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functional determinant and WKB approximation let be a Hamiltonian in one dimension, i would like to evaluate the functional determinant $ det(E-H) $ in onde dimension
i believe that $ det(E-H)= Cexp(iN(E)) $ here $ N(E)$ is the number of energy levels less than a given number 'E'
my steps
*
*i use the identity $ logDet(E-H)=TrLog(E-H)$
*i replace the sum $\sum_{n} log(E-E_{n})$ by an integral over the phase space $ \iint_{D}dpdp log(E-p^{2}-V(x))$
*I take the derivative respect to 'E'
*I use the identity $ \int_{-\infty}^{\infty} \frac{dx}{x^{2}-a^{2}} = \frac{\pi i}{a}$
*I use the Bohr-Sommerfeld quantization condition $\int_{C}dx (E-V(x))^{1/2} = (n(E)+1/2)h$
*i use integration respect to 'E' again
7 i take the exponential
is this semiclassically correct :) thanks.
| The formula doesn't work. Most of the manipulations are formally ok, although it is probably best to start right at step 3--- the derivative of the logarithm of the determinant is the (trace of the) Green's function, which is better behaved than the determinant itself.
Step 5 is incorrect--- there is no reduction using the WKB condition, because the quantity $\sqrt{E-V}$ is in the denominator, and the integration is unbounded. The correct semiclassical expansion for the Green's function is given by the Gutzwiller trace formula.
The best way to check all this is to try it out on the Harmonic oscillator. The formula you give doesn't work, although the semi-classical bit is nice. The semiclassical HO green's function is
$$\int dp dq {1\over E - p^2 - q^2} $$
Which is elementary (up to being divergent--- you can move E a little), and evaluates to $log(E) \pm i\pi (E>0)$ , where $\pm$ means either add, or subtract, or ignore depending on how you deal with the divergent point. plus a divergent constant, which is irrelevant.
| {
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Can a photon passing by an open space barycenter of a system of masses be modeled as if all the system's mass is at the barycenter? To be clear, this example can't apply to the Solar System, since the barycenter is within the Sun, similarly the Earth/Moon system's barycenter is within the Earth.
But, given a system of gravitationally attacting masses revolving about a barycenter that is not contained within any of the masses, i.e. a barycenter in open space, would a photon passing near that point behave in the same manner as a photon passing a point mass of the same mass as the system itself?
| The barycenter has no mass and therefore no forces emanating. This is evident by your example of the moon earth barycenter which continually moves in the mantle 1700 km down or so. If it had any effect it would be working as a whip in cream, generating from quakes to volcanoes!
It is just a geometrical point whose use is to give an observer outside the system a reference point for calculations at large distances from the system.
The photon, or any other particle, will feel the gravity forces of the individual masses according to the laws of gravitation, the distances it has from the individual masses. BTW the forces felt by the photon will be very short and transient.
| {
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About the Ether Theory acceptance Why was the Ether Theory refused by Modern Physics? If you please explain me, I just wanted to understand it more.
| It's because the luminiferous aether was, by definition, composed out of some particles or elementary building blocks with a well-defined location in space. Consequently, it picks a privileged reference frame, the rest frame of the aether. In this rest frame, the speed of light – vibrations of the aether – could be constant, $c$.
However, things moving relatively to this aether by the speed $v$ should detect a different speed of the light relatively to them – the speed would go from $c-v$ to $c+v$, depending on the direction. However, this modification of the light speed, the so-called aether wind, was shown to be non-existent by the Morley-Michelson experiment which measured the speed to be $c$ regardless of the source and the observer. This falsifies the existence of the aether.
The equivalent but even more robust refutation of the aether came from the theory. A physicist named Albert Einstein built a whole new theory of spacetime, the so-called special theory of relativity (a picture of this physicist is often being shown by the ordinary people as well), that also assumes/guarantees that the speed of light is always constant and there can't be any privileged reference frame. Relativity has been backed by the Morley-Michelson experiment as well as hundreds of much more specific experiments.
One of the things it guarantees is that light (electromagnetic radiation) has to be made out of disturbances of the empty space, the vacuum itself, and not a localized material carrier.
| {
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Why is beta negative decay more common than beta positive? In simple terms, why is beta negative decay more common than beta positive?
I know it's something to do with occuring inside/outside the nucleus - but I can't find a simple, easy to understand explanation!
| (The following extends Georg's remarks earlier, where K-capture refers to K-electron capture.)
Beta-plus decay competes with electron capture, but there are few positrons around for beta-minus decay to compete with, so even when beta-plus decay is possible, its branching ratio may be small or overwhelmed by EC.
Moreover, in EC (versus beta-plus decay) the energy difference between initial state and final benefits from the addition of an electron to the reactant side and loss of a positron on the product side (so about 1MeV total). So some nuclei decay by EC that can't decay by positron emission.
| {
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Quality factor for a quantum oscillator? I've been reading papers about nanomechanical oscillators, and the concept of quality factor often pops up. I understand to some extent about Q factor in classical sense, but since nanomechanic oscillators are often treated quantumly, what does Q factor mean then?
| This is just a quick answer that I'm hoping will be superseded, because I'd like to see a good answer to this question. In the context of microwave resonators etched on superconducting chips, I know from experience that the Q factor seems to be basically the $T_1$ time. Sorry I can't be more help -- this is a point of confusion for me too.
| {
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Planetary model of atom still valid? When I was in school, I learned (from Democritus) that an atom was
similar to a solar system, with the nucleus being the sun, and the
electrons being the planets. Of course, there are some differences:
*
*The "sun" isn't a single entity, but a collection of protons and nuetrons.
*Two planets can share an orbit (which might be possible in a solar
system too, but it doesn't happen in our solar system).
Is this model still valid? Here are my problems with it:
*
*In "Surely You're Joking, Mr Feynman", Richard Feynman implies
that electrons are more a theoretical concept than real objects.
*I have trouble understanding atomic bonds (ionic and covalent) in
this model.
*I also have trouble understanding electron "orbit jumping" in this
model, as well as several other things.
Is there a better model for someone learning this for the first time?
| Yes, in some cases.
Nearly a century after Danish physicist Niels Bohr offered his planet-like model of the hydrogen atom, a Rice University-led team of physicists has created giant, millimeter-sized atoms that resemble it more closely than any other experimental realization yet achieved.
Using lasers, the researchers excited potassium atoms to extremely high levels. Using a carefully tailored series of short electric pulses, the team was then able to coax the atoms into a precise configuration with one point-like, "localized" electron orbiting far from the nucleus
| {
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Additional accelerating force during take off of a rocket? During the take off of a rocket, the exhaust produces some pressure below the rocket, which gives an additional force. How large is this force in comparison to the force produced directly by the exhausted gases because of conservation of momentum?
When the rocket is flying through air, is there any measurable (minimal) effect of the additional thrust produced by pressure because the exhaust hits the outside air?
| Any pressure difference along the rocket will result in air motion. Normally the rocket jet entrains some air down and makes a negative pressure difference. However, if you enclose the rocket in a container, you can get a positive pressure difference. Watch videos of rocket launching, especially those vapor clouds along the missile, and you will get an idea how important the pressure difference is in reality.
| {
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The time component is $\gamma m c$, so shouldn't $E=mc$? Basically, the book is Brian Cox's Why Does $E=mc^2$?: (And Why Should We Care?).
I just finished Chapter 5, where we derived the spacetime momentum vector (energy-momentum four vector, as he establishes the physics jargon).
Let $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
So, as we found out, the vector's spatial component is $\gamma m v$, leaving the time component as $\gamma m c$.
He went on, under the guise of making the outcome more intellgible, to saying we could happily multiply the time component by $c$ without changing it's meaning. Cool, I thought, no problem. Next, he pointed out $\gamma\approx1+\frac{1}{2}\frac{v^2}{c^2}$, so $\gamma m c^2 \approx mc^2 + \frac{1}{2}mv^2$. Et voilà, $mc^2$.
Granted, he's obviously trying to simplify things so I can reach an intuitive understanding, but from that point onwards, he uses $mc^2$ as the conversion value.
I'm confused. Could someone explain why we use $mc^2$ and not the version scaled down by a factor of $c$?
| Uh, because $mc$ is not energy? And what do you mean "time component"? Your $\gamma mc$ is momentum.
| {
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Relativistic transformation of the wave packet length Let us suppose we have an excited atom at rest. It has a certain mean lifetime $\tau_0$. If we wait sufficiently long time $t>>\tau_0$, we will find a deactivated atom and a (spherical) electromagnetic wave function of photon with about $\tau_0\cdot c$ long layer with non zero probability to find a photon within. Something like a fast expanding probability "ring" with a $\tau_0\cdot c$ width of the ring.
Now, let us consider this system in a moving reference frame. It seems to me that this width $\tau_0\cdot c$ is relativistic invariant: it is a difference between two "fronts" of electromagnetic wave rather than a length of a material body subjected to the Lorentz contraction. Is it correct? In other words, whether this picture relativistic invariant?
| First, just as a note, I would want to say that the problem will be somewhat ill defined because in QM it may be simple to define transition probabilities for going from one state from another, but, the actual process over time which should give you the Dirac transition current from charge and spin, the current which is responsible for the electromagnetic field, is on a whole different level.
Going back to your question: The circles you are drawing correspond to "Spheres of Simultaneity". In other reference frames they become "Ellipsoids of Simultaneity", so they are not Lorentz invariant.
I did put a significant amount of work in all the 3D images (4.9 through 4.24) in the introductory chapter: "Non-Simultaneity from the wave equation" to visualize why and how this happens.
http://physics-quest.org/Book_Chapter_Non_Simultaneity.pdf
for instance:
Regards, Hans
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Decoherence inside black holes I have a question about decoherence. Assume there is a macroscopic black hole floating around and you have some macroscopic object with you with a huge number of internal degrees of freedom. Conventional decoherence theory predicts decoherence. If this macroscopic object is dumped into the black hole, will decoherence still happen? There are two environments here: the internal degrees of freedom of the macroscopic object which are coarse-grained over, and the exterior of the black hole. Which environmental degrees of freedom should you trace over? If black hole complementarity is right, the internal degrees of freedom and the external degrees of freedom don't commute, so clearly, unless some mathematical generalization of the partial trace can be invented, you can only trace over one or the other. A trace over the former leads to decoherence, but a trace over the latter can't for the simple reason that nothing can escape the black hole. Only thermal Hawking radiation can escape, but it is so scrambled that it carries no decoherence information.
| The question by the OP is misguided as it stands. Both choices of partial traces end up with decoherence. That is easy to see when the partial trace is over the internal degrees of freedom. It is also the case when it is over exterior of the black hole precisely because Hawking radiation is thermal and decoherent. Correct me if I am wrong, but what I guess the OP wanted to ask is whether the preferred pointer basis match up. They do not because a pointer inside the black hole can never be compared with a pointer outside the black hole.
| {
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Can Plasma Ignite? This question on Scifi.se: Why does warp plasma burn green? mentions a fictional type of plasma called 'Warp Plasma' that is capable of igniting and when it does it sprays plasma flames/gas out of pipes as if it were oil or some other form of fuel.
Wikipedia says that the sun consists of hot plasma interwoven with magnetic fields. I couldn't find anything about plasma ignition (This is what I originally thought might be a reason why stars are so bright).
Is plasma capable of igniting?
| Plasma is a kind of matter very similar to gas in which atoms have been ionized. It is very hot and lets off radiation (also in the visible spectrum) due to recombination so it might already be considered flame. Thus it isn't clear what "igniting plasma" really means.
In cold plasma where only a small fraction (few percent) of atoms have been ionized you can have normal chemical reactions (aided by plasma's high temperature). Example of such a process occurs in plasma created by lightening and produces ozone (O3). Thus mixing in the right chemicals can result in oxidation occurring within plasma which could be considered burning. For example plasma made up of element which easily oxidizes could be affected by mixing in gas oxygen. I guess it could be considered ignition, but it seems a stretch relative to the ordinary meaning of the word since even cold plasma is already very hot (few thousands degrees Celsius) and causing oxidation to occur inside such plasma would not add a lot of heat (in relative terms).
As for the color, careful engineering can produce plasma where recombination energy is such that light of a given color is emitted.
| {
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What is the force between two magnetic dipoles?
*
*What is the force between two magnetic dipoles?
*If I have two current loops parallel to each other with currents $I_1$ and $I_2$ and radii $R_1$ and $R_2$ a distance $z$ from each other, what is the force between them?
*What would change if they were two solenoids instead of current loops?
*Would the same hold if it was two magnets?
*Or a magnet and a solenoid?
| The field of a magnetic dipole is, at long distances (in some naturalized unit convention), equal to
$$ B_i = \partial_i ({\mu \cdot r \over 4\pi r^2})$$
where $\mu$ is the magnetic moment vector (the current times area of the loop, in the direction perpendicular to the area of the loop, times N for a solenoid with N windings, or just a given value for a fixed magnet). This form is universal for all magnetic dipoles at long distances, so it's the same for small loops, for small magnets and for small solenoids.
The form is easy to understand, because it's the form of the field for an electric dipole
The response of a second rigid magnetic dipole to this field is a torque which tends to align the magnetic moment along the direction of the field, plus a force proportional to the field gradient. These are also determined by the magnetic dipole of the magnet.
$$ T = \mu \times B $$
$$ F_j = - \mu^i \partial_i B_j $$
The first tends to align the magnetic dipole with the field, and this tends to happen quickly, the second moves the aligned dipole to the region of stronger field, and this happens more slowly.
These equations give a complete determination of the forces and torques acting between two dipoles, but if you substitute in the dipole field, the expressions become complicated and unilluminating. The picture is that there is a $1/r^3$ magnetic field which will align magnets so that their magnetic moment points along it (in the presence of dissipation), and then will lead to an attractive force as the aligned magnet drifts to a region of stronger field.
| {
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Hamiltonian with position-spin coupling I am solving a Hamiltonian including a term $(x\cdot S)^2$.
The Hamiltonian is like this form:
\begin{equation}
H=L\cdot S+(x\cdot S)^2
\end{equation}
where $x$ is the position operator, $L$ is angular momentum operator, and $S$ is spin operator. The eigenvalue for $L^2$ and $S^2$ are $l(l+1)$ and $s(s+1)$.
If the Hamiltonian only has the first term, it is just spin orbital coupling and it is easy to solve. The total $J=L+S$, $L^2$ and $S^2$ are quantum number. However, when we consider the second term position and spin coupling $(x\cdot S)^2$, it becomes much harder. The total $J$ is still a quantum number. We have $[(x\cdot S)^2, J]=0$. However, $[(x\cdot S)^2,L^2]≠0$, $L$ is not a quantum number anymore.
Anybody have ideas on how to solve this Hamiltonian?
| This problem appears interesting for the following reason. Let us write it down in Cartesian coordnates:
$$-\frac{1}{2}\left(\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}\right)+\frac{1}{2}(x\cdot S)^2\psi+L\cdot S\psi=E\psi$$
where I have introduced a 1/2 factor for later convenience. Now, I concentrate on x and I consider the operator
$$-\frac{1}{2}\frac{\partial^2}{\partial x^2}+\frac{1}{2}(x\cdot S)^2$$
One can introduce creation and annihilation operators in a similar way as for the harmonic oscillator
$$A_S=\frac{1}{\sqrt{2}}\left(\frac{\partial}{\partial x}+xS\right)$$
and the corresponding eigenvectors will be labeled as $|n,S\rangle$. The next step is to write down $L\cdot S=\frac{1}{2}(J^2-L^2-S^2)$ and we can restate this problem in the form
$$\left(A_S^\dagger A_S+\frac{1}{2}\right)\psi-\frac{1}{2}\left(\frac{\partial^2\psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}\right)+\frac{1}{2}(J^2-L^2-S^2)\psi=E\psi$$
| {
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Renormalization Group: Different fixed points Extending the Gaussian model by introducing a second field and coupling it to the other field, I consider the Hamiltonian
$$\beta H = \frac{1}{(2\pi)^d} \int_0^\Lambda d^d q \frac{t + Kq^2}{2} |m(q)|^2 +
\frac{L}{2} q^4 |\phi|^2 + v q^2 m(q) \phi^*(q)$$
Doing a Renormalization Group treatment, I integrate out the high wave-numbers above $\Lambda/b$ and obtain the following recursion relations for the parameters:
$$\begin{aligned}t' &= b^{-d} z^2 t & K' &= b^{-d-2}z^2 K & L' &= b^{-d-4}y^2 L \\
v' &= b^{-d-2}yz v & h' &= zh
\end{aligned}$$
where $z$ is the scaling of field $m$ and $y$ is the scaling of field $\phi$.
One way to obtain the scaling factors $z$ and $y$ is to demand that $K' = K$ and $L' = L$, i.e., we demand that fluctuations are scale invariant.
But apparently, there is another fixed point if we demand that $t' = t$ and $L' = L$ which gives rise to different scaling behavior, and I wonder
a) why I can apparently choose which parameters should be fixed regardless of their value ($K$ and $L$ in one case, $t$ and $L$ in the other case)
b) what the physical meaning of these two different fixed points is...
(My exposure to field theory/RG is from a statistical physics approach, so if answers could be phrased in that language as opposed to QFT that'd be much appreciated)
| The short answer is that if you have coefficients for all the terms, you have two independent exact fake scale invariance for the field $\phi$ and $m$ which just rescales the fields and the coefficients appropriately to keep the Hamiltonian exactly the same. This is not a real invariance of the action, since it changes the parameters of the action, it is best thought of as choosing the dimensional scale of the two fields. You usually do this by fixing the terms "L" and "K", but you get a different scaling if you fix the "t" and "L" terms, which is physical in different limits.
I should point out that this model is exactly solvable, there are no real interactions in this model, so the Renormalization Group analysis is just dimensional analysis in disguise. There are two rotated q-modes mixing $m(q)$ and $\phi(q)$ which are completely free.
| {
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Why can't $ i\hbar\frac{\partial}{\partial t}$ be considered the Hamiltonian operator? In the time-dependent Schrodinger equation, $ H\Psi = i\hbar\frac{\partial}{\partial t}\Psi,$ the Hamiltonian operator is given by
$$\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2+V.$$
Why can't we consider $\displaystyle i\hbar\frac{\partial}{\partial t}$ as an operator for the Hamiltonian as well? My answer (which I am not sure about) is the following:
$\displaystyle H\Psi = i\hbar\frac{\partial}{\partial t}\Psi$ is not an equation for defining $H$. This situation is similar to $\displaystyle F=ma$. Newton's second law is not an equation for defining $F$; $F$ must be provided independently.
Is my reasoning (and the analogy) correct, or is the answer deeper than that?
| Asking why
$$i \hbar \frac{\partial}{\partial t}$$
isn't the hamiltonian operator in QM is the same as asking why the time derivative isn't the hamiltonian in Hamilton's equations:
$$
\frac{d p_i}{dt} = -\, \frac{\partial H}{\partial q_i}, \qquad \dots
$$
| {
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Radar Frequency Bandwidth I've come across an interesting question in the course of doing some exam review in a quantum mechanics book and thought I'd share it here.
"What must be the frequency bandwidth of the detecting and amplifying stages of a radar system operating at pulse widths of 0.1usec? If the radar is used for ranging, what is the uncertainty in the range?"
I don't know the solution; although I'd guess that it could involve a Fourier transform (a generic first crack at anything with waves) and the uncertainty relation (as the problem does call out uncertainty in range; which could be interpreted as uncertainty in x).
| The first part of the question deals with the so called "Transform Limited" pulses. A transform limited pulse is one where the time-bandwidth product is a minimum (unity). This can be loosely thought of as a manifestation of the energy time uncertainty relationship.
So:
$\Delta t\times B=1$. From this you can find the bandwidth.
The next part of the question seems to be related to the coherence length of the radiation, which is defined as:
$L_{coherence}=\frac{c}{\pi \Delta f}$
Note that $\Delta f$ is the bandwidth B from the first part of your question. So, anything beyond this range will have an associated error in position measurement.
Edit: Also take a look at Wiki, where there is an expression for coherence length in terms of wavelength. What you need is the $\Delta\lambda$ term.
| {
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What are some interesting calculus of variation problems? That I could create as a classical mechanics class project? Other than the classical examples that we see in textbooks (catenary, brachistochrone, Fermat, etc..)
| Here is one I just made up, but it has a nice flavor--- suppose you have a 2-d bullet going very fast through a 2-d gas. The gas molecules reflects specularly off the bullet, making glancing collisions. What shape of bullet of a fixed area has the least drag?
This problem gives
$$\int {1\over 1+y'^2} + \lambda y dx $$
And the equation for y' you get is
$$ y' = \lambda x (1 - 2 y'^2 - y'^4) $$
or
$$ y = \int \lambda x ( 1 - 2 y^2 - y^4) $$
Which you can solve in a series by plugging in $y={\lambda x^2\over 2}$ and iterating a few times using the relation above as a recursion.
| {
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What is the physical sense of the transition dipole moment? So if the states are the same we achieve the expectation value of the dipole moment for a given state. I mean
$ \langle \mathbf{\mu} \rangle = \langle \psi \vert \hat{\mathbf{\mu}} \vert \psi \rangle$
But I don't feel the physical sense in the case of transition dipole moment when psi-functions on both sides are different
$\langle \psi_{1} \vert \hat{\mathbf{\mu}} \vert \psi_{2} \rangle$
Help me to understand, please.
| Actually the name transition dipole moment pretty much contains its definition. It describes the possibility of a transition between states $\psi_1$ and $\psi_2$.
Physically this can be understood as well. An atom in a particular state may, by absorbing say an electromagnetic wave, change its state. The probability for this change between states is described by the transition dipole element. Therefore, if an atom has a $0$ transition dipole moment, it cannot change its state by absorbing an electromagnetic wave.
| {
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This Expansion-of-the-Universe-Diagram Confuses Me The following blue-cone Wikipedia diagram confuses me.
At any point of cosmological time the encircling horizontal lines in the diagram are of finite circumference. That is indicative of a closed model of the universe.
Queries:
1. Why does the author use a closed model of the universe to explain his point?
2.Can we conclude "It is also possible for a distance to exceed the speed of light times the age of the universe, which means that light from one part of space generated near the beginning of the Universe........." if we draw the same diagram on a flat sheet of paper[instead of using the cone we take a flat surface],remembering that the null geodesics are always straight lines in the flat spacetime context?
| The author of the diagram explains very well how and why he drew the diagram this way:
http://en.wikipedia.org/wiki/File:Embedded_LambdaCDM_geometry.png#Mathematical_details
In particular it states:"I deliberately cut off the embedding short of a full circle to emphasize that space does not loop back on itself (or, if it does, not at a distance governed by the arbitrary parameter R)".
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Why do 3d spheres and gravity tend to rotating discs on one plane? Whether is it our solar system or a whole galaxy, there is usually a massive object (star or black hole) at the centre with gas and objects rotating around it.
The gravitational effect of the star/black hole extends uniformly (more or less) in every direction in 3d. Why does matter tend towards a single plane?
Furthermore, what happens to matter that approaches after the "disc" is formed when it is pulled in from anywhere off the plane, why does it join the plane rather than forming another plane?
I suspect angular momentum has something to do with it, but would appreciate a "pop science" explanation.
Many thanks
Andrew
| It's not true that everything tends towards a single plane.
A counterexample are the dark matter halos of galaxies : See for instance http://news.bbc.co.uk/2/hi/8444038.stm .
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What is escape velocity? In reality, how can something no longer be under the gravitational influence of something else? Isn't G a continuous function and although you leave the immediate vicinity of the earth with an escape velocity won't it always exert a force, however small it may be. Won't that force eventually pull the object back to the earth (assuming the absence of other objects)
| Yes the Earth will always exert a force, but that force will get smaller and smaller as the object get further and further. Even if we assume Earth and the object are the only 2 masses in the universe, the gravity from earth will always slow down the object, forever (ie until its distance is infinite), but the velocity of the object will always have a radial positive component, even if it tends towards 0 at infinity.
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Physics Paradox about Newtons Second Law $F=ma$
*
*If force equals mass times acceleration, wouldn't a basketball dropped from the top of the Eiffel tower exert the same force on the ground as a basketball dropped a foot off the ground? They both have the same mass, and they both are accelerating towards the ground at a rate of $g = 9.81\,{\rm m/s^2}$. (I don't know what terminal velocity is that well as I'm only in physics 1 in high school, but just assume that air drag is not important and the ball doesn't reach terminal velocity.)
*Also, if a ball is dropped high enough to reach terminal velocity, then it accelerates at $0\,{\rm m/s^2}$, so it has a force of ZERO when it hits the ground?
| Force does equal mass times acceleration. However $9.8~\text{m/s^2}$ is the acceleration of the ball imposed by gravity. The acceleration that the ball experiences upon impact with the ground is instead proportional to its current velocity.
Upon impact, $a=-\frac{v}{t}$, where v is current velocity and t is the time impact lasts.
If the ball were travelling $100~\text{m/s}$ and impact lasts 2 seconds, the acceleration upon impact would be $-50~\text{m/s^2}$.
The time impact lasts is related to the properties of all the materials in question. Striking a solid concrete surface, a much shorter impact, striking a pool of water, longer. And the related deceleration is not necessarily distributed evenly over time, it can be much higher in the beginning of impact and lesser near the end, etc, but that's beyond the scope of the question.
Finally, even if the ball has stopped accelerating just before it hits the ground (reaches terminal velocity), sudden impact with the ground will have the ball rapidly decelerate, which is the definition of Kinetic Force.
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How does gravity work underground? Would the effect of gravity on me change if I were to dig a very deep hole and stand in it? If so, how would it change? Am I more likely to be pulled downwards, or pulled towards the edges of the hole? If there would be no change, why not?
| Acceleration due to gravity at depth d below the earth's surface is given by:
$g(d) = G M_e \dfrac{R_e - d}{R_e^3}$
Where,
G = Universal gravitational constant
Me = Mass of the earth
Re = Radius of the earth
d = depth below the earth's surface
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Difference between coordinate and proper distance in Schwarzschild geometry I'm trying to understand the difference between proper distance $d\sigma$ and coordinate distance $dr$ in Schwarzschild geometry. The bottom bit of the diagram represents flat space, the upper bit curved space. The inner circles represent Euclidean spheres of radius $r$, the outer circles radius $r+dr$.
Is the proper radius of these circles the same as $r$? I think I mean if I measured the radius of these circles with a real ruler would I get the coordinate distance $r$ of the Schwarzschild metric.
The more I think about this the more confusing I find it.
Thank you
| In the Schwartzchild coordinates the r co-ordinate is the value you get by dividing the circumference of the circle by 2$\pi$. That is, it's the radius of the circles you've projected onto the base of your diagram.
As others have mentioned in the comments, deciding what you mean by the "real" distance from the circle to the singularity is a vexed issue. Far away from the singularity r agrees with what we think of as the radius, but of course that's only because space is (nearly) flat far away from the singularity. Once the curvature becomes significant r will not be the same value as you get by integrating the metric from the singularity out to your circle.
JR
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What does a unitary transformation mean in the context of an evolution equation? Let be the unitary evolution operator of a quantum system be $U(t)=\exp(itH)$ for $t >0$.
Then what is the meaning of the equation
$$\det\bigl(I-U(t)e^{itE}\bigr)=0$$
where $E$ is a real variable?
| Just a mathematical note in response to the previous answer:
$e^{i \hat{H} t /\hbar}$ is not defined as the exp-series, although it is common to define it so in physics textbooks. But it is not possible to do this as the series is generally not converging (in the operator norm).
One has to use the spectral calculus, in which the "calculation"
$$
e^{i \hat{H} t /\hbar} |n\rangle = e^{i E_n t /\hbar} |n\rangle
$$
becomes (some kind of) a definition.
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Effects of a Coiled Cable Okay, I've got a little bit of a layman's question here.
We're doing a bit of spring cleaning in our office and we've found a cabinet with boxes upon boxes of stored wires, so naturally, this discussion arose...
Picture a normal, bog-standard wire, with a plastic outer coating. Now, quite often when these wires are stored, they will wrapped up and twisted, to effectively make a coil.
I was just wondering what the effects of this type of storage would have.
What if you had a 15m wire and only used the each end to cover about a single meter (leaving 14m still twisted and wrapped in the middle), what the effect of the electrical current running through this have?
Thanks for helping us settle a mild dispute!
| Sorry for necromancing but someone on the internet is wrong.
Nothing major happens with regard to magnetic fields since the cable hold a wire pair carrying equal and opposite current, thus creating two magnetic fields that almost completely cancel each other out.
The most notorious feature of loaded coiled cables is that they potentially generate a lot of heat in a tight space. In most cases it's not an issue, but at high load with little cooling such a coil could be a fire hazard.
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Why are the quarks so named? Quarks have a variety of names (or flavours):
*
*Up
*Down
*Strange
*Charm
*Bottom or Beauty
*Top or Truth
Why do they have such odd names?
| The up/down top/bottom should be self evident: in a matrix representation the vector is written that way
+1/2>
-1/2>
So in the isotopic spin space (the SU(2) of the SU(2)xSU(3)xU(1) of the Standard Model) according to the charge the one on top was called the up and the one on bottom, the down.
The strange got its name from the strange mesons, which when discovered were behaving strangely, with respect to pions, needing a new quantum number because they were generated in pairs ( Lamda K) and the quantum number became the "strange" one.
Charm was a whim as , they were charmed by its existence since it had been predicted to exist given the quark model expectations; from the quark entry in wikipedia:
Glashow, who coproposed charm quark with Bjorken, is quoted as saying, "We called our construct the 'charmed quark', for we were fascinated and pleased by the symmetry it brought to the subnuclear world
Top and Bottom, again because of the position in the vector, and Beauty instead of Bottom out of whimsy again, keeping the B, and Truth keeping the T.
Who said that physicists are not having fun?
For completeness, the name "quark" has the origin in Finnegan's Wake of James Joyce:
For some time, Gell-Mann was undecided on an actual spelling for the term he intended to coin, until he found the word quark in James Joyce's book Finnegans Wake:
Three quarks for Muster Mark!
Of course our Germanic language friends say that Quark is a type of cheese!
Actually the quark article in wikipedia has an etymology section for the quarks, to be read for completeness.
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Spin angular momentum of a system of particles : Is there any energy associated with it? Consider a system of point particles , where the mass of particle $i$ is $μ_i$ and its position vector is $\vec{r}_i$. Let $\vec{r}_\text{cm}$ is the position vector of the center of mass of the system. Considering the system from a reference frame attached to the center of mass, the system may have a spin about the center of mass and it is given by the spin angular momentum $\vec{L}_{spin}$. It is given by the expression
$$\vec{L}_{spin} = \sum_i \mu_i \Bigl[(\vec{r}_i - \vec{r}_\text{cm}) \times (\dot{\vec{r}}_i - \dot{\vec{r}}_\text{cm}) \Bigr]$$
The rate of change of this spin angular momentum is the total torque acting on the system about the center of mass in the center of mass reference frame.
My question is, is there any (spin kinetic (may be)) energy associated with the spin angular momentum in the center of mass reference frame ? How is it defined ?
| Similar to the derivation of separation of angular momentum into $L_{CM}$ and $L_{internal}$, one can derive similar expression for Energy as
$E = \frac{1}{2}M_{total}v_{CM}^{2} + \frac{1}{2}\sum \mu_{i} v_{i}^{'2}$.
Proof:
$$E = \frac{1}{2}\sum \mu_{i} v_{i}^{2}$$
$$v_{i} = v_{CM} + v_{i}^{'}$$
$$E = \frac{1}{2}\sum \mu_{i} v_{CM}^{2} + v_{CM}\sum \mu_{i} v_{i}^{'} + \frac{1}{2}\sum \mu_{i} v_{i}^{'2}$$
Since in CoM frame $\sum \mu_{i} (r_{i}-r_\text{cm}) =0 \to \sum \mu_{i} v_{i}^{'}=0$.
$$QED$$
L and E within Com frame can be related only if body is rigid. One can refer Klepner & Kolenkow Classical Mechanics.
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Elasticity of Space; How does the expansion of Space affect gravity? Does space have an elastic quality?
What I was thinking about was if space is expanding, is it being 'stretched', like a balloon being blown up, and if so, is this causing gravity to weaken? Imagine space as a 2 dimensional sheet (got this from one of Brian Greene's books) with planetary bodies resting on it and causing a depression in it, if you were pulling this sheet from all sides over a period of time, you would cause the depression of the planetary body to decrease and eventually become flat, which if we go back to reality, would mean that the gravitational 'constant' had changed to the point where the planetary body had no influence on those objects which were previously orbiting around it (or even residing on it's surface).
Is this the case in reality? Or does space not have an elastic quality? If not, can you explain to me what exactly it means for space to be expanding?
In case you didn't notice, I'm a layman (hence the Brian Greene books :p), so try to keep your answers/explanations conceptual if possible.
| I remember that Prof. Susskind said in the cosmology course of his "Stanford ongoing studies series" (it must have been somewhere in the first part of this 8 Lecture course)
http://www.newpackettech.com/Resources/Susskind/PHY28/Cosmology_Overview.htm
that space is continuously created in the course of the expansion such that the energy density keeps constant. This can be described by Hook`s law with a negative spring constant but it is not really a "rubber sheet". He mentioned to have derived a theoretical model to describe this process which is called "Newton Hooke cosmology".
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Simple applications of group theory which can be understood by a senior undergrad I am looking for references (books or web links) which have "simple" examples on the use of group theory in physics or science in general.
I have looked at many books on the subject unfortunately they usually require extensive technical coverage of the basics, i.e. the 1st 100 pages or something, to be able to start discussing applications.
I believe that there is an easy way to explain anything (it's just hard to find it).
| My two favorites for group theory in physics are:
*
*Lie Groups for Pedestrians
for a general introduction to Lie groups, mostly in a particle physics context.
*Levine's Quantum Chemistry, for an introduction to group theory in molecules.
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What frequency photons are involved in mediating physical force? If the force felt when pushing an object is mediated by the electromagnetic interaction and hence photons, what is their frequency?
| Their frequency is determined with the velocities. For example, if you have a heavy, moving with $\vec{v}= const$ source charge that creates a time-dependent potentials $\phi(\vec{x},t)$ and $\vec{A}(\vec{x},t)$, then the Fourier expansions of those potentials will contain the following time-dependent exponentials: $\phi_{\vec{k}}, \vec{A}_{\vec{k}}\propto exp(-i\vec{k}\vec{v}t)$, i.e. the frequencies $\omega=\vec{k}\vec{v}$ are related to the wave vectors in a different way than for the true photons. Any probe charge will feel these "frequencies" in a superposition, as a unique time-dependent force $e\vec{E}(t)+\frac{e}{c}[\vec{V}_p ,\vec{B}(t)]$.
Also, the number of those "photons" is uncertain, that makes the notion of virtual photons a useless concept.
Finally, the transferred energy for a given scattering angle is determined solely with kinematics (the same for any kind of interaction potential), so the notion of the virtual photon is worthless.
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What fundamental principles or theories are required by modern physics? We have been taught that speed of light is insurmountable but as we know an experiment recently tried to show otherwise.
If the experiment did turn out to be correct and confirmed by others, would it make physics to be rethought of? What other concepts are fundamental to physics, which, if disproved would need radical rethinking?
If this sounds too juvenile and/or misinformed, please understand that I am a layman, having nothing, professionally or academically to deal with science, directly, and this question is out of curiosity. I have developed a liking to "science stuff" and been reading popular science variety of literature lately. This question was also prompted by what Sheldon Cooper had to say in one of the episodes (I was watching a rerun).
| Most progress in physics is incremental to start with, as far as data and experiments go. Theories change following new data but on the whole they change by incorporating the old theories as limiting cases for certain parameters of the new theories, or convolutions over the variables of the new theories.
@Ronmaimon's list is valid , and if an experiment violates one of these conditions the theories would have to be reshuffled/reformulated or, as has happened in the past, the phenomenon explained by new particles. I remind that the neutrino was discovered because energy and momentum conservation had to hold, for example.
The Standard Model of particle physics has to be incorporated in any new theory because it is a shorthand for all the data up to now with very few dark spots ( CP violation comes to mind). Incorporated does not preclude new ways of looking at the data, just that there should be consistency with the old.
If strings are the theory of everything, on the other hand they bring us many unexplored dimensions, and if we have managed to have such complicated theories with 3+1 dimensions, God knows what smart theorists can come up with trying to accommodate violations, and they are already exploring theories to fit these superluminal neutrinos if they turn not to be a systematic error.
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Why can't echoes be heard inside a room? If I go camping and shout anywhere, in the forest or on a cliff, I usually hear the echo of my voice.
Why when I shout in my room I do not hear any echoes?
| Sound does echo inside a room but you might not notice it much for two reasons.
Firstly the time for the echo to return is very small so you will not hear a long sound repeated as an echo, instead you will get a resonance type echo, like when you sing in the bathroom.
Secondly most rooms are full of soft furnishings that quickly absorb the sound and damp the resonance.
If you are in a large room with no furnishings and you clap once loudly you will certainly hear the echo rebounding rapidly off the walls.
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Why doesn’t gravity break down in a large black hole? By popular theory gravity didn’t exist at the start of the Big Bang, but came into existence some moments later. I think the other forces came into existence a little latter.
When a black hole crushes matter to a singularity (infinite density), at some point shouldn’t the forces cease to exist including gravity?
Kent
| You are right. Both the Big Bang and a black hole are what is called a space-time singularity and the physics of these object is, to put it simply, poorly understood. We know a lot about a black hole seen from outside, or about what happened some ridiculously small after the Big Bang, but it is unknown if the laws of physics as they are currently understood cease to be valid before the Planck epoch ($10^{-43}$ seconds after the Big Bang).
The same applies for a black hole: laws of physics describe extremely well what happens outside the black hole, and, to an extent, even in proximity of the center of a black hole but the closer you get to the singularity, the higher the curvature and we simply don't know if Einstein field equations (the cornerstone of General Relativity) are valid for higher curvature, or are just a low-curvature approximation or the exact laws.
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Physics related Podcasts
Possible Duplicate:
Are there any good audio recordings of educational physics material?
In the same way that was already asked about good books of Physics in this StackExchange, I would like to know good physics podcasts! What are the most informative and enjoyable to hear?
| I don't know of any regular podcasts, but there are lots of Physics lectures on Youtube. Just search for something obvious like "physics".
If you follow any of the Physics blogs you'll often see links to Youtube videos posted.
JR
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Could gravity hold electron charge together? Could the gravitational force be what holds the charge of the electron together? It seems to be the only obvious possibility; what other ideas have been proposed besides side-stepping the issue and assuming a "point charge"? How would this affect the electron "self-energy" problem? The question is related to the idea of geons.
| The short answer... we do not know. (Were 'we' is humanity or physicists - take your pick.)
A more interesting answer is...
The electron size is known to be 10^-18 meters or smaller. If gravity was holding it together then it might be at the Schwarzschild radius.
Rs = 2GM/c^2
so with values substituted it would be
2 x 6.67300x10^-11 x 9.10938291x10^−31 / (3x10^8)^2 = 1.35x10^-57 meters
However, this is less than the Plank length (10^-33 m). Therefore, if it is held by gravity then it would likely have a radius near the plank length. Supersymmetry (SUSY), for example, has gravity increasing to have all forces equal at the plank length.
If you check out Lenard Susskind's lectures on ER=EPR and GR you will find that he thinks that physics is leading us towards the idea that elementary particles and black holes might be related. Black holes have only three properties angular momentum (spin) mass, and charge. Sounds like an elementary particle.
This is a hint not a theory. It is very early to say.
EDIT:
It would have been nice to know why this was downgraded with the addition of a comment. There is nothing wrong with the physics. If it was downgraded because it does not answer the question then that does not make sense because there is no known answer to the question.
Black holes look like macroscopic elementary particles but elementary particles do not look like black holes.
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Why is Mendel Sachs's work not taken seriously? Or is it? Back in college I remember coming across a few books in the physics library by Mendel Sachs. Examples are:
General Relativity and Matter
Quantum Mechanics and Gravity
Quantum Mechanics from General Relativity
Here is something on the arXiv involving some of his work.
In these books (which I note are also strangely available in most physics department libraries) he describes a program involving re-casting GR using quaternions. He does things that seem remarkable like deriving QM as a low-energy limit of GR. I don't have the GR background to unequivocally verify or reject his work, but this guy has been around for decades, and I have never found any paper or article that seriously "debunks" any of his work. It just seems like he is ignored. Are there glaring holes in his work? Is he just a complete crackpot? What is the deal?
| Mendel Sachs may have been blacklisted, which would certainly be wrong. But his theory has a fatal error. His derivation depends on the assumption that certain 2x2 complex matrices, standing for quaternions, approach the Pauli spin matrices in the limit of zero curvature. This is impossible; the Pauli matrices are not quaternions and the argument collapses.
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how to represent the effect of linking rigid-bodies together? I have 2 rigid-bodies (b1,b2) if i linked one to the other (as if they are conjoined together) , how to represent b1 effect on b2 and b2 effect on b1
Is there any LAW that affect the position/orientation of the other body ?
notes :
*
*i am using Quaternions for orientations
*i don't want to treat them as one body
*i have only primitive shapes (box,sphere,..) to link.
| The open-source physics engine ODE allows you to connect two bodies using any of a number of different joints. One of those joints is the "Fixed" joint. It's much more stable, in the physics engine, to represent the two bodies as a single body but maintain two separate geometries for collision purposes. However, ODE probably handles collision detection/resolution differently from what you have in mind. It only detects collision after one frame of interpenetration and then constrains the relative velocity of the colliding bodies in such a was as to force them apart on the next time step. That type of constraint is much easier to satisfy for a single rigid body than two, but perhaps you're actually preventing penetration and so need a different technique.
The fixed joint simply constrains the two bodies to have zero relative angular velocity and zero relative linear velocity (and also has an error correction term to eliminate small numerical drift). After that, the LCP solver handles the rest.
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What is the connection between Poisson brackets and commutators? The Poisson bracket is defined as:
$$\{f,g\} ~:=~ \sum_{i=1}^{N} \left[
\frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}} -
\frac{\partial f}{\partial p_{i}} \frac{\partial g}{\partial q_{i}}
\right]. $$
The anticommutator is defined as:
$$ \{a,b\} ~:=~ ab + ba. $$
The commutator is defined as:
$$ [a,b] ~:=~ ab - ba. $$
What are the connections between all of them?
Edit: Does the Poisson bracket define some uncertainty principle as well?
| I don't know any link between Poisson bracket and anti-commutator, but I do know the link between Poisson bracket and commutator.
$$[\hat a,\hat b]=i\hbar\{a,b\}_\text{Poisson}$$
Subtleties
As the operator $\hat a$ and $\hat b$ are counterparts to classical dynamical variable, they must be ①functions of canonical coordinates and momenta (ruling out spin, which cannot be put in a Poisson bracket) ②Hermitian operators (try $[\hat{x}\hat{p},\hat{p}\hat{x}]$).
In addition, the equality sign isn't really an equality, because r.h.s. are commutative numbers while l.h.s are non-commutative operators, so you must be careful relating two sides. For example, the quantum analogy of $xp$ is neither $\hat{x}\hat{p}$ or $\hat{p}\hat{x}$, but $\frac{1}{2}\left(\hat{p}\hat{x}+\hat{x}\hat{p}\right)$.
| {
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What is the meaning of speed of light $c$ in $E=mc^2$? $E=mc^2$ is the famous mass-energy equation of Albert Einstein. I know that it tells that mass can be converted to energy and vice versa. I know that $E$ is energy, $m$ is mass of a matter and $c$ is speed of light in vacuum.
What I didn't understood is how we will introduce speed of light?
Atom bomb is made using this principle which converts mass into energy; in that the mass is provided by uranium but where did speed of light comes into play? How can speed of light can be introduced in atom bomb?
| c: “Light Velocity” (c: celeritas, Latin: “swift”),
is the speed of photons in a straight, flat vacuum:
299, 792, 458 meters per second ( 3e^8 mps).
This reference is a constant that never varies.
(Does NOT mean mass is moving at light speed!)
c^2 = 89, 875, 517, 873, 681, 764 ( 9e^16 ), where
“9e^16” becomes the “mass to energy ratio”
(nucleons to photons) but NOT a velocity.
Each kg of matter is composed of
6e^26 nucleons (N) and every nucleon contains
1.5e^−10 joules (J) of energy.
The total energy (E) in a mass (m) of 1 kg
is equal to the number of nucleons (N) per kg
multiplied by the energy (J) per nucleon (J/N):
E: 9e^16 J = 6e^26 N/kg x 1.5e^−10 J/N
Compare this to:
E=mc^2: E: 9e16 J = m: 1 kg x c^2: 9e^16
m=E/ c^2: m: 1 kg = E: 9e^16 J / c^2: 9e^16
This energy is released by atomic annihilation
(using anti-matter). The energy released
would be Gamma Rays.
If this energy is released all at once:
9e^16 J would equal approximately
21.5 Mt (Megatons of TNT).
This is 1,000 times greater than either
the Trinity or Nagasaki devices.
| {
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Where to find cross section data for $e^{-}$ + $p$ $\longrightarrow$ $p$ + $e^{-}$? Where to find cross section data for $e^{-}$ + $p$ $\longrightarrow$ $p$ + $e^{-}$ ?
PDG's cross-section data listing does not include it.
| The answer is highly dependent on the scale of the momentum transfer.
The figure of merit is $Q^2$ is the squared momentum transfer, and in some regimes the missing energy $\omega = \epsilon' - \epsilon$. The formalism is usually developed in the lab frame with a stationary proton target and a energetic electron beam. We write $Q = \mathbf{k}' - \mathbf{k}$ and $\mathbf{k} = (\epsilon,\vec{k})$ and $\mathbf{k}' = (\epsilon',\vec{k}')$ are the four momentum of the incident and scattered electron respectively.
*
*If $Q^2 \ll m_p^2$ then you can treat the proton as a point particle to first order and you can simple look this up. In the upolarized case you use the Mott cross-section.
*For Q^2 on the same order of magnitude at the proton mass squared the problem is complicated enough that one typically uses a parameterized experiment results in the shape of a set of "form factors" (note that the formalism typically used at medium energies is different from but equivalent to that used at high energies). The JLAB results that luksen links to are among the highest precision available at this time.
In nuclear physics parlance you get
$$\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \left( \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} \right)_{\text{Mott}} \frac{Q^2}{\left|\vec{q}\right|^2} \left[ G^2_\mathrm{E}(Q^2) + \tau \epsilon^{-1}G^2_\mathrm{M}(Q^2)\right] .$$
To a first approximation you can use the dipole form for the form factors $$G_\mathrm{M} \approx \mu G_\mathrm{E} \approx \mu \left( 1 + \frac{Q^2}{0.71\text{ GeV}^2}\right)^{-2} ,$$
where $\mu$ is the magnetic moment of the proton.
*For $Q^2 \gg m_p^2$ you are in the deep inelastic scattering regime and can treat the proton at a collection of bound partons. The phrase you're looking for is "structure functions" (PDF link).
| {
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Does the scientific community consider the Loschmidt paradox resolved? If so what is the resolution? Does the scientific community consider the Loschmidt paradox resolved? If so what is the resolution?
I have never seen dissipation explained, although what I have seen a lot is descriptions of dissipation (i.e. more detailed pathways/mechanisms for specific systems). Typically one introduces axioms of dissipation for example:
entropy $S(t_1) \geq S(t_0) \Leftrightarrow t_1 \geq t_0$ (most often in words)
These axioms (based on overwhelming evidence/observations) are sadly often considered proofs. I have no problem with useful axioms (and I most certainly believe they are true), but I wonder if it can be proven in terms of other (deeper and already present) axioms. I.e. is the axiom really independent? or is it a corollary from deeper axioms from say logic (but not necessarily that deep).
(my opinion is that a proof would need as axioms some suitable definition of time (based on connection between microscopic and macroscopic degrees of freedom))
| *
*The Second Law has nothing to do with time, and it isn't formulated in terms of S. Instead, it is used to define entropy. See Caratheodory and Born.
*Loschmidt's objection concerns the trivial fact: irreversible macroscopic behaviour cannot be derived using reversible equations of motion. Boltzmann obviously hailed. His H-theorem is just a mathematical exercise. Later, Ehrenfest attempted to do the job using corse-graining of the phase space, but also failed.
*The reversibility problem isn't properly defined. What is irreversible is thermodynamic state. A mechanical state is perfectly reversible by time inversion in the equations of motion. But a thermodynamic state is not. Any adiabatic transformation is adiabatically irreversible.
| {
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Can the charge of particles spontaneously flip from positive to negative or vice versa? I'm thinking of matter antimatter annihilation, are there reactions where normal matter converts to antimatter?
| There are reactions where normal matter converts to antimatter. For example, in neutral kaon oscillations, a beam of kaons (or rather, what are created as kaons) will appear to convert to antikaons after some distance, then back to kaons, and so on.
However, a key property that allows this to happen is that the kaons are neutral. This wouldn't happen with charged particles, because electric charge is conserved. The flavor identity that distinguishes (neutral) kaons from antikaons, on the other hand, is not conserved.
So no, the charge of a particle cannot spontaneously flip from positive to negative or vice versa. But it is possible for matter to turn into antimatter.
| {
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How Does Dark Matter Form Lumps? As far as we know, the particles of dark matter can interact with each other only by gravitation. No electromagnetics, no weak force, no strong force. So, let's suppose a local slight concentration of dark matter comes about by chance motions and begins to gravitate. The particles would fall "inward" towards the center of the concentration. However, with no interaction to dissipate angular momentum, they would just orbit the center of the concentration and fly right back out to the vicinity of where they started resulting in no increase in density. Random motions would eventually wipe out the slight local concentration and we are left with a uniform distribution again.
How does dark matter form lumps?
| My problem is with the time scale of the phenomenon. It has been proposed that the universal network of filaments and concentrated lumps of dark matter is the framework upon which ordinary matter condensed to form our present day clusters of galaxies. If this is so, it would seem reasonable that there be some rapid mechanism for its formation early in the development of the universe. With only gravity to draw it together and mechanisms like the three body interaction expelling one of the bodies to permit coalescence into higher density lumps, how did the dark matter network manage to form first? Ordinary matter has the same gravitational means of aggregating plus the electromagnetic interactions for reducing angular momentum. Why didn't the ordinary matter condense first?
| {
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Why does electron-positron annihilation prefer to emit photons? If gravitons are also massless, and neutrinos nearly so, why aren't pairs of either of them normally expected outcomes of electron-positron annihilations? Are they possible but simply unlikely, or is there actually some conserved quantity prohibiting their creation?
Edit: I'm talking about the low-energy limit, not in accelerator beam collisions.
| It is my understanding that electron-positron annihilation with neutrino-antineutrino production is possible at any energy, but the cross-section of such reaction is extremely low at low energy. While electron-positron annihilation with two photon production requires a virtual electron, electron-positron annihilation with neutrino-antineutrino is only possible due to weak force, so it requires a virtual W or Z boson (Phys. Rev. D, D.A. Dicus, v.6, p.941 (1972)). The masses of these bosons are five orders of magnitude greater than that of electron, so the cross-section is extremely small at low energies. Or, in other words, the reaction is possible, but very rear, as weak force is much weaker than electromagnetic force at low energy.
| {
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How are the Pauli matrices for the electron spin derived? Could you explain how to derive the Pauli matrices?
$$\sigma_1 = \sigma_x =
\begin{pmatrix}
0&1 \\ 1&0
\end{pmatrix}\,, \qquad
\sigma_2 = \sigma_y =
\begin{pmatrix}
0&-i\\ i&0
\end{pmatrix}\,, \qquad
\sigma_3 = \sigma_z =
\begin{pmatrix}
1&0\\0&-1
\end{pmatrix}
$$
Maybe you can also link to an easy to follow tutorial ?
| This link seems to be along the required path of thought.
Please note the "axiomatic" facts: experimental inputs in value of $S^2$, raising and lowering operators, desirability of hermitian operators... that go inside the derivation.
Also, once having chosen them, note that the 3 Pauli matrices along with the 2d identity matrix can be used as a basis to write down any 2x2 matrix as has been mentioned here.
| {
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Calculating lagrangian density from first principle In most of the field theory text they will start with lagrangian density for spin 1 and spin 1/2 particles. But i could find any text where this lagrangian density is derived from first principle.
| Seek for 'Klein-Gordon equation' and 'Dirac equation' - they can be found in any textbook concerning basic relativistic quantum mechanics (such as, e.g. Landau ). Klein-Gordon (spin=0 and any natural spin after modifications) comes directly from the energy-momentum conservation of special relativity $p_\mu p^{\mu} = -m^2$, whereas Dirac equation for fractional spins is guessed as 'square root' of Klein-Gordon (in certain sense).
| {
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What are the practical applications of decoherence? Let me clarify this question somewhat. I know decoherence is ubiquitous in nature and explains the emergence of a classical world from quantum physics. My question is really about how a knowledge of how decoherence actually works can be put to use in a practical application. An application we can't design in the absence of such a knowledge, even though decoherence is still happening all the time.
Thanks
| Quantum key distribution comes to mind. It is used to ensure that there are no eavesdroppers on the key distribution channel because they would decohere it measurably.
| {
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Energy in an EM wave should depend on frequency I just finished reading Feynman's Lectures on Physics vol.I, §34-9: "The momentum of light". The author explains that there is a relation between the wave 4-vector $k^{\mu}$ and the energy-momentum 4-vector $p^{\mu}$ of an EM wave, namely
$$p^{\mu}=\hbar k^{\mu}, $$
or equivalently
$$\tag{deB}W=\hbar \omega, \mathbf{p}=\hbar \mathbf{k},$$
and those equations are called de Broglie relations.
However, as I learned in my classical electromagnetism course, flux of energy in such a wave is quantified by Poynting's vector, yielding formulas such as the following:
$$\tag{1} I=\frac{1}{2 \mu_0 c} E_0^2, $$
where $I$ stands for "average intensity" of the wave and $E_0$ for "maximum amplitude of electric field".
Question Where is $\omega$? It does not appear in formula (1) nor in any other formula based on Poynting's vector. But as of equations
(deB) it should do so. Am I wrong?
Thank you.
| The section you are referring to clearly states that those equations do not apply to the wave, but to the "particles" of light, the photons. The resolution is that two waves of the same amplitude but different frequencies contain different numbers of photons. This has interesting consequences, for instance it means it is possible to communicate via radio waves carrying miniscule amounts of power while an optical signal of similar intensity would be drowned out by shot noise.
| {
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Physically what happens during Avalanche breakdown to the pn junction? What does breakdown mean physically?
I saw this in wikipedia:
The avalanche process occurs when the carriers in the transition
region are accelerated by the electric field to energies sufficient to
free electron-hole pairs via collisions with bound electrons.
Say I have a lightly doped Germanium pn junction, will I be able to use it again or throw it away once Avalanche breakdown take place?
And also what happens in case of Zener breakdown? I know it requires highly doped pn junction and not so much pot diff..
| The avalanche breakdown itself doesn't cause destruction, it just moves charge carriers around within the material.
But the breakdown results in extremely high currents through the semiconductor, which results in high temperatures. In particular with avalanche breakdown the higher the temperature gets, the higher the current becomes, which raises the temperature, etc. This results in a runaway thermal effect. The thermal stress is what causes physical destruction, e.g. melting.
| {
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How is gradient the maximum rate of change of a function? Recently I read a book which described about gradient. It says
$${\rm d}T~=~ \nabla T \cdot {\rm d}{\bf r},$$
and suddenly they concluded that $\nabla T$ is the maximum rate of change of $f(T)$ where $T$ stands for Temperature. I did not understand. How gradient is the maximum rate of change of a function?
Please explain it with pictures if possible.
| The equation
${\rm d}T~=~ \nabla T \cdot {\rm d}{\bf r}$,
says that the change in T, namely ${\rm d}T$, is the scalar product of 2 vectors, $\nabla T$ and ${\rm d}{\bf r}$, which can also be written as the magnitude of the 1st vector times the magnitude of the 2nd vector times cosine the angle between them.
${\rm d}T~=~ |\nabla T| |{\rm d}{\bf r}|\cos\theta$.
Now assume that we are fixing the length of the infinitesimal displacement vector but we can move it around changing its direction, and hence changing $\theta$. You notice that $dT$ is maximum if $\theta$ is $0$.
$\theta=0$ means both vectors have the same direction, and since $d{\bf r}$ is the displacement vector then in this case you move along the same direction of $\nabla T$ that makes $dT$ maximum.
Hence you can interpret $\nabla T$ as the vector whose direction is the direction along which the change of the function $T$ is maximum.
| {
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How can a human eye focus on a screen directly in front of it? I am asking this question here because I think the answer has something to do with the way light is bent as it's captured through the eye.
I saw a show a while ago about tiny screens on contact lenses to pull up data on objects you see in the real world, I also just saw this article about Google testing the same idea with screens in the lenses of sunglasses.
The part I do not understand is how your eye would focus on a screen that is so close. My eyes (and I believe most others) cannot focus on anything closer than a couple inches away. Yet the lenses on glasses are much closer to the eye than a couple inches.
So I would guess they would have to use some special technology to separate the light rays in a way that your brain could make an image from it. If that's correct, how would it work? if not, how would they get your eye to focus on a screen so close?
| In theory, an ultra ultra high resolution display could act as an holographic filter and produce an image that seems to originate from far away.
Or you can project an image from a single point straight into the eye. That does not require focus to be sharp. But it will be very challenging to align the virtual image with the real image (you need to anticipate any focus change of the eye)
Or maybe you can put a lens in front, and another lens behind a transparent display. One lens to allow you to focus on the screen and the other one to cancel out the effect. But it will not be practical (thick construction, loss of periferic vision, ...)
Or, use a semi-transparent mirror to combine the normal view with a lens-modified display. So far the ony practical idea.
| {
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Why can't we think of free fall as upside down rocket? /\
/ \
| |
| m |
| |
------ <--- floor (Rocket A)
This rocket is accelerated (g) upwards then mass(m) falls on the floor.
------ <--- floor
| |
| m |
| |
\ /
\ /
\/ (Rocket B)
This rocket is accelerated (g) downwards then mass(m) falls on the floor.
----- <--- ceiling
| |
| m |
| |
----- <--- floor (Elevator E)
This elevator is falling freely on the earth. Acceleration due to gravity is g. The mass stay in the midair.
Why? Why can't we think of the elevator as upside down rocket? Why doesn't mass go to the ceiling of the falling elevator?
NOTE: Principle of equivalence of this document is what I am trying to understand.
| You can't.Cause the rocket is in a gravity free zone.When it accelerate downwards the mass inside the rocket feels a force proportional to it's acceleration in opposite direction.Which causes the mass inside the rocket moves towards the floor.
But for a free falling elevator in gravity field,the mass inside it feels no force hence it will stay in rest.
| {
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What is the physical meaning of a "complete" Hilbert space in QM? What does the word "complete" means from the physical point of view?
I do not understand what it physically means to say that a Hilbert space is a complete vector space.
| I think your question is "why wasn't quantum mechanics formulated on normed vector spaces?" i.e "why was the completeness criterion required?"
I don't know a rigorous answer, but it seems reasonable for the following reason:
Completeness means that every Cauchy sequence of elements of H converges to an element of H. The QM postulate says that physical states are represented by vectors (strictly speaking rays) in H, so if I had an infinite sequence of physical states which were getting "physically" closer and closer together - in the sense that the characteristics of the physical quantities encoded in the states were converging, then it seems reasonable to require that the thing they're converging to is also a physical state. Mapping this over to H, then the Cauchy completeness criterion will take care of this.
The reason I'm worried that this is a bit of a weak answer, however, is that not all elements of H necessarily represent realizable physical states. For example the sums of vectors in different superselection sectors certainly doesn't. So maybe the Hilbert space criterion where every Cauchy sequence converges to an element of H is sufficient but not necessary.
| {
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Finding distance when the force is a function of time I'm having trouble with this homework question
A mysterious rocket-propelled object of mass 49.0 kg is initially at rest in the middle of the horizontal, frictionless surface of an ice-covered lake. Then a force directed east and with magnitude $F(t) = (16.3\text{ N/s})t$ is applied.
How far does the object travel in the first 5.50s after the force is applied?
For some reason I'm not getting the correct answer. I think maybe I'm not understanding how to use the magnitude of the force they are giving me. I know how to use a constant force, but is this different because the force is a function of time?
I tried starting it like this:
$$\begin{align}F &= ma\\
16.3(t) &= (49)(a) \\
16.3(5.5) &= (49)(a) \\
89.65 &= (49)(a) \\
a &= 1.82959\ \mathrm{m/s^2}\end{align}$$
So now we know that:
$$\begin{align}t &= 5.5\text{ s} \\
a &= 1.82959 m/s^2 \\
V_o &= 0\end{align}$$
So I plug it into my equation:
$$\begin{align}\Delta X &= V_o t + 1/2 a t^2\\
\Delta X &= (0)(5.5) + (1/2)(1.82959)(5.5)^2\\
\Delta X &= 27.7\text{ m}\end{align}$$
But that's not the right answer.
| If the force is constant with time, then the distance is a polynomial with time of order 2. If force varies linearly with time, then the distance is a polynomial of order 3.
$$ x(t) = C_0 + C_1\, t + C_2\, t^2 + C_3\, t^3 $$
So there are four (4) unknown coefficients to the expression for distance. Two of them are given from the initial conditions ($x=0$ and $v=0$ at $t=0$), leaving only the coefficients for $t^2$ and $t^3$ to be determined.
The other two are found from the equations of motion
$$ F(t) = m \frac{{\rm d}^2 x(t)}{{\rm d}t^2} $$
given that $F(t)=k\,t$ and $x(t)=\ldots+C_2\,t^2+C_3\,t^3$
I hope you have had calculus and knows how to do a 2nd order derivative of a polynomial. This will give you the coefficients $C_2$ and $C_3$ since the above needs to be solved for ALL values of time $t$.
In the end plug $k=16.3$ and $t=5.5$ and whatever values the initial conditions give you and you will get your answer.
| {
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Calculating Impact Velocity Given Displacement and Acceleration
Assume a car has hit a wall in a right angled collision and the front bumper has been displaced 9 cm. The resulting impact is 25g. Also, it is evident by skid marks that the car braked for 5m with an acceleration of 1.5m/s^2. What is the impact velocity in this collision?
Here's what I get out of it.
$$\begin{align}\Delta d &=0.09\text{ m}\\
a &= 196\ \mathrm{m/s^2}\\
V_2 &= 0\text{ m/s}\\\end{align}$$
Then I determine $V_1$ by:
$$\begin{align}V_2^2 &= V_1^2 + 2a\Delta d\\
0 &= V_1^2 -35.28 \\
5.94\text{ m/s} &= V_1\end{align}$$
My textbook does not give this answer. Could anyone please explain why. I have been looking at it for hours.
| Why is $a=196 m/s^2$? It's 20g, not 25g. And it looks like you did not take into account the deceleration before impact.
| {
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What determines color -- wavelength or frequency? What determines the color of light -- is it the wavelength of the light or the frequency?
(i.e. If you put light through a medium other than air, in order to keep its color the same, which one would you need to keep constant: the wavelength or the frequency?)
| Actually, there is something important all these answers are missing. Color is determined by the response of the human eye, not by energy or frequency. In order to get the full range ('gamut') of colors, I need a mix of red, green and blue light (hence the RGB displays) and the primaries can themselves all be different frequencies. That is, one RGB system can have one frequency for the red, while another has a somewhat different frequency for red, the only hard and fast requirement being that both of them choose that frequency from somewhere in the red range. But the choice affects the gamut.
Now I said "human eye", but of course, other animals see colors, too. Bees see colors into the ultraviolet. But of course, we have no idea what the ultraviolet colors look like to them, only that they do see them, and can distinguish shades of them.
Wikipedia has a lot of good further info on this, but it is scattered among several articles. Probably http://en.wikipedia.org/wiki/Color_theory#Color_abstractions is the best starting point. For something much more thorough and technical, see Poynton's excellent Color FAQ at http://www.poynton.com/ColorFAQ.html
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Does sending data down a fiber optic cable take longer if the cable is bent? Ok, so, my simplified understanding of fiber optics is that light is sent down the cable and it rebounds off the sides to end up at its destination. Which got me thinking, if it has to bounce more times (and having a shorter travel between each bounce), does the light (data) take longer to get to the other end of the cable? Like this:
http://i.imgur.com/pCHUf.jpg
A part of me is saying no, because it's still the same distance to travel and bouncing doesn't take up any time, but another part of me is saying yes because the light will have further to travel the more times it bounces, and thus will take longer to get to its destination. I'm swaying towards it taking more time.
Thanks!
| That picture is only really true for lightguide type large plastic fibres. For single mode fibre used in communication the wave is essentially directly down the centre
| {
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The force of a phase transition At standard temperature and pressure, I fill a bottle to capacity with $N$ liters of water, then place a weight of mass $M$ kg on its opening to serve as a lid. What values of $N$ and $T$, where $T$ is the temperature of the bottle, are sufficient to raise the lid?
| Easy! Any value of $T$ will suffice. (unless it's ice in a certain temperature range)
Since it's probably a reasonable expectation that you're talking about liquid, subcooled, water for the duration of the problem this is nothing more than multiplication. The mass of the water is invariant from state $1$ to state $2$ at a higher temperature.
$$M = V \rho(T) $$
Then compute the difference in volume, here $\rho_f$ is the density of saturated fluid. That is an approximate way to find the density of water by neglecting the compression effect due to pressure.
$$\Delta V = M_2 - M_1 = V \left( \rho(T_1) - \rho(T_2) \right) \approx \left. V \frac{d\rho_f}{dT} \right|_{T_1}$$
Divide by area to find the distance it rises.
$$\Delta z = \left. \frac{V}{A} \frac{d\rho_f}{dT} \right|_{T_1} $$
This change will be positive provided that the derivative is positive. The derivative is positive for the vast majority of materials and regions. A notable exception is where the density vs. temperature for ice reverses for a small temperature region.
| {
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What happens when a supersonic airplane flies through a cloud? What happens when a supersonic airplane flies through a cloud? Will it punch a hole or is it more like a bullet through water (= hole closes immediately after the aircraft has passed)?
Is there some special effect because of the supersonic speed? Or maybe the question should be: Does the airflow around an airplane change when the sound barrier is broken?
| It creates shock waves, which under the right conditions like a supersonic rocket did in this picture, causes concentric cloud rings. Clouds are essentially just volumes where the humidity, temperature, and pressure are such that the air is locally supersatured with water. The craft passing through the cloud will send out waves that disturb the pressure, changing the saturation and causing visible ripples.
EDIT: For a diagram of what is happening here, see this image.
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In a Sterling Engine, does heat from the hot side transfer to the cold side? A Sterling Engine is a closed system. The "hot" side oscillates between higher temperature with higher pressure and lower temperature with lower pressure. Does Nature switch back and forth between one temperature/pressure combination and another temperature/pressure combination without loss of energy within the closed system?
The International Sterling Engine Society says they can get 60 Watts from a 50 degree temperature differential. Is there the equivalent of 60 Watts of power pulled from the hot tank? If not, where does the 60 Watts come from?
| A Stirling engine moves a fluid from a hot end to a cold end - extracting mechanical work as it does so. The power input comes from maintaining the temperature of the hot end - usually by burning some fuel externally.
The 'clever' part of a Stirling engine, and the thing that gives it it's high efficiency, is that the hot end of the mechanism stays hot and the cold part stays cold, so you don't waste energy constantly re-heating a piston
image from http://en.wikipedia.org/wiki/Stirling_engine - It's an alpha type engine, less common in practice but easier to understand
| {
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Newton's corpuscular theory Where did Newton get the idea that light had a particle nature and not a wave nature? At those times, AFAICT there were no phenomena that showed particle nature. But wave nature is much easier to detect. So, why didn't Newton change his theory to a wave theory? I recall something about "motion like Eels", but by Occam's razor, it makes more sense to just call it a wave and not a wavelike particle.
Was it just his arrogance? Or did he have some reason to stick to particle nature?
Edit: I actually wanted to ask why he didn't change his theory after wave nature was discovered. He instead complicated it with the Eels. I failed to see what supported the particle point. But it's answered now :D.
| Newton gives two main arguments for a corpuscular view of light in his Opticks:
(1) Light consists of rays of inherent and inalterable dispositions (as regards colour, refrangibility, etc.). This is argued for throughout, but see esp. the classic prism experiments in props. I and II. Wave theorists, on the contrary, base their explanations on modifications of rays.
(2) The law of refraction "may be demonstrated upon this Supposition. That Bodies refract Light by acting upon its Rays in Lines perpendicular to their Surfaces" (p. 79 of Dover ed.), i.e., the law of refraction follows by assuming a gravity-like force attracting particles of light toward the heavier medium. (Of course this implies that light speeds up when it is refracted towards the normal, i.e., that light is slowest in vacuum and fastest in dense materials; a fact not experimentally disprovable at the time.)
(1) and (2) are elegantly combined if rays of different colours consist of particles of different sizes.
| {
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Expected Energy Production From High Efficiency Solar Cells First, a bit about my thoughts. I believe we have the capability today to provide energy, water, food, education, and transportation to every man woman and child on the planet. To that end, I would like to become a force that brings about this change.
In trying to meet the first goal, which is to provide energy, I have come across two technologies which greatly interest me, the first of which must be in place to begin the second.
The first is the high efficiency solar cells developed by Patrick Pinhero at the University of Missouri. Assuming that said solar cell captures 80% of available light, how much energy can I expect them to produce per meter of cell? How would this vary betweeen environments such as the Nevada desert and central Florida, how did you come to these conclusions, and is there any formula I can use to calculate an expected energy output?
| A fantastic, free, book for calculations of this kind is provided by David McKay in Renewable energy without the hot air.
| {
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Prove that negative absolute temperatures are actually hotter than positive absolute temperatures Could someone provide me with a mathematical proof of why, a system with an absolute negative Kelvin temperature (such that of a spin system) is hotter than any system with a positive temperature (in the sense that if a negative-temperature system and a positive-temperature system come in contact, heat will flow from the negative- to the positive-temperature system).
| Negative temperature - yes I encountered that once: I seem to recall that it's the state that arises when, say, you have a system of magnetic dipoles in a magnetic field, and they have arrived at an equilibrium distribution of orientations ... and then the magnetic field is suddenly reversed and the distribution is momentariy backwards - basically the distribition given by substituting a negative value of T. Other scenarios can probably be thought of or actually brought into being that would similarly occasion this notion. I think possibly the answer is that the system is utterly out of thermodynamic equilibrium, whence the 'temperature' is just the variable that formerly was truly a temperature, and is now merely an artifact that gives this non-equilibrium distribution when rudely plugged into the distribution formula. So heat is transferred because you now have a highly excited system utterly out of equilibrium impinging upon a system that approximates a heat reservoir. I think there's no question really of accounting for the heat transfer by the usual method, ie when both temperatures are positive, of introducing the temperature difference as that which drives the transfer.
And would it even be heat transfer atall if the energy is proceeding from a source utterly out of thermodynamic equilibrium? It's more that the transferred energy is becoming heat, I would say.
| {
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What do the dimensions of circulation mean, and how is circulation related to action? The dimensions of circulation $\int_C \vec{v}\cdot d\vec{r}$ seem strange, but if you include
(even a constant) density $\rho$, then $\int_C \rho\vec{v}\cdot d\vec{r}$ has dimensions
the same as action/volume. Is there any significance to that? Is there any heuristic way to
think about circulation which helps understand the dimensions?
| The question is "What is the unit of circulation?"
Ans: $(m^2/s)$
Explanation: $\text{circulation} = \text{vorticity}\times\text{area} = (1/s)\times(m^2) =(m^2/s)$
| {
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Can someone explain the science behind MIT's 230% efficient LEDs? I was reading Gizmodo the other day and I didn't quite understand the Physics behind this. Could anybody shed some light on how this effect actually works?
| As others have said, a heat pump with an efficiency greater than unity doesn't have any problem with the laws of thermodynamics. There are still a few problems with this which mean that it isn't as good as it seems.
The efficiency of a heat pump depends on the temperature difference - you can get higher efficiencies when you're just increasing the temperature a bit. Now the effective temperature of white light is the temperature of the sun's surface, so there's no way that this can be a light emitting diode. In fact it emits at a wavelength of about 2.5 microns, which is in the infrared.
However, that still corresponds to a temperature of about 1200 K. The device is heated to 135 degrees C, i.e about 400 K. Thats still too much of a temperature difference for a heat pump to have 230% efficiency, so there's clearly something else going on.
Were they adding in the normal black body radiation from the device? No, they were subtracting it, but the interesting thing is that it is much larger than the emission from the device. In the paper they say that the black body emission in the appropriate wavelength range is about 40 nW, while the emission from the device is 69pW.
This explains how they can have a heat pump with such high efficiency. If you think of the effective temperature for the emission, well 40 nW is 135 Celsius, so 40.069 nW will be a bit more than 135 Celsius. So there's no problem in having a heat pump with 230% efficiency.
| {
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Why are there 4 Dimensions and 4 Fundamental Forces? Is it a coincidence that there are four fundamental forces and four spacetime dimensions ? Does a universe with three spacetime dimension contain four fundamental forces? Can magnetism be realized in three dimensional spacetime?
Edit 1:
Thanks alot for the answer. I'm a first year undergrad student so this may seem naive. If there is a symmetry that mixes spacetime symmetries and internal symmetries of the theory (e.g. gauge symmetry) then there might be ways that spacetime geometry have effects on the particle spectrum and interactions of the theory so you can describe things in geometrical manner . The mechanisms that lead to the structure of spacetime may explain why there are four forces.
| Well, I will not join the chorus in agreeing there are four fundamental forces.
Our everyday world depends on two forces, gravity and electromagnetism, and in this sense they would be fundamental to our everyday world.
When we started the scattering experiments we discovered another two forces, strong and weak. The higher we go in energies the more forces may be discovered, because forces are exchanged particles in Feynman diagrams and none would be more fundamental than another one, imo. There are the grand unified theories for example, where there are a lot of carriers of force similar to a photon or a gluon.
If/when string theory becomes evident, even the four dimensions will no longer be true, as string require eleven dimensions.
Thus yes, it is a coincidence that at this point in time we know four dimensions and have two forces well explored, another two under examination at the LHC and an unknown number N for the future.
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Why does optical pumping of Rubidium require presence of magnetic field? The optical pumping experiment of Rubidium requires the presence of magnetic field, but I don't understand why.
The basic principle of pumping is that the selection rule forbids transition from $m_F=2$ of the ground state of ${}^{87} \mathrm{Rb}$ to excited states, but not the other way around ($\vec{F}$ is the total angular momentum of electron and nucleus). After several round of absorption and spontaneous emission, all atoms will reach the state of $m_F=2$, hence the optical pumping effect.
But what does the Zeeman splitting have anything to do with optical pumping? Granted, the ground state, even after fine structure and hyperfine structure considered, is degenerate without Zeeman splitting, but the states with different $m_F$ still exists.
In addition, how is the strength of optical pumping related to the intensity of magnetic field applied?
| In a hyperfine pumping you pump atoms to the other hyperfine level, let's say you apply your laser to the $F_g = 2 \rightarrow F_e = 2$ transition of the $^{87}$Rb D$_1$, then the atoms from the $F_e = 2$ level will decay to both ground state hyperfine levels $F_g=1,2$ and eventually will be pumped into the $F_g = 1$ level.
In a Zeeman pumping scheme the polarization of the exciting laser becomes important. If you apply let's say $\sigma^+$ polarized laser to the same transition $F_g = 2 \rightarrow F_e = 2$, then magnetic quantum number is changed by $\Delta m = 1$ and the $m_g = +2$ sublevel cannot absorb the exciting radiation as there is no $m_e = +3$ sublevel in the excited state, as atoms from the excited state sublevels $m_e = +1,+2$ can and do sponataneously decay to the sublevel $m_g = +2$ atoms that haven't been pumped to $F_g = 1$ level by the hyperfine pumping are pumped to the $m_g = +2$ by the Zeeman pumping.
What's the role of the magnetic field? I would say it's needed only to detect that you have pumped your atoms into $m_g = +2$ sublevel. As long as all the Zeeman sublevels are degenerate you cannot distiunguish the. Once the magnetic field is applied and the degeneracy is removed, you can use radiofrequency to detect the atoms, if you satisfy equality condition for the Zeeman shifts befor neighbouring magnetic sublevels and anergy of the radiofrequency quantum $\Delta E = \mu_BgB = h\nu_{rf}$, then the radio frequency induces transitions of $\Delta m = \pm 1$ within the ground state hyperfine level and atoms are brought back to the sublevels that can absorb light ($m_g \neq +2$) allowing to observe some fluorescence or changes in absorption.
| {
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Piston movements in four stroke cycle? I was reading about a four stroke cycle. Here's what I understood:
*
*In the first stroke, the piston starts at the top and moves down.
*In the second stroke, the piston moves upwards.
*In the third stroke, the piston moves down due to the combustion by spark plug.
*In the final stroke, the piston moves up and the cycle continues.
I can understand why the piston moves down in third stroke due to the gasoline explosion. But, what moves the piston up and down in Step 1, 2, and 4?
| As Manishearth says, for engines with more than one cylinder the firing of the other cylinders rotates the crankshaft. However, as any fan of vintage motorcycles will know, you can have four stroke engines with a single cylinder. In this case the engine has a heavy flywheel attached to the crankshaft and the momentum of the flywheel keeps the crankshaft turning while it's compressing the petrol/air mixture.
| {
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Does it take significantly more fuel to fly a heavier airplane? I was reading in the papers how some-airline-or-the-other increased their prices for extra luggage, citing increased fuel costs.
Now I'm a bit skeptical. Using the (wrong) Bernoulli-effect explanation of lift, I get this:
More luggage$\implies$more lift needed $\implies$ more speed needed$\:\:\:\not \!\!\!\! \implies$more fuel needed.
At this point, I'm only analysing the cruise situation. When the plane is accelerating, this will come into effect, but more on that later.
Now, I know that the correct description of lift involves the Coanda effect and conservation of momentum, but I don't know it well enough to analyse this. Also, there will be drag forces which I haven't (and don't know how to) factored in. I can see that viscosity must be making a change (otherwise planes wouldn't need engines once they're up there), but I don't know how significant a 1kg increase of weight would be.
So, my question is: Are airlines justified in equating extra baggage to fuel?
Bonus questions:
*
*If more baggage means more fuel, approximately what should the price be for each extra kilo of baggage?
*What happens when we consider takeoff and landing? Does a heavier plane have to use a significantly large amount of fuel?
| One thing in your argument is that more lift, means a higher speed. This may not be what airliners do. Airplanes (at long flights) choosse their cruising altitude based on their weight. Higher weight means lower altitude. I think this should be included in the incremental cost calculation of additional piece of luggage.
First, simple Google hit: http://www.ehow.com/about_4572148_why-do-planes-fly-feet.html . Some of the physics aspects are however mentioned here, that can be used in your derivation.
| {
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Faster-than-$c$ photons As far as I know, according to quantum field theory, there are some photons that go faster than c, which is the speed of light in vacuum.
However, there seems to be a paper and a corresponding experiment that show every photon obeys the speed limit of $c$. (http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.106.243602)
So, my question is:
*
*Is this experiment accepted universally?
*Regardless of the acceptance of the experiment, if every single photon is shown to obey the speed limit of $c$, what does this mean for quantum field theory?
| Special Relativity only requires that light travel at c in vacuum. In any normal dielectric the speed of light will be less than c. This is what gives rise to Cherenkov radiation. https://en.wikipedia.org/wiki/Cherenkov_radiation
If you consider phase and group velocity, the issue is really only straightforward in homogeneous dielectrics. Without going into a lot of details, there exist real media in which the permittivity is negative. Near a resonance of the dispersion relation the phase and group velocity can be anything. In metals, below the plasma frequency the permittivity is negative. Finally, in random media quantities like wavevector and velocity are not well defined. You have to look at average phase and group delay, etc. But I suspect, just the first paragraph is what you're after.
| {
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Magnetic force and work If the magnetic force does no work on a particle with electric charge, then:
How can you influence the motion of the particle? Is there perhaps another example of the work force but do not have a significant effect on the motion of the particle?
| The energy of a freely moving particle is its kinetic energy $E=\dfrac{m v^2}{2}$. If the energy of the particle remains unchanged after someone acts on it, it means that no work has been done. However, the direction of ${\bf v}$ could have changed, provided that $v^2$ and hence $E$ is kept the same.
This means that one can affect the motion of a particle by changing the direction of its velocity and making no work for it.
If you do make work, however, you change $E$ and hence $v^2$. Hence, you must change ${\bf v}$, and affect the motion.
To conclude, the work cannot be done without affecting the motion of a particle.
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A question about definition of Fermi energy Wikipedia states the definition of Fermi energy as for "a system of non-interacting fermions". If we have to assume free electrons in a solid behave this way before we are able to calculate Fermi energy, how can Pauli exclusion be justified (because electrons are non-interacting)? Can Fermi energy be similarly defined for electrons confined to a single atom?
| You might say, of course, that in some sense the fermions do interact (and it is even called exchange interaction). However, it is physical forces, like Coulomb ones, that are understood to be absent. A relevant discussion has taken place here: Degeneracy Pressure, What is it?
Concerning the second question about single atoms, the answer is no. Firstly, Fermi statistics, as any statistics, can only be applied to macroscopic objects. Secondly, even if you were able to create a giant nucleus of a large charge and cover it with a macrosopic number of electrons, the electrons would be interacting with each other through Coulomb forces, hence will not represent a degenerate gas and hence will neither follow the statistical distribution of degenerate gases nor possess Fermi energy.
| {
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Difference in timbre between 'quiet' and 'far away' I'd like to know what are the differences in timbre - or the acoustic properties of a sound - that allow us to differentiate between a sound which is quiet (but close-by) and one which is far away.
For example, you can tell when someone near to you is playing an instrument quietly even without looking to see where they are - they don't sound 'far away'.
Hearing a loud gig or a car stereo playing from the next street doesn't sound like it's quiet - it sounds loud, but far away.
But other times we can't differentiate - I sometimes hear a siren on TV and think it's on the street!
I thought only the amplitude (i.e. volume) of a sound wave diminished with distance - does the shape/frequency alter too?
Is this ability just to do with having two ears to locate the source - surely someone who is deaf in one ear can still tell an orchestra is playing a diminuendo and not gradually getting further away?!
| There are different cues to the perception of distance:
*
*intensity,
*direct-to-reverberant energy (D/R) ratio (decreases with distance),
*spectral balance (reverberations have more low frequencies than the source signal, so when D/R ratio decreases, basses increase. Sound going through walls and windows also lose treble),
*interaural coherence (the closer the source is, the more different the signals to left and right ears are).
These different cues (maybe more, and mainly the two first) are weighted in your brain to give a stable assessment of distance. Research shows that the weighs associated with these cues vary with the type of signal (noise or speech), and angular position of the source [2].
As you guessed, the interaural coherence is not a major cue for distance perception, especially for greater distances, see [1].
sources:
1 http://www.acoustics.org/press/160th/lavandier.html
2 http://www.mendeley.com/research/assessing-auditory-distance-perception-using-virtual-acoustics/
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Slit screen and wave-particle duality In a double-slit experiment, interference patterns are shown when light passes through the slits and illuminate the screen. So the question is, if one shoots a single photon, does the screen show interference pattern? Or does the screen show only one location that the single photon particle is at?
| Let me try a slightly different way to answer this (well worn) question.
The photon doesn't have a location, or at least not a well defined location, until you interact with it and cause it to localise.
When the photon hits the photomultiplier, or photographic plate, or whatever you're using as the screen the interaction occurs at a point and that localises the photon. Until then it's somewhat meaningless to talk about the position of the photon. I don't mean the photon has a position but we don't know it, I mean the photon simply doesn't have a position. That's why it doesn't make sense to ask which slit the photon went through. because the photon's position is ill defined it occupies the whole experimental apparatus.
So a single photon does indeed passthrough both slits, but it then interacts with the screen at a point. The point of interaction with the screen is random, with the probability o the position being given by the square of the wavefunction. That's why over time the pattern created by many phtons gives you the interference pattern.
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Flow rate of a syringe Suppose a syringe (placed horizontally) contains a liquid with the density of water, composed of a barrel and a needle component. The barrel of the syringe has a cross-sectional area of $\alpha~m^2$, and the pressure everywhere is $\beta$ atm, when no force is applied.
The needle has a pressure which remains equal to $\beta$ atm (regardless of force applied). If we push on the needle, applying a force of magnitude $\mu~N$, is it possible to determine the medicine's flow speed through the needle?
| The appropriate equation for laminar flow (i.e., not turbulent) of a
liquid through a straight length $l$ of pipe or tubing is:
$$Flowrate = \frac{\pi r^4 (P - P_0)}{8 \eta l}$$
where $r$ is the radius of the pipe or tube, $P_0$ is the fluid
pressure at one end of the pipe, $P$ is the fluid pressure at the other end of
the pipe, $\eta$ is the fluid's viscosity, and $l$ is the length of the pipe or
tube. In your case $P$ is presumably $\mu$ divided by $\alpha$ and $P_0$ is $\beta$. Make sure you keep the units consistent - your question gives $\beta$ in atmospheres.
The equation is called Poiseuille’s law. Google for this for more details.
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Why don’t photons interact with the Higgs field? Why don’t photons interact with the Higgs field and hence remain massless?
| There is an aspect to this question that nobody seems to have addressed and that is, although the higgs (the 'radial' component of the field) is neutral, and therefore doesn't interact with the photon at 'tree level' we still see the decay $h \rightarrow \gamma \gamma$. This is because, roughly, by quantum effects a higgs will fluctuate into a particle/ anti-particle pair (electrons, quarks etc) which can they produce photons. So while the higgs does not strictly interact with the photon, at low energies we can parameterize a low-energy effective interaction where the higgs does interact with the photon. This is diagrammatically expressed in the Feynman diagrams:
which I have borrowed from http://resonaances.blogspot.com/2012/07/h-day-3-how-to-pump-up-higgs-to-gamma.html.
| {
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Where is the flaw in deriving Gauss's law in its differential form? From the divergence theorem for any vector field E,
$\displaystyle\oint E\cdot da=\int (\nabla\cdot E) ~d\tau$
and from Gauss's law
$\displaystyle\oint E\cdot da=\frac{Q_{enclosed}}{\epsilon_0}=\int \frac{\rho}{\epsilon_0}~d\tau$
Hence,
$\displaystyle\int\frac{\rho}{\epsilon_0}d\tau=\int (\nabla\cdot E)~d\tau$
Textbooks conclude from the last equation that
$\displaystyle \nabla\cdot E=\frac{\rho}{\epsilon_0}$
My question is how can we conclude that the integrands are the same? Because I can think of the following counter example, assume
$\displaystyle \int_{-a}^a f(x)~dx=\displaystyle \int_{-a}^a [f(x)+g(x)]~dx$
where $g(x)$ is an odd function. Obviously the 2 integrals are equal but we cannot conclude that $f(x)$ is equal to $f(x)+g(x)$ so where is the flaw?
| Your counterexample is obviously correct: it is not at all true that, if the integral of a function is the same as that of another function, then the two function coincide.
To mathematically prove the differential form of Gauss' law, if you choose the domain of integration as a paralellepiped $P$ whose sides are $[x_0,x_0+h_x]$, $[y_0,y_0+h_y]$ and $[z_0,z_0+h_z]$ and call $\|h\|=\sqrt{h_x^2+h_y^2+h_z^2}$ the length of the diagonal, by applying what is said here, you can see that$$\lim_{\substack{h\to 0\\h_xh_yh_z\ne 0}}\frac{1}{h_xh_yh_z}\int_{x_0}^{x_0+h_x}\int_{y_0}^{y_0+h_y}\int_{z_0}^{z_0+h_z}\frac{\rho(x,y,z)}{\varepsilon_0}dxdydz=\frac{\rho(x_0,y_0,z_0)}{\varepsilon_0}$$and$$\lim_{\substack{h\to 0\\h_xh_yh_z\ne 0}}\frac{1}{h_xh_yh_z}\int_{x_0}^{x_0+h_x}\int_{y_0}^{y_0+h_y}\int_{z_0}^{z_0+h_z}(\nabla\cdot E)(x,y,z)dxdydz=(\nabla\cdot E)(x_0,y_0,z_0)$$Therefore, since $$\int_{x_0}^{x_0+h_x}\int_{y_0}^{y_0+h_y}\int_{z_0}^{z_0+h_z}\frac{\rho(x,y,z)}{\varepsilon_0}dxdydz$$$$=\int_{x_0}^{x_0+h_x}\int_{y_0}^{y_0+h_y}\int_{z_0}^{z_0+h_z}(\nabla\cdot E)(x,y,z)dxdydz$$you have the thesis.
| {
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Is the change in kinetic energy of a particle frame independent? Intuitively, I would expect the change in kinetic energy of a particle to be frame independent. It just doesn't "feel" right that between two points in time-space, one frame should measure a change in kinetic energy of a particle different to another frame.
Is my intuition right?
Is the change in kinetic energy of a particle frame independent?
| The answer is No, the change in kinetic energy depends on the frame of reference. Consider e.g. a single non-relativistic particle measured relative to two different initial frames.
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Neutrino beam energy Neutrino is one of the most mysterious particles in todays physics. Even when values of some parameters like for example mass associated with it are not known (or there is great range of possible values), its existence is not questioned. There is a field of research named neutrino astronomy. Because neutrino interaction with ordinary matter is very weak, sensitivity of detectors is minuscule. The only natural sources of neutrinos we could detect are sun and SN1987A supernova that exploded in year 1987. I am curious what is the total energy of neutrinos flowing to earth from the Sun? For convenience I want this value as power density [W / m2]. For comparison electromagnetic energy flowing to Earth from Sun in upper limits of atmosphere in equator areas is about 1360 W / m2.
| http://www.ncbi.nlm.nih.gov/pmc/articles/PMC33947/table/T1/ gives a table of neutrino number density for various sources and energies. The flux of solar neutrinos is 5 x 10$^{10}$cm$^{-2}$s$^{-1}$, i.e. 5 x 10$^{14}$m$^{-2}$s$^{-1}$, and the energy per neutrino is 10$^7$eV. 1eV is about 1.6 x 10$^{-19}$J, so I make that about 800Wm$^{-2}$.
Plot of the solar neutrino spectrum (courtesy of the KamLAND collaboration):
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Why do books have dog ears? I googled the question and found no explanation. It seems that dog ears are inevitable (for paperbacks, notably) even if you've always been careful. From my experience, they are about equally likely to appear on the top corners as on the bottom corners (for both the beginning pages and the ending ones). Dog ears for the middle pages of the book are less likely but they can also appear in frequently used old books. Can someone explain why?
I apologize if this is not the right kind of question to post here. I can find no other sites on SE for it.
| Paper is made out of cellulose fibers which are bound together by small amounts of glue under significant heat and pressure in the paper mill. Under a scanning electron microscope the surface of a sheet of paper looks like a haystack that has been smashed flat by great pressure.
As long as the glue (called binder) retains its grip on the deformed cellulose fibers, the paper sheet is stable i.e., its dimensions (including its thickness!) do not change. But mechanical flexure gradually breaks down the structure of the paper sheet and the squashed cellulose fibers loosen up and want to relax back to their unsquashed state.
In a book, the portion of the sheets that get flexed the most are the corners, which slowly grow in thickness as the cellulose comes unglued. Moisture and finger grease make this swelling worse, and high quality paper resists this better than low grade paper.
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Why does a glass rod when rubbed with silk cloth aquire positive charge and not negative charge? I have read many times in the topic of induction that a glass rod when rubbed against a silk cloth acquires a positive charge. Why does it acquire positive charge only, and not negative charge?
It is also said that glass rod attracts the small uncharged paper pieces when it is becomes positively charged. I understand that a positively charged glass rod attracts the uncharged pieces of paper because some of the electrons present in the paper accumulate at the end near the rod, but can't we extend the same argument on attraction of negatively charged silk rod and the pieces of paper due to accumulation of positive charge near the end?
| You might know that all matter is made up out of atoms. Now, atoms themselves have a core, or nucleus, and electrons orbiting around the nucleus. The core has positive charge, the electrons have negative charge.
When you are rubbing the glass rod with the silk cloth, electrons are stripped away from the atoms in the glass and transferred to the silk cloth. This leaves the glass rod with more positive than negative charge, so you get a net positive charge.
Why do the electrons go from glass to silk and not from silk to glass? That depends a lot on the minute details of the material. Ultimately, for every two materials there is one of them where electrons are happier. It just turns out that for glass and silk, electrons are happier at the silk cloth.
Now to your second question. Here, the important thing to note is that in your typical solid material, the positive charges, which are the cores of the atoms, cannot move around much. They are locked into a rigid structure. The tiny electrons, however, can move around much better. That's why the glass rod can induce a net negative charge at one end of the paper clips.
EDIT: Let me add that there should also be some attraction between the silk and a bunch of paper: The electrons in the paper will be pushed away by the electrons in the silk, leaving the end of the paper that is closer to the silk with a net positive charge that then gets attracted. However, it might very well be that in your silk cloth the electrons are overall too spread out to have a strong enough attractive effect.
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What would happen to electronic circuits when traveling near the speed of light? Imagine a space ship, loaded with all sorts of computer systems, traveling near the speed of light.
Electricity itself is very fast, and can reach speeds close the speed of light. (up to 99% according to wikipedia). So, what would happen to the electronic circuits in this spaceship?
Will the computers shut down, because electricity can't reach the components? Or are they just not related to each other and will the computers keep working perfectly?
| You're dealing with an incomplete form of relativity.
In the frame of the spaceship, nobody will notice anything different, since all inertial frames are equivalent
In the "ground" frame, electricity would be moving at a different speed, by the relation $$\rm v_{e,ground}=\frac{v_{ship}+v_{electricity}}{1+\frac{v_{ship}v_{electricity}}{c^2}}$$
We cannot simply use relative velocities, we need to use the above equation. If you compare this with the time dilation of the system, the computers will all seem to be working the same, albeit slower, from your POV.
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How to combine the error of two independent measurements of the same quantity? I have measured $k_1$ and $k_2$ in two measurements and then I calculated $\Delta k_1$ and $\Delta k_2$. Now I want to calculate $k$ and $\Delta k$.
$k$ is just the mean of $k_1$ and $k_2$. I thought that I would need to square-sum the errors together, like so:
$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$
But if I measure $k_n$ $n$ times, $\Delta k$ would become greater and greater, not smaller. So I need to divide the whole root by some power of $n$, but I am not sure whether $1/n$ or $1/\sqrt n$. Which is it?
| The formula you've specified
$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$
is the formula to obtain error of quantity $k$, as being dependent on $k_1$ and $k_2$ according to the following expression
$$ k = k_1 + k_2.$$
Generally, to calculate experimental error of a dependent quantity (and the expression stated in your question), you start with the expression for dependent quantity
$$k = f(k_1, k_2, ...)$$
and use statistical expression
$$\Delta k = \sqrt{\sum_i \left(\frac{\partial f}{\partial k_i} \Delta k_i \right)^2}.$$
If
$$k = \frac{k_1 + k_2}{2}$$
then
$$ \Delta k = \frac{\sqrt{(\Delta k_1)^2 + (\Delta k_2)^2}}{2} $$
So the generalized answer might be: you have to divide by $n$ and not $\sqrt{n}$.
However, bear in mind that the statistical expression is applicable only when the measured quantities are "independent" of each other. If $k_1$ and $k_2$ are the same quantity measured in two measurements, this is not exactly true, and the exact statistical expression is much more complicated.
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Is there a general physics simulator for learning purposes? Is there a complete physics simulator that I can use to do general simulations for learning purposes? For example:
*
*Create a sandbox.
*Fill with a gas.
*Load a 3d solid model like this (but 3d).
*Fill it with a dense liquid.
*Load gravity.
*Watch, measure and understand how a barymeter works.
It doesn't need to be precise, just usable, so I guess it is not impossible. The point would be to simulate and visualize any kind of exercise you would find in your physics book. It would be the mother of the learning tools. If it doesn't exist, is anybody interested in programming it?
| The closest program to the description is Phun:
http://phun.en.softonic.com/
Download it, it's a lot of fun.
Oh, I see, you want 3D immediately. Ambitious enough so that I won't erase my answer.
Update, May 2012. You may try to download trial of Wolfram System Modeler,
http://www.wolfram.com/system-modeler/
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How to derive the Manley-Rowe relation for the process of the second harmonics generation? Derive the Manley-Rowe relation for the process of the second harmonics generation.
Manley-Rowe relation: ~ The Manley-Rowe relations are mathematical expressions developed originally for electrical engineers to predict the amount of energy in a wave that has multiple frequencies.
Second-harmonic generation: ~ Second harmonic generation (SHG); also called frequency doubling; is a nonlinear optical process, in which photons interacting with a nonlinear material are effectively “combined” to form new photons with twice the energy, and therefore twice the frequency and half the wavelength of the initial photons. It is example of nonlinear phenomena (I, 3). In the SHG process, the intense wave at the frequency $\omega$ propagates in the medium with second-order nonlinearity ( VI, 1).
| The Manley-Rowe relation arises from conservation of energy and momentum. For the case of SHG, the presence of the nonlinear optical (NLO) material eliminates the conservation of momentum (any momentum difference between the initial and final photons can be provided by the bulk material). So what's left is conservation of energy.
Let $N_\omega$ and $N_{2\omega}$ be the number of the fundamental frequency and the second harmonic. Usually NLO people care about how these sorts of things change with the distance that the wave moves through the material. So let the direction of propagation be $x$. Then, by energy conservation:
$$2\frac{dN_{\omega}(x)}{dx} + \frac{dN_{2\omega}(x)}{dx} = 0.$$
This follows from $E=\hbar\omega$. That is, it takes two of the $\omega$ photons to provide the energy in one $2\omega$ photon.
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Why would it be true that people with longer legs walk faster than ones with shorter legs? When a person walks, the only force acting on him is the force of friction between him and the ground (neglecting air resistance and all). The magnitude of acceleration due to this force is independent of the mass of the object (longer legs have more mass). Hence the person should move with with a velocity independent of the length of his legs.
But I have heard (also observed) that people with longer legs walk faster than ones with shorter legs. If that is true, then why?
One can argue that the torque about the pivot due to friction is more in case of longer legs, But then the torque due to gravity (when one raises his leg to move), which opposes the frictional torque, is also more for longer legs. And why would these torques make a difference anyway, as they have no effect on the acceleration of the center of mass?
| Think about the limiting cases. An ant-sized marching band would take a long time to march the length of a football field. The reason they take so long has nothing to with friction - it's just that their legs are smaller and so each stride moves them a shorter distance.
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The equivalent electric field of a magnetic field I know that Lorentz force for a charge $q$, with velocity $\vec{v}$ in magnetic field $\vec{B}$ is given by
$$\vec{F} =q \vec{v} \times \vec{B}$$
but there will exist a frame of reference where observer move at same velocity with that of charge $q$, so according to him $v=0$. hence he will see no magnetic force is exerted on charge $q$. I have work on this problem for a while and found that the special relativity predicts equivalent electric force will acting upon charge instead. I want to know the relationship between this equivalent electric force and magnetic force.
Thanks in advance
| We can write the Lorentz transform of the fields in a very clean and easy to understand way.
To simplify the expression we use a short hand notation for the various components of the fields parallel and orthogonal to the boost $\vec{\beta}$, further simplified by setting $c$ to $1.$
Lorentz transform of the electromagnetic field
$$
\begin{array}{lclclcl}
\mathsf{E}' & = & \mathsf{E}_\| & + & \mathsf{E}_\bot\ \gamma & + & \mathsf{B}_\otimes\ \beta\gamma \\[5px]
\mathsf{B}' & = & \mathsf{B}_\| & + & \mathsf{B}_\bot\ \gamma & - & \mathsf{E}_\otimes\ \beta\gamma
\end{array}
$$
The parallel and orthogonal components are defined, using the unit vector $\hat{\beta}$, as:
$$\begin{array}{lcll}
\mathsf{E}_\| &=&
\left(~ \hat{\beta}~\cdot~\mathsf{E}~ \right)~\hat{\beta} &
\mbox{parallel component with regard to $\vec{\beta}$} \\[5px]
\mathsf{E}_\bot &=&
\left(~ \hat{\beta}\times \mathsf{E}~ \right)\times\hat{\beta} &
\mbox{orthogonal component with regard to $\vec{\beta}$} \\[5px]
\mathsf{E}_\otimes &=&
\left(~ \hat{\beta}\times \mathsf{E}~~ \right) &
\mbox{$90^o$ rotated orthogonal component}
\end{array}$$
So in words:
*
*The fields parallel to the boost don't change
*The fields orthogonal to the boost are multiplied with $\gamma$
*The $E$ and $B$ fields fields orthogonal to the boost are converted into each other.
For more see this chapter from my book (PDF).
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A charged particle moves in a plane subject to the oscillatory potential A charged particle moves in a plane subject to the oscillatory potential:
$U(r)=\frac{m\omega^2 r^2}{2}$
There is also a constant EM-field described by:
$\vec{A}=\frac{1}{2}[\vec{B}\times\vec{r}]$
where B is normal to the plane.
This produces the Lagrangian:
$L=\frac{m}{2}\dot{\vec{r}}^2+\frac{e}{2}\dot{\vec{r}}\vec{A}-U(r)$
Now my friend says we need to transform this into polar coordinates and that produces:
$L=\frac{m}{2}(\dot{r}^2+r^2\dot{\phi}^2)-mr^2\omega_L\dot{\phi}-U(r)$
where $\omega_L$ is the Larmor precession frequency:
$\omega_L=-\frac{eB}{2mc}$
My question is, How does he get this transformation? I don't really understand where the second term is coming from in the mechanical kinetic energy.
| In polar coordinates $d\vec{r}=\hat{e}_r dr+\hat{e}_{\phi}rd\phi$. Devide it by $dt$ and you will have the particle velocity $\dot{\vec{r}}$. Square the latter and you will get the kinetic energy.
| {
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In the known universe, would an atom not present in our periodic table exist? I have watched this movie Battleship. In it the researchers say this piece of metal is alien because we cant find this metal on earth.
So that would mean somewhere else in the universe any of the following should be true?
*
*Atoms' composition is not similar to that as on earth (nucleus, electrons, anything else)
*Elements with atomic numbers above 120 or 130 are stable (highly impossible without point 1)
*The realm itself is observed by different binding forces (but then, once that elements realm has changed, it should become unstable and collapse)
| Metallic hydrogen is a metal that's not found on earth (but may be present in Jupiter): http://en.wikipedia.org/wiki/Metallic_hydrogen
Wether it does anything but evaporating or burning at ambient temperatures and pressures (or whatever conditions those aliens encountered in this movie), I don't know.
Since metals a generally in the lower left corner of the periodic table, I see no other candidate except transuraniae.
Of course, the alien metal could be handwavium, a material traditionally used for filling up plot holes.
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Why do objects follow geodesics in spacetime? Trying to teach myself general relativity. I sort of understand the derivation of the geodesic equation $$\frac{d^{2}x^{\alpha}}{d\tau^{2}}+\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\tau}\frac{dx^{\gamma}}{d\tau}=0.$$ which describes "how" objects move through spacetime. But I've no idea "why" they move along geodesics.
Is this similar to asking why Newton's first law works? I seem to remember reading Richard Feynman saying no one knows why this is, so maybe that's the answer to my geodesic question?
| It is related to what Einstein called "the happiest thought of my life", that for an observer falling freely from the roof of a house, the gravitational field does not exist.
If we could choose a system of coordinates and a suitable definition of derivative, so that the acceleration (the derivative of velocity) were zero for that case, the feeling of the person in free fall would match the maths.
GR does that job. The system of coordinates is defined by the metric tensor, and the geodesic equation is nothing more than the result of setting the covariant derivative of the 4-velocity to zero.
If we make an analogy from the 4-dimensional spacetime to a 2-dimensional surface of the earth, the change of perspective is similar to realize that the airplane journey from Tokyo to Paris is a curve if observed in a common world atlas of the airline magazine. But it is "straight" if we join the 2 cities by a string in a world globe.
| {
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Relative Change of Volume Simple question, in materials publications I often see the relative change of volume in a system reported as
$$ \Delta \left (V \right )/V $$
is the denominator volume supposed to be initial or the final volume? I would assume it is the final volume as it likely parallels the relative error calculation, but I'd like to make sure.
-- It occurs to me that this is actually very likely dependent on the situation, still input would be appreciated.
| Usually, as for example in the formula that estimates volumetric thermal expansion
$$\frac{\Delta V}{V} = \beta \Delta T$$
$V$ represents initial volume.
Actually, the real definition of volumetric thermal expansion coefficient $\beta$ is stated in the differential form
$$\frac{\text{d} V}{V} = \beta \text{d} T,$$
which means that the first expression is only an integrated version of the second one under an assumption that $\beta$ is temperature independent (and that $V$ is not a variable but initial volume). Since such coefficients are constant only for very small temperature ranges, obviously $\Delta V \ll V$, so it is almost irrelevant whether $V$ represents initial or final volume.
If however, if $\Delta V \approx V$, it would be IMHO appropriate from the author to explicitly specify which volume is represented by $V$.
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How does the process of freezing water remove salt? How does freezing water to make ice remove whatever salts were in the water to begin with?
| In simple terms, there isn't any space in the ice crystal lattice for the extra atoms and there is no way to plug either of the ions (or the whole salt molecule) into the growing pattern.
So more and more water joins the frozen mass, leaving a more and more concentrated brine until essentially all the water is frozen and the salt remains behind. As Manishearth notes in the comments this requires getting things rather colder than the usual "freezing point" of water.
| {
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noise level (units confusion) i had a question in one of my classes regarding SNR in underwater acoustic channels. There are a couple of terms with the unit dB re uPa. I know it stands for dB with reference to uPa but I am not exactly sure what it means. Can I convert it to dB. If yes, how?
Thanks in advance!!
| dB$\mu$Pa would mean "dB's with respect to 1 $\mu$Pa".
If I have a pressure of X Pascals, then to express it in dB$\mu$Pa, I would compute
$$20log_{10}(\frac{X}{1\mu Pa})$$
i.e
$$20log_{10}(10^6X)$$
since there are $10^6$ $\mu$Pa per Pa.
So the rule is: compute how many micropascals you have and take $20log_{10}$ of that number.
For the second part of your question, "converting it to dB", you would have to specify exactly which dB quantity you were interested in. Wikipedia lists a rather large number of options.
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Is the converse of Noether's first theorem true: Every conservation law has a symmetry? Noether's (first) theorem states that any differentiable symmetry of the action of a physical system has a corresponding conservation law.
Is the converse true: Any conservation law of a physical system has a differentiable symmetry of its action?
| I do not know how to prove the following but it should answer your question factually at least. The following I quote from the book 'Classical Mechanics' by Goldstein-
"It should be remarked that while Noether's theorem proves that a continuous symmetry property o a Lagrangian density leads to a conservation condition, the converse is not true. There seems to be conservation conditions that cannot correspond to a symmetry property. The most prominent examples at the moment are the fields that have soliton solutions, e.g. , are described by th sine-Gordon equation or the Korteweg-deVries equation."
I hope this answers your question.
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Worlds branching or diverging in Many-Worlds Interpretation? In recent years there seems to have been a growing discussion surrounding MWI's ontology.
In the 2010 volume "Many Worlds?", Simon Saunders has a chapter dedicated to discussing whether the worlds in MWI branch or diverge.
The branching view states that before a measurement there is 1 observer in 1 world, after the measurement there is 2 observers in 2 worlds.
The divergent view states that there were 2 observers and worlds all along, but they were identical up to the point of differentiation, so there were no way for you to tell which world you were really in.
Alastair Wilson has written several papers on this in the last years: http://alastairwilson.org/
Any MWI proponents got any opinion either way?
| The two interpretations only differ in ontology. Neither is bullshit, rather both are standard MWI. When you have an observer, this is a computational entity, and to map it to a wavefunction is a nontrivial task. Whether you decide to say there were "really" two observers that were exactly the same before the measurement or whether there was "really" only one observer which branched is a meaningless question in the sense of Carnap, and has no answer, not even philosophically.
The important principle is that the "number" of consciousnesses is determined by interaction with other entities. If you have a computer running a simulation of a conscious observer, and you decide to do a double-check step on each computational step, do you get a second observer? If you do the double-checking in a different computer, is it a new observer? Observationally, you only get a new observer when you talk to two different people, and this only happens after the two computations are distinct.
This question is entirely about philosophy of mind, and this is where computational and positivist formulation is most important.
| {
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Will a stone thrown in space move forever? If I throw a stone in space, in a place where gravity is equal zero, and the space had no end, and no objects to collide with, will the stone move forward forever, because no air, so no friction?
| According to Newton's first law, yes. The velocity of any object will remain constant if no forces affect it. That holds in any Inertial frame of reference (if you are accelerating by yourself, then the stone will be accelerating relative to you, even if no forces act upon it).
| {
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How to create unusual sundial? I am considering small "artsy" project. I would like to create sundial by placing gnomon on the window and painting hour lines on the window facing wall.
Since this is to be placed in bedroom I am constrained by my geographic location, wall, window placement and orientation. The esthetic and size (I have one wall only) of the project is gating factor (otherwise known as "the wife" factor).
Due to above I am perfectly OK with the fact that this sundial will "work" by limited time of the day and even limited time of the year. However whenever it will work (that is the shadow of the gnomon will be cast on the "said" wall) I would like it to be as accurate as possible.
Also all of the above make the calculations for creating hour lines quite challenging (at least for me) and to be honest I do not know where to start.
Could you point me the resources that would help in calculating hour lines (software, tutorials, books, math equations)?
Could you describe how would you approach the task of calculating hour lines?
(Resources that are little heavy on math side are OK for me. I am also capable of wring software on my own.)
| Look at this web page, it makes something similar to what you are looking for, http://sundial.damia.net
| {
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What causes millisecond pulsars to speed up? Millisecond pulsars are supposed to be old neutron stars. However, they are spinning even more rapidly than newly formed pulsars. Since pulsars slow down as they age, something must have caused these older pulsars to "spin up" and be rotating as fast as they are. What is the mechanism for doing so?
| They accrete gas from a disk, fed by either a wind or Roche lobe overflow from their companion. Almost all known millisecond pulsars are in binary systems, but I think some in globular clusters may have been disrupted by three-body encounters, so appear to be isolated.
| {
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Optimal Angular Field of View (AFOV) Given the rather huge price differences between eye pieces at the same focal length. How exactly does the AFOV affect the view seen through the eyepiece?
Are higher / lower AFOV better for certain situations? or is higher always better?
| Simply, a larger FOV provides a wider hunk of sky in the eyepiece. This is useful for capturing entire objects at higher magnifications than could be obtained with cheaper eye-pieces. It doesn't provide more light, just a wider view. As for better, if you've gone to the expense of buying one for a given focal length, it'll still work just as well as something cheaper so may as well use it.
| {
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