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Why can't "missing mass" (=dark matter) be photons? After a star lives and dies, I assume virtually all of its mass would be photons. If enough stars have already lived and died, couldn’t there be enough photon energy out there to account for all the "missing mass" (=dark matter) in the universe?
And if there were enough photons to account for all the missing mass, what would it look like to us?
| Photons are easily detectable. We can count how many photons are there at any distance of us by just counting the photons reaching us from there. It is impossible that the hidden photons ramble the whole universe but mysteriously avoid us.
| {
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Why is temperature constant during melting? This is an elementary question but I do not know the answer to it. During a phase transition such as melting a solid to a liquid the temperature remains constant. At any lower temperature the heat provided went to kinetic energy and intermolecular potential energy. Why is it that at the melting point no energy goes into kinetic (that would increase the temperature)?
| Imagine a container containing just ice at $-1^\circ \rm C$. When you heat it, the energy goes into kinetic motion of the molecules, and its temperature increases. Similarly, if the container is filled with liquid water at $1^\circ \rm C$ its temperature will increase for the same reason.
But now imagine the container is filled with 90% ice and 10% water at $0^\circ \rm C$. If you heat the water part up, it's temperature will temporarily increase a little. But now the water is hotter than the ice, so heat will be transferred from the water to the ice. When the ice is heated above $0^\circ \rm C$ it melts, and this uses up some energy, cooling the water. This will continue until the ice and the water are the same temperature again, so you'll end up back at $0^\circ \rm C$, but with a higher proportion of liquid water and less ice.
This is why, if you heat a mixture of the two phases slowly enough, all the energy will go into melting the solid rather than increasing the temperature. It continues until all the solid has melted, which is when the temperature starts increasing again. The same thing happens in reverse if you decrease the temperature.
| {
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Other Gross-Neveu like theories? By "Gross-Neveu like" I mean non-supersymmetric QFTs whose partition function/beta-function (or any n-point function) is somehow exactly solvable in the large $N_c$ or $N_f$ or 't Hooft limit.
(..supersymmetric examples would also be helpful to know of if in case there are no other theories like the above..)
| There are many quantum field theory models which are exactly solvable in the Large $N$ limit, such that the $\mathbb{C}P^N$ model, the Thirring model, the $O(N)$ vector model etc. Please see the following review by Moshe Moshe and Jean Zinn-Justin covering many of these models. The main idea is that Feynman diagrams (for example the vacuum diagrams in the case of the partition function) are proportional to certain powers of $N$ depending on the number of vertices, lines and loops, and the leading order diagrams can be summed. There are other methods which lead to the same results such as variational computations. When the fields in the model belong to the fundamental vector representations of $O(N)$ or $U(N)$, the computation of the large $N$ limit (summation of the leading diagrams) is quite easy, however, when the fields belong to the adjoint representation (for example, the gluons in QCD), the analysis becomes more complicated. The case of large $N$ QCD was solved by t'Hooft, please see the following review by Aneesh Manohar.
| {
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Effect of gas or liquid within a compound lens system Hi my question is if a compound lens system if filled with gas or a liquid how does it affect the system when compared to the lens system being separated by air alone. Does this affect the focal power of the system or the effective power at all.
| One of the Gaussian equations for a compound lens system describe this :
$$
\phi_{\text{tot}}=\phi_1 + \phi_2 - \phi_1\phi_2\tau
$$
where $\tau = \frac{t}{n}$ is the reduced distance between the lenses (or lens systems), $n$ is the index of refraction of the medium between them (the gas or liquid which you describe), and the $\phi$s are the powers where
$$
\phi = \frac{1}{f_e}
$$
where $f_e$ is the effective focal length of each respective system or lens.
You can see from the above, for a compound system with two positive powers, the power of the combined system will increase if you increase $n$.
| {
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Do new universes form on the other side of black holes? I have four questions about black holes and universe formations.
*
*Do new universes form on the other side of black holes?
*Was our own universe formed by this process?
*Was our big bang a black hole seen from the other side?
*Are there solid reasons why this might not be the case?
| It's not exactly a duplicate, but have a look at my answer to Entering a black hole, jumping into another universe---with questions.
For certain types of black holes it's possible to find a trajectory that takes you inside the event horizon then back out again, but when you emerge you'll find there is no way (without travelling faster than light) to get back to where you started.
The question is whether this counts as another universe (I would say not) or indeed whether the trajectory is physically realistic or not (Luboš would say not and I don't know enough about the subject to comment!).
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Evaluating propagator without the epsilon trick Consider the Klein–Gordon equation and its propagator:
$$G(x,y) = \frac{1}{(2\pi)^4}\int d^4 p \frac{e^{-i p.(x-y)}}{p^2 - m^2} \; .$$
I'd like to see a method of evaluating explicit form of $G$ which does not involve avoiding singularities by the $\varepsilon$ trick. Can you provide such a method?
| Before answering the question more or less directly, I'd like to point out that this is a good question that provides an object lesson and opens a foray into the topics of singular integral equations, analytic continuation and dispersion relations. Here are some references of these more advanced topics: Muskhelishvili, Singular Integral Equations; Courant & Hilbert, Methods of Mathematical Physics, Vol I, Ch 3; Dispersion Theory in High Energy Physics, Queen & Violini; Eden et.al., The Analytic S-matrix. There is also a condensed discussion of `invariant functions' in Schweber, An Intro to Relativistic QFT Ch13d.
The quick answer is that, for $m^2 \in\mathbb{R}$, there's no "shortcut." One must choose a path around the singularities in the denominator. The appropriate choice is governed by the boundary conditions of the problem at hand. The $+i\epsilon$ "trick" (it's not a "trick") simply encodes the boundary conditions relevant for causal propagation of particles and antiparticles in field theory.
We briefly study the analytic form of $G(x-y;m)$ to demonstrate some of these features.
Note, first, that for real values of $p^2$, the singularity in the denominator of the integrand signals the presence of (a) branch point(s). In fact, [Huang, Quantum Field Theory: From Operators to Path Integrals, p29] the Feynman propagator for the scalar field (your equation) may be explicitly evaluated:
\begin{align}
G(x-y;m) &= \lim_{\epsilon \to 0} \frac{1}{(2 \pi)^4} \int d^4p \, \frac{e^{-ip\cdot(x-y)}}{p^2 - m^2 + i\epsilon} \nonumber \\
&= \left \{ \begin{matrix}
-\frac{1}{4 \pi} \delta(s) + \frac{m}{8 \pi \sqrt{s}} H_1^{(1)}(m \sqrt{s}) & \textrm{ if }\, s \geq 0 \\
-\frac{i m}{ 4 \pi^2 \sqrt{-s}} K_1(m \sqrt{-s}) & \textrm{if }\, s < 0.
\end{matrix} \right.
\end{align}
where $s=(x-y)^2$.
The first-order Hankel function of the first kind $H^{(1)}_1$ has a logarithmic branch point at $x=0$; so does the modified Bessel function of the second kind, $K_1$. (Look at the small $x$ behavior of these functions to see this.)
A branch point indicates that the Cauchy-Riemann conditions have broken down at $x=0$ (or $z=x+iy=0$). And the fact that these singularities are logarithmic is an indication that we have an endpoint singularity [eg. Eden et. al., Ch 2.1]. (To see this, consider $m=0$, then the integrand, $p^{-2}$, has a zero at the lower limit of integration in $dp^2$.)
Coming back to the question of boundary conditions, there is a good discussion in Sakurai, Advanced Quantum Mechanics, Ch4.4 [NB: "East Coast" metric]. You can see that for large values of $s>0$ from the above expression that we have an outgoing wave from the asymptotic form of the Hankel function.
Connecting it back to the original references I cited above, the $+i\epsilon$ form is a version of the Plemelj formula [Muskhelishvili]. And the expression for the propagator is a type of Cauchy integral [Musk.; Eden et.al.]. And this notions lead quickly to the topics I mentioned above -- certainly a rich landscape for research.
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Canonical Commutation Relations Is it logically sound to accept the canonical commutation relation (CCR)
$$[x,p]~=~i\hbar$$
as a postulate of quantum mechanics? Or is it more correct to derive it given some form for $p$ in the position basis?
I understand QM formalism works, it's just that I sometimes end up thinking in circles when I try to see where the postulates are.
Could someone give me a clear and logical account of what should be taken as a postulate in this regard, and an explanation as to why their viewpoint is the most right, in some sense!
| The choice of postulates is somewhat arbitrary in the sense that given a set of postulates you almost always can find an alternative set. The choice is guided by subjective criteria such as simplicity, closeness to experiment, or theoretical elegance.
However there are situations where some postulates/theorems do not make sense. For instance, $[\hat{x},\hat{p}] = i\hbar$ makes no sense in the Wigner & Moyal formulation of quantum mechanics, neither as postulate nor as theorem, because this formulation of quantum mechanics does not use operators:
The chief advantage of the phase space formulation is that it makes quantum mechanics appear as similar to Hamiltonian mechanics as possible by avoiding the operator formalism, thereby "'freeing' the quantization of the 'burden' of the Hilbert space.
Although the phase space formulation of quantum mechanics does not use commutation relations, them can be still obtained as a theorem when one makes the transition from the general phase space state to the configuration space wavefunction: $W(p,x;t) \rightarrow \Psi(x;t)$. Precisely, an explicit derivation of the $[\hat{x},\hat{p}] = i\hbar$ is given in my paper Positive definite phase space quantum mechanics
| {
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How can Sub-Atomic Particles be Visualized? Can you see or accurately visualise sub atomic particles or are they known only by maths and/or inference?
| We can image the sub-structure of nucleons by a number of different techniques involving high energy scattering. The results are generally presented in terms of "parton distribution functions" or "structure functions".
One such experiment that I had some small relationship with (though not enough to be an author) was NuSea (E866) at Fermilab in the mid 1990s, which used muon Drell-Yan as the a probe to image the QCD sea in particular.
| {
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Averaging decibels Wikipedia:
The decibel (dB) is a logarithmic unit that indicates the ratio of a
physical quantity (usually power or intensity) relative to a specified
or implied reference level.
If I measure some physical quantity in decibels, then what is the preferred way to calculate the mean of the measured values? Is it enough to simply average them, or should I convert them back to linear scale, calculate the average, and convert it back to decibels (example)? When should I use which approach, and why?
| The "physically natural" quantity to average is the actual power, or energy, but it depends exponentially on the number of decibels. So if you were averaging the power or energy, the result would be pretty much equal to the power or energy of the largest (loudest) reading in decibels.
So even though it's physically less natural, you probably want to compute the average number of decibels itself. But as you said, it's abuot "preferences". Your question isn't a question about observables, it's about subjective choices, so there can't of course be any "only correct and objective" answer. For various applications, various averages may be more or less useful or representative.
| {
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Conservation of Energy in a Capacitor Consider a parallel-plate capacitor in free space. A negatively charged point particle with initial velocity $v$ passes through the space between the pair of parallel plates (with an initial path perpendicular to the normal vector of the plates).
The point particle accelerates towards the positively charged plate but passes beyond the edge of the plate.
How is energy conserved, given that the capacitor does work on the particle by accelerating it in the direction towards the negatively charged plate?
EDIT: Was reminded by Art Brown that a negatively charged particle accelerates towards the positive plate.
| The point charge $q$ moves in a potential field $\phi$ (generated by the capacitor), so the point charge has potential energy $U=q \phi$. It is accelerated by a force $\boldsymbol{F}$ along the gradient of that potential ($\boldsymbol{F}=q \boldsymbol{E}= -q \boldsymbol{\nabla} \phi$). For any such situation, you can show from the equation of motion ($\boldsymbol{F}=m\boldsymbol{a}$ in Newtonian dynamics) that the sum of the point charge's kinetic and potential energies is constant at all times.
In particular, if the potential field goes to 0 at infinity in every direction, a particle starting at infinity with speed $v$ ($v = $ magnitude of the velocity) that eventually escapes to infinity must end up with the same speed (assuming no collisions en route, as assumed by the problem).
(By the way, a negatively charged particle will be attracted to the positively charged plate.)
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Beyond WKB approximation for energies In the first term the energies are given by the Wentzel–Kramers–Brillouin (WKB) formula
$$ \oint p dq = 2\pi \left( n+\frac{1}{2} \right) $$
However, can this formula be improved to include further corrections? For example the wave function in the WKB approach can be evaluated to several orders of magnitude in $ \hbar $
| In the following, I'll describe to you how in principle one can compute higher order corrections to the Bohr-Sommerfeld condition.
In order to find higher order corrections to the Bohr-Sommerfeld formula, we need to include higher order corrections of the wavefunction of the form:
$\Psi(x) = \sum_n \hbar^n a_n(x) exp(\frac{i}{\hbar} \int^x \sqrt{2m(E-V(y)) }dy)$
Inserting in the Schrodinger equation and requiring fullfilment to each
order of $\hbar$, we obtain:
$a_0 = \frac{Const.}{(2m(E-V(y)))^{\frac{1}{4}} }$
and
$ a_k \Delta S + 2 \nabla a_k. \nabla S = i \Delta a_{k-1}$
where:$ S= \int^x \sqrt{2m(E-V(y)) } dy$. In one dimensions these equations can solved in closed form.
The quantization condition is achieved through the requirement that the
wave function phase does not change over a closed loop:
$\oint^x \frac{1}{\Psi} \frac{d\Psi}{dx} dx = 2\pi i n\hbar $
Please see, the following article by A. Voros where this procedure was
applied to the anharmonic oscillator. Equation (4.4), describes an
explicit solution of the wave function to the first higher order.
Remark: Voros uses the Bargmann representation of the phase space instead of the usual coordinate-momentum representation.
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Is the Lagrangian density in field theory real? As the Lagrangian in classical mechanics corresponds to energy, it must be real. But is that the case in quantum field theory? I mean, it should still correspond to some sort of energy, but what about all the "$i$"s here and there, such as in the Dirac Lagrangian $i\bar{\psi}\gamma^\mu\partial_\mu \psi$ and the current density $J_\mu = ie[\dots]$ (see Griffiths for example)?
Another question is, how can it be hermitian, $\mathcal{L} = \mathcal{L}^\dagger$, when we have those "$i$"s? Wouldn't I get a minus sign if I complex-conjugated the interaction term and the Dirac field term? I'm really confused and hope someone can help
| In quantum field theory, the Lagrangian density is an operator, not a number. So it doesn't make sense to say it has to be real; "real" is a term that applies to numbers, not operators.
What does have to be true is that $\mathcal{L}$ has to have real expectation values in all physical states, and that in turn means it has to be hermitian (what mathematicians call self-adjoint). But hermiticity is not just a matter of being real. You can have other non-hermitian factors besides $i$. In particular, the derivative $\partial_\mu$ in the Dirac Lagrangian is antihermitian, and so the combination $i\partial_\mu$ as a whole is hermitian.
| {
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When is the right ascension of the mean sun 0? I understand that the right ascension of the mean sun changes (at least over a specified period) by a constant rate, but where is it zero? I had naively assumed that it would be zero at the most recent vernal equinox, but when I try to calculate the equation of time using this assumption and true sun positions, all my values are about 7.5 minutes larger than they should be.
When (at what date and UT time) is the right ascension of the mean sun 0? And why?
| No, the right ascension of the mean Sun is NOT zero at the vernal equinox. It is in fact nearly identical to the ecliptic longitude of the mean Sun (the difference is due to UT vs ephemeris time), and this is defined such that it coincides with the ecliptic longitude of the apparent Sun when the Earth is at perihelion. So that should be the starting time to calculate the equation of (ephemeris) time.
Using an ephemerides site like this one, one can check that the Earth was at perihelion on Jan 2 2013, at 4:37 UT, and will return to perihelion on Jan 4 2014, at 11:58 UT. The difference between these is called an anomalistic year. The same site shows that on Jan 2 2013, at 4:37 UT, the Sun had an apparent right ascension
$$
\alpha_p = 18^\text{h}51^\text{m}56^\text{s},
$$
and the corresponding ecliptic longitude is
$$
\lambda_p = \tan^{-1}(\tan\alpha_p/\cos\varepsilon) = 18^\text{h}47^\text{m}46^\text{s},
$$
where $\varepsilon=23^\circ 26' 21.4''$ is the obliquity of the ecliptic. The equation of ephemeris time is then
$$ \Delta t = M + \lambda_p - \alpha,
$$
where $M = 2\pi(t-t_p)/t_Y$ is the mean anomaly, $t_p$ is the moment of perihelion and $t_Y$ is the length of the anomalistic year. The quantity $M + \lambda_p$ is the ecliptic longitude of the mean Sun. The apparent right ascension $\alpha$ can be calculated from
$$
\begin{align}
M &= E - e\sin E,\\
v &= 2\tan^{-1}\left[\sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}\right],\\
\lambda &= v + \lambda_p,\\
\alpha &= \tan^{-1}(\tan\lambda\cos\varepsilon),
\end{align}
$$
with $E,e,v,\lambda$ the eccentric anomaly, orbital eccentricity, true anomaly and apparent ecliptic longitude (see also wiki). This should give you an approximation of the equation of time, accurate to a few seconds. If you want a really detailed calculation, have a look at this paper.
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Internal energy according to the van der Waals equation I am trying to derive the internal energy of a gas which obeys the van der Waals equation.
I have however encountered some problems. I calculate the integral of $dU$ from $V=0,T=0$ to $V=V, T=\infty$ to $V=V,T=T$.
I can calculate the work:
$$\left(p+\left(\frac{a}{n}\right)^2\right)(V-nb)=nRT \implies p=\frac{nRT}{V-nb}-\frac{an^2}{V^2}$$
For the second part of the path $V$ is constant so $W=0$.
$$W=-\int\limits_\infty^V p\textrm dV=\int\limits_V^\infty \frac{nRT}{V-nb}-\frac{an^2}{V^2} \textrm dV\\=nRT\ln(V-nb)|_V^\infty+\frac{an^2}{V}|_V^\infty\\=
\infty-\infty +\frac{an^2}{\infty}-\frac{an^2}{V}=-\frac{an^2}{V}$$
I know I haven't been mathematically rigorous but that is not really important to me at the moment. I think this is right.
I can't however think of how i should calculate the heat involved in following this path.
Any help on how to do this is appreciated.
EDIT: I see now that the work I calculated is wrong as well as $$nRT\ln(V-nb)|_V^\infty\neq\infty-\infty$$
| In case you want to take it from the partition function,
$U = k_B T^2 \frac{\partial}{\partial T} \left( \ln Q \right)$
and
$Q(N, V, T) = \frac{1}{N!} \left( \frac{2 \pi m k_B T}{h^2} \right)^{\frac{3N}{2}} \left( V - Nb \right)^N e^{\frac{a N^2}{V k_B T}}$,
which gives the same result as the other answers.
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The meaning of imaginary time What is imaginary (or complex) time? I was reading about Hawking's wave function of the universe and this topic came up. If imaginary mass and similar imaginary quantities do not make sense in physics, why should imaginary (or complex) time make sense?
By imaginary I mean a multiple of $i$, and by complex I mean having a real and an imaginary part, i.e., $\alpha + i\beta$, where $\alpha, \beta \in {\mathbb R}$.
| Another way to look at it is to imagine that time is a curved dimension, in sense it will be cyclic.
To visualize that imagine a plane of two dimensions, then the third usual dimension will be a perpendicular line to this plane.
Now consider this line to be bent in a circle, thus this dimension will go around and around, in sense it will have maximum value after which you will come back.
For information, is approach called Compactification, and if you familiar with complex numbers, remember that purely imaginary exponential is cyclic.
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Is it possible to reduce the sound, when two metal objects collide (perhaps with some coating) without reducing the rigidity of the surface? I have a system, where there are ball bearings on the pistons that clamp the metal plate with special dents for ball bearings. The system should be precise, because it is used for microscopy. It also should be as noiseless as possible. It also should be fast, so the impact at high velocity is inevitable.
I've thought of introducing some resin coating, but it will reduce the rigidity. Are there any solutions for this problem? Is there any strong relationship between sound and rigidity? I believe that there may be some rigid materials that somehow don't favor phonons.
| Unfortunately, anything that reduces noise without slowing the system down is going to do it by sacrificing rigidity.
If it is within your engineering parameters, you might introduce a damper on the pistons that engages within a millimeter or so of the ultimate extension of the piston. By slowing the piston down before impact, you can reduce the noise, without compromising the rigidity of the surface. It would have to be just before impact, slowing over a very short distance, in order to keep your system from being too slow. (I'm thinking of something akin to the dampers that are used on kitchen drawers, that slow the drawer down just before the drawer slams shut.)
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Who used the concept of symmetries first? Who "invented" the concept of symmetries? This article is quite extensive, but it blurs the history with the modern understanding.
https://plato.stanford.edu/entries/symmetry-breaking/
Some of the concepts can be traced to Galileo and Newton, but I'm quite certain the modern notion is incompatible with their view of the world. Does the notion come from group theory specifically? Can the first mention be traced accurately?
Although the spatial and temporal invariance of mechanical laws was
known and used for a long time in physics, and the group of the global
spacetime symmetries for electrodynamics was completely derived by H.
Poincaré [7] before Einstein's famous 1905 paper setting out his
special theory of relativity, it was not until this work by Einstein
that the status of symmetries with respect to the laws was reversed.
| For any application of symmetries throughout history, one can always find an earlier, more rudimentary application, so there is no first person, at least not as far as records go. The very first mathematics/physics/natural sciences results were geometric in nature, so the symmetries of the objects played a crucial role, if often implicitly.
The modern framework of symmetries as concerns physics is that of group theory, of course. A particularly important class of groups is that of Lie, so a natural candidate is Sophus Lie. He was definitely not the actual first, but he did formalize and generalize many notions of the time. In physics, a symmetry is often synonymous with a Lie group, so this seems a reasonable person to pick, if we must.
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Energy from man-made tornadoes Peter Thiel just paid $300,000 to Canadian inventor Louis Michaud who is working to construct useful "man-made tornadoes" or "atmospheric vortex engines" which could be components of future power plants.
Less ambitiously, they could replace chimneys and reduce the losses. More ambitiously, the surface-higher-atmosphere temperature difference could be enough to drive these vortices, i.e. give us energy almost from nothing.
See some very short introduction of mine.
Questions:
*
*Isn't it a perpetual motion machine? If it isn't, where and how does it take energy for operation?
*Is there a physical upper bound on the amount of energy one could construct in this way? Does the hypothetical engine reduce the lapse rate or otherwise modify the atmosphere?
| I accidentally stumbled over this question and wanted to post an answer, after some brief reading on http://vortexengine.ca/english.shtml, despite it was asked already over a year ago.
I think the basic idea of the concept is founded on a major confusion between cause and effect regarding meteorological phenomena. While certain conditions on a meteorological scale (ie. large scale) may lead to the formation of energetic vortices, it is in reverse silly to assume that generating a vortex (by means of a realtively tiny energy input) would create the necessary (large scale) conditions to invoke a stationary vortex, that could be tapped.
An other way of thinking about the problem is that any continuous buoyancy driven updraft process of this kind is a thermal process and therefore limitted by Carnot Efficiency:
$\eta_c=1-\frac{T_{cold}}{T_{warm}}$
so
$P_{mech}<\dot{Q}\eta_c=\dot{m}_{turbine}c_p(T_{warm}-T_{cold})\eta_c$
for any real process. This is true for any tower height or medium and the temperatures refer to ground conditions. So by any practical means the power that could be generated would always be a rather small fraction of the thermal power fed into the tower (and of course even less, if the hydrostatic pontential is used up to generate a vortex).
| {
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Why does the nature always prefer low energy and maximum entropy? Why does the nature always prefer low energy and maximum entropy?
I've just learned electrostatics and I still have no idea why like charges repel each other.
http://in.answers.yahoo.com/question/index?qid=20061106060503AAkbIfa
I don't quite understand why U has to be minimum. Can someone explain?
| consider a system of gas particle inside a perfume bottle. Now lets say you spray the gas out, and the system of particle now gets spread all around in a Random manner. Well,now lets see this particles as waves(matter waves) ,just like the electron clouds. I admit, I don't have any deep understanding of quantum mechanics but from what I have learnt from my intermediate chemistry is that more the electron clouds spread, lesser is their energy. So , when the particles spread uniformly everywhere with maximum entropy, the electron cloud spreads all through out, As such they attain a stable and lower energy state. Now lets come to your next question of lower energy. It isn't so that nature or systems of nature prefer lower energy states, but rather they can't proceed further when they have the lowest possible energy. Like consider a ball thrown up from the ground when it comes down to ground it lacks any more energy or any external force to be back to air. Now it still remains a mystery to me as to how they can attain this stable state in finite amount of time and can there exist systems which has a stable state but it surpasses it. So ,this was all my audacious attempt to explain this tendency of nature with intuitions of quantum mechanics.
| {
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experimental technique for measuring temperature of an ant I am taking a course on thermodynamics. I have a question from my text(halliday & resnick,physics-1). They asked me to measure temperature of an ant or an insect or a small body,like a small robot. If I build a thin thermometer then it is probable that surface tension would have greater influence than thermal expansion. Then How can I measure the temperature of an ant?
| I am not sure, but maybe you can use a micro-thermocouple (https://www.variohm.com/images/datasheets/Micro-Thermocouples.pdf) For example, 44 AWG wire is 0.05 mm diameter, if I am not mistaken.
| {
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Separation between slip rings of DC generator
This is a DC generator which converts mechanical energy to Direct Current. The commutators in a DC generator are separated (as you can see in the image). It is explained in our book that it prevents the change in direction of current. But I don't understand.
Even if we don't keep them separated, how will the current change direction?
In short, what is the function of the gap between the two split rings?
| Imagine a that one of the carbon brushes is connected to a positive terminal of a DC voltage source (eg: battery) and the other brush to the negative terminal.
A current, $I$, will flow through the rotor winding from the positive terminal to the negative terminal. As the current flows through the magnetic field $B$ set up by the permanent magnet, the wire carrying the current experiences a force, given by $F=IL\times B$, where L is the length of the wire segment in the magnetic field. So for the wire segment a-b, the current is flowing into the page so the force on the wire due to the magnetic field will be upwards, whereas for segment c-d, the current is flowing out of the page (back towards the negative terminal) so the force on the wire will be downwards. The net result is a torque (spinning force) on the rotor, causing it to spin clockwise.
When the rotor spins $90^{O}$, the commutator also spins $90^{O}$ (since it is connected to the rotor shaft). At this point, there is no torque on the rotor. The rotor however will continue past the $90^{O}$ point under its own inertia. The split in the commutator now causes the current to flow in the reverse direction through the winding. Since the rotor has moved past $90^{O}$, the current through d-c is now into the page and the resulting force is upward, whereas the current through section b-a is out from the page and the resulting force is downward. So the torque on the rotor is in the same direction as before.
So the commutator produces a 'reversal' of the applied voltage is required every time the rotor winding goes past the $90^{O}$ and $270^{O}$ point (when the torque on the rotor is zero). The voltage must be reversed at these points to ensure the current in each segment of the loop has the same 'sense' with respect to the permanent magnetic field, as the rotor spins, to ensure the net torque is always in the same direction (clockwise). Otherwise, the rotor will simply 'wobble' about the $90^{O}$ point and not spin as a proper motor should!
| {
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Is the superposition principle universal? In David J. Griffiths' Introduction to Electrodynamics, he claims that the superposition principle is not obvious but has always been found to be consistent with the experiments. So I was wondering have we found some physics quantities which do not follow superposition principle? If we have not till now why can't we generalize and make it into a law?
More specifically: Griffiths was talking about electromagnetic force. My question is about the existence of something like mass or charge and which doesn't follow this superposition principle.
| Any physical quantity that can be organized as a vector space obeys the superposition principle. I would go as far as to say that the superposition principle arises from the fact that a vector space is closed under the weak operation $+$ of the field $\mathbb{(R,+,\cdot)}$.
| {
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Can one heat up a vacuum? I've got a question about heating a vacuum. If there were, say, a container in space, at 2.7 degrees kelvin (the typical temperature of space, if I'm not mistaken) and as empty as space (as close to a vacuum as space allows), how would one go about pressurizing and heating that container? If a gas such as oxygen were introduced, would it freeze due to the temperature or would it sublimate due to the vacuum? If the former, I don't understand how heat could be introduced because heat needs a medium to heat. If the latter, once the vacuum was overcome, and a sufficient pressure was acquired, wouldn't the oxygen freeze and re-create the vacuum? Would both heat and pressure need to be introduced at the same time?
Thank you.
| If you inject the gas very slowly, the system will remain in thermodynamic equilibrium
source
By looking at the diagram above, you can heat up the gas as you inject it and keep it on the bottom right of the quadrant, thus gaseous.
| {
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What is the current radius of cosmological event horizon? Doing some crude calculations (using the value of $H_0$ at this point of time only, since it is time dependent but not distance dependent thanks to Johannes answer) what is the radius of cosmological event horizon at this point of time? (not looking for the changes of CEH through time)
From here we have for $H_0 $:
$$H_0 = 73.8 \pm 2.4 (\frac{km}{s})\frac{1}{Mpc}\tag{I}$$
We are seeking the distance $L$ s.t. $H_0L = c = 3\times 10^6 \frac{km}{s}$
$$L=\frac {c}{H_0} = \frac {3\times 10^6}{73.8 \pm 2.4} Mpc \tag{II}$$
Where 1 pc = 3.26 light years ($ly$),
$$ L=\frac {c}{H_0} = \frac {3\times 10^6\times10^6\times3.26}{73.8 \pm 2.4} ly \tag{III}$$
$$ L= \frac {9.78\times 10^{12}}{73.8 \pm 2.4} ly \tag{IV}$$
$$L=1.3\pm0.1\times 10^{11} ly \tag{V}$$
Is this calculation correct?
Would the correct calculation make sense? (By making sense I mean it would seem in accordance with some observation and not in contradiction to some other observations? Or results like this are unconfirmable, just mere flights of fancy were they do not relate to anything physical?
The only thing I could use to see it is not invalid (yes double negative, I cannot say it was valid) is the fact that observable universe is $45.7×10^9 ly$ but then again by that account $L=10^{123}ly$ would seem just as valid.
| The Hubble length $c/H_0$ does not coincide with the radius of the observable universe.
Your calculation assumes a Hubble parameter that doesn't change over time. This is not correct: the Hubble parameter $H$ changes over time, and $H_0$ (the Hubble constant) indicates the current value of $H$. To refer to $H_0$ as a 'constant' is a bit of a misnomer, it is effectively a constant in space, not in time.
Also note that if $H$ would have been constant over time, the Hubble time $1/H$ would be the time taken for the universe to increase in size by a factor of $e$. It is a coincidence that the current value for $H$ leads to a Hubble time very close to the current age of the universe.
| {
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Reasons for violation of universality in statistical mechanics The Universality in statistical mechanics is nicely explained by the renormalization group theory. However, there are fair amount of numerical and theoretical studies show that it can be violated in models such as Ising model, spin glass, polymer chain and percolation (Example references: 1, 2, 3, 4, 5, 6, 7) and some violations are pretty strong. So, my questions are:
*
*What are the general reasons of the violation of the universality?
*Also, why can't they be explained by the renormalization group theory and scaling hypothesis?
*Is it always true that there are some "core elements" for each universality class? Whenever they are missing, we can definite say they are not in the same universality class.
*Is there an example of commonly recognized "universality class" (such as 2D Ising model) that show different critical exponents by changing the microscopic details? Note, it should not happen by the definition of universality class, but there is no proof for all universality class.
I would accept answers discussing solid examples for some universality class about these questions.
Note: Universality refers to a large class of different systems exhibiting the same properties regardless of its microscopic details such as the lattice type (square, hexagonal, triangular, and kagome lattices). In statistical mechanics, the universality class are usually determined by the symmetry of the order parameter (such as up-down symmetry in Ising model and spherical symmetry in Heisenberg model) and the topological structure (such as dimensionality).
| Your reference $3$, for instance, is an Ising model modified by randomness. It is of course not clear how universality works in such complex systems. Another link studies (finite size) clusters, far from the ideal infinite system of the analytical solution.
In general, recall that universality is a feature of phase transitions with respect to a parameter of the system. Models are typically constructed using a lattice Hamiltonian and phase transitions derived from the partition function, with $1/T$ as the crucial parameter. The (solved) 2D Ising model illustrates the importance of conformal symmetry and related field theories for typical critical phenomena. But few systems are exactly solved, and systems may be studied away from criticality.
Renormalisation a la Wilson turns the critical point into fixed points for RG flows. Again, the process only works for relatively simple Hamiltonians. (Such techniques are now used even in the 4D case)!
| {
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Change in intensity of electric field with constant velocity Consider a +Q charged particle is travelling towards another test charge +Q. Now what would be the difference in electric field experienced by the test charge(avoid the gradual decrease in distance between them)? Would the field lines look compressed and effective field strength increased for the test charge?
| A test charge "feels" only an external field. If another charge is approaching, it will experience a stronger field due to $1/R^2$ field strength dependence.
| {
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Don't understand the integral over the square of the Dirac delta function In Griffiths' Intro to QM [1] he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being
$$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right)$$
(cf. last formula on p. 101). He then says that these eigenfunctions are not square integrable because
$$\int_{-\infty}^{\infty}g_{\lambda}\left(x\right)^{*}g_{\lambda}\left(x\right)dx
~=~\left|B_{\lambda}\right|^{2}\int_{-\infty}^{\infty}\delta\left(x-\lambda\right)\delta\left(x-\lambda\right)dx
~=~\left|B_{\lambda}\right|^{2}\delta\left(\lambda-\lambda\right)
~\rightarrow~\infty$$
(cf. second formula on p. 102). My question is, how does he arrive at the final term, more specifically, where does the $\delta\left(\lambda-\lambda\right)$ bit come from?
My total knowledge of the Dirac delta function was gleaned earlier on in Griffiths and extends to just about understanding
$$\tag{2.95}\int_{-\infty}^{\infty}f\left(x\right)\delta\left(x-a\right)dx~=~f\left(a\right)$$
(cf. second formula on p. 53).
References:
*
*D.J. Griffiths, Introduction to Quantum Mechanics, (1995) p. 101-102.
| You need nothing more than your understanding of
$$
\int_{-\infty}^\infty f(x)\delta(x-a)dx=f(a)
$$
Just treat one of the delta functions as $f(x)\equiv\delta(x-\lambda)$ in your problem. So it would be something like this:
$$
\int\delta(x-\lambda)\delta(x-\lambda)dx=\int f(x)\delta(x-\lambda)dx=f(\lambda)=\delta(\lambda-\lambda)
$$
So there you go.
| {
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Do particle pairs avoid each other? Please end my musings Can you explain what happens when a particle and its antiparticle are created. Do they whiz away from each other at the speed of light or what? I suppose that they don't because otherwise they would never meet and annihilate each other, but then, if I had just been created with an antiparticle I would do all I could to stay away from him/her. On the other hand, for the sadistic/suicidal type, they might actually be attracted to one another.
[the question is serious, even though it's written light-heartedly. Please explain.]
| This is a clarifying answer.
Here are particle antiparticle pairs in a bubble chamber photo from Fermi lab, created by the reaction
gamma + proton -->electron positron pair + proton
The gamma is unseen before it hits the proton ,because it does not ionize and when it hits the proton, most of its energy is taken up by the new pair production.
As one can see the products do not have the velocity of light but maybe a few hundreds to a few MeV each. The momentum can be calculated from the curvature given the value of the magnetic field, which is perpendicular to the page. The tracks end in spirals because the electrons ( positrons) ionize the fluid, and so we can see the tracks, but there is loss of momentum and energy due to this ionization, creating the spiral.
In the center of mass of the initial state "proton gamma" the three particles acquire their momentum from the conservation of momentum and energy laws. The matrix element kinematics are such that the target proton is a spectator, very seldom seen with a mm track in the chamber and the electron positron pair take most of the energy of the gamma..
| {
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When are leap seconds added in various time zones? I understand that technically when a leap second is added, it is added after midnight UTC, but I'm unclear how the addition is handled in other timezones. For precise reckoning of course (e.g. astronomical ephemeris), it must be added at the same instant (e.g. at 1 AM in France, or 7 PM in in New York), but for many locations doing so could be disruptive and potentially confusing. Even in Britain, the addition at midnight on New Year's Eve could be confusing, since it would either extend the countdown by a second, or require it to be started a second "late".
Are leap seconds in fact added everywhere, in all time zones, at the same UTC instant, or do different time zones (and perhaps even different countries and organizations) account for them at different, more "convenient", times.
| Yes, the leap second are added at the same "instant" in the whole world, whatever is the part of the day on a given place. This is documented by a rather typical screenshot of some "clock" in 2008 according to Chicago's U.S. Standard Central Time:
The confusing second was added right before 6 p.m. – in as big a city as Chicago and similarly in the whole world. More comments about leap seconds and the plans to abolish them, see:
http://motls.blogspot.com/2012/01/leap-seconds-may-be-abolished-in-2015.html?m=1
In general, leap seconds add some problems which is why there are efforts to abolish them. Still, one shouldn't overstate the problems. The amount of activities that require this 1-second precision in "absolute time" is rather limited, and even if we count trains into this set (which is a big exaggeration), the 1 second may usually be easily "lost" during the journey. More high-tech applications – like the communication with NASA's satellites etc. - are so professional and administered by so advanced people that they're expected to deal with all potential subtleties here.
| {
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Total Energy of the Universe? I've heard the total energy is zero, but I've also heard it cannot be said to be zero since there's so much unknown stuff in the universe. Is that true?
| I think it can be safely stated that no one really knows the answer to this, because like you've said, there's so much unknown stuff in the Universe. There's that dark matter and dark energy stuff that's pretty mysterious as of today.
Here's a Wikipedia article on the zero-energy-universe, btw.
http://en.wikipedia.org/wiki/Zero-energy_universe
I think it would make sense for the total energy to be zero. Because, really, if its not, then how can that be? Where did the energy come from? Its pretty safe to say that the conservation of Energy is something that's widely accepted and held to a pretty high degree. The reason why it would be zero is because all of the positive energy equals the amount of negative energy the Universe has.
http://en.wikipedia.org/wiki/Conservation_of_energy
I hope someone else can answer this better than I can.
| {
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Conformal Quantum Mechanics I heard the term Conformal Quantum Mechanics used today.
*
*What exactly does this mean?
*Why would one want to study this?
| It's slang for conformal field theory for manifolds of $D=1$ dimensions. (If you take the dimension to be time, 1-D QFT describes the time evolution of a system living in zero spatial dimensions, i.e. at a single point, so it's not really field theory but QM.)
It is special in the sense that normally, CFT representations are specified by two labels (the operator dimension $\Delta$ and spin $l$), but in 1D there is no spin, so you only have scaling dimensions.
Also, all 1-D manifolds are trivially conformally flat (= they can be rescaled to obtain a Euclidean line). This fails in $D \geq 3$ dimensions. In 2D, this point is a bit subtle (you can have conformal manifolds with a boundary, such as the half plane with $y > 0$).
I've seen some papers about the subject, and it seems to be that a small group of people study conformal QM, probably because it's relatively easy to obtain exact results, such as closed-form expressions for conformal blocks. (This is a hard problem in general dimensions - I can give references if you're interested.) Mathematically there's no hope of extending these results to higher dimensions, but maybe these people try to get some intuition from this simple case.
You can also informally rephrase the subject by thinking of conformal QM as the study of quantum systems with no characteristic timescales, where all temporal correlation functions behave as power laws. (Of course, this is very crude, since scale invariance is less restrictive than conformal invariance.)
| {
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Why doesn't my pinhole camera work? We all know that light travels in straight a line, which can be proved by pinhole imaging as in the picture shown :
But when I'm doing this little experiment with an apple, no matter how I change the distance between the object and the pinhole, an image can never be observed on the cardboard behind. So what's going wrong? Please help!
| It actually is providing an image, you just cannot see it because of contrast. The light you have in the picture is flooding the room with enough ambient light that the image that is actually formed has no contrast.
*
*Pinhole camera image quality heavily depends on the pinhole size.
However, the smaller the pinhole, the "dimmer" the image will be,
this is the problem that you have.
*In order to be able to view the image you will need to make the
pinhole basically in a box. The box could be an entire room (e.g.
cover all windows in a room with aluminum foil so it is very dark,
then prick a pinhole in the foil on one of the windows) or a smaller
box with a "viewing screen." The key thing is that whatever "box"
you use needs to be relatively dark inside.
A viewing screen can be something clear but diffuse, like parchment paper or privacy glass. Even with the viewing screen, you need to somehow make it dark enough so that your eyes can see it.
You could also make an enclosure large enough for you head to fit into, but still maintain a good level of darkness.
If you have an old webcam, this might work inside of an enclosed box, then you can view the image on your computer.
This is one of the things that makes using pinhole cameras in demonstrations difficult, you want the object being imaged to be well lit, but you need the viewing screen to be dark.
| {
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Earth moves how much under my feet when I jump? If I'm standing at the equator, jump, and land 1 second later, the
Earth does NOT move 1000mph (or .28 miles per second) relative to me,
since my velocity while jumping is also 1000mph.
However, the Earth is moving in a circle (albeit a very large one),
while I, while jumping, am moving in a straight line.
How much do I move relative to my starting point because of this? I
realize it will be a miniscule amount, and not noticeable in practise,
but I'd be interested in the theoretical answer.
| In the rotating frame of the earth's surface, there are two fictitious forces acting on you, a centrifugal force and a Coriolis force. These are both quite small in absolute terms, and the centrifugal force is also not noticeable because it simply feels equivalent to a slight change in the over-all direction and magnitude of the gravitational field. The Coriolis force has a horizontal component and will therefore in theory cause you to come down in a slightly different place then the one from which you jumped up. In practice, this effect is much too small to observe in this situation due to all the other systematic and random errors.
There is identically zero effect due to the earth's motion around the sun, since the earth is free-falling and therefore constitutes an inertial frame in the general-relativistic sense.
| {
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Charge conjugation in Dirac equation According to Dirac equation we can write,
\begin{equation}
\left(i\gamma^\mu( \partial_\mu +ie A_\mu)- m \right)\psi(x,t) = 0
\end{equation}
We seek an equation where $e\rightarrow -e $ and which relates to the new wave functions to $\psi(x,t)$ .
Now taking the complex conjugate of this equation we get
\begin{equation}
\left[-i(\gamma^\mu)^* \partial_\mu -e(\gamma^\mu)^* A_\mu - m \right] \psi^*(x,t) = 0
\end{equation}
If we can identify a matrix U such that
\begin{equation}
\tilde{U} (\gamma^\mu)^* ( \tilde{U} )^{-1} = -\gamma^\mu
\end{equation}
where $ 1 =U^{-1} U$.
I want to know that, why and how did we do the last two equation. More precisely, I want to know more details and significance of the last two equations.
| The key here is that the gamma matrices are given by their commutation relationships and those do not determine a unique representation for the matrices.
If you start from the Dirac equation
$$\gamma^\mu (i\partial_\mu - e A_\mu) \Psi = m \Psi$$
and make the following generic transformation $\Psi = U \Psi'$ with $U$ a constant matrix with inverse $UU^{-1}=1$ the equation becomes
$$\gamma^\mu U (i\partial_\mu - e A_\mu) \Psi' = m U \Psi'$$
multiplying by the inverse matrix
$$U^{-1} \gamma^\mu U (i\partial_\mu - e A_\mu) \Psi' = m \Psi'$$
this is equivalent to the original equation if $\gamma^\mu{'} = U^{-1} \gamma^\mu U$. This relation guarantee that the new matrices satisfy the same commutation relationships than the original.
Regarding the specific case of charge conjugation I think the most fundamental approach uses the CPT theorem. In this case parity is trivial therefore it remains charge conjugation (C) and time reversal (T). The Dirac equation is invariant if both the sign of the charge and time are reversed. This is the basis for Stuckelberg Feynman interpretation of antiparticles as particles travelling backward in time.
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Phase difference of driving frequency and oscillating frequency Suppose a mass is attached to a spring and is oscillating (SHM). If a driving force is applied, it must be at the same frequency as the mass' oscillation frequency. However I'm told that the phase difference between the driving frequency and the mass's frequency must be $\frac{\pi}{2}$.
Why is that? I would have thought they should have to be in phase to be in resonance?
| For resonance to occur (at $\gamma$=0 and $\omega=\omega_0$), the system must be able to absorb all the energy input coming from the external driving force, and this results in incessant growth in the amplitude. For this to happen, the external force, which does not have to be in sine/cosine form, should be positive (pushing forward) while the oscillator is traveling from $-A$ (amplitude) to $+A$, and the external force should be negative (pulling back) while the oscillator is traveling from $+A$ to $-A$.
The plot below shows the solution of $\ddot{x}(t)+\omega_0^2x(t)=f(\omega_0 t)$ where $f(\omega t)$ is a square-wave of which frequency $\omega$ here matches the natural frequency of the harmonic oscillator $\omega_0$. The amplitude grows in time due to the resonance ($\omega=\omega_0$) as expected.
One can observe from the plot above that the force is positive while the oscillator is traveling in the forward direction (green shaded region), and the force is negative while the oscillator is traveling in the backward direction (pink shaded region); this leads to a complete absorption of the energy from the driving force.
Now let's have a sine-wave external force instead of a square wave. The result is the same, but this time we can discuss the phase difference between the two sine waves (the amplitude and the external force). From the plot, one can see how $x(t)$ and the force have a phase difference of $\phi=\pi/2$.
In layman terms, the force should be positive (forward) while the oscillator is moving forward, and vice verse. (Note that this statement does not say the force should be positive when the oscillator amplitude is positive.)
Finally, that the phase $\phi = \arctan \left( \frac{2 \xi \omega \omega_0}{\omega^2 -\omega_0^2} \right)$ becomes equal to $\pi/2$ when $\omega \rightarrow \omega_0$ (here $\omega_0=3$) and $\xi \rightarrow 0$ can be seen from the plot below. For $\omega \rightarrow \omega_0=3$, we get $\phi=\pi/2$ for all nonzero values of $\xi$. When $\xi=0$, we get a step function (not shown in the plot), but we can consider $\phi=\pi/2$ for any value of $\xi$ that is arbitrarily close to zero.
Want to see all of these in action? Here's a video from MIT: https://www.youtube.com/watch?v=aZNnwQ8HJHU
Hope all these help.
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Showing that position times momentum and energy times time have the same dimensions I've been asked to show that both the position-momentum uncertainty principle and the energy-time uncertainty principle have the same units.
I've never see a question of this type, so am I allowed to substitute the units into the expressions and then treat them as variables?
If so, here's my attempt. Forgive me if I've done something silly, as I'm no physicist.
Starting with the position-momentum uncertainty principle:
$$\Delta{}x\Delta{}p \geq h / 4\pi$$
Substituting the units into the expression (at this point, diving by $4\pi$ won't necessarily matter):
$$(m)\left(kg \cdot \frac{m}{s}\right) \geq J \cdot s$$
Combining $m$ and bringing $s$ to the other side:
$$\frac{kg \cdot m^2}{s^2} \geq J $$
Knowing that $J = kg \cdot m^2/s^2$:
$$J \geq J$$
Now, for the energy-time uncertainty principle:
$$\Delta{}E\Delta{}t \geq h / 4\pi$$
Substituting the units into the expression (again, diving by $4\pi$ won't necessarily matter):
$$J \cdot s \geq J \cdot s$$
Diving by $s$:
$$J \geq J$$
Is this valid? Or could I not be more wrong?
| It's just that magnitudes can't be ordered, I mean: things like $meter>second$ don't make sense.
In every physical equation, or inequation, both sides must be in the same unit, so what you're doing is actually checking that the units at the right are the same as the ones in the left. That must be true, as you're saying a numerical product of momentum and time (a real value) will be higher than another product of momentum and time, that is, the units of $h$. The $>$ sign doesn't mean that units must be higher, such a thing doesn't exist, it means that quantities measured in the same units (quantities of the same kind of thing) will behave like that, one will bi bigger than the other.
I don't know if I'm explaining it properly.
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Rigorous approaches to quantum field theory I have been reading Quantum Mechanics: A Modern Development by L. Ballentine. I like the way everything is deduced starting from symmetry principles.
I was wondering if anyone familiar with the book knows any equally elegant presentation for quantum field theory. Weinberg's books start off nice with the irreducible representations of the Poincare group/algebra but the later chapters lose me with the notation. Also, most books I've read on QFT (Srednicki, Peskin and Schroeder, Mandl and Shaw) make a valiant initial attempt at a nice consistent framework but end up being a big collection of mathematical recipes and intuitive insights that seem to work but the overall structure of the theory seems to be sewn up. The relativistic equations crop out of rather flimsy arguments, the canonical anti/commutation relations are imposed out of density indeterminate air, functional methods are developed cause we know no better, infinities come about with renormalization theory to the rescue but it seems very alien from the initial context.
Is there any approach that ties all these things together in an elegant mathematical framework which accounts for all the patch up work that is needed? I am not talking about axiomatization just a global point of view that encompasses all the issues.
| I have not read it myself, but did you take a look at Zee's book? I heard that while it is not very technical, it is great on a conceptual level.
| {
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Why do we weigh less when falling?
I don't want to go to science world to find out because it would be a long round-trip.
I understand that acceleration/deceleration would effect the weight and I can also imagine that someone at terminal velocity would weigh nothing but I can't get my understanding in terms of forces and how that would effect the weight. For example, what would a formula relating mass, speed (in direction of gravity), gravitational force and weight look like?
| As an alternative example, think about how it feels to drive a car.
*
*When the car is standing still, you will sink into the seat and feel the seat pressing against you (= normal weight).
*When you press the throttle and the car accelerates, you will be pressed further back into the seat and will in turn feel the seat pressing harder against you (= increased weight).
*When the car has reached cruising speed and is not accelerating any longer, you will not get further pressed into the seat any longer (= normal weight again). There are still forces acting on the car (engine is driving it, wind resistance and friction are slowing it), but they are balanced and results in no acceleration.
*When you press the brake and the car decelerates, you will be lifted out of the seat, which will in turn lessen it's pressure on you (= decreased weight).
In other words, your weight will only change when the car is accelerating/decelerating.
| {
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Cyclic Coordinates in Hamiltonian Mechanics I was reading up on Hamiltonian Mechanics and came across the following:
If a generalized coordinate $q_j$ doesn't explicitly occur in the
Hamiltonian, then $p_j$ is a constant of motion (meaning, a constant, independent
of time for a true dynamical motion). $q_j$ then becomes a linear function of time.
Such a coordinate $q_j$ is called a cyclic coordinate.
The above quote is taken from p. 4 in Ref. 1.
What I don't understand is why $q_j$ is a linear function of time if $p_j$ is constant in time. In other words, why does $p_j$ constant in time imply partial $\frac{\partial H}{\partial p_j}$ is a constant? (In particular, $\frac{\partial H}{\partial p_j}$ could depend on any of the other coordinates or momenta.)
Reference:
*
*Patrick Van Esch, Hamilton-Jacobi Theory in Classical Mechanics, Lecture notes. The pdf file is available from the author's homepage here.
| It's an old question and already has a fine answer, I just wanted to quickly add a more physical example.
Take a 2D particle in a central potential in polar coordinates, the Lagrange function is
$$ L(r,\dot r,\theta,\dot\theta) = \frac 1 2 \left( \dot r^2 + r^2 \dot\theta^2 \right) - V(r) \;. $$
Now the angle $\theta$ is a cyclic variable, which gives us the conservation law
$$ \frac{\partial L}{\partial \dot\theta} = r^2 \dot\theta = \text{const} \equiv c \;. $$
However,
$$ \theta(t) = \theta_0 + \int_{t_0}^t \frac{c\, \mathrm dx}{r^2(x)}$$
is not linear in time in general.
| {
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Do perfect spheres exist in nature? Often in physics, Objects are approximated as spherical. However do any perfectly spherical objects actually exist in nature?
| If we are considering the states of matter we are familiar with (fermions), since these structures are inherently discretized (solids, liquids, gases), they will not exhibit perfect spherical symmetry.
We can loosen the definition of perfect and construct a cutoff where variations in the radius are negligible below some length scale and call that perfect.
Replying to wouter, the black hole will have an event horizon that is spherical if and only if the black hole has zero angular momentum. It ultimately depends on the interpretation of Cactus' question about what an "object" constitutes and if an event horizon is considered a valid answer.
| {
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What happens to heat waste produced by energy generation? What happens to heat waste produced by energy generation on earth that must be there according to the laws of thermodynamics? So, it never dissipates and remains on earth?
| Take for example a conventional coal fired power plant as used widely in most countries.
About 40% of the energy from burning the coal is converted to electricity, and the remaining 60% is lost as heat to the environment.
The 40% of the energy that gets converted to heat is used by us consumers and eventually gets converted to heat and is lost as heat to the environment.
So all the energy used by burning coal (or gas, or from nuclear energy) eventually ends up in the environment, where it heats up the air/rivers/seas/whatever. However the amount of heat lost to the environment is small compared to the amount received from the Sun and generated inside the Earth.
| {
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What is the electrical conductivity (S/m) of a carbon nanotube? I have been searching around for a while for this but I am having trouble finding any actual figures, all I can seem to find is that it is "very high".
So I am wondering, does anyone have any figures of what the electrical conductivity of a carbon nanotube is, a theoretical or estimated answer is fine. I am preferably looking for the answer in $Sm^{-1}$.
| The conductivity of a nanotube depends on the chirality / diameter of the individual nanotube and the conductivity. The chirality determines whether a nanotube is classified as either metallic or semiconducting.
The highest conductivity that has been measured for a carbon nanotube is ~100 MS/m, which is equivalent to a resistivity of 1 uOhm-cm. This result was reported by McCuen and Park (2004) for a metallic carbon nanotube.
https://doi.org/10.1557/mrs2004.79
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why is orbital moment quenched while atoms forming solid atom has well defined spin(up and down) and orbital(s,p,d,etc) momentum, but when forming crystals, why the spin degree continues to be good quantum number while orbital momentum is quenched?
| Orbital angular momentum is a good quantum number for the atomic problem because the Coulomb potential between the electron and nucleus is rotationally invariant, but the potential an electron feels in a crystal is not. A non-spherically-symmetric potential can couple states with different $l_z$, and so if $\psi_{l_z}$ were the eigenstate of the spherically symmetric problem with angular momentum $l_z$, then the correct atomic eigenstates in, for example, a cubic or tetragonally symmetric potential separate into linear combinations like $\psi_{l_z} \pm \psi_{-l_z}$ which measures out to total $l_z=0$.
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Force and Torque Question on an isolated system If there's a rigid rod in space, and you give some external force perpendicular to the rod at one of the ends for a short time, what happens?
Specifically:
What dependence does the moment of inertia have?
If it rotates, what is the center of rotation?
Does it matter that the rod is rigid? What happens if it's "springy", say a rubber rod instead.
Is there a difference between exerting a force for a short period of time, and having an inelastic collision (say a ball hits the end of the rod instead of you pressing).
| The moment of inertia of a rod with a perpendicular axis of rotation through the center of mass is $$I=\frac{1}{12}ml^2$$ where m is the mass and l the length.
If you apply a torque $$\tau=r\times F$$ (all vectors), then $$\tau=I\alpha$$ lets you calculate the angular acceleration alpha.
If the rod deforms, energy will be needed for deformation and creation of heat. If the force F is exerted for a longer time, the total work on the rod has to be integrated over time. No. F has to be integrated to get the work.
| {
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Why does smoke go out the window of the car - and what if there's wind blowing instead of the car moving? When driving a car while smoking with the window open (safety and legal issues aside), I've noticed that the smoke tends to go outside the window.
*
*Why does the smoke go outside?
*If the car is standing still and there is wind blowing at the same velocity the car was going - will the smoke behave the same?
| To the first question: no idea, honestly. My intuition says that air should come by the front windows and out from the rear ones. Probably air creates difference of pressures in the neighboorhood of the window and suck air out, just like a vaccum cleaner: it sucks air by making it go out... Making theoretic studies may be very difficult, but if we could see a wind tunnel simulation with windows opened we may see what air is doing there.
To the second question: definitely yes. The two situations are actually the same, that's why they use wind tunnels to test aerodynamics of cars instead of getting them out in tracks (they also do it, but for driving stuff rather than aerodynamics).
| {
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Crystal momentum and the vector potential I noticed that the Aharonov–Bohm effect describes a phase factor given by $e^{\frac{i}{\hbar}\int_{\partial\gamma}q A_\mu dx^\mu}$. I also recognize that electrons in a periodic potential gain a phase factor given by $e^{\frac{i}{\hbar}k_ix^i}=e^{\frac{i}{\hbar}\int k_idx^i}$. I recall that $k_i$ plays a role analogous to momentum in solid state physics. I also recall that the canonical momentum operator is $P_\mu=-i\hbar\partial_\mu-qA_\mu$. Notice that when you operate with the momentum operator on a Bloch electron, $\psi(x)=u(x)e^{\frac{i}{\hbar}k_ix^i}$, you get $e^{\frac{i}{\hbar}k_ix^i}(-i\hbar\partial_i+k_i)u(x)$.
My question is whether a parallel can be drawn between the crystal momentum, $k$, and the vector potential $A$. It seems they play a similar role quantum mechanically, but I have never seen Bloch's theorem described in terms of vector potentials. I suppose one does not even need a nontrivial vector potential for Bloch's theorem to hold. Still, crystal momentum and the vector potential play very similar roles in phase factors and I wonder whether there is any deeper meaning to that.
| Just as V is the potential energy, $\vec{A}$ is the potential momentum. This is also true in a crystal.
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What is the result of a classical collision between THREE point particles at the same precise instant? Classical Mechanics is said to be deterministic, a statement that nearly always is followed by that quote from Laplace, something like
If at one time, one knew the positions and velocities of all the
particles in the universe, the laws of science should enable us to
calculate their positions and velocities at any other time, past or
future.
I always scratch my head after hearing/reading that. If 3 or more rigid point particles happen to collide elastically at the same precise instant, is it not impossible to predict the resulting trajectories? If it is possible, how?
| Taking the case of point particles and "contact" collisions seriously actually causes trouble even in the two dimensional case: the instantaneous forces are necessarily infinite even if the impulses remain finite.
The solution to that problem--to recognise that all real particles interact via fields over non-zero distances--solves the three particle problem as well. You just integrate the equations of motion (possibly numerically as this may not be easy in closed form).
This isn't necessarly in the 2 particle elastics case because conservation of energy and momentum fully constrain the outcome allowing us to elide this question in a introductory presentation.
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What is the current through the lamp? We have the following circuit:
A neon lamp and a inductor are connected in parallel to a battery of 1.5 $V$. The inductor has a 1000 loops, a length of $5.0 cm$, an area of $12cm^2$ and a resistance of $3.2 \Omega$. The lamp shines when the voltage is $\geq 80V$.
*
*When the switch is closed, $B$ in the inductor is $1.2\times 10^{-2} T$.
*The flux then is $1.4 \times 10^{-5} Wb$
(calculated myself, both approximations).
You open the switch. During $1.0 \times 10^{-4} s$ there is induction. Calculate how big the current through the lamp is.
My textbook provides me with the following answer:
$U_{ind} = 1000 . 1.4 \times 10^{-5} / 1.0 \times 10^{-4} = 1.4 \times 10^{2} V$.
$ I = U/R_{tot} = 1.4 \times 10^{2} / (3.2+1.2) = 32A$
My concerns:
*
*How do we know that $1.4 \times 10^{-5}$ is $|\Delta \phi|$? This is the flux in the inductor while the switch is closed, but when you open it doesn't induction increase/decrease the flux? Or will the flux just become 0 and hence give us $1.4 \times 10^{-5}$ ?
*Why do we have to take the $R_{tot}$? What does the resistance of the inductor have to do with the lamp?
p.s. - This question can't be asked on electronics SE, since their site doesn't allow for such a question.
| I think the other answers cover your question about the change in flux. As to why $U/R_{tot}$ is used instead of $U/R_{lamp}$... it's not quite clear whether you're expected to find the current at a particular point on the circuit, or what point that might be. If I had to pick one current to characterize the circuit, it would be the current through a bit of wire that didn't have any parallel counterparts (or equivalently the sum of the currents through a set of parallel wires). The voltage drop has to be equal across both the lamp and the inductor, so the current is just $I_{tot}=V/R_{tot}$. If after this you're interested in the current through the lamp, you can find it easily by calculating the current through the inductor individually ($I=V/R_{ind}$), then $I_{lamp}=I_{tot}-I_{ind}$.
In your original question you don't say anything about needing to find the current through the lamp, but in one of your comments you imply that that's the quantity you're after - is that actually what the original problem statement asks for?
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Why does unbroken supersymmetry imply the absence of tachyons? Just a quick question, same as in the title. I'm trying to understand stable D-branes.
| In theories with unbroken supersymmetry, the energy can be written as
$$E=\sum_i c_i Q_i Q_i $$
where $Q_i$ are some Hermitian supercharges and the coefficients are positive. This is a sum of squares of Hermitian operators which is why it's positively semidefinite. It can't be negative.
Tachyons obey $E^2-p^2=m^2$ for a negative $m^2$ so the 4-momentum (or $d$-momentum) is a spacelike vector, assuming it is real. For spacelike vectors, one may always change $E$ to a negative number by a Lorentz transformation, but no such states with $E\lt 0$ may exist due to the positivity above, which means tachyons can't exist at all.
Spacetime supersymmetry was of course the feature that eliminated the spacetime tachyons of bosonic string theory when it was superseded by superstring theory in the early 1970s.
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What exactly are we doing when we set $c=1$? I understand the idea of swapping from unit systems, say from $\mathrm{m\ s^{-1}}$ to $\mathrm{km\ s^{-1}}$, but why can we just delete the units altogether?
My question is: what exactly are we doing when we say that $c=1$?
| Not much, although in a given reference frame, velocity being equal to zero is (usually) a fixed point on the scale. So you still have two points, which is enough to define this scale.
When looking at the Lorentz Transformations for something like $SO(1,3)^{\uparrow}$ (the space that special relativity takes place in), you'll see the term $\beta$ (defined as $v/c$) equate to the magnitude of $v$, as c is treated as an identity element. The domain of $v$ is therefore placed between zero and $c$, which reveals the primary reason people may set $c=1$: to represent all $v$-s as a fraction of $c$
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How to guarantee that a kilogram of antimatter will quickly annihilate another kilogram of matter? What I mean is, suppose we could somehow get a kilogram of antimatter and contain it safely. Now lets say we want to make a bomb using this kilogram, now, we have two ways, either store another kilogram of matter inside the bomb itself and let the matter and the antimatter touch each others when we want to bomb to detonate, or just expose the kilogram to the air and it will explode. But, my question here is simple, either of the previously mentioned ways will just allow the first particles touching each others to annihilate and sending the rest matter and antimatter in opposite ways making the reaction harder and slower to continue. I know eventually the whole kilogram will be annihilated, but it's all about reaction speed in explosives and that's the main difference between nuclear reactors and nuclear weapons.
So now, is there a way to ensure that the matter and antimatter will completely annihilate each others with a high rate of reaction ?
| You face exactly the same problem as the makers of the first explosives. Although these days we use explosives that react intramolecularly, the original explosives like gunpowder were made from a mixture of an oxidising agent (potassium nitrate) and a reducing agent (sulphur and charcoal). To get the gunpowder to go bang rather than fizz you had to mix the reagents extremely intimately so that the transport of the reacting molecules/atoms was faster than the explosion.
So basically you need an extremely intimate mixture of the matter and antimatter. That is going to make containment fun!
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How do transformers work? A transformer is basically a primary inductor connected to a voltage $U_P$ which you want to transform. You also have an iron rod and a secondary inductor. So when there is a current $I_P$ the iron rods becomes magnetic. When you connect the primary inductor to AC, that means that you'll have a changing current, which causes a change in flux which causes induction. My question is, is $U_S$ just the induction voltage created by the iron rod?
| It's always handy to have some background information:
In Europe the mains voltage is 230 Volts, which is too much for a lamp for example, so it needs to be lowered to for example 12 Volts. This is done by using a transformer.
The primary winding is connected to the mains voltage of 230 Volts. The AC in this primary winding causes a varying magnetic flux in the iron rod (core) which on its turn creates a varying magnetic flux through the secondary winding. Because of electromagnetic induction a voltage is induced in the secondary winding. The primary winding has more turns than the secondary winding which causes the secondary voltage to be lower than the primary voltage:
$$ \dfrac{N_P}{N_S} = \dfrac{U_P}{U_S} = \dfrac {I_S}{I_P}$$
You can see that be decreasing/increasing the number of turns in the windings you can control the voltage created by electromagnetic induction.
Here is an illustration with an example ($U_S = 220V, U_P = 110 V$)
You can see how simple it really is.
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What is $v \, dp$ work and when do I use it? I am a little confused, from the first law of thermodynamics (energy conservation)
$$\Delta E = \delta Q - \delta W $$
If the amount of work done is a volume expansion of a gas in, say a piston cylinder instrument at constant pressure,
$$\Delta E = \delta Q - p \, dv$$
Here $p$ is the constant pressure and $dv$ is the change in (specific) volume.
So, when do I take into account
$$\delta W = d(pv) = p \, dv + v \, dp$$
I am assuming that for cases of boundary work, at constant pressure, the $v \, dp$ term is zero.
So under what conditions should I consider the $v \, dp$ term?
| I hate answering my own question but I'd like to share this with you as it is definitive:
$$\mathrm{d}h = \mathrm{d}u + \mathrm{d} (p v)$$
$$h_2 - h_1 = u_2 - u_1 + p(v_2 - v_1) + v (p_2 - p_1)$$
Now, for a pump working in the compressed liquid (subcooled liquid zone), it is noted that the change in specific volume $v$ is minimal when the pump adds pressure energy to the liquid. It is minimal because liquid is incompressible.
This leaves us with the following terms:
$$h_2 - h_1 = u_2 - u_1 + v (p_2 - p_1)$$
$$\Delta h = C (T_2 - T_1) + v(p_2 - p_1)$$
An other assumption that is made that the fluid flow through the pump does not raise the temperature much which would allow us to drop the $C (T_2 - T_1)$ term where $C$ is the specific heat of the liquid.
*
*This leaves us with the equation that was troubling me:
$$h_2 - h_1 = v (p_2 -p_1)$$
So this is valid for a pump working in the compressed liquid region of the $P-v$ or $T-v$ diagram.
If you'd like to comment further, please do so. This has been quite enlightening for me already! :)
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The viscous force between the layers of liquid is same, then why there is variation in the velocities of its layers? I have learned in my textbook that when the liquid flows the bottom layer of the liquid never moves because of friction, but the upper layers move with increasing velocities how it is possible if the viscous force between all these layers is same
| Consider that
$$ \mbox{Force} \propto \mbox{Velocity Gradient} $$
Equal force means, the same velocity gradient, i.e. linear distribution of velocities across the flow. The flow near the boundary has zero velocity and so velocity increases linearly the further away you go from the boundary.
| {
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what is use of relativistic action? this is relativistic action:
$$S=\int_C \mathcal {L}dt$$
where the $\mathcal{L}$ is $-m_oc^2\gamma^{-1}$
what is use of relativistic action!?
| The use of an action is do derive all the dynamical equations of a theory from the least-action principle, $\delta S=0$ (action is minimized along the right path). Quantum mechanically, the use of an action is to derive the transition amplitudes from an initial state to the final state by summing over histories weighted by $\exp(iS/\hbar)$ in a quantum theory.
Similarly. The use of a relativistic action is do derive all the dynamical equations of a relativistic theory from the least-action principle, $\delta S=0$ (action is minimized along the right path). Quantum mechanically, the use of a relativistic action is to derive the transition amplitudes from an initial state to the final state by summing over histories weighted by $\exp(iS/\hbar)$ in a relativistic quantum theory.
Yes, I used the copy-and-paste and added the adjective "relativistic" to the second paragraph. ;-)
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Evaporating Black holes and entropy
*
*I'm told that for Black Holes, when they radiate (Hawking radiation) particles and anti-particles) split at the event horizon, one going to infinity, the other into the BH. They then lose mass. How is that possible? Wouldn't their masses increase, since they are absorbing particle or anti-particles?
*Also, their entropy varies as their masses, and since entropy increases, wouldn't their masses also be increasing, along with their area? How does that reconcile with case 1?
| 1) In the tunneling picture, there are two scenarios:
i)a virtual particle/antiparticle pair is created just outside of the horizon. The negative energy particle then tunnels into the horizon and the positive energy particle is radiated away.
ii) a virtual particle/antiparticle pair is created just inside of the horizon. The positive energy particle then tunnels out of the horizon and the negative energy particle remains inside.
In (i) although a particle is "falling in" to the horizon, it's a negative energy particle and hence results in a mass loss.
There are equivalent ways to derive the results without using these virtual particles though. These other ways involve working out the vacuum state of the asymptotic observer, who sees a flux of particles coming out of the hole. Either way the mass decreases.
2) The entropy of a closed system must not decrease. The BH on its own doesn't constitute a closed system. You need also to consider the states of the radiation. See the discussion of the "Generalized Second Law" here.
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Coulomb gauge fixing and "normalizability" The Setup
Let Greek indices be summed over $0,1,\dots, d$ and Latin indices over $1,2,\dots, d$. Consider a vector potential $A_\mu$ on $\mathbb R^{d,1}$ defined to gauge transform as
$$
A_\mu\to A_\mu'=A_\mu+\partial_\mu\theta
$$
for some real-valued function $\theta$ on $\mathbb R^{d,1}$. The usual claim about Coulomb gauge fixing is that the condition
$$
\partial^i A_i = 0
$$
serves to fix the gauge in the sense that $\partial^iA_i' = 0$ only if $\theta = 0$. The usual argument for this (as far as I am aware) is that $\partial^i A'_i =\partial^iA_i + \partial^i\partial_i\theta$, so the Coulomb gauge conditions on $A_\mu$ and $A_\mu'$ give $\partial^i\partial_i\theta=0$, but the only sufficiently smooth, normalizable (Lesbegue-integrable?) solution to this (Laplace's) equation on $\mathbb R^d$ is $\theta(t,\vec x)=0$ for all $\vec x\in\mathbb R^d$.
My Question
What, if any, is the physical justification for the smoothness and normalizability constraints on the gauge function $\theta$?
EDIT 01/26/2013
Motivated by some of the comments, I'd like to add the following question: are there physically interesting examples in which the gauge function $\theta$ fails to be smooth and/or normalizable? References with more details would be appreciated. Lubos mentioned that perhaps monopoles or solitons could be involved in such cases; I'd like to know more!
Cheers!
| It means that the gauge ambiguity is practically removed in the Coulomb gauge if you deal with a "nice" $\mathbf{A}$ (which is your purpose).
However, it does not mean you only deal with the radiation (propagating solutions). Transversal $\mathbf{A}$ is different from zero for a uniformly moving charge too.
| {
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Is it possible for a physical object to have an irrational length? Suppose I have a caliper that is infinitely precise. Also suppose that this caliper returns not a number, but rather whether the precise length is rational or irrational.
If I were to use this caliper to measure any small object, would the caliper ever return an irrational number, or would the true dimensions of physical objects be constrained to rational numbers?
| The hypotenuse of a right angled triangle with legs 1 is irrational.
Alternatively, consider a pyramid. As you take measurements of the 'base length' towards the apex, you get a continuous sets of values. One of these must be irrational.
Of course, you can then start an argument about what 'physical' object is, and if length is truly continuous, or it has to be discrete because it is constructed by atoms.
| {
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Exact diagonalization of graphene's tight binding Hamiltonian While directly diagonalize graphene's tight binding Hamiltonian, which is numerical. We have to use a finite-sized graphene.
So how to deal with boundary conditions? The usual solutions are zigzag or armchair condition, but to make our model more realistic to real infinite graphene plane, how about using periodic condition at boundaries while exactly diagonalizing the tight binding Hamiltonian?
| You're mixing two things here. One is what the structure of the boundary is, e.g. armchair or zigzag. The other is what the wavefunction does at the boundary.
For your finite size cluster of carbon atoms, you have to decide what shape it has, which basically means deciding how many lattice sites you include and where you put them. This would decide whether your finite cluster has armchair or zigzag boundaries.
If you treat these boundaries as "real" system boundaries, then you have closed boundary condition. But if you want periodic boundary conditions, you have to decide on a way to map one boundary to the other, i.e., how does everything get "wrapped around".
E.g. for a rectangular lattice this is very easy, you just declare that electrons leaving the lattice to the left re-enter it on the right, basically stating that site $0$ and site $N$ are the same.
For your graphene lattice, you can do the same, you just have to be careful that the boundary of your finite sized cluster is such that this wrapping actually works. That means: For every atom on the boundary of your finite cell, there must still be a way to identify its 6 unique "nearest neighbors", some of which are "actual" neighbors, and some appear on the opposite site of the cluster.
| {
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Resonant inductive coupling and Schumann resonances I was reading about WiTricity (http://en.wikipedia.org/wiki/WiTricity) a technology developed by MIT to wirelessly transmit electricity through resonance, and I have this question:
Given the phenomenon of resonant inductive coupling which wikipedia defines as:
the near field wireless transmission of electrical energy between two
coils that are tuned to resonate at the same frequency. http://en.wikipedia.org/wiki/Resonant_inductive_coupling
And the Schumann resonances of the earth ( ~7.83Hz, see wikipedia), would it be theoretically possible to create a coil that resonates at the same frequency or one of it's harmonics (7.83, 14.3, 20.8, 27.3 and 33.8 Hz) to generate electricity?
I have a feeling that these wavelengths may be too big to capture via resonance (they are as large as the circumference of the earth if I understand it correctly), so alternatively would it be possible to create a coil that resonates with one of the EM waves that the sun sends our way?
| You Must reflect the signal, "back upon itself"for it to amplify.Like a reflective inner surface of a tube surrounding it.Like a Laser does.Then you must pin the charge, in Maxwell Ultra Capacitors.The Resonant coupling is a sustained resonance.A resonant tank circuit would work.Positive to North, and Negative to South.For the transfer of wireless energy, you must create a powerful (beam) again like a laser/Maser beam, "focused via a lens".They are more or less broadcasting it, propagating it, which is not the answer for powerful beam energy,and long distance transfer.
| {
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Can low-gravity planets sustain a breathable atmosphere? If astronauts could deliver a large quantity of breathable air to somewhere with lower gravity, such as Earth's moon, would the air form an atmosphere, or would it float away and disappear? Is there a minimum amount of gravity necessary to trap a breathable atmosphere on a planet?
| The gravity of a planet holds the atmosphere in place. The moon doesn't have enough mass / gravity to do so. If you moved air to the moon there's so little gravity the air would simply float away.
| {
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Equivalence principle question I understand the equivalence principle as "The physics in a freely-falling small laboratory is that of special relativity (SR)." But I'm not quite sure why this is equivalent to "One cannot tell whether a laboratory on Earth is not actually in a rocket accelerating at 1 g".
| The way it makes sense to me is that if you shoot a ray of light across a room inside an accelerating rocket ship, the ray will fall some distance (relative to the room). Therefore on Earth the ray of light must also fall (ostensibly "due to gravity"), otherwise you would know whether you are in a ship or not based on whether light does or does not fall. And I imagine there would be some kind of physical consequence to that. The following video seems to imply that the consequence would be that inertial mass and gravitational mass would not be the same, which makes intuitive sense to me.
This is a video from "The Mechanical Universe." It's a good overview of the deduction.
http://www.youtube.com/watch?v=VbOKxkj0_wc&feature=youtu.be&t=13m24s
| {
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What would be the effect of a slanted muzzle on the trajectory of a bullet? Let's say I cut off the end of a gun barrel at a 45° angle: What would the effect be on the trajectory of a bullet fired through that barrel?
Would the bullet be less stable (I guess)? Would it make the gun fire with an angle, and would that be toward the "small" end?
| I think the main effect would be that the muzzle velocity will be lower.
This is because the expanding gases of the gun power will have a shorter distance along which they push the bullet. Once the back of the bullet emerges out the short side of the barrel, the gases will escape and depressurize, loosing the ability to keep pushing the bullet.
The main effects of the lower muzzle velocity will be lowering the range of the bullet (i.e. it will fall to the ground after traveling a shorter horizontal distance), and lowering the penetration capability of the bullet.
I don't think this will make the gun fire with an angle.
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When do I apply Significant figures in physics calculations? I'm a little confused as to when to use significant figures for my physics class. For example, I'm asked to find the average speed of a race car that travels around a circular track with a radius of $500~\mathrm{m}$ in $50~\mathrm{s}$.
Would I need to apply the rules of significant figures to this step of the problem?
$$ C = 2\pi (1000~\mathrm{m}) = 6283.19 $$
Or do I just need to apply significant figures at this step?
$$ \text{Average speed} = \frac{6283.19~\mathrm{m}}{50~\mathrm{s}} = 125.664~\mathrm{m}/\mathrm{s} $$
Should I round $125.664~\mathrm{m}/\mathrm{s}$ to $130~\mathrm{m}/\mathrm{s}$ since the number with the least amount of significant figures is two?
| I have been taught and continue to teach that you should record (and use) the answers to your intermediate calculations (and/or conversions) with one extra significant digit beyond the number of significant digits that will be in your final answer.
| {
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Which theory predicts that dubbed tires have more friction? The physical reasoning I suppose could be that more contact areas mean more and higher friction. But is there an actual formula or a more mathematical explanation?
Application for bicycling was my question https://bicycles.stackexchange.com/questions/14319/could-i-switch-tires-on-an-mtb-to-tires-with-less-friction-if-i-mostly-use-the-m
| For a rolling tyre the resistance (it's not really friction) is due to viscous losses in the rubber as the tyre is deformed.
If you take a piece of rubber and put work in to stretch it then get work out as it relaxes, the work you get back is less than the work you put in. The balance goes into heating the rubber. If the piece of rubber is part of a tyre, this loss of energy means you have to put energy in to keep your vehicle moving at a constant speed i.e. you have to apply a force. That force is what you feel as rolling resistance. If you touch your car tyres after you've driven the car you'll find they are warm because they've been heated by the losses within them.
The more the tyre deforms, the more energy is lost, and therefore the greater the resistance. So if you use a smooth tyre and pump it to a very hard pressure the rolling resistance will be low. If you use a knobbly tyre and run it at a low pressure the resistance will be high.
You ask if there's a way to describe this mathematically, and yes there is assuming you know the material properties of the rubber you're using. However you'd need to use a finite element analysis.
| {
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London into Australia in 90 minutes Me and my friend are having a debate on whether it would be possible for a human to travel at 15,000 miles an hour from London to Australia in the matter of 90 minutes. Would a human be able to survive travel in such at fast speeds knowing he will have to overcome immense amount of g's. Basically is it possible for a human? or will he suffer death in the process?
| Gagarin flew around the world in 90 minutes 50 years ago, apparently without serious health problems.EDIT: OK, it was 108 minutes.
| {
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Is time dilation an illusion? It is said that we can verify time dilation by flying a very accurate clock on a fast jet or spaceship and prove that it registers less time than the clocks on earth. However, the clocks on earth would be moving relative to the clock on the spaceship, and since time always dilates and never goes faster regardless of the direction of relative motion, the clocks on earth should register less time than the clock on the spaceship.
Is this true? Whenever there is a fast-moving object such as a rocket do all clocks on earth really become slow?
If the rocket with the clock landed after moving at relativistic speed, would its clock and the earth's clock again show the same time since during its travel both appeared slow to each other?
Or is all this just an illusion, ie. the clocks just appear to be slow to each other but in actually run at normal speed, and neither is behind when the rocket actually lands?
| You're referring to what is commonly known as the twin paradox. The Wikipedia page provides several different ways of analysing the situation, but one way to look at it is this -
When the clock on the spaceship leaves earth, it'll experience an acceleration (even if it's a really small acceleration for a long time, or a huge acceleration for a small time) to reach relativistic speeds, and will experience an acceleration again when it has to turn around to return to earth.
Special relativity only claims that inertial frames of reference are equivalent. Since one clock experiences acceleration, and hence is in a non-inertial frame momentarily, the two situations aren't equivalent.
If you view the time dilation due to acceleration as a gravitational acceleration (principle of equivalence) and then do the calculations, the results obtained for both the clock on the earth as well as the clock on the spaceship agree. Again, the wikipedia link contains in detail this argument, as well as other arguments.
| {
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Thought experiment regarding an object approaching a mirror Here's a thought experiment I came up with in class today when my mind drifted (I however highly doubt I'm the first to think about this since it is pretty rudimentary) :
Let's say superman approaches a (flat) mirror at 200,000 km/s. At what speed does superman's image approach him?
This has given me a big headache. I have pondered a lot of solutions, but none of them really convinced me. What is the correct solution to this thought experiment?
| Superman's image approaches him at c, it will be blue shifted a bit towards xrays.
Luckily Superman has xray vision so he should be able to see himself.
For the side question of what you would see at c, try this link.
http://www.space.com/19268-star-wars-hyperspace-physics-reality.html
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What grounds the difference between space and time? We experience space and time very differently. From the point of view of physics, what fundamentally grounds this difference?
Dimensionality (the fact that there are three spatial dimensions but only one temporal) surely cannot be sufficient, as there are tentative proposals among string theories for models with multiple spatial dimensions, and two time dimensions.
One of the most lauded answers in the philosophy of spacetime has to do with the fact that our laws predict temporally, rather than spatially. That is to say, if we are given enough information about the state of the world at one moment, we can predict (quantum considerations aside) the future state of the world. However if we reverse the roles of time and space here, and instead give information about a single point of space for all of time, it seems we cannot predict spatially.
Are there equations in physics that can be considered to predict across space (for a given time)?
| What makes the difference is that the order of Time is made by the presence of memory. That is, consciousness creates the "arrow of Time".
| {
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How does the correlation length of weather emerge? The question is pretty simple: If I know the weather where I stand, I can estimate the weather 5 meters or 1 km away away pretty well, but I'll have a hard time guessing what the weather is, say, 50 km away.
Therefore, it seems that the climatic system has a length-scale. Where does it come from? Navier-Stokes equations do not feature an internal length scale, and it doesn't seem that the scale comes from earth's radius either.
| Part of your scale comes from the observer, the time and space averaging of the observations.
You say the weather 10 meters distant is pretty similar to here, but look a bit closer, and it is not. The wind 10 meters from where you are can be quite different from where you stand. A sunny rock can be dry and hot 1 meter away from a shady moist spot with moss growing.
| {
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How accurate are our calculations about distant stars keeping in mind their enormous distances? Since many stars are hundreds of light years away from the Earth and therefore, what we observe of them today is really their distant past, how can we say anything with certainty about their composition, size and nature? Betelguese, for example, is said to be in the last stage of stellar evolution, but taking into consideration its approximately 640 light years of distance from the Earth, aren't we actually making assertions about the star that are 640 years old? If that is the case, how can we claim to know the actual present status of the star?
| Well, we don't claim to know the status of the star at the present time. If the star is close enough (within our galaxy, or local neighbourhood), we measure the distance in light years, or parsecs. If they're further away than that, it's easier to quantify their distance in terms of a redshift.
When we say that a star has some properties (size, temperature etc.) in that statement is the knowledge that the light received is delayed, so it won't reflect the state of the star right now.
Even though knowing the state of a star as it was several thousand years ago may not seem like it's a useful thing, it's the only tool we have to probe the evolution of stars, and the evolution of the universe. Since the further away something is, it will reflect conditions of the universe at earlier and earlier times. Which is where we can test cosmological theories of the evolution of the universe, or our theories of the evolution of galactic or stellar systems.
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What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass? Say I have a rigid body in space. I've read that if I during some short time interval apply a force on the body at some point which is not in line with the center of mass, it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass.
What is the proof of this?
|
I've read that if I during some short time interval apply a force on the body at some point which is not in line with the center of mass, it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass.
To my understanding, your question is flawed. If a single force is applied to a rigid body under the influence of no other forces, either:
*
*The line of action of the force passes through the center of mass, causing a pure translation and no rotation
*The line of action of force does not pass through the center of mass, in which case you end up with a pure rotation about an axis which does not pass through the center of mass. In other words the instantaneous axis of zero velocity induced by a single force can never be the center of mass.
If you apply an eccentric force, the center of mass of the body will undergo a linear acceleration, and the body itself will undergo an angular acceleration. In a fixed reference frame, this can be viewed as a pure rotation about a certain point, but this point will never be the center of mass of the body.
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Why is ski jumping not suicidal? At least on television, ski jumpers seem to fall great vertical distances before they hit the ground - at least a few dozen meters, though I couldn't find exact distances via a quick search. And yet they almost always land on their feet as if they just fell two or three meters. (Here's a whole lot of footage from the Vancouver Olympics if you need to refresh your memory.)
Without going into the level of equations (which I wouldn't understand), why are ski jumpers able to fall such great heights without seriously injuring themselves?
| A main part is because they are landing on a slope. It means that for most, if not all, of the jump they are travelling parallel to the ground. The jump looks impressive because a) it is and b) the camera angles don't show this clearly.
This slope means that when the jumper lands they can carry on moving forwards with virtually the same velocity they had when they were in the air. The slope isn't constant, so if you don't jump far enough or jump too far the landing can be be dangerous.
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Confused over complex representation of the wave My quantum mechanics textbook says that the following is a representation of a wave traveling in the +$x$ direction:$$\Psi(x,t)=Ae^{i\left(kx-\omega t\right)}\tag1$$
I'm having trouble visualizing this because of the imaginary part. I can see that (1) can be written as:$$\Psi(x,t)=A \left[\cos(kx-\omega t)+i\sin(kx-\omega t)\right]\tag2$$
Therefore, it looks like the real part is indeed a wave traveling in the +$x$ direction. But what about the imaginary part? The way I think of it, a wave is a physical "thing" but equation (2) doesn't map neatly into my conception of the wave, due to the imaginary part. If anyone could shed some light on this kind of representation, I would appreciate it.
| The trick is in hiding information about phase of wave in this kind of representation. There is a nice appendix from a book about holography :
http://onlinelibrary.wiley.com/doi/10.1002/9783527619139.app1/pdf - part A.3
It stays that:
For general wave function $ y = A · cos (\omega t − kr + \alpha) $, $kr$ and $\alpha$ can be combined in one phase $\phi$, so that $ y = A · cos (\omega t - \phi) $. Here function explicitly depends on the time and phase. It can be transformed such that, it explicitly depends on one of these parameters.
By formula for cosine of difference of arguments :
$ y = A · \cos \phi · \cos ωt + A · \sin \phi · \sin ωt $
or
$ y = A_1 · \cos \phi + A_2 · \sin \phi$.
Using represenatinon of complex number, we can rewrite equation above as
$ y = A · \cos \phi + i · A · \sin \phi$
and by Euler's law we get :
$ y = A · e^{i · \phi}$.
| {
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What is enstrophy? In turbulence, the enstrophy of a flow in a domain $\mathcal{D} \subset \mathbb{R}^{D}$
$$
\mathcal{E} = \int_{\mathcal{D}} |\vec{\nabla} \times \, \vec{v}|^2 d^{D}x
$$
appears sometimes, it's cool and has cute properties in 2d (see this answer about turbulent flows), etc.
Now, what does it represent physically? Okay, it's the integral of the vorticity squared, but has it a physically understandable interpretation? Why is it interesting to study that?
| From mathematics point of view, it is a surface integral of the scalar quantity |Curl v $|^2$. The physical meaning of it, in the context of fluid dynamics in 2-D or 3-D, is that it has the units $(m/s)^2$ which when multiplied by the density of a fluid represents some form of energy. As for the meaning of the intgral in a D-dimentional space, the problem is more of mathematics than physics. For a physical interpretation of this integral in D-dimensions, one needs to know first what the Curl v itself could represent, in a similar context to that of fluid flow in D-dimensional space, and then wonder what the physical meaning of its intgral on D-dimensional surface could mean. Its dimensionality, $(m/s)^D$, does not seem to relate to something physically meaningful for arbitrary values of D. I am not aware of a good physical analogy of this in D dimensions.
As for the the word itself, it derives from the Greek $\epsilon \nu - \sigma \tau\rho o\phi\eta$. It means: "during the rotation",in a state of rotation or "while turning". In other words, the word makes reference to turning, rotation. I hope this helps you visualise the meaning of the word in relation to the turbulant motion of a fluid.
Similar to the way Entropy is derived: $E\nu\tau\rho o\pi\eta$ which means: in a state of change, during a change.
| {
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Identity as a trivial reducible representation In particle physics, I was taught that a representation of a group is a function $r: group \rightarrow matrices\,(n\times n)$ such that $r(g_1)r(g_2)=r(g_1g_2)$ and $r(e)=I_{n\times n}$. Then, that a representation is reducible when you can find a matrix $A$ such that $Ar(g)A^{-1}$ is in diagonal-block form for every element of the group.
Then the professor tried to find in complicated ways reducible representations of $SO(N)$, $SU(N)$ and so on. But the trivial function that assigns $I_{n\times n}$ to every value of $g$ is not already a reducible transformation? I know it must be somehow useless, but what did I lose?
| Well if you're just looking for one example of a reducible representation of a group, then that's a fine one. However, representation theory of groups has a bunch of wonderful complexity that is illuminated by studying other examples of reducible representations with a lot more structure.
For example, addition of angular momentum in quantum mechanics involves writing reducible representations of $\mathrm{SU}(2)$ as directs sums of irreducible ones.
Cheers!
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Why does the presence of a battery change whether a circuit is in parallel or series? If I take a closed circuit with two capacitors and a voltage difference, the circuit is apparently in parallel, but if I introduce a battery, the circuit is in series. Why does the presence of the battery make a difference?
Here's a diagram of what I mean:
According to my book, the left circuit is in parallel, but the right circuit is in series. I don't see why—after all, for any voltage to run through the circuit on the left, it has to pass through C1 before C2 (or vice versa).
| If you look at the structure of the circuit on the left, and where voltage must be, you will see that the wire connecting C1 and C2 ensures that the plates connected to point A are at one voltage, let's call it VA, and the other plates connected to voltage VB.
So this can be thought of as two batteries , each with voltage differential of VA-VB in parallel.
For the second circuit, the two capacitors are in series with respect to the battery (as mentioned by @FrankH)
| {
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Maxwell-Boltzmann velocity PDF to CDF I asked on Math.SE and was advised to try here instead.
I need to draw from a Maxwell-Boltzmann velocity distribution to initialise a molecular dynamics simulation. I have the PDF but I'm having difficulty finding the correct CDF so that I can make random draws from it.
The PDF I am using using is:
$$f(v)=\sqrt \frac{m}{2\pi kT} \times exp \left( \frac{-mv^2}{2kT}\right) $$
I am told that to find the CDF from the PDF we perform:
$$CDF(x)= \int_{-\infty}^x PDF(x) dx $$
After integrating $ f(v) $ I get:
$$ CDF(v)= \sqrt \frac{m}{2\pi kT} \times \left( \frac{\sqrt\pi\times erf \left( \frac{mv}{2\pi kT} \right) }{2\times \left( \frac{m}{2kT} \right)} \right) $$
$$CDF(v)= _{-\infty} ^{x} \left[ {\sqrt \frac{m}{2\pi kT} \times \left( \frac{\sqrt\pi\times erf \left( \frac{mv}{2\pi kT} \right) }{2\times \left( \frac{m}{2kT} \right)} \right)} \right] $$
*
*After I reach this point I am unable to proceed as I do not know how to evaluate something between $x$ and ${-\infty}$.
*I am also concerned that I have not done the integration correctly.
*I want to implement the CDF in C++ in the end so I can draw from it. Does anyone know if there will be a problem with doing this because of the erf, or will I be alright with this GSL implimentation ?
Thanks for your time.
@bryansis2010 on Math.SE says that I can evaluate in the range $x$ to $0$ instead of $-\infty$ as we do not drop below 0 Kelvin.
Would this then make the CDF:
$$ CDF(v)= \sqrt \frac{m}{2\pi kT} \times \left( \frac{\sqrt\pi\times erf \left( \frac{mv}{2\pi kT} \right) }{2\times \left( \frac{m}{2kT} \right)} \right) $$
as $erf(0)=0$
Is this correct?
| The solution is to realise that that function is merely a Gaussian. In fact Each component of the velocity vector has a normal distribution with mean =0 and st-dev $\sqrt {kT/m}$.
All that is left to do at that point is to get the Gaussian CDF (well known) and sample from it, making sure to plug in our mas and temperature.
$$CDF(x)=\frac{1}{2}\times \left[ 1+ erf\left( \frac{x-\mu}{\sqrt{2{\sigma}^2}} \right) \right]$$
GSL implements gsl_cdf_ugaussian_P (double x) here
| {
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What would the collision of two photons look like? Could someone explain to me what the collision of two photons would look like? Will they behave like,
*
*Electromagnetic waves: they will interfere with each other and keep their wave nature
*Particles: they will bounce like classical balls
I assume that energy of that system is too small to make creation of pairs possible.
| A lowest order QED Feynman diagram for the process photon + photon $\rightarrow$ electron + positron looks like shown below (the time axis is the horizontal axis).
From the point of view of energy conservation, this process is only possible if sum of the energy of the photons is above twice the electron mass. In the center of mass frame of the di-photon system, the photons need to have at least 511 keV.
| {
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Difference between torque and moment What is the difference between torque and moment? I would like to see mathematical definitions for both quantities.
I also do not prefer definitions like "It is the tendancy..../It is a measure of ...."
To make my question clearer:
Let $D\subseteq\mathbb{R}^3$ be the volume occupied by a certain rigid body. If there are forces $F_1,F_2,....,F_n$ acting at position vectors $r_1,r_2,...,r_n$. Can you use these to define torque and moment ?
| Moment is bending due to linear force and the distance from the axis is perpendicular whereas in torque rotation takes place beyond 360 degrees.
| {
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Can we make images of single atoms? I was wondering how far in imaging physics had gotten. Do we hold the technology to actually take a picture of, say, an alpha particle, or even a single atom?
I realise we aren't talking about camera pictures, so what kind of imaging techniques have taken images/is most likely to be the technique to take an image of a single atom?
| Yes - depending on what you mean by image (if you squint you can read IBM in individual atoms)
Atomic Force Microscopy measures the position of individual atoms.
There is a presentation of IBM's nantotech research
| {
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Is cosmic background radiation dark-matter and/or dark-energy? Dumb question alert: Is it possible that the cosmic background radiation might be the source of dark-matter and/or dark-energy? What is the mass of the background radiation in the known universe?
| I can only elaborate on what's already been said, but no, dark matter and dark energy do not exist as part as the CBR. The CBR we conventionally think of was constructed based on observed EM radiation. Dark matter and energy are only detected currently through their gravitational effects, and won't show up as radiation in the conventional effects, (although I am no expert, I am led to understand that under some models of the theoretical graviton particle, dark matter might radiate gravitons, but don't cite me on that).
| {
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Proper notation for normalized scalar? I have not been able to find a resource to tell me the standard notation for a normalized scalar value. Normalized vectors (i.e. unit vectors) are typically denoted by placing a hat over the variable, something like:
$${\bf \hat{e} = \dfrac{e}{||e||} }$$
However, does the same apply to normalizing (and nondimensionalizing) a scalar? Would it be correct to write:
$$ \hat{L} = \dfrac{L}{L_0} $$
This is assuming that $L$ and $L_0$ are just scalar lengths. If I am defining my own notation, is it verboten to call this something like $\bar{L}$ (with a bar)?
If it makes any difference, I am a mechanical engineer and this will be going in my thesis.
| There is not, to my knowledge, a uniform standard on this subject. I have seen normalized quantities expressed by adding a twiddle (tilde) over the character (as in $\tilde{A}=A/A_0$), but I this notation is often reserved to indicate a time-varying quantity, instead. Unless someone knows of a strong standard in this regard, I'd say your best bet is to simply define the notation you will be using.
| {
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Is light affected by gravity? Why? I would like to know if light is affected by gravity, also, I would like to know what is the correct definition of gravity:
"A force that attracts bodies with mass" or "a force that attracts bodies with energy, such as light"?
Is light massless after all?
| I'd like to answer the first question "Is light affect by gravity? Why?" because I remembered an adorable thought experiment.
In the spirit of the equivalence principle consider an observer in a closed box. As we all know, the observer inside that box would be unable to tell (neglecting tidal effects) the difference between a uniformly accelerated box (think rockets) or gravity (think the Earth).
Let's say that the observer is in outer space and he's being accelerated with some rockets. Inside the box, he shoots a beam of light from one wall to the other across the rocket, in the direction perpendicular to the direction of perceived inertial acceleration/direction of travel. It's obvious that the light will have a curved trajectory because as it was travelling from one wall to the other, the rocket moved a bit.
By the equivalence principle, the light should also bend if the observer was in a box on the surface of Earth, because otherwise, he would have a way to distinguish a uniformly accelerated frame from a frame in a gravitational field.
What did I tell you, adorable, right? :)
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Why is electric potential scalar? I can't conceptually visualize why it would be so. Say you have two point charges of equal charge and a point right in the middle of them. The potential of that charge, mathematically, is proportional to the sum of their charges over distance from the point ($q/r$). But intuitively, my thought process keeps going back to the concept of direction and how the electric field at that point would be zero. So why would the electric fields cancel while the electric potentials just add up algebraically?
| When we bring a test charge let us say (+q) to a certain point, we exert some force. Our exerted force has to be equal or greater to the force exerted by the electric field in order to overcome it. Thus the forces being opposite and equal cancel each other and hence the direction is undetermined.
Since it is a compulsion for vector to have both magnitude and direction, hence here there is no direction but magnitude and hence we call it a scalar quantity.
| {
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How is possible for current to flow so fast when charge flows so slow? How is it possible for current to flow so fast when charge flows so slowly?
We know electrons travel very slowly while charge travels at ~the speed of light.
| As you may've noticed, the drift velocity of electric charges is $\sim {10}^{-5} \frac{\mathrm{m}}{\mathrm{s}} ,$ which is much slower than the flow of electric current.
So, how could electric current travel near the speed of light? It's because the charges' electric field is propagating near the speed of light.
| {
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Can a black hole bounce? Is there a limit to the amount of matter that a black hole can accrete per second and if so could a certain sized black hole bounce off a dense enough surface?
| Let's answer the first part of your question first. In general, astrophysical objects are limited in the amount they can accrete by the Eddington limit.
What happens is as matter is accreted onto the object (black hole in our case), it heats up due to conversion of gravitational potential energy into kinetic thermal energy. This hot matter emits photons whose total energy per time is called the luminosity of the object. If this luminosity is high enough (and you should get a higher luminosity for more accreted matter) then the outward pressure of the photons on the matter actually can overcome the gravitational inward pull. This point (rate of matter accretion) is called the Eddington limit.
To partially answer the second part of your question, if the black hole had non-zero total net electric charge, then it could certainly bounce off another material of sufficient density. However if the charge on the black hole was identically zero, there would be no repulsive force to create a bounce.
| {
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Schrodinger equation in term of Fokker-Planck equation From Wikipedia on the Fokker-Planck equation:
$$\tag{1}\frac{\partial }{\partial t}f\left( x^{\prime },t\right) ~=~\int_{-\infty}^\infty dx\left( \left[ D_{1}\left( x,t\right) \frac{\partial }{\partial x}+D_2 \left( x,t\right) \frac{\partial^2}{\partial x^2}\right] \delta\left( x^{\prime}-x\right) \right) f\left( x,t\right).\qquad $$
Integrate over a time interval $\varepsilon$,
$$f\left( x^\prime ,t+\varepsilon \right) $$
$$~=~\int_{-\infty }^\infty \, dx\left(\left( 1+\varepsilon \left[ D_{1}\left(x,t\right) \frac{\partial }{\partial x}+D_{2}\left( x,t\right) \frac{\partial^{2}}{\partial x^{2}}\right]\right) \delta \left( x^\prime - x\right) \right) f\left( x,t\right)$$
$$\tag{2}+O\left( \varepsilon ^{2}\right).\qquad $$
OK, but Fokker-Planck equation for one dimension is usually
$$\tag{0} \frac{\partial}{\partial t}f(x,t) = -\frac{\partial}{\partial x}\left[\mu(x,t)f(x,t)\right] + \frac{\partial^2}{\partial x^2}\left[ D(x,t)f(x,t)\right].$$
I was not able to understand how one gets from the original equation (0) to the above (1) and how does the first equation (1) lead to the second equation (2). Can anyone explain this?
| Hints:
$\underline{(0) \Rightarrow (1)}$: Don't try to accomplish everything at once. Do it slowly in as many steps as you need to be sure that you are calculating correctly and understand everything. The trick is to integrate by part. Be very careful to keep track of what depends on $x$ and what depends on $x^{\prime}$.
$\underline{(1) \Rightarrow (2)}$: Use Taylor series
$$f( x^{\prime} ,t+\varepsilon )
~=~f( x^{\prime} ,t) +\varepsilon\frac{\partial }{\partial t}f\left( x^{\prime },t\right) +{\cal O}\left( \varepsilon ^{2}\right).$$
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Why does electromagnetic waves travel slower in well isolated conductor? Wikipedia writes, that
Propagation speed is affected by insulation, so that in an unshielded copper conductor ranges 95 to 97% that of the speed of light, while in a typical coaxial cable it is about 66% of the speed of light
Why does insulation affect this speed?
| That's because adding shield means that there is now distributed capacitance between core & shield in addition to the inductance of the central wire.
Current/voltage transitions gets slower because charging/discharging "capacitor" on each segment through "inductor" of the wire takes time -> propagation speed decreases.
| {
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How to choose a solution from all possible solutions of general relativity So there are so many solutions for general relativity - then how does one "choose" the solution that is right one? By checking with observation? (though I also know that it is currently unknown which one is the correct solution.)
| Generally speaking we start with a known stress-energy tensor and boundary conditions and look for solutions for the curvature. When doing this we're not usually overloaded with possible solutions, and it's normally pretty obvious which solutions are physically relevant.
Where multiple physically relevant solutions exist we select the appropriate one by comparing with experiment. For example we select the FLRW rather than the Gödel metric because experiment suggests the universe has no net rotation. Likewise the value of $\Lambda$ has to be fixed by experiment, as was done in 1998 by Perlmutter and Riese's groups.
| {
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Estimate number of hairs on human head A technique of vital importance at all levels in physics is estimation. This is obvious from the first chapter in any introductory physics textbook, but is also related to the working physicist. Checking orders of magnitudes during research presentations is common practice - I've seen many good questions with good followup answers that started with "If I estimated that value I would get something much different". In general, the actually result is not the interesting thing - it's what individual things will affect the result. There are even famous examples of this: Fermi's piano tuner problem, and the Drake equation. Apparently, Fermi was so good at this that he estimated the size of the Trinity nuclear bomb test to within a factor of 2 (see the wikipedia article for a discussion of that).
In this spirit, I would like to see someone try and estimate the number of hairs on the human head. The answer must include the basic assumptions so we can see where the major unknowns lie, and the best answer is one which requires no specific knowledge
| First estimate roughly the no. of hairs in 1mm^2 and consider the distance between two hairs is uniform all over the head and calculate the area of the whole head and subtract the area of the head having no hair.then multiply that with the hair contained in 1mm^2.
hair is supposed to distribute uniformly.
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How to evaluate commutator with angular momentum? I need to evaluate the commutator $[\hat{x},\hat{L}_z]$. I believe the $L_z$ is referring to the angular momentum operator which is:
$L_z = xp_y - yp_x$
using this relationship i end up with:
$[x,L_z] = x(xp_y - yp_x)-(xp_y - yp_x)x$
my next step is substituting in for the p operator but i still dont get anywhere. any suggestions???
| Usually I find it easiest to evaluate commutators without resorting to an explicit (position or momentum space) representation where the operators are represented by differential operators on a function space.
In order to evaluate commutators without these representations, we use the so-called canonical commutation relations (CCRs)
$$
[x_i,p_j] = i\hbar \,\delta_{ij}, \qquad [x_i, x_j]=0,\qquad [p_i, p_j]=0
$$
Now, in order to evaluate and angular momentum commutator, we do precisely as you suggested using the expression
$$
L_z = x p_y - y p_x
$$
and we use the CCRs
\begin{align}
[x, L_z] &= [x, xp_y-yp_x]\\
&= [x,xp_y] - [x,yp_x]\\
&= x[x,p_y]+[x,x]p_y-y[x,p_x]-[x,y]p_x \\
&= -i\hbar y
\end{align}
In the last step, only the third term was non-vanishing because of the CCRs. I have also used the fact that the commutator is linear in both of its arguments,
$$
[aA+bB,C] = a[A,C] + b[B,C], \qquad [A,bB + cC] = b[A,B] + c[A,C]
$$
where $a,b,c$ are numbers and $A,B,C$ are operators, and the following commutator identity that you'll find useful in general:
$$
[AB,C] = A[B,C] + [A,C]B
$$
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Why can't we store light in the form of light? We can store cold (ice), heat (i.e. hot water bag) and electrical charge (batteries). We can even "store" a magnetic field in a magnet. We can convert light into energy and then, if we want, back to light. But we can't store light in form of light in significant amounts. What is the explanation of that in physics terms?
| Isn't that what we call "heat" is just a form of electromagnetic waves in the infrared spectrum? So we may say that we can store an electromagnetic wave in a certain media, like water in a thermos. But the first thing, we do not use a thermos filled with hot water as an energy storing container, we just use it for a very simple purpose, to store hot water! But if we were using it as, let's say, a kind of a battery, with a some kind of devise inside that would convert heat into electricity or a mechanical power, to move a vehicle, that would be a different story. So, may be, the very first thing that we need is to find such a media to store the sun light, as that hot gas containing atoms of rubidium or may be that should be some sort of a solid matter, and a second step is to create a sort of a convertor to transform that collected energy into a mechanical or electrical power. By the way, filling a car tank with hot boiling water instead of gas sounds like not a bad idea!))
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A Book about the Bohr-Einstein debate? A book about the Bohr-Einstein debate?
Is there any book that details the correspondence between the two? The only books I could find are popular science books, I wonder if there is a book that lists the correspondence in a more 'raw' form.
| The following essay by N.P. Landsman: "When champions meet: Rethinking the Bohr–Einstein debate." Studies In History and Philosophy of Science Part B: Studies In History and Philosophy of Modern Physics 37.1 (2006): 212-242” contains an extensive bibliography on the debate.
In particular the main references containing or discussing the letters are listed in footnote 1 on page 213.
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distance of electron from proton An electron is projected, with an initial speed of $1.10 \times 10^5 \text{m/s}$, directly towards a proton that is essentially at rest. If the electron is initially a great distance from the proton, at what distance from the proton is its speed instantaneously equal to twice its initial value?
i know it can be solved by equating the total energy as
K.E+P.E(electron)=K.E+P.E(Proton)
kinetic energy is $ \frac{1}{2}mv^2$. how can i find out potential energy of each particle?
| You have written down the right equation. Use the electric potential energy between the electron and the proton and write:
$\frac{1}{2}mv_1^2-\frac{e^2}{4\pi\epsilon_0}\frac{1}{R} =\frac{1}{2}mv_2^2-\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}$
Put $v_2=2v_1$, ignore the $1/R$ term since $R$ is very large, so that the electron is in the classical energy region (i.e. in the continuous energy spectrum), and solve the resulting equation for $r$. You should find that the distance $r$ for which the speed of the electron doubles is inversely proportional to the initial speed. Substitute the physical constants into the equation you found to calculate the value of $r$. I think it should come out to be $\sim 1.4\times 10^{-8}$m.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/56109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Absolute zero and Heisenberg uncertainty principle I got to read Volume I of Feynmann's lectures. It said that at absolute zero, molecular motion doesn't cease at all, because if that happens, we will be able to make precise determination of position and momentum of the atom. We do know that Heisenberg uncertainty principle holds for microscopic particles in motion. But what then is wrong to consider that all molecular motion ceases at absolute zero? In other words, does the uncertainty principle not hold when there is no motion?
Need some help!
| The uncertainty principle is a fundamental property of quantum systems, and is not a statement about observational success. No particle either free or in crystal can have zero momentum otherwise a nonsensical infinity is required for the standard deviation of position $\Delta x$, in the uncertainty principle $\Delta x \Delta p \geq \hbar / 2$.
$0 \cdot \infty$ is undefined, so breaks the principle if the product is interpreted as zero. More likely the product should be interpreted as also undefinable, in which case the principle itself becomes undefinable. So all particles must move at least to some extent at all times or the uncertainty principle itself breaks down.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/56170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Influence of charged particle's own electric field on itself I read this in my textbook: A charged particle or object is not affected by its own electric field.
Since I find this completely unintuitive and my mind is yelling "wrong! wrong! how could a particle even distinguish between its own field and the external fields?" I would really like to hear an explanation about why this is true, if it is.
(A corollary question that comes to my mind after thinking about this is... What about gravity and mass considering the analogous situation? And the other two forces?)
| A correct treatment of the self-force problem, 100% rigorous and free of the paradoxes associated with the conventional textbook treatments is given here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/56233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
How to determine the direction of a wave propagation? In the textbook, it said a wave in the form $y(x, t) = A\cos(\omega t + \beta x + \varphi)$ propagates along negative $x$ direction and $y(x, t) = A\cos(\omega t - \beta x + \varphi)$ propagates along positive $x$ direction. This statement looks really confusing because when it says the wave is propagating along $\pm$ x direction, to my understanding, we can drop the time term and ignore the initial phase $\varphi$ while analyzing the direction, i.e. $y(x, 0) = A\cos(\pm\beta x)$, however, because of the symmetry of the cosine function, $\cos(\beta x)\equiv \cos(-\beta x)$, so how can we determine the direction of propagation from that?
I know my reasoning must be incorrect but I don't know how to determine the direction. So if we don't go over the math, how to figure out the direction of propagation from the physical point of view? Why $-\beta x$ corresponding to the propagation on positive x direction but not the opposite?
| $y(x,t)=A\cos(\omega t+\beta x+\phi)$ in this equation $\omega t$ and $\beta x$ symbols of the coefficient are same i.e( ++ or --) then the wave is negative direction travelling wave.
$y(x,t)=A\cos(\omega t−\beta x+\phi)$ in this equation $\omega t$ and $\beta x$ symbols of the coefficient are alternative i.e( +- or -+) then the wave is positive direction travelling wave.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/56338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 2
} |
Calculating the mass equivalency of a song? I've recently become fascinated with the idea of sound energy having a theoretical equivalent mass. I've read over this thread:
Do light and sound waves have mass
I understand this part: $m_{eq}=E/c^2$ and $E=A\rho \xi^2\omega^2$
Where I am getting tripped up is the way to measure $E$.
Essentially, I want to measure $E$ and eventually $m$ of a song (for it's entire duration). The only equipment I have is an iPhone app that measures intensity (dB) (apparently it's fairly accurate for levels below 100 dB). I plan to play the "music" at about 80dB over speakers in a room that is 50'x30'x15' and I will assume the temperature is at room temperature. If your wondering, it's for a conceptual audio art piece.
| A song has no mass equivalence. The sound waves from playing a song do---to the extent that they carry energy, and you can relate an equivalent mass1 to energy.
To cut to the chase, it would probably be easiest to look up how much power (energy per unit time) your speakers produce, then multiply that by the duration of the song.
If you measure the 'intensity' of the song with your iphone, that tells you the power-density at the location of the phone---while the speakers are actually generating sound-waves in all directions. Through an elaborate series of calculations and approximations, you could convert this to an energy. For completeness:
The number of decibels actually measures the pressure of the sound waves, which you can convert to an amplitude. You can estimate the frequency ($\omega$) as, something like, the higher frequency end of the human audible range (possibly adjusted for the type of song). The total power is then computed by finding the surface-area of a sphere around the source ($A$), at the distance you are measuring.
Again, using the power of the speakers will be more accurate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/56589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Tensor perturbation inflation During inflation the metric is de-Sitter so $dt^2-d\underline{X}^2 $.
I know that the eqn.motion governing GW's from inflation (tensor perturbations) is
$$2H\dot{h}+\ddot{h}-\nabla^{2}_{i}h~=~0,$$
derived from varying
$$S^{(2)}~=~\int \frac{a^{2}(t)}{2}~\partial^{\mu}h ~\partial_{\mu}h ~d^{4}x.$$
See paper 1 and paper 2. Surely this isn't valid during inflation because it isn't de-Sitter. I have read lots of papers/lectures where quantum fluctuations are found by deriving equations of motion from a metric with an $a(t)$ and then Fourier decomposed. Surely during inflation the metric can't contain an $a(t)$. So why do all these lectures/papers use it?
| De Sitter space is a special case of the Robertson-Walker spacetime. If you want, you can even work out what the coordinate transformation is.
If your question isn't answered by this, could you please amend it to make it clearer exactly where your hangup is?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/56677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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