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Decoherence inside black holes I have a question about decoherence. Assume there is a macroscopic black hole floating around and you have some macroscopic object with you with a huge number of internal degrees of freedom. Conventional decoherence theory predicts decoherence. If this macroscopic object is dumped into the black hole, will decoherence still happen? There are two environments here: the internal degrees of freedom of the macroscopic object which are coarse-grained over, and the exterior of the black hole. Which environmental degrees of freedom should you trace over? If black hole complementarity is right, the internal degrees of freedom and the external degrees of freedom don't commute, so clearly, unless some mathematical generalization of the partial trace can be invented, you can only trace over one or the other. A trace over the former leads to decoherence, but a trace over the latter can't for the simple reason that nothing can escape the black hole. Only thermal Hawking radiation can escape, but it is so scrambled that it carries no decoherence information.
| Grow up and be a no-nonsense physicist. Translation - become a positivist. Ask what are the empirical experimental observations that can be made here. Any more metaphysics is mere sophistry and illusion, and has to be committed to the flames.
Any experimenter jumping into the black hole to measure the macroscopic object can never deliver us any empirical data. No experiment performed inside a black hole can ever have any empirical results. Commit what is inside the black hole to the flames.
The only things which are empirically observable are the Hawking radiation. They do not decohere what is inside the hole.
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Spin Echo Experiment I'm doing an spin echo experiment for different dilutions of Glycerol as an undergraduate physics lab experiment. At the end of my experiment I will need to extend it and do some initiatives! I have a few potential extensions in mind. But I was wondering what other cool stuff one can do with typical equipments for such experiments (I've got things like giant magnets, RF oscillators, probes and other measuring devices, etc)
Thanks.
| You could try bilateral experiments opposed to mechanical ones which are more fun if you into watching things happen by themselves without computing the control mechanisms. Start with some type of "conversion plane" which could be any material that interacts with the experimental magnets, then modify the levels of planar differentials according to the most dependent part which is finding properties with patterns that respond, also adjusting the plane could be difficult.
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What type of substances allows the use of the Ideal Gas Law? I know that I can use the ideal gas law with pure gases or pure liquids. But can I also use the ideal gas law at saturated gases and saturated liquids as long as they aren't two phase substances?
| Aside from the classical cases where the ideal gas law applies, it also applies to describe the exact entropy of a dilute solution, even if that solution is in a dense liquid. The reason is that the entropy of a dilute solution in a dense liquid is exactly the same as the entropy of a dilute gas, the number of possible positions for the solute particles is the same as the number of possibilities for the gas.
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How would nucleosynthesis be different if the neutron were stable? If the strong nuclear force were just 2% stronger, the neutron would be a stable particle instead of having a half life of about 13 minutes. What difference would that have made to Big Bang nucleosynthesis, to the growth of structure, to the formation of stars, nucleosynthesis in stars?
| Assuming that the proton is heavier than the neutron, by more than the mass of the electron (plus the mass of a neutrino, plus the ionization energy of hydrogen), this is easy to answer, it would just make hydrogen unstable to decay to a neutron an an electron positron pair, so that a mostly hydrogen universe will decay into neutrons and electron-positron pairs, which wil annihilate into photons. So I will assume that the difference between neutron mass and proton mass is less than the mass of the electron, so that both the proton and the neutron are stable.
The most drastic effect of this is on big-bang nucleosynthesis, where two new stable species can be created, neutrons and tritium, and He-3 would be unstable to inverse beta-decay into tritium. So you would produce hydrogen, deuterium, tritium, helium, lithium, and neutrons. The initial conditions are mostly neutrons, not mostly protons, because the mass is inverted, and so you would get a lot of He-4, very little H-1, and most of the universe's mass would consist of stable neutrons and alpha-particles. These neutrons might collide to form neutron clusters, which would then beta-decay to protons once the binding energy was greater than the neutron proton mass difference.
There wouldn't be stars, but there might be gravitationally bound neutron clusters. Neutrons are neutral, and find it hard to dissipate energy, but the time scales are long, so they might be able to eventually settle down into neutron-star-like objects.
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Why can't $ i\hbar\frac{\partial}{\partial t}$ be considered the Hamiltonian operator? In the time-dependent Schrodinger equation, $ H\Psi = i\hbar\frac{\partial}{\partial t}\Psi,$ the Hamiltonian operator is given by
$$\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2+V.$$
Why can't we consider $\displaystyle i\hbar\frac{\partial}{\partial t}$ as an operator for the Hamiltonian as well? My answer (which I am not sure about) is the following:
$\displaystyle H\Psi = i\hbar\frac{\partial}{\partial t}\Psi$ is not an equation for defining $H$. This situation is similar to $\displaystyle F=ma$. Newton's second law is not an equation for defining $F$; $F$ must be provided independently.
Is my reasoning (and the analogy) correct, or is the answer deeper than that?
| Answer to the top question is actually very short. Time is an external parameter in conventional QM; parametrizing a unitary evolution. It, as well as $i\partial_t$, nothing has in common with operators, observables etc. In other words $t$ in $\psi(t)$ does not enumerate some basis vectors of an observable like $x$ does in $\psi(x)$.
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Radar Frequency Bandwidth I've come across an interesting question in the course of doing some exam review in a quantum mechanics book and thought I'd share it here.
"What must be the frequency bandwidth of the detecting and amplifying stages of a radar system operating at pulse widths of 0.1usec? If the radar is used for ranging, what is the uncertainty in the range?"
I don't know the solution; although I'd guess that it could involve a Fourier transform (a generic first crack at anything with waves) and the uncertainty relation (as the problem does call out uncertainty in range; which could be interpreted as uncertainty in x).
| Since you're talking about a quantum mechanics book rather than a radar design engineering book, I assume that they're talking about a basic textbook radar system, rather than one with all the advanced features.
The receiver needs to have a bandwidth great enough to receive most of the energy in the reflected pulses. Since they're sine waves modulated by a pulse shape, they will have spectral splatter and the receiver needs to be able to process most of this energy for each pulse, which is now spread over a range of frequencies. You're right in that, to compute this spectral spread, you need to do a Fourier transform - the spread you'll get depends on the pulse shape.
The longer the pulse, the narrower the bandwidth of the spectrum - sometimes people use a "rule of thumb" whereby the bandwidth is estimated as the reciprocal of the pulse duration.
With respect to range, I suppose if you imagine two targets close to each other(let's assume same azimuth), one behind the other, then you can resolve them if the reflected pulse from the nearest has finished being received before the reflected pulse from the furthest starts to be received. It's fairly easy to see that this implies that the distance between the targets must be greater than 0.5*PulseTime*c
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What is the physical sense of the transition dipole moment? So if the states are the same we achieve the expectation value of the dipole moment for a given state. I mean
$ \langle \mathbf{\mu} \rangle = \langle \psi \vert \hat{\mathbf{\mu}} \vert \psi \rangle$
But I don't feel the physical sense in the case of transition dipole moment when psi-functions on both sides are different
$\langle \psi_{1} \vert \hat{\mathbf{\mu}} \vert \psi_{2} \rangle$
Help me to understand, please.
| I never thought of it this way, but off-diagonal components of an operator means that this operator is... Well, non-diagonal. Which means that neither of $| \psi \rangle$ realizes a definite value of this operator.
The dipole operator comes from the term which couples charged particle with electric field. It would be better to write some math here, but I just appeal to quantum approach to Larmor precession. The same situation there (we start from the state which is not the eigenstate of the operator and see how it changes) leads to periodic change of the wavefunction from one state to another where off-diagonal elements are related with the probability of those transitions.
From some other arguments we know that the particle may go down, but can not go up. So, this probability not a probability of some quantum beats, but the probability to go from upper to lower state.
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Is it possible to recover Classical Mechanics from Schrödinger's equation? Let me explain in details. Let $\Psi=\Psi(x,t)$ be the wave function of a particle moving in a unidimensional space. Is there a way of writing $\Psi(x,t)$ so that $|\Psi(x,t)|^2$ represents the probability density of finding a particle in classical mechanics (using a Dirac delta function, perhaps)?
| You can recover Schroedingers equation from the path integral formulation of Quantum mechanics by Feynman. In the path integral picture the classical trajectories are the stationary points of the integrand. So in the stationary phase approximation, they are the contribution of $0$-th order in $\hbar$. Of course that is not a direct relation between the Schroedinger equation and classical trajectories.
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Can stable nuclei theoretically fission through quantum tunneling? As I understand it, an unstable nucleus is going to randomly fission because the forces binding it together are momentarily weaker than the electrostatic repulsion of the protons.
Given that some nuclei are really unstable, and others are stable for billions of years, is it true that so-called stable nuclei are actually theoretically able to fission through quantum tunnelling, just unlikely to do so? Have stable atoms ever been observed to spontaneously fission?
Additionally, will bombarding stable nuclei with particles split them? Can you hit a Carbon atom with a neutron and split it, or will it always add on or miss? Can a photon split a nucleus?
| Stability of nuclei is a matter of the binding energy and conservation of quantum numbers: number of baryons is conserved. If ( and it is a large if) the proton decays, i.e. baryon number has a small but finite probability of not being conserved, then also stable nuclei might decay spontaneously. At the moment the lifetime of the proton has a limit of larger than 10^32 years.
will bombarding stable nuclei with particles split them?
yes, again depending on the energies involved
Can you hit a Carbon atom with a neutron and split it,
yes depending on the isotope and the energy
Can a photon split a nucleus?
Yes, if it has MeV energies, i.e gamma rays.
"The height and shape of the fission barrier are dependent on the particular nucleus being considered. Fission can be induced by exciting the nucleus to an energy equal to or greater than that of the barrier. This can be done by gamma-ray excitation (photofission) or through excitation of the nucleus by the capture of a neutron, proton, or other particle (particle-induced fission). The binding energy of a particular nucleon to a nucleus will depend on—in addition to the factors considered above—the odd–even character of the nucleus. Thus, if a neutron is added to a nucleus having an ... (100 of 9031 words) "
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Is there a limit to the resolving power of a mirror telescope? Like, if you grabbed the asteroid 16 Psyche and hammered it out into a disc of 1 mm thick iron foil and curved it into a telescope mirror with 2.4x the radius of the Sun, could you resolve details on the surface of an exoplanet? At what resolution?
Could you make the mirror arbitrarily bigger and continue to get better resolution?
What are the formulas for calculating this and what are the limits, if any?
If you make a big enough mirror could you see individual houses on Proxima b? Could aliens with big telescopes have videos of the formation of the moon in their libraries?
| The resolving power of a device is limited by the so called Rayleigh criterion (commonly known as the diffraction limit). This is applicable for "classical" states of light (i.e light that has a coherent state representation). For non-classical light (squeezed light or entangled light), you can beat the diffraction limit.
A few interesting articles What Diffraction Limit?, Resonant Lithography,Quantum Limits on Optical Resolution,Quantum Imaging--pdf file. An interesting experimental overview provided by Boyd's group at Rochester.
The point is, for non-classical light the diffraction limit is $\lambda(N)=\frac{\lambda(1)}{N}$ where N is the number of photons in an entangled state (say a two mode state such as: $|N,0>-|0,N>)$ and $\lambda(1)$ is the wavelength dependence you would expect for classical light.
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What is escape velocity? In reality, how can something no longer be under the gravitational influence of something else? Isn't G a continuous function and although you leave the immediate vicinity of the earth with an escape velocity won't it always exert a force, however small it may be. Won't that force eventually pull the object back to the earth (assuming the absence of other objects)
| It's true that it doesn't matter how far away you get from the Earth, there will always be an attractive force, and that if you are at rest gravity will pull you towards the Earth.
However, if you start at Earth's surface with that speed, you can show that your trajectory will never take you back to Earth. It will be a hyperbola. The reason is that your speed in the direction opposite to the Earth never goes down to zero. It gets smaller and smaller as you get further (because gravity pulls you back), but it never gets to zero.
A vary nice way to see it (if you are familiar with potential energy) is the definition from wikipedia: the escape velocity is the speed you need so your kinetic energy plus the gravitational potential energy add up to zero. This means that the object has enough energy to get out of the gravitational well. It doesn't matter if the gravitational well extends to infinity, the object will not return to the Earth because it will always have nonzero speed.
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When has the speed of light been measured, recently? Yes, it is weird, absurd, but I can't stop thinking that the would-be superluminal
neutrino speed has been computed by an arithmetic operation (space/time) and not by direct comparison with a simultaneous light ray running in "parallel".
So the "unspeakable", outrageous question: is the speed of light increasing in the last months? When it has been measured the last time?
Also: is the speed of light measured in a one-way or two-ways (forward and back) method?
It is a "politically incorrect" question, but its logic is rock-solid, I think.
| There has been a lot of coverage of the faster-than-light neutrinos and therefore reasonable to think about the subject. There are some theories that suggest that the speed of light is not a constant, but that change would be over the lifetime of the Universe. In any case, there is no evidence for such changes in the speed of light. As for how often it is measured, well, all the time. GPS navigation makes use of Special Relativity all the time which has the speed of light built into it. It is good to question long standing theories in light of new evidence, but it is going to take a lot to dethrone Relativity; it has been tested far too rigorously for us to question it on the basis of single statistical result.
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Do multiple permanent magnets aggregated together approach the same strength as a single magnet of the same size? Here's an applied physics question. ;) If I buy some cube or sphere magnets like these, can I aggregate them together to create a stronger magnet (almost as strong as a single magnet)?
| To expand on Omar's Answer: KJMagnetics has an FAQ that answers this question: http://www.kjmagnetics.com/FAQ.asp#stack.
If you wander round the KJMagnetics site, they give a lot of technical information (I bought some magnets off them 2-3 years ago).
Their FAQ answer describes the situation for stacking flat magnets vertically on top of each other. Note that if you place multiple magnets next to each other horizontally, instead of in a vertical stack, they will still act in the same way as a larger magnet having the same area. However, in the horizontal case, the magnets will repel each other because like poles are next to each other (N next to N, S next to S). So you would have to impose an equal and opposite force to keep them together. That's not easy when the magnets get large.
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Define Pressure at A point. Why is it a Scalar? I have a final exam tomorrow for fluid mechanics and I was just looking over the practice exam questions. They do not provide solutions. But pretty much I have to define pressure at a point and also say why pressure is scalar instead of a vector.
I am thinking pressure at a point is $P=\lim_{\delta A \to 0} \frac{\delta F}{\delta A}$. Please let me know if I am wrong.
But I do not know at all why pressure is a scalar instead of a vector. I know it has something to do with $d \mathbf{F}=-Pd \mathbf{A}$
| Because it doesn't behave like a vector -- for example, the standard vector norm of the pressure doesn't mean anything, but the sum of pressure components does. Just because something has a bunch of components doesn't mean it's a vector. Pressure refers to the diagonal components of the stress tensor (actually their sum, the trace) -- this arises naturally in terms of momentum fluxes.
As an analogy, consider the fact that the divergence of a vector field is not a vector. Analogously, divergence is just the diagonal components of the tensor derivative. One level further, the dot product is not a vector, for the same reason.
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Heat transfer, Cold vs. Hot If you boil water inside a pot the outer rim bubbles first, I imagine because its hotter. Does that same concept apply for the inside of a refrigerator for example. Is the outer rim always more extreme, or at least at first?
| When you are boiling water bubbles typically form first at the bottom of the pot, and then at the walls. As you guessed, this is because the bottom is usually at higher temperature than the walls, since it's in direct contact with the heat source. The reason that bubbles don't appear sooner in the bulk of the water is that nucleation (the initial formation of tiny bubbles) is easier at surfaces and at their irregularities. To check this, you can use any pot with a big scratch, and you'll see bubbles forming at the scratch first.
In a refrigerator, the conditions are quite different. In a typical refrigerator there's a coiled set of pipes inside the freezer, near the top of the unit. This is the evaporator, and it's where the heat is extracted, and this is where you'll have the lowest temperatures. If the leaks are small, you'll have the lowest temperature at the top and the highest temperature at the bottom. Usually, the back of the refrigerator is colder than the front because heat leaks mainly through the door crack.
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Physics Paradox about Newtons Second Law $F=ma$
*
*If force equals mass times acceleration, wouldn't a basketball dropped from the top of the Eiffel tower exert the same force on the ground as a basketball dropped a foot off the ground? They both have the same mass, and they both are accelerating towards the ground at a rate of $g = 9.81\,{\rm m/s^2}$. (I don't know what terminal velocity is that well as I'm only in physics 1 in high school, but just assume that air drag is not important and the ball doesn't reach terminal velocity.)
*Also, if a ball is dropped high enough to reach terminal velocity, then it accelerates at $0\,{\rm m/s^2}$, so it has a force of ZERO when it hits the ground?
| This is the case of jerk in physics , rightly pointed above, so when the ball hits ground it change in acceleration is 2a in very less time. This is the reason why you get more injured when you fall from high tower.
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How does gravity work underground? Would the effect of gravity on me change if I were to dig a very deep hole and stand in it? If so, how would it change? Am I more likely to be pulled downwards, or pulled towards the edges of the hole? If there would be no change, why not?
| The other answers provide a first-order approximation, assuming uniform density (though Adam Zalcman's does allude to deviations from linearity). (Summary: All the mass farther away from the center cancels out, and gravity decreases linearly with depth from 1 g at the surface to zero at the center.)
But in fact, the Earth's core is substantially more dense than the outer layers (mantle and crust), and gravity actually increases a bit as you descend, reaching a maximum at the boundary between the outer core and the lower mantle. Within the core, it rapidly drops to zero as you approach the center, where the planet's entire mass is exerting a gravitational pull from all directions.
The Wikipedia article on "gravity of Earth" goes into the details, including this graph:
"PREM" in the figure refers to the Preliminary Reference Earth Model.
Larger versions of the graph can be seen here
And there are other, smaller, effects as well. The Earth's rotation results in a smaller effective gravity near the equator, the equatorial bulge that results from that rotation also has a small effect, and mass concentrations have local effects.
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Deterministic quantum mechanics I came across a very recent paper by Gerard 't Hooft
The abstract says:
It is often claimed that the collapse of the wave function and Born's rule to interpret the square of the norm as a probability, have to be introduced as separate axioms in quantum mechanics besides the Schroedinger equation. Here we show that this is not true in certain models where quantum behavior can be attributed to underlying deterministic equations. It is argued that indeed the apparent spontaneous collapse of wave functions and Born's rule are features that strongly point towards determinism underlying quantum mechanics.
http://de.arxiv.org/abs/1112.1811
I am wondering why this view seems to unpopular?
| One place to look is the homepage of Antony Valentini now at Clemson University. He claims that Born's probability rule is only an approximation. David Bohm first made this claim. One can show that entanglement can be used for faster-than-light and even retro-causal back-from-the-future delayed choice signaling once the shackles of Born's rule are cast away.
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Do interaction-free measurements require a physical collapse or splitting in order to be truly interaction free? Interaction-free quantum experiments like Renninger's experiment or the Elitzur-Vaidman bomb tester are often taken to be examples of interaction-free measurements of a system. Unfortunately, such assumptions presuppose the ability to post-select in the future just to make sense. Interpretationally speaking, it is hard to see how post-selection can possibly be made without some form of physical collapse of the wave function or a preferred physical splitting of the wave function into branches. Without either a physical collapse or splitting, is it possible to gain information about a system of which we are totally ignorant about the preparation of its properties we are interested in without an actual interaction with it? Basically, does the idea of interaction-free measurements only make sense within some interpretations of quantum mechanics and not others? Is there a philosophical reading of the two-state formalism which does not presuppose a collapse or a splitting? In the two-state formalism, does the necessity of normalizing the overall probability factor to 1 entail the ontological reality of the other outcomes because we have to sum up over their probabilities to get the rescaling factor? The other outcomes where the bomb did in fact go off?
|
The Elitzur-Vaidman bomb tester isn't really an interaction free measurement. Analyze it using consistent histories. Suppose initially, for the three possible bomb states, we start off with the mixture diag(p, 1-p, 0) for dud, workable but unexploded, and exploded respectively. Let $P_c$ correspond to the projector of a photon detected at C. Let $P_l$ correspond to the projector the photon traveled along the lower path, and $1−P_l$ it travelled along the upper path. Consider the chain operators $C_1≡P_c P_l$ and $C_2≡P_c (1−P_l)$.
Note that $Tr[C_1 ρ C_2^\dagger]=p/4≠0$. The consistency conditions are not satisfied. In a realm where we know the photon was detected at C, we can't say whether or not it took the lower path. It's not really interaction free after all. That presupposes the photon didn't take the lower path.
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Complex numbers in optics I have recently studied optics. But I feel having missed something important: how can amplitudes of light waves be complex numbers?
| With any simple harmonic oscillator there are two quantities we are interested in, the phase and the amplitude. Complex numbers are an easy way to represent both of these in a single value, especially as a complex number can be written in the form $Ae^{i\theta}$ where $A$ is the amplitude and $\theta$ is the phase. This doesn't mean light has some "imaginary" component. The complex number is just a mathematical model for the light.
There are lots of ways in which complex numbers make it easier to construct mathematical models. If you have access to a copy of Roger Penrose's book "The Road to Reality" have a look at chapter 4 where he discusses this aspect of complex numbers.
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Do a 1-dimensional conformal theory exist? can we have in physic or can we speak about 1-d conformal theory in physics ??
for example in this one dimensional theory what would be the generators $ x \partial _{x} $ or $ \partial _{x} $ ??
| Yes, such theories exist. They're known as "conformal quantum mechanics". See
http://scholar.google.com/scholar?q=%22conformal+quantum+mechanics%22&hl=en&lr=&btnG=Search
http://scholar.google.com/scholar?q=ads2-cft1&hl=en&lr=&btnG=Search
There is an $SL(2,{\mathbb R})$ symmetry in them, or its (e.g. supersymmetric) extensions. The potential $1/r^2$ may occur in such QM models. Because there is only time dimension, there is no $\partial_x$, just $\partial_t$, and similarly there is only $t \partial_t$, formally speaking.
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How do we visualise antenna reception of individual radiowave photons building up to a resonant AC current on the antenna? I am a chemical/biological scientist by trade and wish to understand how quantum EM phenomena translates to our more recognizable classical world.
In particular, I want to get a mechanistic picture of what is going on when a tuned antenna is interacting with a photon of the desired frequency? I believe an individual electron on the antenna (many electrons) accepts a photon; but how does the eventual process of a measurable AC current build up on the dipole (or 1/4 wavelength, for example) to be fed with no reactance onto the transmission line?
"When photon meets antenna" is a great meeting ground for a quantum/classical bridge.
Unfortunately, I do not have a serious maths background, but will try anything suggested. I have read and listened to many of the Feynman's popular quantum discussions which only increases my thirst for a better understanding of how quantum EM translates to our more visible world.
| You cannot understand how a radio antenna works by counting the number of photons that strike the copper wire. This number is many orders of magnitude too small to account for the actual power absorbed by an antenna. An antenna would not work if it depended on physically intercepting photons. I explain all this in my blog post "The Crystal Radio".
In fact it is much more useful to analyze atoms in terms of antenna theory than to analyze antennas in terms of atomic theory. I elaborate on this question in my follow-up blog article, "How Atoms are Tiny Antennas".
| {
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Reducing General Relativity to Special Relativity in limiting case I understand that general relativity is applicable to gravitational fields and special relativity is applicable to case when there is no gravity. But is there a derivation on how to reduce General Relativity to Special Relativity in limiting case, much like how General Relativity is reduced to Newtonian gravity in weak-gravity case?
Edit: By reducing I mean, how can we derive the Lorentz transformation from General Relativity under appropriate limits?
| Other answers have already addressed the relationship between General Relativity and the Minkowski metric, but it seems you are most interested in getting from the Minkowski metric to the Lorentz transformation. So let's do that.
Given a set of coordinates in which the metric takes the standard Minkowski form
$ds^2 = dt^2 - dx^2 - dy^2 - dz^2$
you want to find another set of coordinates in which the metric also takes the same form
$ds^2 = d \bar{t}^2 - d \bar{x}^2 - d \bar{y}^2 - d \bar{z}^2$
Consider a linear boost along the x axis. We want to choose coordinates such that $\bar{x} = 0$ where $x - vt = 0$; the general solution is:
$\bar{x} = \gamma(x - vt)$
$\bar{t} = at - bx$
where $\gamma$, $a$, and $b$ are unknowns.
Then
$dt^2 - dx^2 = d\bar{t}^2 - d\bar{x}^2$
$ = (a dt - b dx)^2 - \gamma^2 (dx - v dt)^2$
$ = (a^2 - \gamma^2 v^2) dt^2 - (\gamma^2 - b^2) dx^2 + (\gamma^2 v - ab) dx dt$
so
$a^2 - \gamma^2 v^2 = 1$
$\gamma^2 - b^2 = 1$
$\gamma^2 v - ab = 0$
so
$a = \sqrt{1 + \gamma^2 v^2}$
$b = \sqrt{\gamma^2 - 1}$
so
$\gamma^2 v = \sqrt{\gamma^2 v^2 + 1} \sqrt{\gamma^2 - 1}$
$\implies \gamma^4 v^2 = (\gamma^2 v^2 + 1)(\gamma^2 - 1)
= \gamma^4 v^2 + \gamma^2 - \gamma^2 v^2 - 1$
$\implies \gamma^2(1 - v^2) = 1$
$\implies \gamma = \sqrt{\frac{1}{1-v^2}}$
Calculating a and b (exercise left to the reader) gives $a = \gamma$ and $b = \gamma v$, so that
$\bar{t} = \gamma(t - vx)$
completing the Lorentz transformation as expected.
| {
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Software to simulate and visualize atoms? Not sure if this is a physics or chemistry question. But if the motion of atoms and it's particles can be described by quantum mechanics, then is there a software that simulate full atoms and it's boundings, in a way you can visualize them, and that can be used, for instance, to throw 2 molecules together and watch them reacting?
| quantum espresso is a package that contains many different packages
for electronic structure calculations.
It is now available in a docker container and therefore very easy to use..
Installed docker on linux, windows or mac,
you just need to run the docker image that is available on dockerhub :
docker run -d rinnocente/qe-full-6.0
more info at https://hub.docker.com/r/rinnocente/qe-full-6.0 ...
| {
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What frequency photons are involved in mediating physical force? If the force felt when pushing an object is mediated by the electromagnetic interaction and hence photons, what is their frequency?
| The electrostatic force is mediated by virtual photons and one could say that they are not physical photons and they do not have a frequency, i.e. one could claim your question is invalid.
However, one may also determine the frequency from the energy $E$ of the virtual photons (via $E=hf$) which actually is determined in each Feynman diagram, although the energy-momentum doesn't satisfy $E^2-p^2 c^2 = m_0^2c^4$ in general. The energy is equal, by energy conservation law, to $E_{1,{\rm final}}-E_{1,{\rm initial}}$. For the electrostatic force that only marginally changes the velocity of the charged particles, you have
$$E=0$$
| {
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How much energy is in a lightning strike? According to Wikipedia an average lightning strike has $1$ TW power, the whole world used $16$ TW of power in 2006.
The lightning strike lasts for $30$ microseconds. Does this mean that you get $100$ TW of energy in just $30$ microseconds?
| You are confused about units. Watt is a unit of power (energy/time), Watt-hour is a unit of energy. 16TW is an estimate of the continuous average power usage of the world - which is about 140,000 TWh each year. If the lightning has a peak power of 1TW for 30 microseconds, this corresponds to an energy content of about 8000 Watt-hours.
| {
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What fundamental principles or theories are required by modern physics? We have been taught that speed of light is insurmountable but as we know an experiment recently tried to show otherwise.
If the experiment did turn out to be correct and confirmed by others, would it make physics to be rethought of? What other concepts are fundamental to physics, which, if disproved would need radical rethinking?
If this sounds too juvenile and/or misinformed, please understand that I am a layman, having nothing, professionally or academically to deal with science, directly, and this question is out of curiosity. I have developed a liking to "science stuff" and been reading popular science variety of literature lately. This question was also prompted by what Sheldon Cooper had to say in one of the episodes (I was watching a rerun).
| Not very much !
The rest of physics still has to work, finding that the speed of light can be exceeded in certain circumstances doesn't suddenly change the results in other experiments or allow perpetual motion machines to start working.
There have been some discoveries where things that were classically 'impossible' were found to work in quantum theory - which have led to practical discoveries (like SQUIDS or even GMR hard drives). Although it's hard to see how you could practically speed up the internet using oscillating neutrinos.
edit: The actual Opera experiment looks like a mistake.
But imagine if it was discovered that (for example) you could send a signal faster than light by some QM effect - but over a distance <0.1nm and only below a temperature of 1mK that would invalidate relativity but have no affect on the day-day use of relativity in physics or on the structure of the universe.
In exactly the same way that a tiny difference in the orbit of mercury overturned Newtonian mechanics and led to GR but had no effect on the day-day use of Newtonian mechanics for calculating the flight of cannon-balls!
| {
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Why can't echoes be heard inside a room? If I go camping and shout anywhere, in the forest or on a cliff, I usually hear the echo of my voice.
Why when I shout in my room I do not hear any echoes?
| You just haven't tried a big enough room...try a large, empty gymnasium or something similar (but not a concert hall as they are usually designed to suppress echos.).
The speed of sound is roughly $v_s = 340\text{ m/s}$ (1100 feet per second), and hearing an echo requires at least
*
*a perceptible time between the end of your shouting and the onset of the returning wave (not sure how long, but lets say $t_c = 0.1\text{ s}$ as a guess (a lot of human perception works on time scales not to far from that))
*that the returning wave be sufficiently loud to be distinguished above the background noise
*that the returning wave be sufficiently distinct from the returning waves from other surfaces
*a surface that gives a good reflection
So lets think about how a indoor situation might fail:
*
*The room is too small. If the longest dimension is much less than $v_s*t_c \approx 34\text{ m}$ then the maximum delay between the end of a loud shout and the onset of the echo may be too short for you to distinguish the echo
*The multiple returns from the several walls are overlapping and preventing you from picking out one echo.
*The ventilation system and other ambient noises are comparable in volume to the echo.
*The walls may be non-flat or made of materials that absorb much of the sound energy (this is often a design goal for large spaces)
| {
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If microwave ovens and WiFi both operate on the same frequency, why doesn't WiFi cook things? If we ignore 5GHz WiFi, then both microwaves and WiFi create photons at ~2.4GHz but one of them will boil water in a few seconds but the other doesn't have any effect. So what's the difference?
Is it simply the number of photons created? Is that what the wattage of a microwave measures? If so, what would be the wattage of a wireless router?
Does the enclosed space have anything to do with it?
If it all has to do with power output could I put enough WiFi routers together in a room to cook a turkey (from microwaves and not waste heat)?
| Power - your wifi router puts out about 0.1 - 1.0 W, your microwave oven puts out 1000W.
It would take a lot of wifi routers to cook a turkey - more than you think because the antennea on the router is designed to spread the power evenly around the room rather than concentrate it on the center of the oven.
There is a danger of being 'cooked' from being close to very high power transmitters such as some warship's radar while they are operating.
ps. It's the same reason your laser pointer can't be used to cut steel plates (or James Bond) in half!
| {
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Physics related Podcasts
Possible Duplicate:
Are there any good audio recordings of educational physics material?
In the same way that was already asked about good books of Physics in this StackExchange, I would like to know good physics podcasts! What are the most informative and enjoyable to hear?
| I could collect the sources from various websites and place it here, browse through their content to get what you want
Are there any good physics podcasts?
Physics General Interest Seminar Podcasts
Physics Podcasts
Podcasting the mysteries of the universe
The first link has some good collection . Enjoy
| {
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Could gravity hold electron charge together? Could the gravitational force be what holds the charge of the electron together? It seems to be the only obvious possibility; what other ideas have been proposed besides side-stepping the issue and assuming a "point charge"? How would this affect the electron "self-energy" problem? The question is related to the idea of geons.
| Yes, it can. Here is a toy model using Newtonian gravity.
V/c^2 = e^2/mc^2r - /\r^2
e^2/mc^2 = rc (classical electron radius)
with SSS metric
g00 = 1 + 2V/c^2 = - 1/grr
g/c^2 = -dV/dr = + rs/r + 2/\r
We can get g = 0 with /\ < 0 i.e. AdS metric
In a vacuum where the w = -1 virtual electron positron pairs surrounding the bare charge have higher density than the w = -1 virtual photons, we can have /\ < 0.
The equilibrium will also be stable looking at d^2V/dr^2.
This neglects spin, but we can model that with the centrifugal potential
http://en.wikipedia.org/wiki/Effective_potential
| {
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how to represent the effect of linking rigid-bodies together? I have 2 rigid-bodies (b1,b2) if i linked one to the other (as if they are conjoined together) , how to represent b1 effect on b2 and b2 effect on b1
Is there any LAW that affect the position/orientation of the other body ?
notes :
*
*i am using Quaternions for orientations
*i don't want to treat them as one body
*i have only primitive shapes (box,sphere,..) to link.
| The laws you are looking for are conservation of momentum and conservation of angular momentum. If you stick the two bodies together both laws still must be fulfilled (inelastic effects neglected). In the end you will have a compound single object. With the parallel axis theorem (wikipedia) you can calculate the mass moment of inertia and together with the centre of mass the whole motion of your compound object. You do not have to change the collision logic completely, create a sphere/box that includes your whole object and use that to test for collisions.
An alternative to connect the two rigid bodies is via springs as dmckee pointed out and this approach is quite successful in a lot of physics engines (bridge building games, World of Goo). Even liquids can be modeled with a few hard drops connected via springs.
| {
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What is the connection between Poisson brackets and commutators? The Poisson bracket is defined as:
$$\{f,g\} ~:=~ \sum_{i=1}^{N} \left[
\frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}} -
\frac{\partial f}{\partial p_{i}} \frac{\partial g}{\partial q_{i}}
\right]. $$
The anticommutator is defined as:
$$ \{a,b\} ~:=~ ab + ba. $$
The commutator is defined as:
$$ [a,b] ~:=~ ab - ba. $$
What are the connections between all of them?
Edit: Does the Poisson bracket define some uncertainty principle as well?
| Regarding the significance of the observables momentum and position there are many similarities between Classical and Quantum mechanics. Some of the algebraic relations have been pointed out.
In the end, there is still an important difference, which is obvious by the fact that the function algebra generated by classical quantities is commutative
$$q·p=p·q,$$
and the other is not
$$Q\ P\ne P\ Q=Q\ P-[Q,P\ ].$$
One might ask if there is a structure for the classical function algebra of $q$ and $p$ with a product, which resembles the quantum mechanical algebra $Q$ and $P$. I.e. is there a product, let's denoted it by $\star\ $, for which
$$q\star p-p\star q=[q\ \overset{\star}{,}\ p]\ \ \Longleftrightarrow\ \ [Q,P\ ]=Q\ P-P\ Q.$$
More on questions in this spirit can be found under Weyl quantization.
The most investigated star product is the Moyal product, which per definition fulfills
$$[f\ \overset{\star}{,}\ g]=i\hbar\ \{f,g\}+\mathcal O(\hbar^2).$$
Fields medals are won for this kind of stuff.
| {
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What is the meaning of speed of light $c$ in $E=mc^2$? $E=mc^2$ is the famous mass-energy equation of Albert Einstein. I know that it tells that mass can be converted to energy and vice versa. I know that $E$ is energy, $m$ is mass of a matter and $c$ is speed of light in vacuum.
What I didn't understood is how we will introduce speed of light?
Atom bomb is made using this principle which converts mass into energy; in that the mass is provided by uranium but where did speed of light comes into play? How can speed of light can be introduced in atom bomb?
| c is a priori not the speed of light. It is the speed of massless particles. The way it comes about is as follows: You construct the Lorentz-transformations as the symmetry transformations of Minkowski space. The group has one parameter, that's c. You have to fix it by physical means. You can look at the dynamics of massive particles and massless particles and find that massive particles will approach c asymptotically only at infinite energy, and massless particles always move with c.
Since to our best knowledge photons are massless, c is also the speed of light. Also, that was historically Einstein's motivation, which is why it's usually motivated in textbooks this way. However, should it turn out one day that photons do have very tiny masses, then c will still be there, it will just no longer be called the speed of light.
| {
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Rotating fluid under gravity, fluid dynamics question An incompressible inviscid fluid is rotating under gravity g with constant angular
velocity $\Omega$ about the z-axis, which is vertical, so that $u = (−\Omega y, \Omega x, 0)$ relative
to fixed Cartesian axes. We wish to find the surfaces of constant pressure, and
hence surface of a uniformly rotating bucket of water (which will be at atmospheric
pressure).
Bernoulli's equation suggests that $$p/\rho+|u|^2/2+gz=\text{constant. So,}$$
$$z=\text{constant}-\frac{\Omega^2}{2g}(x^2+y^2)$$
But this suggests that the surface of a rotating bucket of water is at its highest in
the middle, where is this going wrong?
Many thanks
| In the Physics Forums you should find the answer to your exact problem 'Fluid dynamics - finding pressure for a rotating fluid'. Another answer is here (the point is that Bernoulli's law is applicable only along a streamline so that we must use a rotating frame and add the centripetal acceleration).
From another point of view see Newton's 'Bucket argument'.
| {
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For someone who only studied electromagnetism, what is the modern way to explain electromagnetic fields? After reading most of the electromagnetism chapters of Feynman's lectures on physics, I would like to understand in more detail, at least an idea, of what causes the electromagnetic fields. Not sure where to go.
For instance, an electric field is caused by a charged particle. The presence of another particle would mean a force would act upon it. But is there a better explanation for this force to occur? By wikipedia browsing, photons are "the force carrier for the electromagnetic force". What does this mean exactly?
| The force acts between bodies, the electric field determines the force into the equations of motion of bodies. The body interaction is generally retarded so the field depends on time in a retarded way. The total electric field of a given charge is a sum of a "near field" that depends on time, but decays with distance as $1/R^2$ (they say it is "attached" to the charge), and a propagating field decaying as $1/R$. The latter corresponds to real photons (waves propagating with light velocity) whereas both are involved into the charge interaction. Depending on the value of $R$, one or another term may dominate.
| {
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Why don't electrons crash into the nuclei they "orbit"? I'm having trouble understanding the simple "planetary" model of the atom that I'm being taught in my basic chemistry course.
In particular,
*
*I can't see how a negatively charged electron can stay in "orbit" around a positively charged nucleus. Even if the electron actually orbits the nucleus, wouldn't that orbit eventually decay?
*I can't reconcile the rapidly moving electrons required by the planetary model with the way atoms are described as forming bonds. If electrons are zooming around in orbits, how do they suddenly "stop" to form bonds.
I understand that certain aspects of quantum mechanics were created to address these problems, and that there are other models of atoms. My question here is whether the planetary model itself addresses these concerns in some way (that I'm missing) and whether I'm right to be uncomfortable with it.
| Think a little further. When the electrons get accelerated closer to the nucleus they radiates some energy away which fills the vacuum and get scattered by other electrons, accelerating them. This finally becomes an equilibrium condition. This was calculated by Puthoff in 1987.
Ground state of hydrogen as a zero-point-fluctuation-determined state
H. E. Puthoff
Phys. Rev. D 35, 3266 – Published 15 May 1987
| {
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Can the charge of particles spontaneously flip from positive to negative or vice versa? I'm thinking of matter antimatter annihilation, are there reactions where normal matter converts to antimatter?
| Short, short answer: no.
Short answer: That would violate various conservation laws.
Discussion:
To begin with I want to exclude from discussion those particles that are their own anti-particles--the force carrying bosons and the compound particles like the neutral pion that are symmetric under the charge-exchange operator $\mathrm{C}$--or those like the neutral kaon's that David discusses which are connected by weak rotations in flavor space. I'll discuss one exception to this exclusion below
The rest of the charged fundamental particles (i.e. quarks and leptons) all carry conserved quantum numbers that are different for the positively and negatively charged versions.
*
*Each quark and charged lepton carries charge (yeah, duh) which is conserved by the electromagnetic interaction.
*Each quark (down, up, strange, charm, bottom, top) carries at least a color (red, green, or blue for the matter quarks and anti-red, anti-green, or anti-blue fr the anti-quarks) which is conserved in strong interaction.
*Each charged lepton (electron, muon, tau) carries the (imaginatively named) lepton number (+1 for the negative leptons, -1 for the positive leptons) which is conserved in the weak interaction.
But what about the neutrinos?
That's the rub. As thing stand they may or may not be Majorana particles and therefore may or may not change their matter-ness when they change helicity. The primary method in use to try to sort out the Majorana/Dirac nature of the neutrinos (neutrinoless double beta decay) is only possible if a neutrino effectively flips it's matter-ness. An answer should be forthcoming soon as saerches are on.
| {
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How Does Dark Matter Form Lumps? As far as we know, the particles of dark matter can interact with each other only by gravitation. No electromagnetics, no weak force, no strong force. So, let's suppose a local slight concentration of dark matter comes about by chance motions and begins to gravitate. The particles would fall "inward" towards the center of the concentration. However, with no interaction to dissipate angular momentum, they would just orbit the center of the concentration and fly right back out to the vicinity of where they started resulting in no increase in density. Random motions would eventually wipe out the slight local concentration and we are left with a uniform distribution again.
How does dark matter form lumps?
| In fact, three and more body purely gravitational interactions can form clusters and clumps by concentrating some particles and expelling others, as mentioned in a previous answer. Astronomers have discovered that this seems to happen quite rapidly. They call it "violent relaxation". Google "violent relaxation" for more info.
| {
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Physics book for 15 year old boy
Possible Duplicate:
List of good classical physics books
my name is Bruno Alano. As stated in the title, I'm 15 years old (I'll do 16 on 7 of Feb) and much love Computer Science (C, C++), Mathematics and Physics.
Some information may have been unnecessary, but my question is: What is the suggestion of a good physics book for a teenager of my age? I know basic things (speed, shoveller these issues and basic primary and secondary).
A good reason for this is my Awe in mathematics and physics. Besides that maybe one day be useful in what I really want a career (science or computer engineering).
And another question: It is interesting physics in the area I want to go? I'm at an age that would be good to learn beyond what is taught in common schools?
| It is not a physics books, though this book will make you a physics lover.
Surely You are joking Mr. Feynman
If you can get books for highschool that would be a good start, don't rush yourself, learning physics takes time and hard work.
| {
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How are the Pauli matrices for the electron spin derived? Could you explain how to derive the Pauli matrices?
$$\sigma_1 = \sigma_x =
\begin{pmatrix}
0&1 \\ 1&0
\end{pmatrix}\,, \qquad
\sigma_2 = \sigma_y =
\begin{pmatrix}
0&-i\\ i&0
\end{pmatrix}\,, \qquad
\sigma_3 = \sigma_z =
\begin{pmatrix}
1&0\\0&-1
\end{pmatrix}
$$
Maybe you can also link to an easy to follow tutorial ?
| Yet another way is from almost purely experimental considerations. Consider a Stern-Gerlach set up; the identity is
$$\mathbb{I}=|\uparrow\rangle \langle \uparrow |+|\downarrow\rangle \langle \downarrow | $$
Now experimentally we can only get two different numbers, $\hbar/2$ and $-\hbar/2$, and regardless of how we are oriented, in the natural basis corresponding to our orientation the spin operator must take the form
$$S=\frac{\hbar}{2}(|\uparrow\rangle \langle \uparrow |-|\downarrow\rangle \langle \downarrow |$$
So now lets fix a basis, say the z, and hit the identity from one side with a spin vector in a different basis, say the x
$$|\uparrow\rangle_{x}=|\uparrow\rangle\langle \uparrow |\uparrow\rangle_{x}+|\downarrow\rangle \langle \downarrow |\uparrow\rangle_{x}$$
Through various arguments, the two inner-products can be evaluated to give $\pm1/\sqrt{2}$ to get
$$|\uparrow\rangle_{x}=\frac{1}{\sqrt{2}}|\uparrow\rangle-\frac{1}{\sqrt{2}}|\downarrow\rangle $$
to see the arguments see Ballentine, or Sakurai, or Schwinger. Now to get $S_x$ we use the expression for $S$ from above
$$S_x=\frac{\hbar}{2}[(|\uparrow\rangle \langle \uparrow |)_x-(|\downarrow\rangle \langle \downarrow |)_x]$$
where my subscript x means in the x basis, but we know how to express those vectors in the "known" basis, the z basis. Substitution yeilds
$$S_x=\frac{\hbar}{2}(|\uparrow\rangle\langle\downarrow |+|\downarrow\rangle\langle\uparrow |)$$
Consider the following equal signs, begotten with entirely using the identity symbol
$$S_x=\sum_{n=1}^{2}\sum_{m=1}^{2}|n\rangle\langle n|S_x|m\rangle\langle m|$$
The $n$ and $m$ run from 1 to 2 standing for spin up and down. Now evaluate the guy in the middle. this is a two by two matrix whos elements are
$$\langle \uparrow |S_x|\uparrow\rangle\quad\langle \uparrow|S_x|\downarrow\rangle\quad\langle\downarrow|S_x|\uparrow\rangle\quad\langle\downarrow|S_x|\downarrow\rangle$$
to see what these equal just hit the $S_x$ buy above with the two up/down vectors in the z basis and use the orthogonality condition to make them either 1 or zero. The result is one of the Pauli matrices.
Hope this helps.
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Would a thinner atmosphere (such as high altitude) yield better solar panel energy generation? I am curious. If you were to put a solar panel up in high altitude or on the surface of the moon, both of which have much less atmosphere to reflect/refract light, would that solar panel produce more energy?
I am curious because an article comments discussion got me thinking about setting up a low Earth orbit solar collection platform as a power plant for a city or nation. Then I got thinking about how to bring it down to the surface (the energy that is). Which had me thinking about using really long cables stretched from the satellite to a high-altitude dirigible platform that would then allow it to more easily bring the energy to our planets surface.
From there I started wondering if the high altitude dirigible platform would instead be up high enough to garner any added energy collection benefits. It would certainly help with space saving on the ground and could be rather quickly moved into a disaster area for emergency power purposes. Being a dirigible it would also be relatively easy to keep it aloft. With it far above the clouds it should be able to avoid many weather issues as well (as far as I am aware).
So does anyone know if there would be any added benefits from solar collection in a thinner atmosphere environment from a power generation perspective? What pitfalls can you think of? The biggest one I can think of is the efficient transport of the energy.
| There is a real problem operating equipement at high altitude , such as observatories in the Chilean high deserts and in Antarctica (> 15000ft) - UV radiation destroys most plastics very quickly. I don't know what it would do to silicon solar panels but making anything else work there is a real pain
| {
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What is the physical definition of causality? Maxwell's equations give a physical relationship between the electric and magnetic fields $\vec E$, $\vec B$ at the same time, which some interpret as changes in one causes changes in the other etc. I find this confusing because to me, the cause of both is charge and cause should precede effect.
Therefore, how do physicists determine if there is a causal relationship between two physical quantities?
| I may add to the above already-excellent answer, that the reasoning behind the introduction of magnetic field can be explained a little less mathematical.
Basically, first thing to consider is an electric charge. When it does not move in our reference frame it does not produce any magnetic field, just electric one. Now, if it starts to move, we percieve its width (and width of spacetime) parallel to its moving direction to contract so that for every $dl$ we have $dl' = \sqrt{1-v^2} dl$ - Lorentz contraction. Now, the field lines of electric field also contract - and the closer field lines are - the stronger the field. This added influence is simply called 'magnetic field' but it is not a separate entity. So, magnetic field is a relativistic shadow of electric field. This is reflected in Maxwell equation, where charge density only apperas with the electric field, not the magnetic one. So, charge causes electric field and our relative movement w.r.t. charge causes our perception of this field to change.
This, however, is only half of the truth. This is because changes in electric fields have been found to propagate in vacuum - these are electromagnetic waves. In vacuum, Maxwell equations become completely symmetric - and one can not distinguish cause and effect when it comes to electric and magnetic fields. This is where 'relativistic reasoning' has to take precedence over our classic intuition and we have to accept that those fields - however distinct they might seem - are just one entity - the electromagnetic field.
| {
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Calculating lagrangian density from first principle In most of the field theory text they will start with lagrangian density for spin 1 and spin 1/2 particles. But i could find any text where this lagrangian density is derived from first principle.
| Try Steven Weinbergs comprehensive The Quantum Theory of Fields (Vol. 1, "Foundations"). He follows a very systematic approach from "first principles", i.e. from Wigner's classification of unitary irreducible representations of the Poincaré group, over free fields for different mass/spin configurations (including spin 1 and 1/2, which in different formulation lead up to the Klein-Gordon and Dirac equations) to perturbation theory and Lagrangian densities (and lots more).
If you're interested in a more compact treatment of the "first principles" part only (but not Lagrangian densities!), plus theorems that can be proven as a direct consequence of them, such as PCT or spin/statistics, the standard textbook/primer of mathematical QFT is Streater/Wightman, PCT, spin and statistics, and all that.
| {
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What are the practical applications of decoherence? Let me clarify this question somewhat. I know decoherence is ubiquitous in nature and explains the emergence of a classical world from quantum physics. My question is really about how a knowledge of how decoherence actually works can be put to use in a practical application. An application we can't design in the absence of such a knowledge, even though decoherence is still happening all the time.
Thanks
| Since very weak interactions are sufficient to significantly decohere a quantum system, quantum systems can potentially be used as very sensitive force sensors if their decoherence is monitored. This monitoring can take the form of interferometric measurements in which the fringe visibility is measured as a function of time or some experimental parameter. The main challenge would presumably be to isolate the system well enough from all the other interactions which also cause decoherence but which you don't want to measure.
A google search for "decoherence microscopy" will bring up some proposals. Another good starting point are the papers of the group of Markus Arndt, e.g. "Quantum interference of clusters and molecules" in Rev.Mod.Phys., in particular the section on "interference-assisted measurements".
| {
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Static plane in an inertial frame of reference Suppose we are given a mechanical frame consisting of two points. How can we prove that assuming any initial conditions there is an inertial frame of reference in which these points will be in a static plane?
| Here is one interpretation of the question(v1). In the framework of non-relativistic Newtonian mechanics, let us consider an initial frame and two point point masses with initial positions ${\bf r}_1(0)$ and ${\bf r}_2(0)$; and initial velocities ${\bf v}_1(0)$ and ${\bf v}_2(0)$. (By a translation of the initial frame one may assume ${\bf r}_1(0)={\bf 0}$.)
Now perform a Galilean transformation with relative velocity ${\bf v}$ such that
the new initial velocities are
$${\bf v}^{\prime}_1(0)~=~{\bf v}_1(0)-{\bf v},$$
$${\bf v}^{\prime}_2(0)~=~{\bf v}_2(0)-{\bf v}.$$
Next find all solutions ${\bf v}$ such that the three initial vectors ${\bf r}_2(0)-{\bf r}_1(0)$, ${\bf v}_1(0)-{\bf v}$, and ${\bf v}_2(0)-{\bf v}$ are linearly dependent. The three linearly dependent vectors therefore span a plane, a line, or a point. (The line and point case will lie inside infinitely many planes.)
Consider finally the full solution of ${\bf v}$'s and planes. Now there should exists at least one such plane so that the total forces ${\bf F}_1(t)$ and ${\bf F}_2(t)$ on each point particle lie in this plane for all times $t$.
This plane is going to play the role of OP's static plane, which can be proved with the help of Newton's 2nd law.
| {
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What is the velocity area method for estimating the flow of water? Can anyone explain to me what the Velocity Area method for measuring river or water flow is?
My guess is that the product of the cross sectional area and the velocity of water flowing in a pipe is always constant. If the Cross sectional area of the pipe increases at a particular point, then the velocity decreases so that the product $AV$ is a constant. Am I right?
If so, how can we extend this to pipes where the water is accelerating & does not have a constant velocity? For example, the system may be under the action of gravity & hence the acceleration of the water is $g$, the acceleration due to gravity?
| I am assuming that you want the relation for a noncompressible fluid ($\rho$ is constant).
Well, what we are doing is conserving the volume(i.e. mass) of fluid entering and leaving the pipe in unit time. Thus, $\frac{dV}{dt}=const$ (more correctly $\frac{\partial V}{\partial t}$). As $V=Ax$, $\frac{dV}{dt}=A\frac{dx}{dt}=Av$. So, The area at a point(in the pipe) times the velocity at that point is constant.
Actually there's a small flaw in this argument. This will only work if the water is not in a pipe but is instead in freefall (or something similar). Also, this will only work at steady state, when the "shape" of the flow is constant. Otherwise, there will be loss of volume of water in changing the shape too.
If the water was in a pipe with varying area, the principle would be the same (conservation of mass), but you would have to take change of density into account (i.e. an incompressible fluid is no longer possible).
| {
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Could a Dyson sphere destroy a star? Freeman Dyson proposed that the power needs of an advanced civilisation would eventually require the entire energy output of a star to be collected, so that the star would end up surrounded by a dense network of satellites extracting power from the radiation. In science fiction this idea has often mutated into a solid shell that completely surrounds the star. My question is, if it were possible to build such a shell, what effect would it have on the dynamics of the star inside?
I suspect the effect would be catastrophic, but I'd like more details. Here's my reasoning: Wikipedia implies that the temperature gradient between the core and outer layers of a star plays an important role in its stability. The shell around the star would reflect or re-emit a lot of the star's radiation. (I'm assuming the shell isn't composed of perfect solar collectors that would simply absorb all the radiation.) This reflected radiation would reduce the loss of heat from the outer layers of the star. This would reduce or even eliminate the temperature gradient, which I guess would cause the star to expand. My question is whether this is correct, and if so whether it would be enough to disrupt the process of fusion in the star's core. Or would there be some other, less obvious effect on the star's dynamics?
A closely related question is, would stars be stable in a static or contracting universe? In this case all of space would be filled with the radiation emitted from other stars, and I'd be interested to know what effect this would have on stellar dynamics.
| Of course not. A star doesn't care what energy is collected from its
radiation.
| {
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How can a human eye focus on a screen directly in front of it? I am asking this question here because I think the answer has something to do with the way light is bent as it's captured through the eye.
I saw a show a while ago about tiny screens on contact lenses to pull up data on objects you see in the real world, I also just saw this article about Google testing the same idea with screens in the lenses of sunglasses.
The part I do not understand is how your eye would focus on a screen that is so close. My eyes (and I believe most others) cannot focus on anything closer than a couple inches away. Yet the lenses on glasses are much closer to the eye than a couple inches.
So I would guess they would have to use some special technology to separate the light rays in a way that your brain could make an image from it. If that's correct, how would it work? if not, how would they get your eye to focus on a screen so close?
| Sadly I don't have a pair of the Google glasses to play with, but I have seen similar things over the years, and these worked as projectors not screens.
You're quite right that it would be impossible to focus on a screen in the lens of your spectacles as it's far too close to the eye. However a projector mounted on the glasses can project the light into your eye as if it was coming from further away. Then your eye can focus it just as it would any distant object. If you can find any more info about the Google glasses it would be interesting to see if they use this technique.
| {
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Electrodynamics textbooks that emphasize applications Please recommend undergraduate-level textbooks on electrodynamics which emphasize practical applications and real life examples. Please describe the book's level and contents and its intended audience in as much detail as possible.
Please provide both applications from everyday life for curious students as well as engineering applications from a physics perspective.
| A text that I use as a supplement for my undergraduate E&M class is:
Zangwill, Modern Electrodynamics, 2013
This is definitely a more advanced textbook than the usual upper-class E&M texts; I'd put it somewhere between Griffiths and Jackson. Where it excels is in the range of applications provided for the upper-level material. A sample of these:
*
*Electrostatic potential energy in the nuclear liquid-drop model
*Electrostatics of ion channels in cells
*The four-point resistance probe and the method of images
*Focusing by axially symmetric magnetic fields in an electron microscope
*Whistlers
*The classical Zeeman effect
*CMB polarization
Also, it contains the funniest E&M problem I've ever encountered: "Find the lowest resonant frequency $\omega_0$ and the exact half-width $\Gamma$ of the resonant cavity God instructed Moses to build in Exodus 25:10–11."
| {
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What formulations of QM are there? It is usually said that there are different formulations of QM, for example historically there was Schrodinger's (wave mechanics), and Heisenberg's (matrix mechanics), then Dirac's (which showed they are equivalent)
Since they are all physically equivalent I have a few questions:
1-Is Dirac's formulations considered more fundamental in the sense that it can be reduced to one of the 1st two?
2-I also hear about the path integral and density matrices, are they another formulations?
3-Are there more formulations less known to undergraduate students but known by researchers because they are technically advanced?
| 2) Path integral formulation sometimes called another (Fynman) formulation of quantum mechanics, as opposed to Dirak and Heisenberg ones.
Density matrices are to my knowledge never called another formulation.
AFAIK, other more advanced approaches are not called another formulations partly due to the fact that they are too advanced, partly because they were formulated later. Maybe also due to the fact that they are never used for QM itself, but for QFT only.
| {
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Does EM radiation (any, i.e. RF), or sound, radiate everywhere at once? I am having trouble understanding electromagnetic radiation (or waves in general, be it EM or sound). If I have a 1 Watt speaker, is it infinitely divided and spread out so that everyone in every direction around the speaker can hear it?
I do not believe they have "height", to reach more than one person at once, but if they did they would probably collide at one point. How do sound waves travel "backwards" (i.e. you are behind a speaker), are they scattered by air particles or itself so that people behind could hear it (at reduced amplitude)?
I am just unsure how to wrap my head around it.
| Electromagnetic or acoustic waves cannot be "infinitely divided", but the minimal "portions" of the waves (quanta - photons or phonons) typically have very low energy. Radiation from an acoustic speaker or an electromagnetic antenna propagates in all directions, but if the speaker or the antenna is directional, the so called intensity of radiation strongly depends on the direction. In some directions the intensity may be extremely small. Propagation of radiation, including backward propagation (say, behind the speaker), is determined by so called diffraction (you may wish to google this term).
| {
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Degree of freedom paradox for a rigid body Suppose we consider a rigid body, which has $N$ particles. Then the number of degrees of freedom is $3N - (\mbox{# of constraints})$.
As the distance between any two points in a rigid body is fixed, we have $N\choose{2}$ constraints giving $$\mbox{d.o.f} = 3N - \frac{N(N-1)}{2}.$$ But as $N$ becomes large the second term being quadratic would dominate giving a negative number. How do we explain this negative degrees of freedom paradox?
| You're double counting here. Lets take three particles. You're counting $\binom{3}{2}=3$ DOFs, right? But fixing the vector distance between particle 1 and two, and then fixing it between 2 and 3 includes fixing it between 1 and 3. Mathematically, $\vec{d}_{1,3}=\vec{d}_{1,2}+\vec{d}_{2,3}$
The easier way to count DOFs is like this. For a molecule with N particles, number of DOFs is $3N$. Out of these, 3 will be translational. For a point molecule (i.e, a single atom), subtract 3 as it has 0 rotational DOFs. For a perfectly linear molecule, subtract 1, as it has 2 rotational DOFs (Rotation along its axis is irrelevant). Now, we usually neglect vibrational DOFs (at normal temperatures). Vibrational DOFs are whatever DOFs are remaining. Thus, we always have a total of 3N DOFs, out of which we may count only the translational (3) and rotational (2 or 3) DOFs. See the table here.
| {
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Finding distance when the force is a function of time I'm having trouble with this homework question
A mysterious rocket-propelled object of mass 49.0 kg is initially at rest in the middle of the horizontal, frictionless surface of an ice-covered lake. Then a force directed east and with magnitude $F(t) = (16.3\text{ N/s})t$ is applied.
How far does the object travel in the first 5.50s after the force is applied?
For some reason I'm not getting the correct answer. I think maybe I'm not understanding how to use the magnitude of the force they are giving me. I know how to use a constant force, but is this different because the force is a function of time?
I tried starting it like this:
$$\begin{align}F &= ma\\
16.3(t) &= (49)(a) \\
16.3(5.5) &= (49)(a) \\
89.65 &= (49)(a) \\
a &= 1.82959\ \mathrm{m/s^2}\end{align}$$
So now we know that:
$$\begin{align}t &= 5.5\text{ s} \\
a &= 1.82959 m/s^2 \\
V_o &= 0\end{align}$$
So I plug it into my equation:
$$\begin{align}\Delta X &= V_o t + 1/2 a t^2\\
\Delta X &= (0)(5.5) + (1/2)(1.82959)(5.5)^2\\
\Delta X &= 27.7\text{ m}\end{align}$$
But that's not the right answer.
| Your mistake here is that you're using the three equations of motion ($v=u+at$ et al), even though they are only applicable for constant acceleration.
The correct way is this (i'm writing 16.3 as $k$) :$$F=ma=kt$$
$$\therefore a=kt/m$$
$$\therefore \frac{dv}{dt}=kt/m$$
$$\therefore dv=\frac{ktdt}{m}$$
$$\therefore \int\limits_0^vdv=\int\limits_0^t\frac{ktdt}{m}$$
$$\therefore v=\frac{kt^2}{2m}$$
$$\therefore \frac{dx}{dt}=\frac{kt^2}{2m}$$
$$\therefore dx=\frac{kt^2dt}{2m}$$
$$\therefore \int\limits_0^xdx=\int\limits_0^t\frac{kt^2dt}{2m}$$
$$\therefore x=\frac{kt^3}{6m}$$
I had to use a bit of calculus here. Any problem involving rate of change requires calculus. Those three equations of motion are derived from "acceleration is rate of change of velocity" and "velocity is rate of change of displacement" at constant acceleration. If $a$ is not constant, you have to use these equations: $$v=\frac{dx}{dt}$$,$$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$$. A shortcut formula for when acceleration is given in terms of displacement is $$a=v\frac{dv}{dx}$$.
| {
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Looking for the name of a particular device Please move this if it's not in the right location.
I'm looking for the name of a device that I frequently see in many scenarios, specifically that of an office/library which can be described as having multiple rings that rotate in various directions. I was thinking it was a gyroscope or perhaps a celestial globe, but something tells me that it's not quite what I'm looking for. I recall that there is a movie production company which uses this device as their symbolic figure of their logo.
| The device you are describing is a Cardan suspension.
| {
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Calculating Impact Velocity Given Displacement and Acceleration
Assume a car has hit a wall in a right angled collision and the front bumper has been displaced 9 cm. The resulting impact is 25g. Also, it is evident by skid marks that the car braked for 5m with an acceleration of 1.5m/s^2. What is the impact velocity in this collision?
Here's what I get out of it.
$$\begin{align}\Delta d &=0.09\text{ m}\\
a &= 196\ \mathrm{m/s^2}\\
V_2 &= 0\text{ m/s}\\\end{align}$$
Then I determine $V_1$ by:
$$\begin{align}V_2^2 &= V_1^2 + 2a\Delta d\\
0 &= V_1^2 -35.28 \\
5.94\text{ m/s} &= V_1\end{align}$$
My textbook does not give this answer. Could anyone please explain why. I have been looking at it for hours.
| I always look for the easiest way to describe questions like this, and I think the easiest way is to do it in reverse. Start with the car stationary and accelerate it at 25g for 9cm, then accelerate it at 1.5ms$^{-1}$ for 5m. You've correctly identified the relevant equation of motion:
$$v^2 = u^2 + 2as$$
So for step 1 i.e. the bumper deforming:
$$v_1^2 = 2 \times 25g \times 0.09$$
Then for step 2 i.e. the braking:
$$v_2^2 = v_1^2 + 2 \times 1.5 \times 5 = 2 \times 25g \times 0.09 + 2 \times 1.5 \times 5$$
See if that gives you the same answer as your book.
| {
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Mechanism by which electric and magnetic fields interrelate I read that force due to electric field on some particle in one reference frame can exhibit itself as force due to magnetic field in some other reference frame and that electric and magnetic fields are two aspects of same underlying electromagnetic field.
My question is what is the mechanism which can explain how an electric field becomes/creates magnetic field in some other reference frame. Is there any such explanation available in relativity theory? I am not looking for mathematics but a physical explanation.
Wikipedia article http://en.wikipedia.org/wiki/Relativistic_electromagnetism
explains something about origin of magnetic forces in a wire as a consequence of lorentz contraction and motion of electrons in the wire
Calculation of the magnitude of the force exerted by a current-carrying wire on a moving charge is equivalent to calculating the magnetic field produced by the wire. Consider again the situation shown in figures. The latter figure, showing the situation in the reference frame of the test charge, is reproduced in the figure. The positive charges in the wire, each with charge q, are at rest in this frame, while the negative charges, each with charge −q, are moving to the left with speed v. The average distance between the negative charges in this frame is length-contracted to:
where is the distance between them in the lab frame. Similarly, the distance between the positive charges is not length-contracted:
Both of these effects give the wire a net negative charge in the test charge frame, so that it exerts an attractive force on the test charge.
But this still does not explain origin of magnetic field in case when there are no positive charges.
| https://ocw.mit.edu/courses/materials-science-and-engineering/3-a08-attraction-and-repulsion-the-magic-of-magnets-fall-2005/assignments/mag_relativity.pdf
Have a look at this link - it will probably be better than any of our explanation.
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What determines color -- wavelength or frequency? What determines the color of light -- is it the wavelength of the light or the frequency?
(i.e. If you put light through a medium other than air, in order to keep its color the same, which one would you need to keep constant: the wavelength or the frequency?)
| For almost all detectors, it is actually the energy of the photon that is the attribute that is detected and the energy is not changed by a refractive medium. So the "color" is unchanged by the medium...
| {
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What is the origin of the naming convention for position functions? In physics, position as a function of time is generally called $d(t)$ or $s(t)$. Using "$d$" is pretty intuitive, however I haven't been able to figure out why "$s$" is used as well. Is it possibly based on another language?
| As commenters have pointed out, it's German Strecke.
Note that $s$ is for displacement, whereas $d$ is for distance. Distance is the distance along the path traveled by a body, whereas displacement is the birds-eye distance traveled. Displacement can also be negative in 1-D, depending upon your reference positive direction.
For some reason, even though Strecke actually means distance, not displacement, its symbol is used for displacement.
You might want to check out this paper, it's got an analysis of the naming, mainly for electrodynamic units. A few symbols from the table at the end of the paper: $c$ (speed of light) comes from Latin celeritas; $I$ (current) comes from "intensity of current" in French (intensite du courant). The $\mathbf{A}$-potential, $\mathbf{B}$-field, $\mathbf{H}$-field got their symbols from the alphabetic order of the others.
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Example in the book: A simple accelerometer A simple accelerometer
You tape one end of a piece of string to the ceiling light of your car and hang a key with mass m to the other end (Figure 5.7). A protractor taped to the light allows you to measure the angle the string makes with the vertical. Your friend drives the car while you make measurements. When the car has a constant acceleration with magnitude a toward the right, the string hangs at rest (relative to the car), making an angle $B$ with the vertical.
*
*(a) Derive an expression for the acceleration $a$ in terms of the mass m and the measured angle $B$.
*(b) In particular, what is a when $B$ = 45? When $B$ = 0?
I don't care about the answers, the important thing is the following:-
The book says The string and the key are at rest with respect to the car, but car, string, and key are all accelerating in the +x direction. Thus, there must be a horizontal component of force acting on the key.
That's the reason the book decided to consider a force in the $+x$ direction, but I'm looking for a better explanation: how would I find detect the force in the $+x$ direction in another way? To me, when I draw the free body diagram of the string, there looks to be no force acting on the $+x$ direction! I understand it starts with noticing that the string is attached to the ceiling of the car, and that the car has force causing acceleration in one direction, but I don't know how to go further than that.
| Acceleration and force are not the same thing.
So first simplify the problem by just talking about acceleration.
For example, suppose the car is accelerating forward at a rate of 1.0 meter per second per second. It also feels an "acceleration" due to gravity, which is 9.8 m/(s^2).
If you want to know the angle at which the key hangs, it is the angle whose tangent is 1.0/9.8.
Hopefully that is enough for you to run with.
Then if you want to know force, make use of f=ma. i.e. multiply the acceleration by the vehicle's mass.
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Have red shifted photons lost energy and where did it go? I think the title says it. Did expansion of the universe steal the energy somehow?
| The short answer is "yes". The energy lost from the photons is taken up by the energy in the gravitational field. Of course energy is a relative concept but if you take the simplest case of a spatially flat homogeneous cosmology with no cosmological constant then the equation for energy in an expanding volume $V(t) = a(t)^3$ is
$E = Mc^2 + \frac{P}{a} - \frac{3a}{\kappa} (\frac{da}{dt})^2 = 0$
$M$ is the fixed mass of cold matter in the volume, $\frac{P}{a}$ is the decreasing radiation energy in the volume with $P$ constant, and the third term is the gravitational energy in the volume which is negative. The rate of expansion $\frac{da}{dt}$ will evolve in such a way that the (negative) gravitational energy increases to keep the total constant and zero.
For a more general discussion of energy conservation in general relativity see my paper http://vixra.org/abs/1305.0034
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How to find the value of the parameter $a$ in this transfer function? I am given a transfer function of a second-order system as:
$$G(s)=\frac{a}{s^{2}+4s+a}$$
and I need to find the value of the parameter $a$ that will make the damping coefficient $\zeta=.7$. I am not sure how to do this but I might have found something that might have helped so I am going to take a stab at it. I found a transfer function in the book of a second order spring-mass-damper system with an external applied force in the book as:
$$G(s)=\frac{a}{m\omega_{n}^{2}}(\frac{\omega_{n}^{2}}{s^{2}+2\zeta \omega_{n}+\omega_{n}^{2}})$$
I was thinking that I could just write $\omega_{n}^{2}=a$ and $2\omega_{n}\zeta=4$. Then I could just solve for $a$. Is this possible?
| You have the answer.
Consider $2*\zeta*\omega_n = 4$. $\zeta = 0.7$. $\omega_n^2 = a$ What value of $\omega_n$ (or $a$ in your case) satisfies this?
Roughly 8.18 is your answer.
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Expected Energy Production From High Efficiency Solar Cells First, a bit about my thoughts. I believe we have the capability today to provide energy, water, food, education, and transportation to every man woman and child on the planet. To that end, I would like to become a force that brings about this change.
In trying to meet the first goal, which is to provide energy, I have come across two technologies which greatly interest me, the first of which must be in place to begin the second.
The first is the high efficiency solar cells developed by Patrick Pinhero at the University of Missouri. Assuming that said solar cell captures 80% of available light, how much energy can I expect them to produce per meter of cell? How would this vary betweeen environments such as the Nevada desert and central Florida, how did you come to these conclusions, and is there any formula I can use to calculate an expected energy output?
| Have a look at this innovative solar panel technology from CERN.
The ultra-high
vacuum provides the panels' heat chambers with exceptional insulation, vastly reducing heat loss and greatly improving efficiency. "We've had temperatures of 80°C inside the panel when the panels were covered in snow", says Benvenuti.The panels also recover the energy produced by diffuse light more efficiently than traditional panels. The two technologies make them particularly suited to colder, less sunny climates where classic solar panels are less efficient.
In Geneva, the vacuum-sealed solar panels can supply heat at over 300 degrees Fahrenheit with 50 percent efficiency, meaning each square meter of panel generates a half kilowatt of power. Moreover, unlike other solar panels that use parabolic mirrors, Benvenuti’s do not rely only on direct sunlight. Instead, they capture diffuse light, which makes up more than 50 percent of the light in central European climates like Switzerland.
| {
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Irreducible representation in physics I am confused about something.
Group theory books written for physicists say that any reducible representation can be decomposed in terms of irreducible representations (so correct me if I am wrong, to me irreducible representations are like the unit vectors i j k in terms of which any 3D vector can be expanded, or they are like sines and cosines in terms of which any periodic function can be Fourier expanded.)
Now at the same time they say that any bigger representation of a group can be built out of irreducible ones.
What is unclear to me is the physical motivation for each direction. Of course those books contain physical applications but the big picture is never obvious (they get to applications after 200-300 pages of abstract details).
If someone could answer the following questions I would be really appreciated:
1-what is the physical motivation to write a representation in terms of irreducible ones?
2-what is the physical motivation to build bigger representations using irreducible ones?
| The physical motivation is pretty simple. In quantum mechanics this means that if Hamiltonian is invariant under all $g\in G$ you may use the fact that the solutions of this Hamiltonian form a (reducible) representation of this group (as any other full system). Here comes representation theory. It is pretty easy to show that the energies of the states corresponding to the same irreducible representation are equal. Furthermore, there is Wigner-Eccart theorem which allows one to reduce number of values which describe the system by using its symmetry properties.
Thus, knowing irreducible representations of $G$ one may say something about levels degeneracy, selection rules, and so on. As a result, it turns out that it is constructive to classify states of the system in accordance with irreducible representations of this system symmetry group.
It is exactly how energy levels in atoms are classified. This idea may be easily generalized from $SO(3)$ to any other group of symmetry.
For the second question, I do not really get it. Usually one has the number of the states which form a basis of the reducible representation given by number of particles in the system, etc. So it is not very natural to build up reducible representations by purpose.
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What do the dimensions of circulation mean, and how is circulation related to action? The dimensions of circulation $\int_C \vec{v}\cdot d\vec{r}$ seem strange, but if you include
(even a constant) density $\rho$, then $\int_C \rho\vec{v}\cdot d\vec{r}$ has dimensions
the same as action/volume. Is there any significance to that? Is there any heuristic way to
think about circulation which helps understand the dimensions?
| Don't confuse circulation with the vortex that produces it. Circulation is irrotational flow, vortices are rotational and are found in the boundary layer. The flux of the vorticity is circulation. In aerodynamic applications, I find it helps to understand its physical meaning by dividing the circulation by the distance it acts over (i.e. the chord of the wing) to get the imparted velocity to the fluid. This velocity multiplied by the airstream velocity and by air density gives you the pressure difference that lifts the wing.
| {
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Are there more bosons or fermions in the universe? The question is in the title: are there more bosons or fermions in the universe? Or is there the same number of bosons and fermions?
I think there is the same number but I don't know why exactly.
| I suppose that the question is ill-posed for at least three reasons.
Since Fermions and Bosons can be generated from each other, there is a certain dynamics involved. Therefore: When would you count those numbers? The problem I see is connected with defining the (global?) time in which to count - given a lack of simultaneity.
In QFT, there is a particle number operator. This operator depends on the coordinate frame. Which one would you chose? The "number of particles" is not independent of the coordinate system.
Finally, Unruh effect tells us that an accelerating observer sees the vacuum as a heat bath consisting of photons. So, we would need to know the acceleration state as well.
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Why does optical pumping of Rubidium require presence of magnetic field? The optical pumping experiment of Rubidium requires the presence of magnetic field, but I don't understand why.
The basic principle of pumping is that the selection rule forbids transition from $m_F=2$ of the ground state of ${}^{87} \mathrm{Rb}$ to excited states, but not the other way around ($\vec{F}$ is the total angular momentum of electron and nucleus). After several round of absorption and spontaneous emission, all atoms will reach the state of $m_F=2$, hence the optical pumping effect.
But what does the Zeeman splitting have anything to do with optical pumping? Granted, the ground state, even after fine structure and hyperfine structure considered, is degenerate without Zeeman splitting, but the states with different $m_F$ still exists.
In addition, how is the strength of optical pumping related to the intensity of magnetic field applied?
| There are two kinds of optical pumping that are possible. Hyperfine Optical pumping and Zeeman Optical Pumping. The latter is under zero magnetic field and former requires a magnetic field. The spin-wave (ground state coherence) you create depends mainly on the polarization of the light being used and its intensity. If you increase the strength of the magnetic field, the separation between the Zeeman levels increases. For large values, you get a "level-crossing" effect.
What do you mean by "strength of optical pumping?" If you mean the population redistribution, then the strength of the magnetic field alone does not play a role. You have to consider the nature of the light (intensity, polarization,detuning,coherence) and the decoherence of the atomic ensemble as well.
Other applications:
In many experiments that exploit atomic coherence, using the Zeeman levels (i.e a 3-level system that consists of two ground state Zeeman levels of the same hyperfine state and the excited state) alleviates the problem of requiring two identical lasers. You can use one laser (perfectly correlated with itself!) and shift its frequency by a small amount using an AOM (say 80Mhz) as opposed to 6.1 GHz between the ground state hyperfine levels.
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How can people do music with Tesla coils? I saw a lot of videos of Tesla coils doing music on YouTube. And I wonder how can they do that sort of things.
How they can calculate what tone it is going to do? And what are the factors to consider?
| In a simplified picture a Tesla coil is just a transformer which creates a high voltage, high frequency current between the top part and the grounded bottom.
This transformer receives it's energy from an AC source. If you modulate the AC source (amplitude modulation) the sound created from the sparks of the Tesla coil will also be modulated with that frequency and in the end can be used to play music.
| {
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Do neutron stars reflect light? The setup is very simple: you have a regular ($1.35$ to $2$ solar masses) evolved neutron star, and you shine plane electromagnetic waves on it with given $\lambda$. Very roughly, what shall be the total flux of absorbed/scattered EM radiation?
Shall the result change if the neutron star is young and not evolved?
| A neutron star will have a thin layer of normal matter at the surface, and of course this reflects light just like any other normal matter.
But I guess you're really asking if neutronium reflects light, and that's a very good question that a quick Google failed to answer. EM radiation generally interacts with dipoles or scatters off electrons, so I'd guess matter made of neutrons should be transparent.
| {
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Why is the conductor an equipotential surface in electrostatics? Since the electric field inside a conductor is zero that means the potential is constant inside a conductor, which means the "inside" of a conductor is an equal potential region.
Why do books also conclude, that the surface is at the same potential as well?
| This result can be understood mathematically. Suppose the system has reached equilibrium and all charges have stopped moving so that electrostatics applies. Then the potential is a harmonic function $\Delta \varphi = 0$ in $\mathbb{R}^3$ and the conductor is a closed and bounded region in $\mathbb{R}^3$.
A general property of harmonic functions is the maximum principle. In short, if $\varphi$ is harmonic on an open set $\Omega$ and $B \subset \Omega$ is a closed and bounded region, then $\varphi$ has no local minimum or maximum in the interior of $B$ and the absolute maximum and minimum of $\varphi$ occur on the boundary of $B$. In particular, if $\varphi$ is constant on the interior of $B$ it must be (the same) constant on the boundary of $B$.
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Decoherence when no one is looking? I understand that in the single-electron-at-a-time double slit experiment, if a detector is placed before the slit, the interference pattern vanishes.
Suppose I left the detector on, but put a bag over its screen (I can't tell what state the electron is before it passes the slits), does the interference pattern come back?
If so, does that mean the electron "knows" I'm not "looking" and proceeds to interfere with itself?
Edit: I had photons passing through the slit initially. I've updated it to electrons to reflect a more common setup.
| No. The electron doesn't care at all whether you look at the screen or don't. It's a common misunderstanding from taking the notion of "observer" too seriously. In quantum mechanics, the "observer" is anything that causes decoherence. In your example, that's the detector. Putting a bag over the screen doesn't change anything about it.
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What shape is needed to contain a blade made of plasma? Although this may stray into the subject of fiction, this question requires physics expertise.
If one were able to create a strong enough magnetic field to contain a blade of plasma, what shape would be needed to contain it in a loop?
| The problem with using magnetic fields to contain plasma is that the charged particles move at right angles to the field you're applying. This makes it exceedly difficult to contain them. Tokamak reactors manage it by holding the plasma in a loop so that when the particles move at right angles to the field they just go round the loop, but even so current Tokamac's can't hold the plasma for long.
I can't think of any way to constrain a plasma in anything resembling a (lightsabre?) blade, and to be honest if you had that much power available you'd be better off using it in a projectile weapon. Projectiles many not be as sexy as plasma weapons, but they're an exceedingly efficient way of transferring energy to the target.
| {
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Slit screen and wave-particle duality In a double-slit experiment, interference patterns are shown when light passes through the slits and illuminate the screen. So the question is, if one shoots a single photon, does the screen show interference pattern? Or does the screen show only one location that the single photon particle is at?
| The answer is yes to both questions: yes, the screen does show one location for one particle and yes, the accumulated picture after repeating the experiment many, many times does show the interference pattern.
There is a set of beautiful pictures and a video of the double slit experiment in one-particle-per-time mode that can be found here (the experiment is with electron but conceptually there is no difference).
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How can a Higgs decay to heavier products than its mass? How is it possible that a higgs at ~125 GeV can decay into 2 W bosons @ ~ 80 GeV a piece (for example)? Shouldn't a particle only be allowed to decay to lighter particles + energy?
Diagram copied from this question
| The decay channels of the Higgs boson on your graph contain decays to virtual particles as well. In particular, one talks about $ZZ^*$ and $WW^*$ decays, too. The asterisk means that the particular particle which carries the asterisk is virtual. Its being virtual means that its energy and momentum don't have to obey
$$E^2-p^2c^2=m_0^2c^4$$
i.e. they can be "off-shell" but being off-shell requires that the particle fails to exist "permanently" or for extended periods of time. Virtual particles may only exist for a very limited period of time which is even shorter if the particles are "very off-shell".
Because one of the particles is off-shell, its energy may be very small, even as small as zero (or slightly negative), so the total energy of the pair with one real particle and one virtual particle may be "just above" the rest mass of the W-boson or Z-boson for the decay to be sufficiently likely.
The virtual W-bosons and Z-bosons nevertheless decay to similar products as physical (on-shell) W-bosons and Z-bosons. In principle, the energy of a virtual particle may be negative (or there may be decays to two virtual particles) so the decay rate isn't strictly zero even for Higgs masses below the W-boson or Z-boson mass, respectively. Nevertheless, the $WW^*$ or $ZZ^*$ branching ratio rapidly increases above the single W-boson or Z-boson mass.
That's not the case of decays to quark pairs which only becomes significant if both quarks are on-shell. For example, the $t\bar t$ decay channel would only kick for a Higgs mass above 350 GeV, twice the top quark mass. Also, the parametric description of the shape of the branching ratio near the "threshold" (minimum allowed Higgs mass or energy for which the decay is possible) would differ between bosons and fermions.
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Are we inside a black hole? I was surprised to only recently notice that
An object of any density can be large enough to fall within its own
Schwarzschild radius.
Of course! It turns out that supermassive black holes at galactic centers can have an average density of less than water's. Somehow I always operated under the assumption that black holes of any size had to be superdense objects by everyday standards. Compare the Earth to collapsing into a mere 9mm marble retaining the same mass, in order for the escape velocity at the surface to finally reach that of light. Or Mt. Everest packed into one nanometer.
Reading on about this gravitational radius, it increases proportionally with total mass.
Assuming matter is accumulated at a steady density into a spherical volume, the volume's radius will only "grow" at a cube root of the total volume and be quickly outpaced by its own gravitational radius.
Question:
For an object the mass of the observable universe, what would have to be its diameter for it to qualify as a black hole (from an external point of view)?
Would this not imply by definition that:
*
*The Earth, Solar system and Milky Way are conceivably inside this black hole?
*Black holes can be nested/be contained within larger ones?
*Whether something is a black hole or not is actually a matter of perspective/where the observer is, inside or outside?
| Only when you're not looking at it. When you are looking, presuming it's expanding, it is a white hole.
Seriously, if we are inside a true singularity, then all of time is included within it, so issues of redshifting, movement, even gravity, etc., are "red herrings" -- artifacts of the observer's frame of reference that misinform oneself. The real issue, then, is what is the real relationship between the observer and the scale of the universe? Since, there's no Grand Unified Theory, that little factor "G" at the beginning of Newton's formula is pretty fungible (i.e. a lot of degrees of freedom to stipulate what mass is, for example).
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Is chaos theory essential in practical applications yet? Do you know cases where chaos theory is actually applied to successfully predict essential results? Maybe some live identification of chaotic regimes, which causes new treatment of situations.
I'd like to consider this from the engineers point of view. By this I mean results that are really essential and not just "interesting". For example one might say weather calculation effects are explained with chaos theory, but an "engineer" might say: "So what? I could have told you without a fancy theory and it provides no added value since there is nothing I can change now."
| I don't have a very detailed list of applications, but off the top of my head, I would say the Stability of Solar System is an essential prediction. Yes, I understand that it is not extensively essential, because I can't imagine what we would have done if, on paper, the System turned out to be unstable.
As far as the applications of Chaos Theory go, it is more about controlling the Chaos that make more sense. How much noise should you add to the system, how strongly should you couple two systems to make them predictable? These are some generically important question that Chaos Theory tries to answer.
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Where is the flaw in deriving Gauss's law in its differential form? From the divergence theorem for any vector field E,
$\displaystyle\oint E\cdot da=\int (\nabla\cdot E) ~d\tau$
and from Gauss's law
$\displaystyle\oint E\cdot da=\frac{Q_{enclosed}}{\epsilon_0}=\int \frac{\rho}{\epsilon_0}~d\tau$
Hence,
$\displaystyle\int\frac{\rho}{\epsilon_0}d\tau=\int (\nabla\cdot E)~d\tau$
Textbooks conclude from the last equation that
$\displaystyle \nabla\cdot E=\frac{\rho}{\epsilon_0}$
My question is how can we conclude that the integrands are the same? Because I can think of the following counter example, assume
$\displaystyle \int_{-a}^a f(x)~dx=\displaystyle \int_{-a}^a [f(x)+g(x)]~dx$
where $g(x)$ is an odd function. Obviously the 2 integrals are equal but we cannot conclude that $f(x)$ is equal to $f(x)+g(x)$ so where is the flaw?
| The equation
$$\displaystyle\int_{V}\frac{\rho}{\epsilon_0}d\tau=\int_{V}(\nabla\cdot E)~d\tau$$
is true for all region $V$ in space the integration is performed over. That is why it follows that the integrands are equal.
Your counterexample is invalid, because the integrals are equal only when the domain of integration is of the form $[-a,a]$.
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Why do books have dog ears? I googled the question and found no explanation. It seems that dog ears are inevitable (for paperbacks, notably) even if you've always been careful. From my experience, they are about equally likely to appear on the top corners as on the bottom corners (for both the beginning pages and the ending ones). Dog ears for the middle pages of the book are less likely but they can also appear in frequently used old books. Can someone explain why?
I apologize if this is not the right kind of question to post here. I can find no other sites on SE for it.
| I think this is primarily about plastic (non-reversible) deformation. Plastic deformation appears when the stress of material is large enough. Stress is a generally speaking ratio between "force" and "dimension" of the object. At the corners, this ratio is larger, even for the same force, as "dimension" is smaller, that is paper tends to get narrower toward the corner.
Imagine that you want to fold paper through the center or at the very edge. A much larger force will be needed in the former case as the dimension of the fold is much larger.
Part of the problem might be also, that you are usually turning pages by pulling page by its corners, so you usually apply force close to future dog ears.
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Why does a glass rod when rubbed with silk cloth aquire positive charge and not negative charge? I have read many times in the topic of induction that a glass rod when rubbed against a silk cloth acquires a positive charge. Why does it acquire positive charge only, and not negative charge?
It is also said that glass rod attracts the small uncharged paper pieces when it is becomes positively charged. I understand that a positively charged glass rod attracts the uncharged pieces of paper because some of the electrons present in the paper accumulate at the end near the rod, but can't we extend the same argument on attraction of negatively charged silk rod and the pieces of paper due to accumulation of positive charge near the end?
| Well this can be explained by the work function of materials. Due to rubbing, heat is generated which supplies energy for removal of electrons. As the work function of the glass rod is smaller than the silk cloth, it easily loses electrons to the silk cloth which then releases energy (electron gain enthalpy) and thus ensures conservation of energy.
| {
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What is the fate of a 3-Torus universe? Since it is flat, will it expand forever like a flat and open universe or collapse like a closed and curved universe?
| Starting with $\mathbb{T}^3$ with the standard metric, it is just $\mathbb{R}^3/\mathbb{Z}^3$. In particular, taking the FLRW ansatz $\mathrm{d}s^2 = -\mathrm{d}t^2 + a(t)^2 \mathrm{d}\Sigma^2$ where $\mathrm{d}\Sigma^2$ is the flat Euclidean metric, you see that modding out the spatial slice by translations you get immediately a solution with spatial slice being the 3-torus. So the geometry of the universe (locally in space but globally in time) will be identical to that of the flat FLRW solution.
In other words, if you take the flat FLRW solution with Euclidean coordinate system $(x,y,z,t)$ such that the metric is
$$ -\mathrm{d}s^2 = - \mathrm{d}t^2 + a(t)^2 \left(\mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2\right) $$
and restrict the coordinates $x,y,z \in [0,2\pi)$, this will give you a coordinate representation of the universe with flat $\mathbb{T}^3$ slices.
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How to combine the error of two independent measurements of the same quantity? I have measured $k_1$ and $k_2$ in two measurements and then I calculated $\Delta k_1$ and $\Delta k_2$. Now I want to calculate $k$ and $\Delta k$.
$k$ is just the mean of $k_1$ and $k_2$. I thought that I would need to square-sum the errors together, like so:
$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$
But if I measure $k_n$ $n$ times, $\Delta k$ would become greater and greater, not smaller. So I need to divide the whole root by some power of $n$, but I am not sure whether $1/n$ or $1/\sqrt n$. Which is it?
|
$k$ is just the mean of $k_1$ and $k_2$
No, the best value of k is calculated using a weighted mean, weighting by the reciprocals of the squares of the respective individual uncertainty values. An accurate measurement must contribute more to the best value than an inaccurate measurement.
I thought that I would need to square-sum the errors together, like so:
$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$
No, instead $$ (\Delta k)^{-1} = \sqrt{(\Delta k_1)^{-2} + (\Delta k_2)^{-2}} $$
Intuitively, a very uncertain value must make little contribution. The uncertainty in k must always be less than or equal to the smallest of the individual uncertainties. Also, multiple, equally accurate measurements must decrease uncertainty.
see the example at the bottom of page 4 of this reference for the general case of n measurements:
http://www.physics.umd.edu/courses/Phys261/F06/ErrorPropagation.pdf
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Is the EmDrive, or "Relativity Drive" possible? In 2006, New Scientist magazine published an article titled Relativity drive: The end of wings and wheels1 [1] about the EmDrive [Wikipedia] which stirred up a fair degree of controversy and some claims that New Scientist was engaging in pseudo-science.
Since the original article the inventor claims that a "Technology Transfer contract with a major US aerospace company was successfully completed", and that papers have been published by Professor Yang Juan of The North Western Polytechnical University, Xi'an, China. 2
Furthermore, it was reported in Wired magazine that the Chinese were going to attempt to build the device.
Assuming that the inventor is operating in good faith and that the device actually works, is there another explanation of the claimed resulting propulsion?
Notes:
1. Direct links to the article may not work as it seems to have been archived.
2. The abstracts provided on the EmDrive website claim that they are Chinese language journals which makes them very difficult to chase down and verify.
| No. In special relativity, 4 momentum is exactly conserved. The first component of 4 momentum is total mass/energy, but the next 3 are given by:
p = m*γ(v)*v
m is the invariant mass, how much inertia it has when you are moving at the same velocity of it.
This is Newton except now momentum is a non-linear function of velocity. Nonlinearity does not change anything. Mass and momentum still are constant (ignoring leaks),
making γ(v)*v, and thus the center-of-mass velocity v, constant.
So why do we measure force? Possibly currents in the waveguide walls induce currents in the metal support structure which creates small magnetic forces between them.
| {
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Is it possible to mathematically derive the formula for resistance? Resistance is given by $\rho L/A$, where $\rho$ is the material constant, $L$ is the length, and $A$ is the area.
Is there any way that this can be derived mathematically, or is the only way experimentally?
Personally, I think experiment is the only way as I do not know how you would get $\rho$ otherwise.
| The answer is "yes", if you take for granted that $R$ is defined by the relation $\Delta V=IR$. In fact it is derived from (the real) Ohm's Law.
Ohm's law states that, for some materials (the so-called "Ohmic" materials) the current density vector $\vec{J}$ (current per unit area) is parallel to the electric field $\vec{E}$, i.e.,
$$\vec{J}=\sigma\vec{E}=\frac{1}{\rho}\vec{E}\ \ \ \ \ \ \ (1),$$
where $\sigma=1/\rho$ is the conductivity of the material (which is the inverse of $\rho$, the resistivity), which can be considered a constant for some materials (but is not restricted to be constant in general). From here, consider a material of length $L$ which has two extremes of area $A$ where we apply a potential difference $\Delta V$. Using the definition of the potential difference, it is easy to show that
$$|\vec{E}|=\frac{\Delta V}{L}\ \ \ \ \ \ \ (2).$$
On the other hand, we can express the current flowing trough the material, from the definition of current density as
$$|\vec{J}|=\frac{I}{A}\ \ \ \ \ \ \ (3).$$
Using, then, the results of equation $(2)$ and $(3)$ on equation $(1)$, we get
$$\frac{I}{A}=\frac{1}{\rho}\frac{\Delta V}{L},$$
or,
$$\Delta V=I\frac{\rho L}{A}.$$
On the typical relationship, $\Delta V=IR$, then $R=\rho L/A$.
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Why there is a $180^{\circ}$ phase shift for a transverse wave and no phase shift for a longitudinal waves upon reflection from a rigid wall? Why is it that when a transverse wave is reflected from a 'rigid' surface, it undergoes a phase change of $\pi$ radians, whereas when a longitudinal wave is reflected from a rigid surface, it does not show any change of phase? For example, if a wave pulse in the form of a crest is sent down a stretched string whose other end is attached to a wall, it gets reflected as a trough. But if a wave pulse is sent down an air column closed at one end, a compression returns as a compression and a rarefaction returns as a rarefaction.
Update: I have an explanation (provided by Pygmalion) for what happens at the molecular level during reflection of a sound wave from a rigid boundary. The particles at the boundary are unable to vibrate. Thus a reflected wave is generated which interferes with the oncoming wave to produce zero displacement at the rigid boundary. I think this is true for transverse waves as well. Thus in both cases, there is a phase change of $\pi$ in the displacement of the particle reflected at the boundary. But I still don’t understand why there is no change of phase in the pressure variation. Can anyone explain this properly?
| When a transverse wave travel in a medium the particle velocity is in upward direction &wave velocity is in forward direction.When a transverse wave meet the surface of the wall,it exert force in upward direction,because particle velocity of wave is in upward direction.so wall also exert force in downward direction(Newton's third law).so particle the particle velocity get reversed.so if we send crest it reflects as though.ln the case of longitudinal wave velocity& particle velocity both are in forward direction.so when it meet the wall it exert force in forward direction so wall exert force in backward direction. The phase of wave get change by π radian .but compression remain as computation.
| {
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Why would it be true that people with longer legs walk faster than ones with shorter legs? When a person walks, the only force acting on him is the force of friction between him and the ground (neglecting air resistance and all). The magnitude of acceleration due to this force is independent of the mass of the object (longer legs have more mass). Hence the person should move with with a velocity independent of the length of his legs.
But I have heard (also observed) that people with longer legs walk faster than ones with shorter legs. If that is true, then why?
One can argue that the torque about the pivot due to friction is more in case of longer legs, But then the torque due to gravity (when one raises his leg to move), which opposes the frictional torque, is also more for longer legs. And why would these torques make a difference anyway, as they have no effect on the acceleration of the center of mass?
| I think the simplest model that may be useful here is to treat the legs as simple pendula. In "steady state" comfortable walking, it is reasonable to assume that the legs oscillate close to their natural frequency. That is, the forward contacting leg lifts allowing the rear to swing forward freely over the stride. For a (simple) pendulum with:
$$\omega = \sqrt{\frac{g}{l}}$$ the velocity along the ground will be:
$$v \propto l\omega = \sqrt{lg}$$
Note that this result is independent of the mass of the walker and the ground contact forces.
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The equivalent electric field of a magnetic field I know that Lorentz force for a charge $q$, with velocity $\vec{v}$ in magnetic field $\vec{B}$ is given by
$$\vec{F} =q \vec{v} \times \vec{B}$$
but there will exist a frame of reference where observer move at same velocity with that of charge $q$, so according to him $v=0$. hence he will see no magnetic force is exerted on charge $q$. I have work on this problem for a while and found that the special relativity predicts equivalent electric force will acting upon charge instead. I want to know the relationship between this equivalent electric force and magnetic force.
Thanks in advance
| I haven't read them, but this, this, this and this thread (I thank a diligent Qmechanic) are related and clear up the but why-questions you might have.
The transformation of the quantities in electrodynamics with respect to boosts are
$$
\begin{alignat}{7}
\mathbf{E}'&~=~ \gamma \left(\mathbf{E} + \mathbf{v} \times \mathbf{B}\right) &&+ \left(1 - \gamma\right) \frac{\mathbf{E} \cdot \mathbf{v}}{v^2} \mathbf{v} \\[5px]
\mathbf{B}'&~=~\gamma\left(\mathbf{B}-\frac{1}{c^2}\mathbf{v} \times \mathbf{E}\right)&&+\left(1-\gamma\right)\frac{\mathbf{B} \cdot \mathbf{v}}{v^2} \mathbf{v} \\[5px]
\mathbf{D}'&~=~ \gamma \left(\mathbf{D}+\frac{1}{c^2} \mathbf{v} \times \mathbf{H} \right) && + \left( 1 - \gamma \right) \frac{\mathbf{D} \cdot \mathbf{v}}{v^2} \\[5px]
\mathbf{H}'&~=~ \gamma \left(\mathbf{H} - \mathbf{v} \times \mathbf{D}\right) && +\left(1 - \gamma\right) \frac{\mathbf{H} \cdot \mathbf{v}}{v^2}\mathbf{v} \\[5px]
\mathbf{j}' & ~=~ \mathbf{j} - \gamma \rho \mathbf{v} && + \left(\gamma - 1 \right) \frac{\mathbf{j} \cdot \mathbf{v}}{v^2} \mathbf{v} \\[5px]
\mathbf{\rho}' & ~=~ \gamma \left(\rho - \frac{1}{c^2} \mathbf{j} \cdot \mathbf{v}\right)
\end{alignat}
$$where $\gamma \left(v \right)$ and the derivation of the transformation is presented on this Wikipedia page and is most transparent in a space-time geometrical picture, see for example here. Namely, the electromagnetic field strength tensor $F_{\mu\nu}$ incorporates both electric and magnetic field $E,B$ and the transformation is the canonical one of a tensor and therefore not as all over the place as the six lines posted above.
In the non-relativistic limit $v<c$, i.e. when physical boosts are not associated with Lorentz transformations, you have
For the traditional force law, the first formula confirms the prediction that the new $E$ magnitude is $vB$.
Also, beware and always write down the full Lorentz law when doing transformations.
Lastly, I'm not sure if special relativity predicts equivalent electric force will acting upon charge instead is the right formulation you should use, because while the relation is convincingly natural in a special relativistic formulation, the statement itself is more a consistency requirement for the theory of electrodynamics. I'd almost say the argument goes in the other direction: The terrible transformation law of $E$ and $B$ with respect to Galilean transformations was known before 1905 and upgrading the status of the Maxwell equations to be form invariant when translating between inertial frames suggests that the Lorentz transformation (and then special relativity as a whole) is physically sensible.
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Is the Avogadro's constant equal to one? Question: Is the Avogadro's constant equal to one?
I was tasked with creating a presentation on Avogadro's work, and this is the first time I actually got introduced to the mole and to Avogadro's constant. And, to be honest, it doesn't make any mathematical sense to me.
1 mole = 6.022 * 10^23
Avogadro's constant = 6.022 * 10^23 * mole^(-1)
What?
This hole field seems very redundant. There are four names for the same thing! Since when is a number considered to be a measurement unit anyway?!
| Unfortunately, I can't post comments, so I have to write it this way. The statement 1 mole = 6.022 * 10^23 (which you use to show that N_A=1) simply doesn't hold - at least it didn't hold when they taught chemistry in my class.
It's $1 \text{ mol} = 6.022\cdot10^{23}/N_A$, isn't it?
P. S. And what exactly did I do to get my -1?
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Why is there a 90˚ phase angle between particle velocity and sound pressure in spherical waves? My text says that in a plane sound wave (or in the far field), particle velocity and pressure is in phase. As we move closer to the sound source (to near field and more spherical waves), the phase angle between these two quantities will gradually shift towards 90˚.
Why does this phase shift occur? I haven't been able to find an intuitive explanation for why there is a phase shift in spherical waves, but not in plane waves, and vice versa.
| It's a little easier to understand if you look at the electrical case with cylindrical geometry: a long wire carrying AC current. If you are far from the wire, the E and B fields are in phase, because they represent radiated power flowing away from the wire. But close to the wire, you are in the range where inductive current flows, and the energy mostly just washes back and forth in and out of the wire.
The total energy radiated by the wire is only a small fraction of the very large energies that are exchanged inductively in the region near the wire. That's why there has to be a mismatch in the phases...because in the near field, the energy is not radiated but just temporarily stored.
It's the same basic idea with the acoustic case of the spherical source. Near the source, the amplitudes have to fall off as 1/r-squared, from simple geometric arguments. If this represented in-phase acoustic power, your total radiated power would be falling of as 1/r^4, which wouldn't add up.
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Operators Uncertainty $\hat A$ is an operator.
The uncertainty on $\hat{A}$, $\Delta A$ is defined by:
$$\Delta A=\sqrt{\langle\hat A^2\rangle - \langle\hat A\rangle^2}$$
what is difference between
$\langle\hat A^2\rangle$ and $\langle\hat A\rangle^2$
that leads to Uncertainty Relation between two Operators?
more details:
$$
\langle\hat A^2 \rangle=\langle\psi|\hat A^2|\psi \rangle$$
What is the name of difference between absolute value of these two complex conjugates
| *
*$\langle\hat A\rangle$ is the expectation value of $\hat A$.
*$\langle\hat A\rangle^2$ is the square of item 1.
*$\langle\hat A^2\rangle$ is the expectation value of $\hat A^2=\hat A \hat A$.
Item 2 and 3 do not have to be equal.
If $\hat A$ is selfadjoint, then it is possible to show
*
*that the expectation value $\langle\hat A\rangle~\in~\mathbb{R}$ is real, and
*that $\langle\hat A^2\rangle ~\geq~ \langle\hat A\rangle^2$.
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Entanglement spectrum What does it mean by the entanglement spectrum of a quantum system? A brief introduction and a few key references would be appreciated.
| If a system S is composed of two subsystems A and B, then a state of S is a vector $$|\Psi\rangle \in H_A\otimes H_B$$ Tracing over the "B degrees of freedom" allows you to define the reduced density matrix $\rho_A$ The entanglement entropy is defined as$$-Tr(\rho_Aln\rho_A)$$
I believe that the entanglement spectrum just refers to the spectrum of eigenvalues of $\rho_A$. Sorry I don't know any references though.
Edit to add: The entanglement entropy is a fairly crude measure of the entanglement present (just a single number). Knowledge of the entanglement spectrum provides further information on the entanglement properties - it includes much more information about the entire reduced density matrix $\rho_A$. This has been used, for example, in investigations of the scaling behaviour of extended quantum systems.
| {
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In the known universe, would an atom not present in our periodic table exist? I have watched this movie Battleship. In it the researchers say this piece of metal is alien because we cant find this metal on earth.
So that would mean somewhere else in the universe any of the following should be true?
*
*Atoms' composition is not similar to that as on earth (nucleus, electrons, anything else)
*Elements with atomic numbers above 120 or 130 are stable (highly impossible without point 1)
*The realm itself is observed by different binding forces (but then, once that elements realm has changed, it should become unstable and collapse)
| In fact, some nuclear theorists do believe that there will be relatively stable heavy elements, as per your point 2. The so-called Island of Stability is predicted to occur because stability is maximized at certain so-called magic numbers which correspond to especially stable isotopes when the number of protons and/or neutrons matches one of the numbers. In particular, Z=114, 120, and 126 may have long-lived isotopes. These haven't yet been produced because it's difficult to get to the requisite number of neutrons to achieve a stable nucleus.
I should emphasize that this is just a hypothesis with essentially no experimental evidence at the moment. It is, as far as I know, a fairly active area of research. It almost sounds like crackpot science, but it definitely isn't and has a number of notable physicists and chemists connected to the hypothesis.
It is nonobvious whether these would be metals, though. If all you want is exotic metals, you'll have a much easier time just making compounds that haven't been synthesized on earth.
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Proton-Neutron Lattice as a form of matter? Would it be possible for a lattice of protons and neutrons (I'm picturing a plane of hexagons in my head) to exist bound by the strong nuclear force (not gravity)? I know that the strong force losses its power when an atomic nucleus gets to be too large, but in a lattice, it would only have to bond one proton to a few neutrons or one neutron to a few protons at a time. Would this work? If so, what would be the properties of such a material?
| A lone proton in between a crystal lattice could find itself with more negative charge then equilibrium from the neighbouring clouds of electrons. Thus there is a compressive force driven by the lattice and heat, in addition to proton repulsion of the nuclear force.
I expect over time their will be a determination that the coulomb barrier may not be static under all conditions and because it is not yet understood it's difficult to replicate working Low Energy Nuclear reactions. However the anomalous results show something is happening. That can only mean that the understanding is incomplete based on observations to date.
| {
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Ideal gas in a vessel: kinetic energy of particles hitting the vessel's wall Reading Landau's Statistical Physics Part (3rd Edition), I am trying to calculate the answer to Chapter 39, Problem 3.
You are supposed to calculate the total kinetic energy of the particles in an ideal gas hitting the wall of a vessel containing said gas.
The number of collisions per unit area (of the vessel) per unit time is easily calculated from the Maxwellian distribution of the number of particles with a given velocity $\vec{v}$ (we define a coordinate system with the z-axis perpendicular to a surface element of the vessel's wall; more on that in the above mentioned book):
$$
\mathrm{d}\nu_v = \mathrm{d}N_v \cdot v_z = \frac{N}{V}\left(\frac{m}{2\pi T}\right)^{3/2} \exp\left[-m(v_x^2 + v_y^2 + v_z^2)/2T \right] \cdot v_z \mathrm{d}v_x \mathrm{d}v_y \mathrm{d}v_z
$$
Integration of the velocity components in $x$ and $y$ direction from $-\infty$ to $\infty$, and of the $z$ component from $0$ to $\infty$ (because for $v_z<0$ a particle would move away from the vessel wall) gives for the total number of collisions with the wall per unit area per unit time:
$$
\nu = \frac{N}{V} \sqrt{\frac{T}{2\pi m}}
$$
Now it gets interesting:
I want to calculate the total kinetic energy of all particles hitting the wall, per unit area per unit time. I thought, this would just be:
$$
E_{\text{tot}} = \overline{E} \cdot \nu = \frac{1}{2} m \overline{v^2} \cdot \nu
$$
The solution in Landau is given as:
$$
E = \nu \cdot 2T
$$
That would mean that for the mean-square velocity of my particles I would need a result like:
$$
\overline{v^2} = 4\frac{T}{m}
$$
Now, I consider that for the distribution of $v_x$ and $v_z$ nothing has changed and I can still use a Maxwellian distribution. That would just give me a contribution of $\frac{T}{m}$ each. That leaves me with $2\frac{T}{m}$, which I have to obtain for the $v_z$, but this is where my trouble starts:
How do I calculate the correct velocity distribution of $v_z^2$?
| The following calculation gives the correct answer:
$$Z\int_0^{\pi/2}\int_0^\infty 2\pi v \sin\theta\; v\; \mathrm{d}\theta\mathrm{d}v\; e^{-mv^2/2kT}\; v \cos\theta\; \frac{1}{2}mv^2,$$
where $Z$ is such that
$$Z\int_0^{\pi}\int_0^\infty 2\pi v \sin\theta\; v\; \mathrm{d}\theta\mathrm{d}v\; e^{-mv^2/2kT} = n,$$
where $n$ is the particle number density.
The correct answer is
$$\left(\frac{2kT}{\pi m}\right)^{1/2}\; nkT = \left(\frac{2kT}{\pi m}\right)^{1/2}\; p.$$
| {
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Possibility for radiation in dark matter that is not interactive with regular matter? Definition: Radiation in this case does not refer to electromagnetic radiation. It refers to any kind of emission of energy, even energy that does not interact with regular matter.
Just like dark matter does not interact with electromagnetic radiation, could regular matter not interact with "dark matter radiation" (I'm not talking about the usual "really high wavelength radiation" kind)?
| Yes, just as you have bosons mediating interactions in the Standard Model sector, you could have dark radiation mediating interactions in the DM sector. Given that more energy is in the form of DM than SM particles, this won't be surprising. Of course, one needs proof for such a species of particles to exist :) A particle physics approach would be to understand the exact nature of the DM particle and its interactions. A cosmological approach would be to see if there are missing relativistic components, as in this paper.
By the way, by radiation, I hope you mean relativistic particles. It is in that sense that radiation is used in the cosmological context.
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Is there a good explanation for the observation of Martian canals? Martian "canals" have been observed by independent observers after their first description. Now, they are attributed to "optical illusion", but I think that this is not a good choice of word, because an optical illusion should be visible today as well. It would have to be a psychological effect, but it is rather astonishing to have people draw whole maps of non-existent canal systems.
Is there a good explanation (optical or psychological) for the observation of Martian canals?
| The human brain likes to find patterns in what it observes. That's why we see patterns in the stars which we call "constellations." The canals of Mars are similar. For most of them, there is something there, just at the edge of vision, and the human tendency is to "connect the dots" and see lines were only vague streaks exist.
It's worth noting that some experienced observers, notably E. M. Antoniadi, were never able to see anything resembling canals. There's some thought that Percival Lowell, the main proponent of canals, may in fact have had some defect in his vision, because he saw canals on Venus as well, which most observers see as a perfectly plain white ball.
I've observed Mars for decades visually with excellent telescopes, and have never ever seen anything even vaguely resembling a canal.
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What do the colors in false color images represent? Every kid who first looks into a telescope is shocked to see that everything's black and white. The pretty colors, like those in this picture of the Sleeping Beauty Galaxy (M64), are missing:
The person running the telescope will explain to them that the color they see in pictures like those isn't real. They're called "false color images", and the colors usually represent light outside the visual portion of the electromagnetic spectrum.
Often you see images where a red color is used for infrared light and purple for ultraviolet. Is this also correct for false color astronomy images? What colors are used for other parts of the spectrum? Is there a standard, or does it vary by the telescope the image was taken from or some other factor?
| The images that are currently taken with the High End, (read: $$$), Astro-cameras, indeed produce complete full color images. A perfect example is http://www.kevindixon.westhost.com/Deep_Sky_CCD-Siciliano.htm. This is one of many by this particular Astrophotographer. None of the color is false at all. Astrophysicists will use spectral analysis to assign colors to specific elements, thereby creating a "False Color" image. This gives them the ability to 'view' the makeup of an object that they desire to study in detail as to the distribution of the elemental makeup of a specific object of interest. You will find at the above link, all the details regarding the image. How long it took, the camera used, and even the Amateur Astronomer's Telescope that it was taken through.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/24895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 4
} |
What objects look best in an O-III filter? I've heard that an O-III (Oxygen III) filter is great for planetary nebulae.
Is this true for all planetary nebulae, or just some or most?
What other target types are often improved with an O-III filter?
| The dominant OIII emission line is the $^3P-^1D_2$ line, which is at 2.51 eV. If you want to see a pretty picture, you'll want a wide field of emission of this line, which would mean that you need a gas cloud which is exciting the OIII line thermally. The thermal energy required to excite the 2.51 eV line is around 8,000 K, so you need gas clouds with that sort of temperature.
Planetary nebulae and supernovae remnants will of course have temperatures around there. H II regions are excellent sources of O III emission, too.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/24987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Recommend good book(s) about the "scientific method" as it relates to astronomy/astrophysics? I am interested in astronomy/astrophysics, but I am not science major (I am a computer science graduate). Facts and results of the field are presented to the public without showing how these facts/results got known or inferred. And I have that curiosity to know how we know what we know about the universe (either observationally or mathematically).
So my question is, what book(s) do you recommend for someone who has knowledge of
*
*algebra, trigonometry and geometry
*college-level calculus
*classical mechanics
and does not get intimidated by mathematical language?
I expect the book(s) to answer questions like (not necessarily all of the questions, but questions of the same level and kind as these):
*
*How do we know how distant from the earth a celestial body (for example, a star) is?
*How do we know the volume/mass of celestial bodies?
*How do we know the materials that a planet is made of?
*How do we know that our solar system orbits around the center of the galaxy?
*How do we calculate the total mass of the galaxy?
| You can refer :-
Basics of astronomy
It will help you to understand the basics of astronomy.
The Solar system and stars
It will help you to understand our solar system and stars.
From stars to our Galaxy
It will give you the information about stars formation and About Galaxies.
Expanding Universe
It will give you the information about the expanding universe.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/25030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
Are solar physics images of use to the night-time community? I'm actually interested in cases of cross-discipline data re-use.
I know that the SOHO/LASCO coronographs are used for comet finding, that solar telescopes were used to get information about Venus's atmosphere during its solar transit in 2004, and STEREO were rolled to try to look at L4 and L5.
But do the night-time or planetary communities use the images for anything else? For example, does STEREO's separation from Earth make the coronographs or heliospheric imagers useful for any triangulation, or are the exposure times or spatial resolution problematic, or is the relative separation insignificant in the grand scheme of things?
Update: I'm not interested in space weather or for warnings about when to put satellites into 'safe mode'... I'm more interested if the low-level telescope data is useful.
| I've been pondering your question for a couple of days, and the only answer I can think of from my own experience is that Coronal Mass Ejections trigger aurora on Earth, and aurora can seriously interfere with night-time observations of faint objects. I'm thinking of an occasion when I travelled to an extremely dark site in Algonquin Park and set up for a night of deep sky observing, only to be blasted by a brilliant aurora display!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/25118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What is the current status of Pluto? Pluto has been designated a planet in our solar system for years (ever since it was discovered in the last century), but in 2006 it was demoted.
What caused this decision? And is there a chance that it could be reversed?
Edit: well, http://www.dailygalaxy.com/my_weblog/2017/03/nasas-new-horizon-astronomers-declare-pluto-is-a-planet-so-is-jupiters-ocean-moon-europa.html is interesting; this is science, so anything could (potentially) change.
| Pluto has been reclassified as a dwarf planet.
It was reclassified as such because a growing number of objects were found similar to Pluto, which exhibited at least one notable difference from the other planets. The choice would have been to accept these other objects as planets or to develop a new class of object.
The primary features of a planet are that it has an orbital path 'clear' of other debris, orbits around the sun, and is massive enough to maintain hydrostatic equilibrium. Pluto (as well as other dwarf planets) fail to meet the first criterion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/25162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
} |
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