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Sum of all forces Let us glue up these two images, where we get closed loop thrust of water. Force $F_3$ has direction $-x$ and force $F_2$ has $x$ direction. What is the sum of all forces? Can it be more than zero? Speed of water is constant.Angles are the same.Half circle is not exactly circled at the ends due to the angles.
One more subquestion. What if speed of water is very high and we have quite big amount of centrifugal force?
| If I understand the question, you are basically talking about a stream of water flowing in a closed loop. Assuming the density is constant, and the speed of the water at each point is constant, then the total momentum of the water (which is the sum of the momentum of each little piece of water in the stream) is constant, and there must not be any net force on the water (by newton's second law).
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Function for heating of a planet by its star I'm looking for an algorithmic model which will gives rough estimate of the average temperature of a planet's atmosphere (good enough to say whether there will be liquid water anywhere, or if metals and plastics will melt/ignite) based on the luminosity of its star, the radius of its orbit, the constituents of its atmosphere, the mass of the planet, and the surface area of the planet.
Specifically, I figure I need two mathematical models. The first would simulate the incoming thermal energy from the star, and the second would calculate how much is lost as blackbody radiation by the planet, taking into account the effects of the 'greenhouse effect'.
Does anyone know how to model these things?
| I recently wrote a tutorial for my astronomy club which seems to address your question. It includes numerical examples that specifically covers planetary temperatures and greenhouse effects.
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Maximum angular velocity to stop in one rotation with a known torque I have an object I can rotate with a given torque. I would like to stop applying torque once I've reached a defined maximum rotational speed. The maximum rotational speed should be defined so that applying maximum torque will stop the rotation of the object within one rotation. If I know my torque and moment of inertia, how can I find the maximum rotational velocity to allow me to stop the object in one rotation?
Time is whatever is needed.
I've tried finding the angular acceleration required to stop the object, but that leaves me with the time variable. Of the equations I've tried, I'm left with a time variable as well as the maximum angular velocity.
| To stop the object you must do work. For a constant torque perpendicular to the moment arm, the work it does is equal to $\tau\cdot\Delta\theta$, and you want $\Delta\theta\leq2\pi$.
It should be obvious that the greatest angular velocity that a torque $\tau$ can stop will take it the full $2\pi$ radians to stop. In a rotating system, the rotational kinetic energy is given by $E_r=\frac12I\omega^2$ (a direct analogue of $E_K=\frac12mv^2$ ). Now consider work-energy equivalence.
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Where can I find the full derivation of Helfrich's shape equation for closed membranes? I have approximately 10 papers that claim that, from the equation for shape energy:
$$ F = \frac{1}{2}k_c \int (c_1+c_2-c_0)^2 dA + \Delta p \int dV + \lambda \int dA$$
one can use "methods of variational calculus" to derive the following:
$$\Delta p - 2\lambda H + k(2H+c_0)(2H^2-2K-c_0H)+2k\nabla^2H=0$$
But I'm having a lot of trouble tracking down the original derivation. The guy who did it first was Helfrich, and here's his and Ou-yang's paper deriving it:
http://prl.aps.org/abstract/PRL/v59/i21/p2486_1 . However, they don't show an actual derivation, instead saying "the derivation will appear in a full paper by the authors" or something like that. Yet everybody cites the paper I just linked for a derivation. Does anybody know a source that can derive this, or can give me some hints to figure it out myself? To be honest I can't even figure out how to find the first variation.
Edit: So, after some careful thought and hours and hours of work and learning, I realized that the answer that got the bounty was wrong. The author stopped replying to my messages after I gave him bounty.... thanks guys. That said, I've almost got it all figured out (in intense detail) and will post a pdf of my own notes once I'm done!
| The paper you're looking for is probably:
"Bending energy of vesicle membranes: General expressions for the first, second, and third variation of the shape energy and applications to spheres and cylinders"
Ou-Yang Zhong-can and Wolfgang Helfrich
Phys. Rev. A 39, 5280–5288 (1989)
http://pra.aps.org/abstract/PRA/v39/i10/p5280_1
Hopefully you have university access - I couldn't find a free copy anywhere. This is actually a more complicated question than most physics papers like to make it appear! Good luck with your project. Membranes are a very cool subject.
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When are the OPE relevant? I've seen OPEs commonly used in 2d CFT, it's quite apparent to me that, in that case, it dresses a bridge between the algebraic and the operator formalism especially when combined with radial ordering and the use of contour integral. Even more powerful in the minimal models where it leads to the bootstrap equations and the resolution of the 3 pt functions. I also heard that OPE are sometimes used in other circumstances, for instance in the QCD chapter of Peskin & Schroeder's book but I don't recall the motive. I'd be curious to know what generally it is relevant to decompose the products of operators into an "operator basis" ie. associate an algebra to the space of operators.
| My memory of Peskin & Schroder is a little hazy, but they're probably discussing the Shifman-Vainshtein-Zakharov sum rules. The idea is that you can use the OPEs for composite operators representing mesons/hadrons to derive formulae that express meson/hadron n-point functions in terms of the VEVs of various QCD condensates. (Edit: Just discovered that Shifman has some very nice lecture notes on the subject.)
More generally: OPEs are always relevant (if not always easy to use in a given situation) because they carry almost all the information about a field theory. You can actually define a QFT by writing down the set of local observables, the OPEs between them, and the VEVs of the local observables. It's as good a formalism as the Hamiltonian or path integral formalisms -- better in some ways, because it applies when the path integral doesn't.
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Graphene as optical and UV mirrors One usually hears about graphene as a good thermal conductor, and good light absorber due to its tunable bandgap properties. But i haven't heard about its aplicability as an optical mirror. In fact, mostly the opposite is true: the optical transmissivity of monolayer graphene is very high ($\approx$ 98%)
Since there are material engineering tricks to tune the separation between two or more graphene layers, i would expect that multi-layer destructive interference can be achieved with optical wavelengths, enhancing reflectivity
I know that probably graphene is not the best material to do this, but allow me to insist. Why? Well, graphene mantains it's properties well over 3000K, and a couple of layers of it can be quite strong and very light. All these properties make it the ideal material for a laser-pushed sail.
By the way, i'm asking about optical and UV frequencies, i know that monolayer graphene is reflective on the microwave region, and i'm not asking about that. Thanks
| There has been some research in the use of graphene in optics, such as from Berkeley's Ultrafast Nano Optics Group and this article about the use of graphene in digitl communications.
In terms of graphene's UV and visual reflective properties, according to the University of Manchester and "High optical absorbance of graphene", it is graphene's absorbance properties that quite unique - with a one atom sheet having 2.3% absorbance in the visual range. For both the visual and UV range, this OptisInfobase article go into far more detail (with data).
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What could magnetic monopoles do that electrically charged particles can't? I understand the significance to physics, but what can a magnetic monopole be used for assuming we could free them from spin ice and put them to work? What would be a magnetic version of electricity?
EDIT
Sorry this wasn't clear. The question is mixed between the quasiparticle and the theoretical elementary particle based on some similarities between the two. I am more interested in the quasiparticle and if they have properties in some way that are similar to particle version:
There are a number of examples in condensed-matter physics where
collective behavior leads to emergent phenomena that resemble magnetic
monopoles in certain respects, including most
prominently the spin ice materials. While these should not be
confused with hypothetical elementary monopoles existing in the
vacuum, they nonetheless have similar properties and can be probed
using similar techniques.
http://www.symmetrymagazine.org/breaking/2009/01/29/making-magnetic-monopoles-and-other-exotica-in-the-lab/
"The Anomalous Hall Effect and Magnetic Monopoles in Momentum
Space". Science 302 (5642) 92–95.
"Inducing a Magnetic Monopole with Topological Surface States"
"Artificial Magnetic Monopoles Discovered"
and comments in articles about quasi-particles like this:
Many groups worldwide are currently researching the question of
whether magnetic whirls could be used in the production of computer
components.
led me to wonder what application might they have? Mixing these two concepts is probably a bad way to present this question. A true magnetic monopole would effect protons whereas the artificial ones don't.
What I don't understand is what advantages an artificial magnetic monopole would have. And does this relate to some theoretical aspect of a true monopole?
| Partly related to your question, Artificial Magnetic Monopoles Discovered, from an article in ScienceDaily just late last month. The monopoles apparently act the same as the ends of a dipole magnet, as has been suggested by Dilaton in the comment above. A more comprehensive article is from Particle Data Group.
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Can deterministic world view be denied by anything other than quantum mechanics If we ignored quantum mechanics and looked at the world with a deterministic Newtonian view. Does not that mean that there is no randomness and that if all the information of the state of the universe during the big bang is accessible one can predict the state of the universe at any period of time and predict that I am writing this question right now.
Of course something like that denies the free will but I am asking if there is any thing other than quantum mechanics that denies the deterministic world view.
| Norton's Dome is an example of nondeterminism in Newtonian Mechanics. There is a potential $V(r)=mgCr^{3/2}$ (say from a dome shaped like $h=Cr^{3/2}$) that gives multiple solutions with no way to pick when, if, or in which direction the particle moves.
However, this is not a serious blow against determinism because firstly, that is not a fundamental force, and no real force is going to look like that exactly, and secondly the nondeterminism only happens for a perfect exact amount of energy, any less and it can't be/get to the top, and any more and it has to be moving when at the top so it slides down the side in the direction it is going and it does so right away. And even if you had a perfect potential like that, and a perfect energy, the nondeterministic equilibrium is unstable, you'd have to perfectly isolate the system from arbitrarily small perturbations.
But, since that is all from Newtonian Mechanics and just sliding a body along a hill at that, it does tell you that popular theories aren't as deterministic as people like to pretend.
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Is a proton collision (collisions like in the LHC) visible to the human eye? I was curious if a proton collision is visible to the human eye.
(This might sound like a really basic question and forgive me if it is. I am very inexperienced in Physics and just wanted an answer to my curiosity)
| Hmm, interesting question at least. Visible as in see two balls crash into each other; no, visible as in see the light emitted from the collision; probably not.
The emitted photons are of the order of $\rm{}MeV$, around $10^{-13}\rm{}$ joules. The human eye works at around $10^{-5}$. So individual collisions no.
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How to cut a stone on a White Dwarf? I've heard that white dwarf stars are extremely dense and hard. So, if I had a piece of white dwarf matter, would it be possible to cut it (or otherwise) into a custom shape? How could one do that?
| We know that matter in White Dwarfs and Neutron stars become hugely dense, ten or fifteen orders of magnitude more dense than ordinary matter. What we don't know is, if such matter stays stable away from the deep gravity well of the star corpses where they form.
We certainly haven't found a single grain of such matter in nature after a couple of centuries of geological observations. But, such a dense material will not stay afloat in any kind of matter crust of any planet, quickly sinking toward the center of planets that capture such grains gravitationally.
If there is some kind of neutronium or strangelet matter with a stable low-pressure phase, the only three ways we are going to ever find out are:
*
*by recreating those densities and pressures in laboratory (our highest laboratory pressure diamond anvils top out at 40 GPa or something, 7 orders of magnitude less of what you are after, so good luck with that)
*We become extremely good at observing gravitational perturbations and optically check the sources.
*My favorite: We adventure a mining expedition to some old moon on the solar system that we are certain that is cool enough for us to go and drill toward the center of the core, and then we get to see what kind of heavy stuff is frozen in there.
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Why does the Sun feel hotter through a window? I have this big window in my room that the Sun shines through every morning. When I wake up I usually notice that the Sunlight coming through my window feels hot. Much hotter than it normally does when you're standing in it outside. I know if the window were a magnifying glass that it would feel hotter because it is focusing the Sun's rays, but I'm pretty sure that my window doesn't focus the rays, otherwise things outside would appear distorted.
So my question is, why does Sunlight always feel hotter when it shines on you through a window than when it shines on you outside? I thought it might simply be a matter of convection, but anecdotal evidence would seem to say it still feels hotter even if you had a fan blowing on you. Am I just crazy?
| Most likely is its always hotter than the outside. the frames absorb heat up into the 160 plus mark depends on where you live. 100 degrees usually about 150 degrees gained we call it solar gain.
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Why do metal objects in microwaves spark? I heard that electrons accumulate at points on metals, and this clearly explains the arcing phenomenon, but how does a microwave make an electron imbalance on the fork?
| The basic idea is that microwaves are a form of electromagnetic radiation-- light. That light consists of an oscillating magnetic and electric field component. By Faraday's Law of Induction, a magnetic field in flux induces a current in a conductor... so it is the non-conservative, induced electric field in the conductor that produces an emf that moves the electrons in the fork.
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How general relativity gets to an inverse-square law I understand that a general interpretation of the $1/r^2$ interactions is that virtual particles are exchanged, and to conserve their flux through spheres of different radii, one must assume the inverse-square law. This fundamentally relies on the 3D nature of space.
General relativity does not suppose that zero-mass particles exchanged. What is the interpretation, in GR, of the $1/r^2$ law for gravity? Is it come sort of flux that is conserved as well? Is it a postulate?
Note that I am not really interested in a complete derivation (I don't know GR enough). A physical interpretation would be better.
Related question: Is Newton's Law of Gravity consistent with General Relativity?
|
I understand that a general interpretation of the $1/r^2$ interactions is that virtual particles are exchanged [...] General relativity does not suppose that zero-mass particles exchanged.
You don't need quantum field theory for this. In a purely classical field theory, we have field lines, and the field lines from a spherically symmetric source should radiate outward along straight lines. In a frame where the source is at rest, we expect by symmetry that the field lines are uniformly distributed in all directions. The strength of the field is proportional to the density of the lines, which falls off like $1/r^2$ in a three-dimensional space.
This whole description is complicated by the polarization of the field. Gravitational fields have complicated polarization modes. Nevertheless, the $1/r^2$ result is unaffected.
Finally, we have an issue unique to GR, which is that the field is the metric, and this means that the field itself affects the measuring tools that we use to measure things like $r$, the field, and the area of a surface through which we're counting the number of field lines that penetrate. These are all strong-field issues, so for large $r$, they don't affect the $1/r^2$ argument.
Is it a postulate?
No. In the standard formulation of GR, the main postulate is the Einstein field equations. From it, we can prove Birkhoff's theorem, which says that the Schwarzschild metric is the external field of a static, spherically symmetric source. The weak-field limit of the Schwarzschild metric corresponds to a $1/r^2$ field.
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Nuclear Binding energy The nuclear binding energy, is the energy that is needed to seperate the nucleons in a nucleus. And binding energy is also defined as the energy given out when a nucleus forms from nucleons.
Also the larger the nucleus is, the more energy is required to break it apart, so why doesn't that mean that larger nuclei are more stable? I mean Uranium has a lower binding energy per nucleon than Iron, but there are many many more nucleons in Uranium that Iron so the total binding energy is going to be much greater.
Basically I don't understand why whether an element gives out energy by fusion or fission (why the lighter element provide energy by fusion not fission and vice versa for heavy elements) depends on binding energy per nucleon and not "total" binding energy
| Nuclear physics is in the realm of quantum mechanics. To first order one can think of "one nucleon" in the collective left over strong interaction field of all the rest, as a potential. As for strong interactions nuclei are indistinguishable to first order the binding energy per nucleon makes sense. The larger this energy the harder to extract a nucleon from the potential.
The total binding energy will depend on second order effects coming in from the Pauli exclusion principle and the electromagnetic repulsion of protons, from clusterings within large nuclei etc. For example there are a large number of unstable isotopes when the electromagnetic repulsion of the protons is not balanced by the neutrality of neutrons or when there are too many neutrons. A simple curve cannot be defined the way it can with the binding energy per nucleon.
The binding energy per nucleon is a necessary condition to see whether fusion can happen, but it is not sufficient to make sure the reaction will happen as all quantum numbers have to be satisfied . In this article there is a discussion of conditions and possibilities for fusion.
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Why do stars flicker? Why do stars flicker and planets don't? At least this is what I've read online and seen on the night sky. I've heard that it has to do something with the fact that stars emit light and planets reflect it. But I don't get it, isn't this light, just "light"? What happens to the reflected light that it doesn't flicker anymore?
I was thinking that it has to do something with Earth's atmosphere, different temperatures or something (if this has any role at all).
| Here is a nice answer, taken from http://www.enchantedlearning.com/subjects/astronomy/stars/twinkle.shtml
The scientific name for the twinkling of stars is stellar scintillation (or astronomical scintillation). Stars twinkle when we see them from the Earth's surface because we are viewing them through thick layers of turbulent (moving) air in the Earth's atmosphere.
Stars (except for the Sun) appear as tiny dots in the sky; as their light travels through the many layers of the Earth's atmosphere, the light of the star is bent (refracted) many times and in random directions (light is bent when it hits a change in density - like a pocket of cold air or hot air). This random refraction results in the star winking out (it looks as though the star moves a bit, and our eye interprets this as twinkling).
Stars closer to the horizon appear to twinkle more than stars that are overhead - this is because the light of stars near the horizon has to travel through more air than the light of stars overhead and so is subject to more refraction. Also, planets do not usually twinkle, because they are so close to us; they appear big enough that the twinkling is not noticeable (except when the air is extremely turbulent).
Stars would not appear to twinkle if we viewed them from outer space (or from a planet/moon that didn't have an atmosphere).
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Have general relativistic effects of all of the components of the stress-energy tensor been measured? The stress-energy tensor is:
Have general relativisic effects of all of the components of the stress-energy tensor been measured? I've heard that the accelerating expansion of the universe is due to negative pressure, but I don't know which terms are involved in the precession of mercury's orbit or frame dragging.
| Nice question. The gravitational effects of the pressure components have been directly verified in at least two ways that I know of. The early universe was radiation-dominated, so I don't think you can reproduce any of the relevant cosmological data (e.g., big bang nucleosynthesis) if you leave out the pressure terms. Kreuzer 1968 is a laboratory test, and Bartlett 1986 tested it using lunar laser ranging. I've written an exposition here of how Kreuzer can be interpreted as a test of pressure as a source, and this review article by Will summarizes the Bartlett experiment in section 3.7.3.
Others might be able to comment on direct experimental tests of the shear stress and momentum density as gravitational sources, but if they didn't contribute, you'd be breaking Lorentz invariance, since a diagonal stress-energy tensor gains off-diagonal elements under a boost.
The precession of Mercury's orbit can be explained in terms of the motion of a test particle in a Schwarzschild metric, so it doesn't test anything about the non-00 terms in the stress-energy tensor.
It seems likely to me that Gravity Probe B's detection of frame dragging can be interpreted as a direct test of the momentum-density part (since I think it can be interpreted as a gravitomagnetic effect).
Bartlett and van Buren, Phys. Rev. Lett. 57 (1986) 21.
Kreuzer, Phys. Rev. 169 (1968) 1007
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Mathematics for Statistical Mechanics I am studying Statistical Mechanics and Thermodynamics from a book that i am not sure who has written it, because of its cover is not present.
There is a section that i can not understand:
${Fj|j=1,..,N}$
$S= \sum_{j=1}^{N} F_{j}$
$<S>=< \sum_{j=1}^{N} F_{j}> = \sum_{j=1}^{N} <F_{j}>$
$\sigma^{2}_{S} =<S^{2}>-<S>^{2}$
line a:
$=\sum_{j=1}^{N}\sum_{k=1}^{N} <F_{j}F_{k}> - \sum_{j=1}^{N} <F_{j}>\sum_{k=1}^{N}<F_{k}>$
I can not understand why these terms are different
First term is:
$\sum_{i}\sum_{j}A_iA_j=(A_1A_1+A_1A_2+A_1A_3+\dots+A_1A_n)+\dots+(A_nA_1+A_nA_2+\dots+A_nA_{n-1}+\dots+A_nA_n)$
Second term is
$A_1(A_1+A_2+A_3+\dots+A_n)+\dots+A_n(A_1+A_2+\dots+A_n)$
I can see they are same.
But as you can see above it is different, how? Where am i wrong?
The question Math for Thermodynamics Basics 's answer have some clue but i could not understand exactly. Thanks.
Thanks
| The average/expected value of a product is in general not the same as the product of expected values. (The "mean value" function is linear though: a sum of mean values is equal to the mean value of the sum.)
The product of the sum and sum of product are related by the covariance:
$cov(X,Y) = <XY> - <X><Y>$, as you stated yourself.
I hope this helps.
source:
http://en.wikipedia.org/wiki/Expected_value#Non-multiplicativity
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Method of images tutorial I'm having an exam in Electrodynamics soon. I think I have most of it under control, but the method of images I'm not quite sure about.
There is not much in my book about this, so I was thinking some of you might know a website, article or something like that, maybe even with examples, that could help me get that last piece of the puzzle in my head?
I have tried Google, but I didn't find much to be honest. That's why I ask here.
| πFor better understanding I had added an image which shows,
*
*An infinitesimally thin grounded sheet AB (I know the diagram is quite different but just assume it)
*A charge $q$ present at right side at distance $l$ from the sheet
"Let me tell you that our purpose of grounding is just to develop zero potential over the sheet. There is nothing more use of it."
So the Image Method let you assume that in this situation (i.e. In which there is zero potential sheet and there is a charge $q$ present at distance $l$ from it) then you can assume that there will be another charge of same magnitude as of $q$ but will be of opposite sign (i.e. $-q$ here).
I had tried to show that totally 'imaginary' image in the figure boxed inside.
Now the main point,
You must be thinking that what is the use of it.
Well, from image method you can find the net force acting here which will be,
$$F=\frac{1}{4πε_0}.\frac{q.q}{(2l)^2}$$
which implies,
$$F=\frac{1}{4πε_0}.\frac{q^2}{4l^2}$$
Similarly you can find Electric field and other electrostatic terms.
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No well-defined frequency for a wave packet? There are similar questions to mine on this site, but not quite what I am asking (I think). The de Broglie relations for energy and momentum
$$ \lambda = \frac{h}{p},
\\
\nu = E/h .$$
equate a specific frequency and wavelength to a particle, yet we know that a wave packet
is a linear combination of an infinite range of frequencies and wavelengths. How is it that we (or nature) choose one frequency and wavelength out of the range? Does this have to do with the collapse of the wave packet when measured? And if so, is the resulting measured frequency a random outcome? Similarly, when an electron jumps from one energy level to another in an atom, it emits a photon of frequency
$$ \
\nu = \Delta E/h .$$
Since the photon is not a pure sinusoidal wave, how can a single frequency be ascribed to the photon?
| *
*according to de Broglie, the wave packet is moving with the group velocity vg and the phase wave is moving with vp=f*lamda. the de Broglie relation lamda=h/p where p=m*vg but lamda=vp/f. this equation involve both wave-particle duality.(p=m*vg for matter and vp=f*lamda). therefore, in this relation, the frequency is refer to phase wave but not the the particle(wave packet) itself
*each emitted photon consists energy E=hf. getting the energy, we can simply calculate the frequency with f=E/h.
hope this will help and sorry for my broken english
| {
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Can we write the electromagnetic potential covariantly in terms of the four-current? In the Lorenz gauge, we have a beautiful relation between the four-current and the four-potential:
$$\Box A^{\alpha} = \mu_0 J^{\alpha}$$
To get $A$ in terms of $J$, however, we have to use a considerably uglier formula; or, at least, this is the formula presented in textbooks:
$$A^{\alpha}(t, \mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{J^{\alpha}\left(t - \frac{1}{c}\|\mathbf{r} - \mathbf{r}'\|, \mathbf{r}' \right)}{\|\mathbf{r} - \mathbf{r}'\|} d^3\mathbf{r}'$$
The first equation is evidently Lorentz covariant. The second one, on the other hand, doesn't look covariant at all. The integrand has some messy dependence on $(t, \mathbf{r})$ and the integration goes over only the spatial dimensions.
Can we rewrite the second equation in a covariant form? If not, then why not?
| The integral expression is Lorentz-covariant, too, and it may be made manifestly Lorentz-covariant, too.
The integral measure $\int d^3 r' / ||\vec r - \vec r'||$ is equal to and may be rewritten as the four-dimensional integral with a delta-function (and step function) added:
$$2\int {d^4 x'} \cdot \delta[(x-x')^2]\cdot \theta(t-t') $$
It's understood that $J$ is substituted at the point $J(x')$.
Note that the step function $\theta$ (equal to one for positive arguments and zero otherwise) is Lorentz-covariant assuming that the points $x,x'$ aren't spacelike-separated (because the ordering of a cause and its effect is frame-independent), and they're not spacelike-separated as guaranteed by the delta-function that is only non-vanishing near/at the null separation of $x,x'$
The step function guarantees that the cause precedes its effect.
The argument of the delta-function is a Lorentz invariant, $(x-x')^2$ which means $(x-x')^\mu(x-x')_\mu$. The sign convention for the metric doesn't matter becase this invariant is the argument of an even function (delta-function).
Finally, the equivalence of the two integrals may be shown by performing the integral over $t'$. The theta-function implies that we only integrate over the semi-infinite line $t'\lt t$. The delta-function implies that the integral is only sensitive on the value of $J(t')$ where $c|t-t'| = |r-r'|$ where the delta-function vanishes.
Finally, the delta-function also automatically generates the $1/|\vec r - \vec r'|$ factor because
$$\delta(y^2) = \delta (y_0^2-|\vec y|^2) = \delta[(y_0+|\vec y|)(y_0-|\vec y|)]=\dots $$
which is equal, because $\delta(kX) = \delta (X)/ |k|$, to
$$\dots = \frac{\delta(y_0-|\vec y|)}{y_0+|\vec y|}=\frac{\delta(y_0-|\vec y|)}{2|\vec y'|}$$
You see that the factor of two was needed, too.
| {
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Uncertainty in path integral formulation In Feynman's path integral formulation, in order to calculate the probability amplitude, we sum up all the possible trajectories of the particle between the points $A$ and $B$.
Since we know precisely that the particle will be at $A$ and $B$, does it mean that the uncertainty of the momentum is infinite?
| Position/momentum Uncertainty, and path integral formulation are exactly the same thing.
Suppose you cut the time interval in time $t_0= t_A, t_1,....,t_n=t_B$.
At time $t_0$, the particle is at the position $x_0=x_A$. Because we know the position, uncertainty about momentum is infinite, but this simply means, that, at time $t_1$, the particle could be in any position $x_1$.
Now, if the particle is, at time $t_1$ at position $x_1$, we can repeat the same argument as above, and say, that, at time $t_2$, the particle could be in any position $x_2$.
So, we see, that all the intermediary positions $x_1, x_2, ....x_{n-1}$ at times $t_1, t_2, ....t_{n-1}$ could take all values.
(The only constraints are the initial and final values of position $x_A,x_B$.)
That means that all paths, beginning at $t_A,x_A$ and ending at $t_B,x_B$ have to be taken in account.
And this is precisely the definition of the path integral formulation.
| {
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Is my boss wrong about our mechanical advantage from our pulley system? I work on a drilling rig as a roughneck and we had a lecture today (at the office) about mechanical advantage in pulley systems. Now, I know that my boss is well educated in oil drilling, but my instincts tell me that he may have this one wrong.
A drilling rig works sort of like a crane in that it has a tall structure supporting a pulley system. There is a large winch permanently installed on the base platform and then it goes over the top of the structure (the crown of the derrick) and down through a floating sheave--this has a few wraps to give us more mechanical advantage. I am including pictures to help describe the situation.
Here the picture shows the floating sheave (the blocks) which we use to do most all of our operations. Most importantly, we use it to pick up our string of pipe that is in the ground.
As seen in this picture, the blocks hold the weight of the string of pipe. Now he told us that if the pipe get stuck in the hole (maybe it snags something or the hole caves in), that we lose all of our mechanical advantage. He said that is why the weight indicator will shoot up and go back down after it is freed. He said that because when the pipe is snagged in the hole then we are not dealing with a free floating sheave anymore and that is what is required to have a mechanical advantage.
I disagree with this because even if it is not free, there is still a mechanical advantage such that (say the normal mechanical advantage is 6 to 1) our pulling force is multiplied by 6. I would like somebody to confirm this for me.
First picture taken from www.worldoils.com on June 21, 2013 Second picture taken from www.PaysonPetro.com on June 21, 2013
| You are right. The simple way to understand mechanical advantage with pulleys is to count the number of ropes that "come out" of a particular block. With the same tension on each of the ropes, the total force is just the tension in each rope times the number of ropes.
Example (advantage of 1, 2, 3):
Whether the weight at the bottom is fixed or moving does NOT change the mechanical advantage (the balance of forces).
| {
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Elastic vs Inelastic vs isospin violating scattering particle physics models I'm looking for a nice paper that explains the difference between three particle physics models for spin-independent dark matter interaction with nuclei: elastic, inelastic and isospin violating scattering. I've found a nice paper that is giving a nice summation of the current results in direct dark matter searches ( http://arxiv.org/abs/1210.4011 ) and want to read up some more.
Cheers,
Adnan
| "Elastic" and "inelastic" mean the same things they do in an introductory mechanics class. In elastic collisions the total mechanical energy is the same after the interaction as before (in a dark matter interaction this means effective the total kinetic energy), while in inelastic interaction that equality can not be counted on. Either some mechanical energy goes into another channel in the final state or some energy that was in another form in the initial state appears in the final state.
Typical possibilities for inelastic interaction would be particle creation, excitation of the struck nucleus, or break up of the target.
"Isospin violating" means that the (approximate) isospin quantum number of the final state is different from that in the initial state. Isospin is the quantum number that distinguishes protons from neutrons (and can be applied to other hadrons as well). A collision that transforms a proton into a neutron without changing the isospin quantum number (if any: it probably doesn't have one) of the dark matter component would be isospin violating (and inelastic because some kinetic energy from the initial state went into raising the mass of the final state).
| {
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If the multiverse theory is true, can there be a Universe where there are different laws of physics? This is probably a very difficult question. But my question is essentially this, if there are other Universes can different laws of physics exist in those Universes and if so, can't there be a Universe where the laws of physics are so different that the multiverse theory has to come out as false?
| If u mind the MWI interpretation of QM, then the answer is: there are no universes where 1st law ofb thermodynamics does not work, but there are universes where the 2nd law of thermodynamics does not work. So some laws work everywhere while others do not.
| {
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Why can't we obtain a Hamiltonian by substituting? This question may sound a bit dumb. Why can't we obtain the Hamiltonian of a system simply by finding $\dot{q}$ in terms of $p$ and then evaluating the Lagrangian with $\dot{q} = \dot{q}(p)$? Wouldn't we obtain then a Lagrangian expressed in terms of $t$, $q$ and $p$?
Why do we need to use
$$H(t, q, p) = p\dot{q} - L(t, q, \dot{q})?$$
Or is it that whatever the Lagrangian is the method of finding $\dot{q}=\dot{q}(t,q,p)$. Will give us that equation for $H$?
| What you wrote is a variable change in the frame of the same Lagrangian. It remains a Lagrangian.
| {
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Can isotropic states have bound entanglement? Let us consider the maximally entangled state
\begin{equation}
|\psi\rangle=\frac{1}{\sqrt{n}}(|0,0\rangle+\cdots+|n-1,n-1\rangle)
\end{equation}
and construct the pseudo-pure state
\begin{equation}
\rho_\lambda=(1-\lambda)|\psi\rangle\langle\psi|+\lambda\frac{I_{n^2}}{n^2},
\end{equation}
where $I_{n^2}$ is the identity matrix and $0\leq\lambda\leq1$. I was told that for any dimension $n$ and for any $\lambda$, $\rho_\lambda$ is either separable or entangled which can be determined by partial transpose. It can be rephrased as
There does not exists any dimension $n$ and any $\lambda$ such that
the corresponding $\rho_\lambda$ is a bound entangled state.
I could not find out the proof and could not make one by myself. Can someone give me a proof? Advanced thanks for any suggestion.
ADDITION: Also the same question can be asked for the case, when the coefficients of $|j,j\rangle$ are non-uniformly distributed nonzero complex numbers (such that the sum is $1$).
| First of all, note that you can extend this class a little bit:
$$
\rho_\mu^\prime=(1-\mu)|\psi\rangle\langle\psi|+\mu\frac{I_{n^2} - |\psi\rangle\langle\psi|}{n^2-1},
$$
where $0 \le \mu \le 1$. You can check that $\rho_\lambda = \rho_\mu^\prime$, with $\lambda = \frac{n^2}{n^2-1}\mu$.
So $\rho_\lambda$ for $1 < \lambda \le \frac{n^2}{n^2-1}$ is also a correct state. It is not a mixture of $|\psi\rangle\langle\psi|$ and $I$, but a mixture of $|\psi\rangle\langle\psi|$ and $I-|\psi\rangle\langle\psi|$.
It is proved in http://arxiv.org/abs/quant-ph/9708015 that $\rho_\mu^\prime$ is separable if and only if $(1-\mu)\le\frac{1}{n}$, and partial transpose is indeed non-positive in the entangled case.
| {
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Why do some air-conditioned stores blast you with jets of air as you enter? I went to a grocery store on a hot day that was very well air-conditioned, and I noticed as I went through the open entrance that there seemed to be a very powerful downward air current right at the doorway. After crossing the invisible threshold, the temperature immediately dropped a good 15 degrees or so.
How does this process work?
| Most shopping malls have this kind of air door that blast you with high velocity air flow when you enter. As you might have noticed, shopping malls are always the cleanest places of them all. And I am not just talking about cleanliness of the floor area but the entire atmosphere in a mall. The reason they keep it clean is obvious. The method which they use to keep it clean is the answer to your question. The cleaning process in a shopping mall is modelled on the same process which is used in an IC fabrication facility, only that the constraints are less stringent. The pressure inside the mall is relatively higher than outside. Because of this airflow is always outward. The air comes from the ceilings & is collected at the floor. There is something called ACH(Air Changes Per Hour). In a fabrication facility, its about 100ACH. If you ever happen to enter a Fab facility, you always have to go through a series of doors, in each you are blasted with high velocity clean air to remove dust particles on your clothes & body. The constraint is less stringent in malls, purpose is same i.e. to aid the system set up in the mall to keep it clean. Also it acts as an air door since it would be irritating for a customer to have an actual door.
| {
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Chemical Potential of Ideal Fermi Gas In Wikipedia's article on Fermi Gases, they have the following equation for the chemical potential:
$$\mu = E_0 + E_F \left[ 1- \frac{\pi ^2}{12} \left(\frac{kT}{E_F}\right) ^2 - \frac{\pi^4}{80} \left(\frac{kT}{E_F}\right)^4 + \cdots \right]$$
where $E_0$ is the potential energy per particle, $k$ is the Boltzmann constant and $T$ is the temperature.
I don't understand how they get the third term, particularly the 1/80 factor. I've often seen this equation expressed just to the tau^2 term, and I understand how to get the 1/12 factor from a binomial expansion of
$$\left( 1 + \frac{\pi^{2}}{8} \left(\frac{\tau}{E}\right)^{2}\right)^{-2/3} $$
However, I've tried continuing the binomial expansion and cannot figure out why the factor is 1/80.
| I not sure how you obtained the last expression. The standard Sommerfeld expansion (for details, see e.g. Ashcroft & Mermin) gives a slightly different result, which is
$$
E_{F} \approx \mu\left[1+\frac{\pi^{2}}{8}\left(\frac{k_{B}T}{\mu}\right)^{2}\,\right]^{2/3} \approx \mu\left[1+\frac{\pi^{2}}{12}\left(\frac{k_{B}T}{\mu}\right)^{2}\right]
$$
to leading non-trivial order in $k_{B}T/\mu$, i.e., $O\big((k_{B}T/\mu)^{2}\big)$.
(I have set $E_{0}=0$ in the expression from Wikipedia.)
We can invert this relation by substituting $\mu = E_{F}\left[1 + c_{2} (k_{B}T/E_{F})^{2} + \cdots \right]$ into the above. That is,
$$
E_{F} = E_{F}\left[1 + c_{2} (k_{B}T/E_{F})^{2} + \cdots \right]\left\{1+\frac{\pi^{2}}{12}\left(\frac{k_{B}T}{E_{F}}\right)^{2}\left[1 + c_{2} (k_{B}T/E_{F})^{2} + \cdots\right]^{-2}\right\}.
$$
Comparing the zeroth order terms in $k_{B}T/E_{F}$ on both sides of the above equation, we simply obtain $E_{F}=E_{F}$. Comparing the second order terms, we have $0 = \frac{\pi^{2}}{12} + c_{2}$. Hence
$$
\mu = E_{F}\left[1-\frac{\pi^{2}}{12}\left(\frac{k_{B}T}{E_{F}}\right)^{2}\right]
$$
up to $O\Big((k_{B}T/E_{F})^{2}\Big)$.
To determine the next-order correction to $\mu$, you should include one higher order term in the Sommerfeld expansion, which gives
$$
E_{F} \approx \mu\left[1+\frac{\pi^{2}}{8}\left(\frac{k_{B}T}{\mu}\right)^{2}+\frac{7\pi^{4}}{640}\left(\frac{k_{B}T}{\mu}\right)^{4}\,\right]^{2/3}.
$$
Expanding this up to $O\big((k_{B}T/\mu)^{4}\big)$, then substituting $\mu = E_{F}\left[1 + c_{2} (k_{B}T/E_{F})^{2} + c_{4} (k_{B}T/E_{F})^{4}+\cdots\right]$ (where we have already determined $c_{2}$ before) into it, and then matching the coefficients of both sides up to $O\Big((k_{B}T/E_{F})^{4}\Big)$ will lead to the desired result.
| {
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A water pipe from sea level to beyond the atmosphere If a pipe extended from just above the ocean floor to outside the atmosphere, would water be sucked up it by the vacuum beyond the atmosphere? If a hole was made in the pipe, above sea level, how would that affect the flow of water? Would it stop it completely?
| No, the water would not be sucked up.
Even if you take a pipe with vacuum, closed the top and dipped the open end of that pipe in water then the water would only rise 10 meters. After that the 'pull' from your vacuum is in balance with the force of gravity acting on a 10 meter water column.
Maybe needless to say: The top of the atmosphere is way higher than 10 meters.
| {
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Is there a name for a quantity that represents (volumetric) flow per unit of mass? In a lot of medical literature about blood flow in the brain, researchers denote the amount of volumetric blood flow that passes through a certain amount of brain tissue as "cerebral blood flow". However, its associated unit is volumetric flow per unit of mass, e.g. 50 mililiters per minute per 100 grams of brain tissue.
However, I don't want to mix up actual volumetric flow (e.g. in mililiters per minute) with this volumetric flow per mass. Can you suggest a name for the latter quantity?
| The word "specific" (as in specific gravity or specific heat capacity) means per unit mass, so you could call it "specific flow" or "specific volumetric flow".
| {
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Why don't planets have Circular orbits? This might be a completely wrong question, but this is bothering me since many days ago. Given the mass (Sun) curves the space around it, gravitation is the result of such curved space (Correct me if I am wrong, source: A Documentary film). Given any point on a circle with center same as center of the mass, curvature in the space should be equal (Intuition).
Planets rotate around the Sun because of the curve in the space they should follow a circular path and the distance between planet and Sun should be at a distance.
Given the fact that earth has a elliptical orbit around the sun, and the distance between Earth and Sun varies according to position of the earth. Why do we have a elliptical orbit not a circular orbit.
| Comets are coming from almost infinite distance towards Sun. Due to Sun's Gravitational force they should be pulled into Sun rather than forming Elliptical or Hyperbolic trajectory.
If object is starting travel from closer distance at considerable speed, we can visualize formation of Elliptical orbit. For object coming from infinite distance getting pulled by Sun's Gravity, should end up falling into Sun
| {
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Are Lagrangians and Hamiltonians used by Engineers? Analytical Mechanics (Lagrangian and Hamiltonian) are useful in Physics (e.g. in Quantum Mechanics) but are they also used in application, by engineers? For example, are they used in designing bridges or buildings?
| I'm a electrical engineer, and have never used either one in over 30 years of designing circuits. I vaguely remember that we went over them briefly in school, but since I haven't used them (knowingly) since, I can't tell you what the physical meaning of either is, which of course perpetuates the fact that I'm not going to use them.
| {
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How is energy extracted from fusion? I understand that combining deuterium and tritium will form helium and a neutron. There are three methods to do this (1) tokamak (2) lasers and (3) cold fusion. I would like to know after helium is formed. How is that energy extracted from tokamak and stored?
| The basic idea in all three cases is that the energy becomes heat, and is extracted using turbines, just like in a fossil fuel burning power station. On a microscopic level the energy is released in the form of kinetic energy of the helium nucleus and the neutrons that are given off. These then collide with other particles, so that this kinetic energy rapidly becomes heat.
| {
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What is meant by Vacuum? What is the exact meaning of the word vacuum? Is it just a state of very low pressure or is it nothingness (as in there is nothing)? Also, when we say space is vacuum - it must be referring to pressure as space has light travelling (which means photons) besides the big masses of comets, planets, stars.
| In Newtonian mechanics this term is pretty easy to define - vacuum is a region of space free of matter. In quantum mechanics, things get a little bit complex, for more details I recommend you to read the corresponding section of the Wikipedia article you've put the link to, it explains the term well enough for a beginner.
| {
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Quantization for particle in a box problem Consider the particle in a box problem in QM. The crux of the reason why QM is able to explain the physical phenomenon is not just the theory but also able to impose boundary conditions which eventually result in quantization. Now in the particle in a 1-d box problem, the wave function is assumed to be zero at the boundaries. It has been said that it is imposed, so that the wave function is continuous. Okay, but what about differentiability? In order for the wave function to satisfy Schrodinger equation,we also need differentiability right? Okay if we assume only left (from one side) derivative to exit, we could have as well assumed only left continuity (from one side). For continuity, we assume it should be from both sides, but for differentiability we need only one side? They also say the slope also must be continuous. I don't see any rationale behind these quantizations!
| Differentiability of the wave function is only required for finite changes in the potential. If the your potential is infinite (as it is outside the inifinitely deep potential well which you describe) the Hamiltonial is ill-defined anyways.
An other case where you can have an infinite potential is if you have a $\delta$-distribution as a potential, there again you will find that the wave-function must be continuous but not differentiable (the difference between left-side and right-side derivative is given by the strength of the $\delta$-potential in this case).
| {
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How does one define "Nuclear Harmonicity"? How does one define Nuclear Harmonicity?
Although the title is pretty much the entire case, the main question is, what exactly is "harmonicity" in terms of nuclear physics? What does it mean for a nucleus to be quasi-harmonic or nearly-harmonic? Is it perhaps in the sense that it approaches a harmonic oscillator, and if so, what's the intuition behind this?
| Without more input on your part I will refer you to the shell model of nuclear physics:
In order to get these numbers, the nuclear shell model starts from an average potential with a shape something between the square well and the harmonic oscillator. To this potential a spin orbit term is added. Even so, the total perturbation does not coincide with experiment, and an empirical spin orbit coupling, named the Nilsson Term, must be added with at least two or three different values of its coupling constant, depending on the nuclei being studied.
Nevertheless, the magic numbers of nucleons, as well as other properties, can be arrived at by approximating the model with a three-dimensional harmonic oscillator plus a spin-orbit interaction.
Nuclei that are well described by this model, with the magic number of nucleons, might be characterized with the "harmonic" adjective (or quasi-harmonic etc).
| {
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Calculating the torque at a point when a motor is stopped? So, I'm trying to solve for the torque $\tau_A$ of a motor. I have attached a strong stick to the motor, like so:
I apply a force $F$ on the stick which stops the motor. The distance from the outside edge of the cylinder to the end of the stick is $L$. The torque for the motor is $\tau_A=F(L+r)$, $r$ is the radius.
My friends believe that the torque at point $B$ is $\tau_B=-FL$, but I believe even though the motor is not moving (due to the force), it still applies a torque at point $B$. It would be less than $\tau_A$ since it doesn't push around the point uniformly, but it should be $\tau_B=-FL+\tau_Ac$ ($c$ is a constant). Using their method, they got that $\tau_A=0$, which I believe happened because in calculating the torque at point $B$, they make $\tau_A=0$.
Who is right? How do I calculate how much $\tau_A$ is applied about point $B$ (assuming I'm correct)?
| Well, nothing is starting to move, right? So there's no net torque at all. Neither around point $A$, not around point $B$.
Let me clarify:
Around point $A$, there is the torque of the motor, $\tau_A$, and the torque due to the force at the end of the stick, $-(r+L)F$. So, the net torque is indeed zero (as is should). There is also a force attacking at point $A$ of magnitude $F$ which is pointing up. This is required to cancel the translatory motion that would otherwise be due to the force on the stick. Since it is attacking at $A$, it will not exert a torque around that point.
Also, one can model the inherent torque of the motor by a force of magnitude $F'$ attacking in $B$ and pointing upwards. To get the same torque, we shall require $rF'=\tau_A$ leading to $F'=(1+L/r)F$. Again, we need to cancel that force by one of the same magnitude attacking in $A$ in the downward direction. Note, that the particular direction and position of this force are arbitrary as long as they oppose each other and lead to the same torque on $A$. However, the above choice is most practical.
Now that we know all the forces, we can consider the torque on $B$. Your friends are right in saying that a torque of $-LF$ acts due to the force on the stick. Yet, there are also the forces on $A$, $F$ (up) and $F'$ (down) leading to an additional torque of $+r(F'-F)$. The force $F'$ that attacks in $B$ is now irrelevant for the torque on $B$. Hence, the net torque on $B$ is
$\tau_B=-LF+r(F'-F)=-LF+LF=0$.
(Again, the torque around $A$ is also zero. It is misleading to call $\tau_A$ the torque around $A$ as it is counteracted by the force on the stick!)
| {
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What are the forces acting between two air bubbles in water? The exact question is
Two air bubbles in water
*
*attract each other
*repel each other
*do not exert any force on each other
*may attract or repel depending upon the distance between them.
The chapter is about gravitation.
The given answer is
A lighter body inside a denser medium behaves like negative mass as far as gravitational force is considered. Two air bubbles i.e. two negative masses will attract each other.
What is negative mass in this context and how can it be applied to such macroscopic objects? How would it result in attraction?
My reasoning is:
Consider the bubble A in the above image.
The air particles forming the bubble A would be attracted more to the left(away from B) as there are more dense particles towards that side-the air particles making up bubble B are less dense than the medium and they will attract the air from bubble A to a smaller extent than if the volume of bubble B was filled with the medium.
A similar case would apply to B due to lesser density of particles forming A and the bubbles would be (indirectly) repelled.
So what is happening in this case?
|
What is negative mass in this context
Consider a single bubble in the middle of a glass of water on your desk.
Normally, gravity pulls air above the desk downwards towards the desk surface.
However, in this case the air in the bubble does not move downwards under the influence of gravity. It moves in the opposite direction. It is as if the air in the bubble has negative mass and is repelled by gravity.
In fact the air does not have negative mass and is not repelled but the real situation is more complex to describe.
| {
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How can you weigh your own head in an accurate way? I read some methods but they're not accurate. They use the Archimedes principle and they assume uniform body density which of course is far from true. Others are silly like this one:
Take a knife then remove your head.
Place it on some scale
Take the reading
re-attach your head.
I'm looking for some ingenious way of doing this accurately without having to lie on your back and put your head on a scale which isn't a good idea.
| A more accurate solution than the "relax your neck muscles" and similar ones:
Go to the morgue. Cut off the head of a recently deceased person with a similar body type. Measure the ratio of the weights of the head and body. Measure your weight.
Even more accurate: measure the density of the head you just cut off. Now measure the volume of your own head by immersing it into water. The volume of the head does not vary as much as of the body, where muscle, fat, etc. play a role. For better accuracy, cut off many heads for an average.
| {
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Vectors with more than 3 components
*
*I have some confusion over Vectors, Its components and dimensions. Does the number of vector components mean that a vector is in that many dimensions? For e.g. $A$ vector with 4 components has 4 dimensions?
*Also, how can a Vector have a fourth dimension? How can we graphically represent vectors with more than 3 dimensions? Its hard for me to visualize such a vector, Can anyone point me to some resource where they explain this graphically and in detail?
| In general, a vector in $D$ dimensions will indeed have $D$ components. A vector with $d<D$ components may, however, always be viewed as one in $D$-dimensional space but with $D-d$ of its components equal to zero.
Visualizing higher dimensions is always tough, because it's not something we have any experience with. We only ever see $0$-$3$ dimensional objects. Sometimes I like to just view a higher dimensional vector as an arrow in some abstract space, but really that's not realistic. So most of the time I don't try to visualize the whole vector. I can usually only imagine 3 (or lower) dimensional projections.
| {
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How do ice spikes form? I recently saw this picture posted on Twitter which shows a so-called ice spike rising from an ice cube tray.
I have read the Wikipedia page, but it doesn't mean much to me. My instinct was that it is caused by vibrations from the freezer setting up a resonant frequency on the surface of the water, and the spike has gradually "built up" as the water freezes. But I'm not a physicist, so I don't know if this makes sense. I'm ideally looking for a layman's explanation of why this happens.
| I think what is happening in rough qualitative terms is that the water freezes around the sides and the top first leaving a hole in the centre. Ice expands by 4%-9% when freezing so as the water below freezes it forces the remaining water up through the hole where is freezes around the edge. The hole shrinks as the water freezes and rises around its edge forming the base of the spike. The spike is hollow so water is pushed up to the top where it freezes at the edge making the spoke grow longer
Update: If you search for ice spike on youtube you will find some good timelapse videos showing these spikes forming. I especially like this one because you can see the unfrozen water pushing up inside the spike.
Sometimes in a larger water container you can get an inverted pyramid shape. The explanation is the same and the shape is due to the crystaline nature of the ice. This video shows the phenomena
| {
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Where does $p^i/p^+$ come from in the EOM of an open string? I have a stupid question about Eq. (1.3.22) in Polchinski's string theory volume 1.
In the equation of motion for an open string, Eq. (1.3.22),
$$X^i (\tau, \sigma) = x^i + \frac{ p^i}{p^+} \tau + i \bigl(2 \alpha'\bigr)^{1/2} \sum_{\substack{n= -\infty,\\n\neq0}}^{\infty} \frac{1}{n} \alpha_n^i \exp\biggl( -\frac{ \pi i n c \tau}{ l}\biggr) \cos \frac{ \pi n \sigma}{l} $$
How do I get the factor $ \frac{ p^i}{p^+} $?
| From $(1.3.18$), we have :
$$\Pi^i = \frac{p^+}{l} \partial_\tau X^i$$
The definition of the total momentum is $(1.3.23 b)$ :
$$p^i = \int_0^ld\sigma ~\Pi^i(\tau, \sigma)$$
So, by definition :
$$p^i = \frac{p^+}{l} \int_0^ld\sigma ~\partial_\tau X^i(\tau, \sigma))~~~~~~~~~~~~~~~~(1)$$
Now, considering equation $(1.3.22)$, and taking the $\tau$ derivate, we get :
$$ \partial_\tau X^i(\tau, \sigma) = \frac{p^i}{p^+} + \sum a_n (\tau) \cos (\frac{\pi n\sigma}{l})$$
The integral of space-periodic-excitations on the interval $[0,l]$ is zero, so we get :
$$\int_0^ld\sigma ~\partial_\tau X^i(\tau, \sigma)) = \frac{lp^i}{p^+}~~~~~~~~~~~~~~(2)$$
Obviously, $(1)$ is the same thing as $(2)$, which explain the factor $\frac{p^i}{p^+}$ in $(1.3.22)$
| {
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What is physics behind of explosion under Atmospheric pressure? An explosion is a rapid increase in volume and release of energy in an extreme manner.
A blast wave in fluid dynamics is the pressure and flow resulting from the deposition of a large amount of energy in a small very localised volume.
The equation for a Friedlander waveform describes the pressure of the blast wave as a function of time:
$$P(t)=P_oe^{-\frac {t}{t^*}}(1-\frac {t}{t^*})$$
where P$_o$ is the peak pressure and t$^*$ is the time at which the pressure first crosses the horizontal axis (before the negative phase).
My Question:
Why does the pressure temporarily drop below ambient after the wave passes?
| If you start with a finite amount of gas in the inner sphere and then deposit a massive amount of energy, the molecules of the gas begin moving rapidly outwards and piling up, creating the blast wave. However, the rate at which the gas is moving outwards may not be balanced by the amount of gas molecules being created by the explosive. If this is the case, then the pressure must decrease below ambient as the molecules are pushed outwards with the blast wave.
You can see this in videos of blast waves. The initial wave continues to move outwards, but the smoke/dirt/debris caused by the explosive will move outwards initially, then inwards as the lower pressure region sucks it back in towards the center. There is actually considerably banging that goes on where the low pressure behind the blast wave moves inwards and outwards until it relaxes back to atmospheric pressure.
Here is a great video that shows the blast and resulting banging as the pressure relaxes.
| {
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Is there an EMF in a conductor moving at constant speed across the uniform magnetic field If a conductor - a long rod - moves at constant speed across the "lines" of a uniform magnetic field, is there an EMF within this conductor? Or, if a conducting rod rotates at uniform rate, pivoted in the middle or at one of its ends in a uniform magnetic field perpendicular to the plane of rotation, is there an EMF generated within the conductor?
| An EMF is generated whenever a conductor moves relative to a magnetic field so that the conductor is cutting across the magnetic field lines. The EMF generated is the cross product of the magnetic field and the motion of the conductor (I may have the sign flipped, don't remember off the top of my head).
To answer your rotating rod question: yes EMF is produced inside the rod. However, note that this EMF will not be end to end of the rod, but from each end to its center.
If you attached brushes to the ends of the rod such that they touched the inside of a fixed cylinder, then the whole apparatus would be a generator between this cylinder and the center of the rod. If you connected this generator to a load, current would now flow, which now produces a force on the rod, which would oppose its motion.
| {
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Can I call additional conditions on potentials a Gauge choice? Let's say I have an electromagnetics problem in a spatially varying medium. After I impose Maxwell's equations, the Lorenz gauge choice, boundary conditions, and the Sommerfeld radiation condition, I still have more unknowns than equations and the solution for (say) the magnetic vector potential is not uniquely determined by the above. This does actually happen when you formulate the equations in plane-stratified media in the plane-wave / Fourier domain. The way to proceed that I have seen is to choose that one of the cartesian components of the vector potential is zero, i.e. impose one of the following as an additional condition to ensure uniqueness of the potentials: $A_x=0$, $A_y=0$ or $A_z=0$. We could of course make up an infinite number of other conditions that leave the fields invariant.
My question is, can I call the above arbitrary choice about the vector potential a "gauge choice"? The reason for imposing it seems to be identical to the reasons we normally impose the Lorenz or Coulomb gauge, namely that the field equations don't dictate anything about certain potential quantities, the choice makes solving the equations uniquely possible, and the physical $\mathbf{E},\mathbf{H}$ fields are invariant to the extra condition on the potentials.
| Per your comment to user1504, you are correct: the Lorenz gauge contains considerable arbitrariness.
To wit, suppose potentials $\Phi, \boldsymbol{A}$ satisfy:
$$ \nabla^2 \Phi - \frac{1}{c^2} \frac{\partial^2 \Phi}{\partial t^2} = - 4 \pi \rho $$
$$ \nabla^2 \boldsymbol{A} - \frac{1}{c^2} \frac{\partial^2 \boldsymbol{A}}{\partial t^2} = - \frac{4 \pi}{c} \boldsymbol{J} $$
$$ \boldsymbol{\nabla \cdot A} + \frac{1}{c} \frac{\partial \Phi}{\partial t} = 0 \quad \text{ (Lorenz condition)}$$
Then, if $ \Lambda $ is a scalar that satisfies the wave equation:
$$ \nabla^2 \Lambda - \frac{1}{c^2} \frac{\partial ^2 \Lambda}{\partial t^2} = 0 $$
the restricted gauge transformation (restricted because of the condition on $\Lambda$):
$$ \boldsymbol{A} \rightarrow \boldsymbol{A} + \boldsymbol{\nabla} \Lambda \quad, \quad \Phi \rightarrow \Phi - \frac{1}{c} \frac{\partial \Lambda}{\partial t}$$
preserves the Lorenz condition.
Caveat! This analysis applies in vacuum. Extension to a homogeneous, isotropic medium looks straightforward, but for a spatially varying background like you posit, I don't know...
| {
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What causes a black-body radiation curve to be continuous? The ideal black-body radiation curve (unlike the quantized emission seen from atomic spectra), is continuous over all frequencies. Many objects approximate ideal blackbodies and have radiation curves very similar in shape and continuity to that of an ideal black-body (often minus some emission and absorption lines from the atoms in an object, such as radiation curves seen from stars). I am wondering what exactly gives rise to a basically continuous black-body radiation curve in real objects? Since atomic energy states are quantized, it seems real life black-body curves would have some degree of measurable quantization to them (or perhaps the degree of quantization is so small the radiation curves look continuous).
| This is the second time in only a few days that I've cited Luboš Motl's excellent answer to What are the various physical mechanisms for energy transfer to the photon during blackbody emission?. As Luboš points out, the precise microscopic mechanisms of the radiation are unimportant because the statistical properties ensure that it follows Planck's law.
To get the characteristic black body curve you just need enough ways to generate EM radiation. Typically thermal vibrations in whatever material you're looking at result in accelerated electrons and oscillating dipoles within the material, and both generate the electromagnetic waves. This isn't a resonant process, so you don't get sharp lines but just a continuum of frequencies.
| {
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Understanding the different kinds of mass in gravity On this site, the Phys.SE question Is there a fundamental reason why gravitational mass is the same as inertial mass? has been asked. See also this Phys.SE question. The 'answer' provided on this forum has been that the curvature of spacetime explains both. The answer is still cryptic for me as I am more a concrete thinker.
Newton said $F=ma$. I can use this formula to measure inertial mass. Experimentally I can measure the motion of an object while applying a constant force to it. Newton also said $F=\frac{GMm}{r^2}$. In this case, what simple experiment will allow the measurement of gravitational mass?
| What you're probably looking for is something like the famous Eötvös experiment which used a kind of torsion balance to test the equivalence of gravitational and inertial mass.
| {
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Can this ratio be written any better? This topic is closely related to previous topic where we were to calculate ratio $\lambda_e/\lambda_p$ for proton and electron with same velocities.
This time we I want to know if it is possible to derive a ratio $\lambda_e/\lambda_p$ for proton and electron who have same kinetic energies $(E_{ke} = E_{kp})$. So i write this like:
\begin{align}
\frac{\lambda_e}{\lambda_p} = \frac{\tfrac{h}{p_e}}{\tfrac{h}{p_p}} = \frac{p_p}{p_e} = \frac{\sqrt{{E_{kp}}^2 + 2E_{kp}E_{0p}}}{\sqrt{{E_{ke}}^2 + 2E_{ke}E_{0e}}} \longleftarrow\substack{\text{Here i know that kinetic}\\\text{energies are the same so}\\\text{i have to use $E_{kp}=E_{ke}$}}
\end{align}
I want to know if this can be reduced a bit more or is this the best possible result. If anyone has any idea please do tell.
| It's difficult to simplify much more your last expression, but:
\begin{align}
\frac{\lambda_e}{\lambda_p} = \frac{\tfrac{h}{p_e}}{\tfrac{h}{p_p}} = \frac{p_p}{p_e} = \frac{\sqrt{1 + 2\frac{E_{0p}}{E_{kp}}}}{\sqrt{1 + 2\frac{E_{0e}}{E_{ke}} }} =\sqrt{\frac{1 + 2\frac{E_{0p}}{E_{kp}}}{1 + 2\frac{E_{0e}}{E_{ke}}}}
\end{align}
Now, you can perform some approximations. For example, if $E_{0e} >> {E_{ke}}$:
\begin{align}
\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{1 + 2\frac{E_{0p}}{E_{kp}}}{1 + 2\frac{E_{0e}}{E_{ke}}}} \approx \sqrt{\frac{E_{0p}}{E_{0e}}}
\end{align}
| {
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SL(2,R) to SL(2,Z) in Type IIB String Theory I heard from Prof. Katrin Becker (in her "SUSY for Strings and Branes - Part 1" lecture) that the classical $SL(2,\mathbb{R})$ symmetry in type IIB String theory becomes $SL(2,\mathbb{Z})$ in Quantum because of charge quantization. However, I cannot see how does it work. Is there any rigorous Mathematical derivation for this?
Thank you.
| It's simple. The dilaton-axion (complexified) field in supergravity (and similar classical theories with a noncompact symmetry) is invariant under $SL(2,R)$ transformations
$$\tau \to \frac{a\tau+b}{c\tau+d}, \quad ad-bc=1$$
However, the same transformation must also transform the charges of objects. For example, one-dimensional branes always carry the charge like $m$ fundamental strings superposed on top of $n$ D1-branes. So the general charge (density) of a D1-brane is given by two numbers $(m,n)$. Under the transformation above, they transform to
$$(m,n)\to (am+bn, cm+dn)$$
because the $SL(2,R)$ transformation mixes the two types of one-brane charges (and similarly for other dimensions of branes, including the instantons).
In the classical theory, the charge of a black $p$-brane, including the one-branes above, is (a generalization of the charge of a charged black hole) given by any real numbers (charges) $m$ and $n$. However, quantum mechanically, $m$ and $n$ have to be integers in certain units. Consequently, we only get allowed one-branes after the transformation if the final charges, $(am+bn, cm+dn)$, are integers for all integers $(m,n)$. This requirement of the integrality of charges implies that $a,b,c,d$ have to be integers by themselves and only the $SL(2,Z)$ subgroup of $SL(2,R)$ maps allowed states in the Hilbert space (a superselection sector) to other allowed states in the same Hilbert space.
Note that the quantization (integrality) of the charges such as $m,n$ above is required by quantum mechanics. The one-branes are electromagnetic duals of five-branes that carry their charges, too – a combination of D5-brane and NS5-brane charges (completely analogous to the two one-branes). Because all these four types of charges (D1,F1,D5,NS5) are allowed to be nonzero but quantum mechanics enforces the Dirac quantization rule that the spacing of the D1-brane charge is inverse up to a $2\pi$ factor to the spacing of the D5-brane charge, and similarly for F1 and NS5, it follows that all these four charges must belong to a lattice. In other words, there has to exist a linear redefinition or convention in which all these four charges are integers.
| {
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What is canonical momentum? What does the canonical momentum $\textbf{p}=m\textbf{v}+e\textbf{A}$ mean? Is it just momentum accounting for electromagnetic effects?
| Imagine this situation:
at time t=0, we have a infinite long straight wire with current zero, and a charged particle q with zero velocity.
at time t=T, we make the current to be I, thus we have a $ \mathbf{B}$ field, and $ \mathbf{A}$ field.
during this process, $ \mathbf{A}$ is build up from zero to some value, therefore we have induced electric field $ \mathbf{E}= - \frac{ \partial \mathbf{A}}{\partial t}$
$\Delta (m\mathbf{v})=\int \mathbf{F} dt = \int q \mathbf{E} dt =\int q \frac{ - \partial \mathbf{A}}{\partial t} dt $
assume this process happened very fast, the particle almost stays at the same position, $ \frac{ \partial \mathbf{A}}{\partial t} =\frac{ d \mathbf{A}}{d t}$
then
$\Delta (m\mathbf{v}) =\int q \frac{ - d \mathbf{A}}{d t} dt = - q \int d \mathbf{A} =-q \Delta \mathbf{A} $
$ \Delta ( m \mathbf{ v } +q \mathbf{A}) =0 $
$ m \mathbf{ v } +q \mathbf{A} = constant $
| {
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Hooke's law doesn't work well with my spring? I'm trying to make an application for Hooke's law using a spring, but the law doesn't give any correct result with my spring, because when I hang a $100\,\mathrm{g}$ object on the spring it's elongates about $0.3\,\mathrm{cm}$ and when I hang a $200\,\mathrm{g}$ object the spring elongates about $1\,\mathrm{cm}$ while it should elongates only $0.6\,\mathrm{cm}$ .
I'm sure that the problem is with the spring design.
The spring I'm using is just like in picture below:
| Let me guess: you take the spring as it is and hang your objects, right? Then measure the displacement.
Try to do the following: hang any arbitrary object so that the string will stretch a bit from its initial state. Then add you 100g and 200g objects to the initial mass and measure the difference in spring's length. I will be surprised if you won't get good results.
Explanation: there are other forces involved when the spring is in its initial condition (as in the picture). When you initially stretch it a bit, you neutralize these forces and the only force left is Hooke's one.
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Lorentz homogeneous group and observables For generators of the Lorentz group we have the following algebra:
$$
[\hat {R}_{i}, \hat {R}_{j} ] = -\varepsilon_{ijk}\hat {R}_{k}, \quad [\hat {R}_{i}, \hat {L}_{j} ] = -\varepsilon_{ijk}\hat {L}_{k}, \quad [\hat {L}_{i}, \hat {L}_{j} ] = \varepsilon_{ijk}\hat {R}_{k}.
$$
For the splitting of algebra, we can introduce operators
$$
\hat {J}_{k} = \hat {R}_{k} + i\hat {L}_{k}, \quad \hat {K}_{k} = \hat {R}_{k} - i\hat {L}_{k}.
$$
So
$$
[\hat {J}_{i}, \hat {J}_{j} ] = -\varepsilon_{ijk}\hat {J}_{k}, \quad [\hat {K}_{i}, \hat {K}_{j} ] = -\varepsilon_{ijk}\hat {K}_{k}, \quad [\hat {J}_{i}, \hat {K}_{j}] = 0.
$$
So, each irreducible representation of Lie algebra is characterized by $(j_{1}, j_{2})$, where $j_{1}$ is max eigenvalue of $\hat {J}_{3}$ and $j_{2}$ is max eigenvalue of $\hat {K}_{3}$.
Then I can classify objects that transform through the matrices of the irreducible representations,
$$
\Psi_{\mu \nu}' = S^{j_{2}}_{\mu \alpha }S^{j_{2}}_{\nu \beta}\Psi_{\alpha \beta},
$$
where $S^{j_{i}}_{\gamma \delta}: (2j_{i} + 1)\times (2j_{i} + 1)$.
For $(0, 0)$ I have scalar field, for $\left(\frac{1}{2}, 0\right); \left(0; \frac{1}{2}\right)$ I have spinor, for $(1, 0); (0, 1)$ I have 3-vectors $\mathbf a, \mathbf b -> \mathbf a + i\mathbf b$ creating antisymmetrical tensor etc.
Also, for scalar $j_{1} + j_{2} = 0$, for spinor - $\frac{1}{2}$, for tensor - $1$. So, the question: is sum $j_{1} + j_{2}$ experimentally observed? Is it connected with a spin?
| Yes, a representation labeled by $(j_1,j_2)$ corresponds to the total spin $j_1+j_2$, (rigourously speaking of spin needs that one of $j_1$ or $j_2$ is zero) and if $j_1=j_2$, this is a real representation, but you may have a representation which is a sum or irreductible representations , some examples:
$(\frac{1}{2},0)$ corresponds to a left-handed Weyl spinor
$(0,\frac{1}{2})$ corresponds to a right-handed Weyl spinor
$(\frac{1}{2},0) + (0,\frac{1}{2})$, is the Dirac bi-spinor
$(\frac{1}{2},\frac{1}{2})$ corresponds to a Lorentz vector.
$(1,0) + (0,1)$, is the electromagnetic field representation
More generally, if the complex conjugate of a representation (interverting in all terms $j_1$ and $j_2$) is the same as the representation, then the representation is real.
For instance, the Dirac representation, or the electromagnetic field representation, are real representations.
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Can visible light be emitted from a non-thermal source? I was reading about thermal and non-thermal radiation and I was wondering if visible light can be emitted from a non-thermal source?
| Super-continuum sources are well known non-thermal white light sources. The gist is a laser beam interacts with a specially tailored nonlinear material to generate ultra-broadband coherent light. A few references:
Supercontinuum light
Demonstration of Stimulated Supercontinuum Generation – An Optical Tipping Point
Generation of a 650 nm - 2000 nm Laser Frequency Comb based on an Erbium-Doped Fiber Laser
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Does magnetism affect corrosion? Supposing there is an iron nail that is left to rust, if we compare the time it takes to rust with that of a magnetized iron nail, will there be any difference in the time of corrosion (assuming other environmental factors are constant)?
| Looks like magnetization can accelerate corrosion (http://www.ifw-dresden.de/de/institute/institut-fuer-komplexe-materialien/abteilungen/chemie-funktioneller-materialien/corrosion/corrosion-of-permanent-magnet-materials/impact-of-the-magnetization-state-on-the-corrosion-of-ndfeb-permanent-magnets/ ) (I did not look at the references there).
| {
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What is the sign of the work done on the system and by the system? What is the sign of the work done on the system and by the system?
My chemistry book says when work is done on the system, it is positive. When work is done by the system, it is negative.
My physics book says the opposite. It says that when work is done on the system, it is negative. When work is done by the system, it is positive.
Why do they differ?
| It's just a convention in physics we are more interested in getting some work output say a mechanical device , engine etc while in chemistry we are more concerned with the internal energy things so we do so in both the cases the result is same
physics case : du = dq - dw , doing work on system increases internal energy as dw = negative for work done on system and vice versa
chemistry case : du = dq + dw ; doing work on system will increase the internal energy of the system as dw = positive which is obvious and vice versa.
we can take any sign convention in a given problem but we should be consistent with that throughout the problem to avoid confusion and mistake.
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Overtaking with non-constant acceleration I have tried to solve this problem by adding the sum of the displacements during acceleration, constant velocity and deceleration, but it does not work out.
Question:
A car accelerates from rest to $20~\text{m/s}$ in $12$ seconds ($a =5/3~\text{ms}^{-2}$), it travels at $20~\text{m/s}$ for $40$ seconds, then retardation occurs from $20~\text{m/s}$ to rest in $8$ seconds ($a = -2.5~\text{ms}^{-2}$). As the car accelerates an RC car, moving parallel to the car, is moving at $14~\text{m/s}$. When will overtaking occur and what will the distance be? The RC car passes the car just as it starts to accelerate.
I can do this without a problem if acceleration is a constant. Is there a differential equation I can use to compute this as that is my better area or must I stick with the SUVAT equations?
Again, if I could be pointed in the right direction that way I can learn.
| The car accelerates from 0 to 20 m/s in 12 s. It has an average speed of 10 m/s over that time, and so covers 120 m in that time. In that same time, the RC car goes 14 m/s * 12 s = 168 m. So the car didn't catch up while accelerating. At 12 s, the car is 168 m - 120 m = 48 m behind the RC car.
The car is now going a constant 20 m/s, whereas the R/C car is going 14 m/s. So it can catch up. The difference is 6 m/s. The car has 48 m to make up, so that takes 48 m / 6 m/s = 8 s after it finishes accelerating, or 20 s after it starts. This is long before the car begins to decelerate, which is 40 s after it's done accelerating.
Therefore the car catches up at 20 s, which is at 14 m/s * 20 s = 280 m.
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Steering forces on a bicycle I always notice this weird thing and try to overcome it but cant.
As shown in the image when I ride the bike by just one hand and pull the handle back say from the right side so as commonly the handle should rotate towards right and the bike should turn to right. But that doesn't happen. No matter what I do the handle turns to left and the bike always go to left. And when I try to push it forward so that it turns to left then the handle turns to right and so the bike also turns to right.
As from how much I know about laws of motion I don't know why it happens. Please explain as it is too much weird and interesting for me.
| Pulling the right handlebar towards you causes the front wheel to turn to the right, which causes the entire bicycle to bank to the left, because the support points are moved to the right.
Then, unless you simply let the bike fall over to the left, you unconsciously balance the bike by turning into the bank.
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Can spheres leaking charge be assumed to be in equilibrium? I am struggling with the following problem (Irodov 3.3):
Two small equally charged spheres, each of mass $m$, are suspended from the same point by silk threads of length $l$. The distance between the spheres $x \ll l$. Find the rate $\frac{dq}{dt}$ with which the charge leaks off each sphere if their approach velocity varies as $v = \frac{a}{\sqrt{x}}$, where $a$ is a constant.
This is embarrassingly simple; we make an approximation for $x \ll l$ and get
$$
\frac{1}{4 \pi \epsilon_0} \frac{q^2}{x^2} - \frac{mgx}{2l} = m \ddot{x}.
$$
We can get $\ddot{x}$ from our relation for $v$, so we can solve for $q$ and then find $\frac{dq}{dt}$.
However, in general, $\frac{dq}{dt}$ will depend on $x$ and hence on $t$. The answer in the back of the book and other solutions around the web have $\frac{dq}{dt}$ a constant.
You can get this by assuming that at each moment the spheres are in equilibrium, so that you have $\ddot{x} = 0$ in the equation of motion above.
Does the problem tacitly imply we should assume equilibrium and hence $\frac{dq}{dt}$ is constant, or am I missing something entirely? I.e. why is the assumption of equilibrium justified? I understand reasoning like "the process happens very gradually, so the acceleration is small compared to other quantities in the problem," but I don't understand how that is justified by the problem itself, where we are simply given that the spheres are small (so we can represent them as points) and $x \ll l$ (which we have used to approximate the gravity term in the equation of motion).
| In your equation
$$\frac{1}{4 \pi \epsilon_0} \frac{q^2}{x^2} - \frac{mgx}{2l} = m \ddot{x}$$
$ \ddot{x}$ can be written as $$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}$$
so the first equation can be written in terms of $x$ for all the terms (as you are given $v$ in terms of $x$ in the question).
After doing that and multiplying through by $x^2$, you get
$$\frac{q^2}{4 \pi \epsilon_0} - \frac{mgx^3}{2l} = -\frac{a^2}{2}$$
i.e. $x$ is constant
This means that we must imagine that $x$ can only vary really slowly, that's the justification - from the things given in the question, for deciding we can put $\ddot{x}=0$
so you get $q^2 \propto x^3$
$$2q\frac{dq}{dt} = 3x^2\frac{dx}{dt}$$
putting $q$ and $\frac{dx}{dt} (= v)$ in terms of $x$ gives the result that $\frac{dq}{dt}$ is constant.
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How are low energy effective actions derived in string theory? For example the eq 2.1 here with regards to Type IIB.
Unless I am terribly missing/misreading something Polchinski doesn't ever seem to derive these low energy supergravity actions.
I would like to see a beginner's explanation (maybe together with review paper for further information) to getting these actions from string theory (and hopefully also something about deriving the black-hole and the brane metrics from them)
| I think that the quickest route to the effective action actually isn't through string amplitudes, but through the beta functions. The conditions for worldsheet conformal invariance are equivalent to the spacetime equations of motion, and from these you can infer an on-shell effective action. This is all you can hope for in string theory (or any theory of quantum gravity), which isn't well-defined off-shell.
As for references, Polchinski does the bosonic case. These computations are easiest in dim reg, but his scheme is fine also. I think the original analysis was by Callan et. al, http://adsabs.harvard.edu/abs/1985NuPhB.262..593C.
There are also really fun technicalities that come up in higher genus corrections. These were pointed out by Fischler and Susskind http://inspirehep.net/record/17879, and expanded on in a series of papers by Fradkin and Tseytlin, and Callan et. al.
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Does the termination point(level) influence water flow from a pipe I get water to my home from a nearby Tank A at a certain height above ground level.
I have a 1" pipe through which I get this water to my home.. I leave this water into my well by connecting a 1" tube to this pipe.
Reason for question: I have seen water pressure vary(lower- more flow) depending upon the height I hold my pipe(on the outlet side) when I empty my fish tank, keeping how deep the pipe is immersed on the other side a constant.
Now, if I elongate and leave the pipe from Tank A at deeper level below ground level should the flow rate increase?(will i get more water in the same time) as apposed to a shorter pipe which still goes into my well.
or simply:
Will I empty my overhead tank on 2nd floor quicker if I use the tap on the ground floor instead of the tap on the second floor. The taps being same size.
| A longer (and deeper emerged pipe) will yield a slightly lower flow rate. This is because the longer pipe will yield a higher drop in the loss of pressure, due to friction. A longer and deeper pipe will never increase the flowrate, since what is causing the water to flow is the height difference between the two water surfaces (assuming both are subjected to atmospheric pressure).
For more information lookup Bernoulli's principle:
$$
p_1+\frac{1}{2}{\rho}v_1^2+{\rho}gz_1=p_2+\frac{1}{2}{\rho}v_2^2+{\rho}gz_2+\left(f\frac{L}{D}+{\sum}K\right)\frac{1}{2}{\rho}v_2^2
$$
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Opening the fridge door to cool a room I'm well aware that the default answer to this textbook default question is "it doesn't work", but still, I believe it does.
To cool the insides of the fridge, the compressor must do work, and since the efficiency isn't 100% you are constantly warming the whole room to cool it's insides, the winning move here is simply turning the fridge off. However, let's suppose the fridge must stay on, wouldn't it be better to open the door?
In other words: Isn't opened fridge turned on better than closed fridge turned on for the whole room temperature?
| No, you are making the fridge do extra work, so more energy is coming in (through the plug) as the pump continues to run since it's not reaching it's cold point. A normal operating fridge does not manifest cold air; it just pumps all the heat out of the inside of the fridge.
The action of pumping the heat out also has heat as a byproduct (which is waste heat from the power coming through the plug as it does work with the pump). So you're really just generating more heat and moving heat around.
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Why does the guy moving on spaceship look younger in twin paradox? If there is no particular absolute choice of frame of reference, the guy who sits on Earth is also moving away from the guy on spaceship perspective and hence time on Earth should also dilate when viewed from the guy on spaceship perspective. But why does the guy moving on spaceship look younger in twin paradox? What am I misunderstanding terribly?
ADDED:: Is time dilation symmetric? If one frame of reference are moving with constant velocity w.r.t other, we have to transformation relations
$$\Delta t = \frac{\Delta t'}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Does each one of them see clock on other tick slowly than their own?
| Here's how I think about it.
First of all, twins A and B have identical clocks consisting of two mirrors, where a photon bounces back and forth between them. That's what you call a "tick" and "toc", and it's a nice way to build a clock because the speed of light is always constant no matter who measures it.
What's more, everything depends on it, like a person's age, which is just a certain number of ticks and tocs.
They are not so far apart or going so fast that they can't watch each other's clock as well as their own.
OK, as far as A is concerned, he's standing still, but B is moving.
Since B is moving, the photon has to travel a longer distance between the mirrors, which takes more time.
So as far as A is concerned, B's clock is running slower, ergo B is aging more slowly.
Now switch to B's frame of reference, and he sees A's clock running more slowly.
So you see, it's all relative :)
Of course, the paradox is resolved when one undergoes acceleration to turn around and come back. That one will actually be younger.
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What do you call the period after sunrise when the sky is bright? At sunrise, the sky isn't actually up in the sky yet. Twilight occurs before sunrise, then at sunrise the leading part of the sun crosses the horizon. But, the sky isn't bright yet. It takes some time for the sky to be blue again.
Then, at the closing of the day, the sky darkens before twilight, then sunset.
Basically, there is a narrower time when the sky is bright and blue, rather than being the time between sunrise and evening twilight. Is there a name for these times, or at least a name for its boundaries?
Edit:
Here's a picture to better point out what I mean:
The shaded part after sunrise I label 1 is the part after sunrise, when the sky is still dark but the sun is out in the sky (here's an image).
The shaded part just before sunset I label 2 is the part when the sky is already darkening, but it's not yet sunset as the sun hasn't set below the horizon (an image again).
The large portion of the day I label 3 would be the part when the sky is blue and bright, the term for which I'm asking for, if it exists (here's an image).
| There is daylight, civil twilight, nautical twilight and astronomical twighlight. Each defined by the position of the sun relative to the horizon.
The wikipedia article seems unusually helpful. With diagrams as well.
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Torque on a Box I think I'm missing something with torques. I seem to have gotten myself confused.
I have a box that's centered at ( 0 , 0 , 0 ) with length ( $x$ dimension ) = 1 , width ( $y$ dimension ) = 0.25, and height ( $z$ dimension ) = 0.5. The edges are parallel to the axes. The $x$ axis is left(-) and right(+), the $y$ axis is up(+) and down(-), and the $z$ axis is into(+) and out of(-) the page.
A force [ 0 , 50 , 0 ] is applied at the point ( 0 , 0 , -0.25 ). To find the torque, we would apply
$\tau = r \times F$
and so $r$ is [ 0 , 0 , -0.25 ] and $F$ is [ 0 , 50 , 0 ]. And the crossproduct is [ 12.5 , 0 , 0 ], so the torque is in the $x$ direction and the box should rotate clockwise?
If I were holding that box in my hand, then ( 0 , 0 , -0.25 ) would be on the side facing me. If I were capable of applying an upward force at ( 0 , 0 , -0.25 ), wouldn't the box start spinning away ( into the page ) from me - and not spin clockwise?
Thanks for taking the time to read. I would really appreciate any help with this.
| Firstly, if the $x$-axis is positive to the right, and the $y$-axis is positive upwards, then the positive $z$-axis should point out of the page in order for your coordinate system to be right-handed (recall that $\hat{\mathbf x}\times\hat{\mathbf y} = \hat{\mathbf z}$ and use the right-hand rule), so let's assume that this is the case (because I think this is what's causing the confusion).
As you point out, the torque is in the positive $x$-direction. This means that the box should rotate in such a way that after a $90^\circ$ rotation, the top is facing you, the side facing you is on the bottom, etc.
This makes physical sense since you're applying a force on the edge opposite the one facing you that is upward. In other words, you're holding the box, you tie a rope to the center of the side opposite the one facing you, and you pull up.
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How to block neutrons What is a good way to block neutrons and what is the mechanism that allows this? It's my understanding that polyethylene is somewhat effective. Why?
| Depending on the velocity of the neutron one can combine different materials to shield neutron radiation.
Diagrams for common materials can be found in The ILL Neutron Data Booklet, 4-2.1 (Download PDF 9 MB)
The strategy is to use
dense material => collisions
hydrogen as moderator => makes neutrons slower
absorbers like B,B10,Cd => captures the neutron (slower ones are easy to catch)
lead and heavy concrete => shields gamma rays too
Hydrogen in polyethylene decreases the velocity of neutrons.
Polyethylene is usually mixed with boron as in Borotron(R).
Boron is a very good neutron absorber. Shutters at neutron beam lines are made from boron carbide with B10.
Cadmium is an efficient neutron absorber too, but it emits intensive gamma rays [Motz, Phys. Rev. B 104, 1353].
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Is the symmetrisation postulate unnecessary according to Landau Lifshitz? The symmetrisation postulate is known for stating that, in nature, particles have either completely symmetric or completely antisymmetric wave functions. According to these postulate, these states are thought to be sufficient do describe all possible systems of identical particles.
However, in Landau Lifshitz Quantum Mechanics, in the first page of Chapter IX - Identity of Particles, he comes to the same conclusion without needing to state any ad-hoc postulate.
It goes like this: Let $\psi(\xi_1,\xi_2)$ be the wave function of the system, $\xi_1$ and $\xi_2$ denoting the three coordinates and spin projection for each particle. As a result of interchanging the two particles, the wave function can change only by an unimportant phase factor:
$$
\psi(\xi_1,\xi_2)=e^{i\alpha}\psi(\xi_2,\xi_1)
$$
By repeating the interchange, we return to the original state, while the function $\psi$ is multiplied by $e^{2i\alpha}$. Hence it follows that $e^{2i\alpha}=1$ or $e^{i\alpha}=\pm1$. Thus
$$
\psi(\xi_1,\xi_2)=\pm\psi(\xi_2,\xi_1)
$$
Thus there are only two possibilities: the wave function is either symmetrical or antisymmetrical.
It goes on by explaining how to generalize this concept to systems with any number of identical particles, etc.
Im summary, no symmetrisation postulate was ever stated in this rationale. Is "shifting by an unimportant phase factor" a too strong requirement for ensuring identity of particles?
| The way Shankar addresses the problem (pg. 278) is by introducing an "Exchange Operator" $P_{1,2}$, which would swap your two particles as follows:
$P_{1,2} |\xi_1, \xi_2 \rangle = |\xi_2, \xi_1 \rangle$
I like the operator notation because it makes it clear (to me, at least) that applying the operator twice is just the identity operator, since swapping two particles twice just gets you back to your original state:
$P_{1,2}^2 |\xi_1, \xi_2 \rangle = |\xi_1, \xi_2 \rangle \longrightarrow P_{1,2}^2 = 1$
This shows that the eigenvalues of the swap are $\pm 1$, meaning your wave function is only either symmetric or anti-symmetric, although there is an implicit assumption that the system in question is in fact an eigenvector of the exchange operator. This is true of particles in the Standard Model in three dimensions, but not generally true (see anyons, for example).
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Can inertia be explained by Bremsstrahlung? Considering that on the atomic level objects consists of densely spaced positively and negatively charged particles, does not the acceleration of those objects lead to Bremsstrahlung of those particles? And although the monopole field is zero, couldn't higher order multipole radiation escape and cause the inertia? I would think at least a small effect like this must happen, even if it doesn't explain all of the inertia.
| No, Bremsstrahlung does not cause inertia.
The power radiated by Bremsstrahlung scales as charge but not mass. Also, it scales as acceleration squared.
This does not reduce to the classical inertia $m\vec a$.
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Why do ice cubes stick together or to the edges of a drinking glass? I was drinking iced-water from a drinking glass (made of glass) at a restaurant yesterday when I was taking a drink, I noticed that there is very little ice water coming out and then suddenly, the ice water mixture comes crashing down. With my open mouth overfilled with ice water, my shirt gets wet – I hate when this happens. After drying myself off, I started thinking why this does happen:
*
*How does ice that starts off loose and separated in water fuse together to form one large clump?
*What causes the ice-water mixture to stop flowing momentarily and then suddenly starting flowing again? Is it that the ice fuses together in the glass and forms one large clump, and then the friction with the edges builds a "dam-like" situation? Or are there individual pieces of ice fusing with the edges of the drinking glass damming-up the ice?
| *
*OK as much I think when two ice cube come together and their sides stick to each other the water on the surface between the two cube starts freezing because on both the sides there is only ice which brings down the temperature of the water between the cube to freezing point which causing the water between then to freeze and the two cubes to join together forming a lump. This property of ice is used in many fields like in ice sculptures making competition in which players use water to join two ice slabs or any other part
*I can't say much about your second question but as much I think the ice cubes stick to the glass wall for a moment and when you again hold the glass the connection between the wall of the glass and the ice cube gets weaker due to the heat of your hand and the cubes then slips and jumps onto your T-Shirt, but I don't know much about it so wait other answers will help me too
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How do we know photons have spin 1? Electrons have spin 1/2, and as they are charged, they also have an associated magnetic moment, which can be measured by an electron beam splitting up in an inhomogeneous magnetic field or through the interaction of the electrons's magnetic moment with an external magnetic field in spectroscopic measurements.
On the other hand, a photon is neutral - how can one measure its spin if there's no magnetic moment? How do we know it has spin 1?
| One method is based on the conservation of angular momentum.
The electronic transition must follow the selection rule $\Delta l=\pm 1$. So the first thing to do is to choose an atom with zero total angular momentum, then let the atom absorb a photon and make a transition to $l=1$ state.
Secondly, we use the Stern-Gerlach experiment to detect the magnetic moment of this atom, which are $m=0,\pm 1$ in our case. Repeating experiments with random photons, we should see that there are mainly three bright spots on the screen: (1) non-deflected, (2) up and (3) down. The distance of outer spot from the center can also be calculated from the theory. Make sure the half-life of the excited state is long enough for the experiment.
In this way you can prove that the photon has spin=1. (That is what I thought, no reference for the actual experiment.)
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Why does electricity need wires to flow? If you drop a really heavy ball the ball's gravitational potential energy will turn into kinetic energy.
If you place the same ball in the pool, the ball will still fall. A lot of kinetic energy will turn into thermal energy because of friction, but the gravitational potential energy will still be converted.
Similarly, why doesn't electricity flow without a good conductor? Why won't Electrons flow from the negative terminal to the positive terminal without a wire attaching them?
Electricity flows like a wave and metals have free electrons in the electron cloud that allows the wave to propagate, or spread. But when these free electrons aren't available to propagate the wave, why don't the electrons just "move" like the ball? Why don't the electrons just "move" through the air to the positive terminal?
A slow drift speed means that the electrons most likely will take a long time to propagate the wave of electricity, but they should still get there.
| When you ask: "why don't the electrons just "move" like the ball? Why don't the electrons just "move" through the air to the positive terminal." I think you need to keep in mind that the ball is made up of neutral atoms which are themselves made up of negatively charged electrons and positively charged protons. The electrons are attracted to the positive terminal, but the protons are repelled equally. Therefore, the ball as a whole does not move. That answers one part of your question. You then might ask why the electrons don't move on their own to the positive terminal? And the answer is that they can if the voltage is high enough to overcome the attraction of the electrons to the protons that they are attached to in the ball. Lightning is a good example of this, as is static discharge.
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How mirror equation can explain farsightedness correction? I have a friend who has just show me his medical prescription for hyperopia (farsightedness) correction and he needs glasses with 4,25 diopters for that, which seemed to be weird for me because I had learned, from the mirror equation, that the maximum correction possible for hyperopia is 4 diopters:
$$
\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}
$$
If we have $0.25m$ for the normal eye distant point and more than $0.25m$ for the farsighted eye distant point (negative sign, because it's a virtual image), then we would have:
$$
\frac{1}{f} = \frac{1}{0.25} + \frac{1}{p'} = 4 - \frac{1}{|p'|} \in\quad ]0,4[, \quad\text{since}\quad |p'| \geq 0.25m \quad\text{and}\quad p'<0
$$
I did some google search and find out that, indeed, hyperopia can reach values even greater, such as 20 diopters, but I can't find pages where doctors explain that with equations or physics teachers explain how things really work in ophthalmology.
Either I am doing some terrible mistake, or doctors are doing some terrible mistake, or this equation just don't apply to hyperopia at all... Which one is true?
| Extreme hyperopia would correspond to your eye lens in relaxed conditions being close to an optical flat. In such a case you would need a contact lens with a focal length of about 25 mm (typical human eyeball diameter). This corresponds to a lens of 1000/25 = 40 diopters.
In other words: a farsighted eyeless requiring 20 diopter correction has a focal length at relaxed conditions of about twice the required (25 mm) focal length.
A simple (approximate) means to estimate the diopters of correction needed, is to use:
$$\frac{1}{f_e} + \frac{1}{f_c} = \frac{1}{D}$$
here $1/f_c$ is the optical correction (in diopters when the corresponding focal length $f_c$ is measured in meters), $f_e$ the focal length of the relaxed eye lens, and $D$ the inner diameter of the eyeball. Note that $1/f_c$ is positive when correcting for farsightedness ($D < f_e$), and negative when correcting for nearsightedness ($D > f_e$).
| {
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Why is there a factor of $4\pi$ in certain force equations? I mean to ask why there is $4\pi$ present in force equations governing electricity? Though all objects in universe are not spherical and circular, the constant of proportionality in both equations contain $4\pi$. Why?
| Some people (myself included) would regard field equations like Gauss' Law as more "fundamental" than force equations. The most obvious reason for this is that Coulomb's Force Law only works when the charges in question are held static - it has to be modified once they are allowed to move.
As the others have stated, the $4 \pi$ has to do with the surface area of a sphere. Gauss' Law (I'll use the integral form to make the surface area connection more apparent) for the electric field tells us:
$$\int \mathbf{E} \cdot \mathrm{d} \mathbf{A}= \frac{q}{\epsilon_0}$$
In words, this means that if you enclose some charges in an imaginary surface, then the sum of the electric field sticking out of that surface multiplied by the area of the surface is equal to the total charge enclosed by it, multiplied by some constant factor. If you take a single point charge and enclose it with a spherical surface of radius $r$, then it reduces to:
$$4 \pi r^2 E= \frac{q}{\epsilon_0}$$
Do some rearranging and it's easy enough to see how it relates to Coulomb's Law.
There's also a form of Gauss' Law for (Newtonian) gravity:
$$\int \mathbf{g} \cdot \mathrm{d} \mathbf{A}= 4 \pi GM$$
This field equation actually contains the factor $4 \pi$ already, so when you enclose a mass with a spherical surface the factor cancels on both sides. This is simply because when Newton wrote down his force law for gravity he didn't know about things like Gauss' Law, and so neglected to include the $4 \pi$ in the force equation. And since then the convention has stuck around, so we're left with a slightly confusing hodgepodge of some field equations needing factors of $\pi$ and some not. In general, if you see a factor of $\pi$ in a field equation then you're probably looking at something gravity-related.
| {
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Degrees of freedom of the graviton versus classical degrees of freedom I have a puzzle I can not even understand.
A graviton is generally understood in $D$ dimensions as a field with some independent components or degrees of freedom (DOF), from a traceless symmetric tensor minus constraints, we get:
*
*A massless graviton has $D(D-3)/2$ d.o.f. in $D$-dimensional spacetime.
*A massive graviton has $D(D-1)/2-1$ d.o.f. in $D$-dimensional spacetime.
Issue: In classical gravity, given by General Relativity, we have a metric (a symmetric tensor) and the Einstein Field Equations(EFE) provide its dynamics. The metric has 10 independent components, and EFE provide 10 equations. Bianchi identities reduce the number of independent components by 4. Hence, we have 6 independent components.
However, for $D=4$, we get
*
*2 independent components.
*5 independent components.
Is the mismatch between "independent" components of gravitational degrees of freedom (graviton components) one of the reasons why General Relativity can not be understood as a quantum theory for the graviton?
Of course, a massive graviton is a different thing that GR but even a naive counting of graviton d.o.f. is not compatible with GR and it should, should't it? At least from the perturbative approach. Where did I make the mistake?
| You can think of diff as bianchi id. The additional 4 dof is killed by the fact that 4 of the 4 of the EFE are constraints.
| {
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What are correlated magnetic moments? My book has the following sentence and I don't understand what correlation or lack of correlation means:
At high temperature the magnetic moments of adjacent
atoms are uncorrelated (to maximize the entropy) so the crystal has no net magnetic moment.
The book is touching on second-order phase transitions and it's describing how magnetic transitions are an example of such 2nd order phase transitions.
| Correlation between two variables (or objects) is, very simply put, how much a change in one variable affects or determines a change in the other. Replacing variables with spins, highly correlated spins would mean that, due to some interactions between them, a change in the direction of one spin will cause a change in the direction of the spin it is correlated with. Thus a high degree of correlation would imply that the behaviour of each spin is governed by its neighbours and overall, they tend to cooperate and point in specific directions rather than randomly, giving long-range ordering.
At high temperatures, the energy scale of this correlation or 'talking' between spins is much weaker than the thermal energy scale. So the directions of the spins are thermally fluctuating and point randomly rather than determine the position of the neighbouring spin. So there is no more long range ordering and the system shows no net magnetic moment.
User17338, hope this helps. Others, please correct me if I am wrong. :)
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How far can you scatter light using a prism? If I were to scatter light how far do you think it would disperse? What prism most effectively scatters light?
| Prisms don't scatter light, they refract it. The amount of refraction depends on the material's index of refraction, or the change in the speed of light in that substance (assuming that the prism is surrounded by normal air). The highest index of refraction, as far as I am aware, belongs to diamond, so a diamond prism would cause the sharpest bending of light (unless there's a higher IoR out there that I don't know about).
| {
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Weak interaction and the Chirality of anti-particles Consider a weak current of the form
$
J^{\mu} = \bar{u}_{\nu}\gamma^{\mu}(1-\gamma^5)u_{e}
$
This describes the part of a weak process where a left-handed electron converts into a left-handed neutrino by emitting/absorbing a W boson. Equivalently, it should also describe the same process for a right-handed positron going to a right-handed anti-neutrino. How do you get this second part from the form of $J^{\mu}$, considering that $P_L = 1-\gamma^5$ is by definition the left handed projector? Whatever antiparticle states contained in $u$ and $\bar{u}$ should have eigenvalue $-1$ of $\gamma^5$ in order to be included in $J^{\mu}$, so, aren't they by definition left-handed?
(note: this is all in the massless approximation so that I can equate chirality and helicity/handedness)
| Yes, antiparticle is still involved in left-handed chirality. The right-handed antiparticle that some books mention is actually helicity. I know many textbooks are WRONG by claiming that antiparticle is right-handed chirality. You can read the book Quantum Field Theory for Gifted to clear this common misconception
*** The weak interaction doesn't care about helicity, it just cares about chirality, and left-hand chirality is the one that involved weak interaction (both for particle and antiparticles)
| {
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Does alternating current (AC) require a complete circuit? This popular question about "whether an AC circuit with one end grounded to Earth and the other end grounded to Mars would work (ignoring resistance/inductance of the wire)" was recently asked on the Electronics SE.
(Picture edited from the one in the above link)
Though I respect the AC/DC experts there, I think (with the exception of the top answer) they are all wrong.
My issue is that they all assume that AC requires a complete circuit in order to function. However, my understanding is that a complete circuit is necessary for DC, but not AC. My intuitive understanding is that AC is similar to two gas-filled rooms with a pump between them - the pump couldn't indefinitely pump gas from one room to another without a complete circuit (DC), but it could pump the gas back and forth indefinitely (AC). In the latter case, not having a complete circuit just offers more resistance to the pump (with smaller rooms causing a larger resistance).
Is my understanding correct - can AC circuits really function without a complete loop?
More importantly, what are the equations that govern this?
If larger isolated conductors really offer less AC-resistance than smaller AC conductors, how is this resistance computed/quantified? Would its "cause" be considered inductance, or something else?
| What do you mean by "to work"? If you mean that you can transfer energy, then to where?
If you continuously alternate the potential at one end of a wire, then this creates a wave that propagates to the other end. Say that other end is earth. If the potential is 0 on earth like and ideal ground, this means it cannot sustain a wave passing through. It is like a mirror. The oscillation will reflect back and you obtain a stationnary beating of counter-propagating waves. Just like agitating a string with the other end attached. This does not convey energy. (You can apply the same reasonning on the Mars side).
| {
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Study of Black-body Radiation Why did scientists study black body radiations from something as complicated as a hollow container rather than the radiation from something simple like a thin solid cylinder?
| A black body is a perfect absorber, and in practice it's difficult to make a material that is a perfect absorber. So to make a black body we choose a material that is as good an absorber as we can find, and form it into a hollow sphere with a small hole in it. The black body is then the hole, not the sphere. It's a nearly perfect absorber because any radiation falling on it is multiply reflected within the sphere. Because the material of the sphere is a good absorber and absorbs most of the radiation at each reflection, the chance that any radiation falling on the opening reflects out again is vanishingly small.
I've never heard of a hollow cylinder being used to model a black body, but the hole at the end of the cylinder would be a black body in a similar way to the hole in the sphere.
| {
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Weather forecasting with coffee bubbles The other day I saw this life-hack:
And I was wondering how true it is. First of all, I always thought(listening to weather forecasts) that low-pressure atmosphere is what correlates with rain; although I never learned the argument behind it.
Anyway, even if low atmospheric pressure correlates with storms and rain; how much(and in what sense) does this affect the bubbles in my morning coffee(if any)? My initial guess is that there should be a threshold pressure that will differentiate whether the bubbles will stick to an edge or suspend in the middle. Since this is an everyday-life question, I should say that experimental answers are welcome as well as rigorous theoretical ones.
Also, I am wondering if we can make a barometer based on the bubbles in a mug of coffee? If so, how sensitive it would be?
| This is quite humorous.
In an 1883 offical US military publication, "Weather Proverbs" by 1st Lt. Dunwoody, at page 107 it is stated "When coffee bubbles collect in the centre of the cup expect fair weather. When they adhere to the cup, forming a ring, expect rain." This is the opposite of the lifehack proverb!
In 1997 Dave Thurlow, using a grant from the US National Science Foundation, insisted that the "bubbles in the middle mean sunny" version was supported by theory. http://www.weathernotebook.org/transcripts/1997/01/31.html
However, he bases this on the coffee/air interface switching from concave up to convex with change in atmospheric pressure, which is extremely doubtful. Instead, the material of the cup wall and its interaction with the coffee would determine concavity.
The Young-Laplace equation describes the air/liquid interface.
The fact that the proverb and its opposite are both professed should be a strong indication that there is no truth in the proverb, absent some logical reason based upon the Young-Laplace equation.
| {
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When is temperature not a measure of the average kinetic energy of the particles in a substance? I had always thought that temperature of a substance was a measure of the average kinetic energy of the particles in that substance:
$E_k = (3/2) k_bT $
where $E_k$ is the average kinetic energy of a molecule, $k_b$ Boltzmann's constant, and $T$ the temperature. (I'm not sure of the 3/2 coefficient.) Then I heard from several folks that this is a simplistic notion, not strictly true, but they didn't explained what they thought was flawed with this idea. I'd like to know what (if anything) is objectionable about this idea? Is it that the system must be macroscopically at rest? Is it that it ignores the quantum mechanically required motion of particles that persists at low temperatures? When is it not valid?
| The expression your wrote down for the energy is the expression for the ensemble average kinetic energy of a monatomic ideal gas. Therefore, we see that for this system, the average energy of the system is simply proportional to the temperature.
For a general statistical mechanical system, however, it might not even make sense to talk about the "average kinetic energy" of the system. For example, take a quantum system consisting of a single spin $1/2$ particle interacting with a magnetic field. It is possible to define a temperature for this system when it is in contact with a heat bath even if the particle is not moving around. In such cases, one appeals to more general definitions of temperature such as that to which John Rennie refers in his answer.
| {
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Unpolarized light vs. randomly rotating polarized light? I am confused with physical picture about unpolarized light.
Is unpolarized light very fast rotating polarized light? or co-existing state of two orthogonal polarization? (or something else?)
If there is a linear polarizer which rotates very very fast and randomly (the polarizer in imagine), the output light is same to unpolarized light? I don't think so but I am not sure.
--
or, instead of linear polarizer, a Faraday rotator with magnetic field whose amplitude is randomly chnaged can be considered, I think.
| The picture you have about unpolarized light is correct, I think, but I would try to avoid the idea of "rotating fast", because it gives an idea of continuity, that I think is what you try to avoid in the concept of unpolarized light.
So, in essence unpolarized light is modelled by short wave trains of some arbitrary pure polarization; this is because if you interfere this light with itself, the interference pattern will blur at some point, that correspond to the average length of these trains.
I never thought about the idea of getting unpolarized light from purely polarized light, but, I think what you propose could work in theory. Now, if you see a real Faraday rotator, I don't think it can do the job.
| {
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Intuitive understanding of the irreps like Wigner-$D$ matrix? Wikipedia defines Wigner $D$-matrix as an irreducible representation of groups $SU(2)$ and $SO(3)$. What is a good way to visualize this representation? Is there any physical system which can be kept in mind as a simple example of the same?
A general explanation of the idea of irreps, beyond just the Wigner-$D$ matrix, would be appreciated.
|
Is there any physical system which can be kept in mind as a simple example of the same?
Yes. Consider a single spin $1/2$ particle, like an electron. In this case, the matrix will be $2$-by-$2$ since its a representation of $\mathrm{SU}(2)$ acting on the two-dimensional spin-$1/2$ Hilbert space. The idea here is that when you rotate the physical system by a rotation $R$ say, then the spin state "rotates" as well (the states in the Hilbert space "rotate" into each other if you will) as follows:
\begin{align}
|\tfrac{1}{2},m'\rangle\longrightarrow D^{1/2}_{m,m'}|\tfrac{1}{2},m'\rangle
\end{align}
In fact, for a rotation by an angle $\theta$ about the unit vector $\mathbf n = (n_x, n_y, n_z)$, we have
\begin{align}
(D^{1/2}_{m,m'})=\begin{pmatrix}
\cos\frac{\theta}{2}-in_z\sin\frac{\theta}{2} & (-n_y-in_x)\sin\frac{\theta}{2} \\
(n_y-in_x)\sin\frac{\theta}{2} & \cos\frac{\theta}{2}+in_z\sin\frac{\theta}{2} \\
\end{pmatrix}
\end{align}
so that when $\theta = 0$, this is the identity matrix; nothing happens to the state, while when $\theta = 2\pi$ (a full rotation), this matrix is $-1$ times the identity, and the state therefore rotates into itself multiplied by $-1$; pretty strange isn't it?
A general explanation of the idea of irreps, beyond just the Wigner-D matrix, would be appreciated.
This is an extremely broad question. The general idea behind a representation of a group $G$ is that it is a mapping that assigns an invertible matrix $D(g)$ to each group element $g$ such that the group structure is preserved (the technical term for this is that it is a group homomorphism). The representation is said to be irreducible if it has no nontrivial invariant subspaces. Concretely, this means that there is no similarity transformation that puts all of the representation matrices into block diagonal form.
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Strings and their masses How do strings present in particles give mass to them? Is it only by vibrating? I have been trying to find the answer but could not find it anywhere, can this question be answered?
| While it is true that an excited string (hence one with a vibration mode above the ground state) looks like a massive particle from far away, this is not the effect that is supposed to explain the mass of any particle ever seen. This is because the mass of the first excited mode of the string is already huge as far as particle masses go. So in string phenomenology, instead, all particles are modeled by strings in their massless ground state excitation and the actual observed masses are induced, as it should be, by a Higgs effect.
While the excited string states are not supposed to show up at energy scales anywhere close to what is being observed, their presence is still crucial: it is all these heavy particle excitations whose appearance as "virtual particles" in scattering amplitudes serve to make string scattering amplitudes be loop-wise finite, hence renormalized.
See the nLab String Theory FAQ the entry How do strings model massive particles?.
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How is momentum conserved when a magnet attracts a metal? Suppose your have any magnetic object and no external force acts upon it, and the object comes near a metal which causes an impulse (think that will happen). However, the magnetic force is internal to the object, and momentum should be conserved, so where have I gotten it wrong?
| There is a corresponding reaction force on the metal object, which is attracted to the magnet. If the metal object is large (say, a fridge) then this is hard to notice and you observe the magnet going to the metal, but for smaller objects like paper clips it's the opposite.
| {
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Where does gravity get its energy from? I would like to know where gravity gets its energy to attract physical bodies?
I know that the law of conservation states that total energy of an isolated system cannot change. So gravity has to be getting its energy from somewhere, or else things like hydropower plants wouldn't be able to turn the power of the falling water into a spinning rotor.
Just to be clear, Lets create an example:
Lets say we have two objects with equal mass close to each other. So gravity does its job and it pulls each other closer, this gets turned into kinetic energy. This is where I'm lost. According to the law of conservation energy can't be created or destroyed and the kinetic energy comes from the gravitational pull so where does the gravitational pull gets its energy.
If that energy isn't being recycled from some where else then that means you have just created energy, therefore breaking the law of conservation.
| As per Newtonts law each thing have energy to attract another thing, beacause in every thing there is potential energy. Due to friction that potential enrgy can convert in another form. for eg. our sirt colour get more dirty, because of friction between neck and colour, in anciant period for generating th fire people used to rubs to white stones on each other.
inside of earth there is so much potential energy,and because of heat or pressure or because of any another reason friction come to exists and due to friction earths potential energy converts in gravity (earth get more energy of attraction than another things which are exists on earh)
| {
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Connection between particles and fields and spinor representation of the Poincare group Let's have a definition of massive particle as an irreucible representation of the Poincare group. Then, let's have a spinor field $\psi_{\alpha \alpha_{1}...\alpha_{n - 1}\dot {\beta} \dot {\beta}_{1}...\dot {\beta}_{m - 1}}$, which is equal to $\left( \frac{m}{2}, \frac{n}{2}\right)$ representation of the Lorentz group. There is the hard provable theorem:
$\psi_{\alpha \alpha_{1}...\alpha_{n - 1}\dot {\beta} \dot {\beta}_{1}...\dot {\beta}_{m - 1}}$ realizes irreducible representation of the Poincare group, if
$$
(\partial^{2} - m^{2})\psi_{\alpha \alpha_{1}...\alpha_{n - 1}\dot {\beta} \dot {\beta}_{1}...\dot {\beta}_{m - 1}} = 0,
$$
$$
\partial^{\alpha \dot {\beta}}\psi_{\alpha \alpha_{1}...\alpha_{n - 1}\dot {\beta} \dot {\beta}_{1}...\dot {\beta}_{m - 1}} = 0.
$$
Can this theorem be interpreted as connection between fields and particles?
| The definition is that a particle in Minkowski space is a unitary irreducible representation of the Poincare group. So one needs to see how various P.D.E.s are related to the classification of unitary irreducible representations of $iso(3,1)$ or $iso(d-1,1)$ in the case of $d$-dimensions instead of $4$.
Note that these are all the Poincare-invariant constraints that can be imposed on the given field without trivializing the solution space (one could imposed $\partial \psi=0$ (gradient), which is Poincare-invariant but too strong as the field must be a constant).
The theorem is not hard to prove. One has to know how to construct irreducible representations of the Poincare group, see chapter 2 of the Weinberg's QFT textbook. Then one solves the equations by standard Fourier transform and shows that the solution space indeed equivalent to what is called a spin-$m$ particle in Minkowski space.
There is nothing special about $4d$ in defining spin-$m$ field, so it is simpler to look at arbitrary dimension, where, say for bosons the above equations are equivalent to
$(\square-m^2)\phi_{\mu_1...\mu_m}=0$
$\partial_\nu \phi^{\nu \mu_2...\mu_m}=0 $
$ \eta_{\nu\rho} \phi^{\nu\rho \mu_3...\mu_m}=0$
$\phi^{\mu_1...\mu_s}$ is totally symmetric in all indices.
In $4d$ one can use $so(3,1)\sim sl(2,C)$ and the last algebraic constraint then trivializes - an irreducible spin-tensor is equivalent to an irreducible $so(3,1)$-tensor
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Frequency of small oscillation of particle under gravity constrained to move in curve $y=ax^4$ How to find the frequency of small oscillation of a particle under gravity that moves along curve $y = a x^4$ where $y$ is vertical height and $(a>0)$ is constant?
I tried comparing $V(x) = \frac 1 2 V''(0) x^2 + \mathscr O(x^3) = \frac 1 2 kx^2$ (assuming $V(0)$ is ground state and $V'(0)$(slope) remains is horizontal at extrememum, but unfortunately $V''(0) = 0$. I am pretty much clueless. Thanks for your help.
ADDED:: Letting $m=1$ the Lagrangian is $L = \frac 1 2 (\dot x^2 + \dot y^2) - gax^4 = \frac 1 2 (\dot x^2 + (4 ax^3 \dot x)^2)-gax^4 = \frac 1 2 \dot x^2(1 + 16a^2x^6)-gax^4$
The above Lagrangian gives the equation of motion as
$$\ddot x(1+16a^2x^6)+\dot x^2 96 a^2x^5 + 4 gax^3 = 0$$
Since we are considering the system a small oscillation whose potential is of order 4, $\mathscr{O}(x^{k>4})$ can be ignored which reduces into $\ddot x = -4agx^3$.
To solve this,
$$\frac 1 2 \frac{d}{dt}(\dot x^2) = -\frac{d}{dt}(agx^4)$$
which gives $\dot x = \sqrt{k - 2agx^4}$, Assuming the system begins from $t=0$ at $x=x_0$ with $\dot x = 0$, $k = 2agx_0$, which turns the integral into
$$\int_0^{T/4} dt= \int_0^{x_0} \frac 1 {\sqrt{2agx_0^4-2agx^4}} dx=\frac{1}{\sqrt{2ag}x_0}\int_0^1\frac{1}{\sqrt{1-y^4}}dy$$
Which gives $$T = 2 \sqrt 2 \sqrt{\frac{\pi}{ag}}\cdot \frac{\Gamma(5/4)}{\Gamma(3/4)}\cdot \frac 1 {x_0}$$
which is a dubious result. Please someone verify it. Any other methods are welcome.
| The motion in such a curve is quite hard to calculate, and even more so if you do not want to get into the messy details of Jacobi elliptic functions like $\text{sn}(u|k)$. However, for the case of small oscillations there is a simple scaling argument that lets you calculate the dependence of the period on the amplitude.
Consider, then, the case of small oscillations, which you correctly calculate by neglecting the effect of the vertical kinetic term $\tfrac12 m\dot y^2$, and which gives the differential equation
$$\ddot x=-4 {ag} x^3.\tag{1}$$
A few words on this equation. For one, it can be explicitly solved in terms of the Jacobi elliptic function $\text{sn}$, which is actually quite harmless. It behaves quite differently to a harmonic oscillator, as can be seen from the phase-space portraits in the Wolfram Alpha link - they would be ellipses in the harmonic case, and are much more 'square' here. Most importantly, because it is anharmonic, its solutions will not be isoperiodic, meaning that the period will depend on the amplitude $x_0$.
To find this dependence, you simply need to re-scale time and position: change $x$ for $\lambda x$ and $t$ for $\mu t$, and your equation will transform into
$$\frac{\lambda}{\mu^2}\ddot x=-4\lambda^3 {ag}x^3,$$
and this has the exact same form as (1) if you set $\mu=1/\lambda$. What this means is that if you have some solution with $x(0)=x_0$ (and $\dot x(0)=0$ for simplicity) with period $T_0$, then you can scale-transform it to match the solution to any other initial condition $x(0)=x_1$ (and $\dot x(0)=0$, of course), with scaling $\lambda=x_1/x_0$, with the caveat that you must also transform the time to $\mu t=\frac{x_0}{x_1}t$, which in the bottom line means the new period is $T_1=\mu T_0=\frac{x_0}{x_1}T_0$. Slightly more elegantly, it means $$\text{the product }x_0T_0\text{ is constant for all solutions.}$$
Get the solution for one $x_0$, and you have the numerical constants, but most importantly, you know the crucial thing, which is the scaling:
$$\boxed{T\propto {1}/{x_0}.}$$
Added:
To elaborate on Michael Brown's comment, you can indeed combine this with some dimensional analysis to get the period's dependence on $a$ and $g$. The period $T$ can only depend on $a,g$ and $x_0$, which means that it must be a function of the characteristic time $\sqrt{x_0/g}$ and the dimensionless quantity $ax_0^3$. Coupled with the scaling above, you are forced into the relation
$$T=\frac{C}{\sqrt{ag}x_0},$$
where $C$ is now a dimensionless quantity, probably of order 1.
| {
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"url": "https://physics.stackexchange.com/questions/75411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does boiling water in the microwave make a cup of tea go weird? When I boil water in the kettle, it makes a nice cup of tea. Sometimes I need to use a microwave because a kettle isn't available. I boil the water in the mug and it looks pretty normal, but when I drop in the teabag the water froths up and looks foamy. I don't see what the chemical difference is here, so I assume it must be some physical difference. I have noticed this with multiple types of tea and multiple microwaves, the results being consistent so it's not just a weird microwave or something like that.
What is the reaction here and how/why does it occur?
Here is a photo of the 'fizzy' looking tea just after dunking in the teabag.
| I doubt that it's superheating, as I understand superheating is a rather violent phenomenon. The most likely explanation is dissolved gases; gas solubility in water decreases with temperature. The kettle boiling process is very turbulent and so can release all the dissolved gases, resulting in pure hot water. The microwaved water is heated in a very gentle way, and so the gases do not yet have a chance to escape from the surface... not until they can start nucleating on your tea bag.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
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Why can't we destroy energy? From a wikipedia article:
In physics, the law of conservation of energy states that the total energy of an isolated system cannot change—it is said to be conserved over time. Energy can be neither created nor destroyed, but can change form; for instance, chemical energy can be converted to kinetic energy.
What is the reason behind this Law: is there a proof that we can't destroy energy? I mean if we couldn't destroy energy in our universe maybe we can do it in other universes the point is why is it a law and not a theory
| The point of defining energy is to obtain a conserved scalar quantity -- as an example of how this works out, have a look at the derivation of the Newtonian formula for kinetic energy, where energy conservation is disguised as "kinetic energy is equal to the heating produced when it's destroyed".
There are plenty of conserved scalar quantities of course -- charge, each component of momentum, whatever -- so the real definition of energy is as the conserved quantity mandated by time-translational invariance through Noether's theorem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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View from a helicopter rotor: why is the horizon distorted? This video ("rotor panorama") was captured by a camera attached to the rotor head of a radio-controlled helicopter, with the frame rate set to the rotor's frequency. During a long segment of the video, the horizon looks distorted:
What causes this distortion?
| This is most likely caused by wobble in the motion of the rotor, compounded with the fact that pixels are captured linearly, probably from left to right (this is what the description refers to as "scanning shutter"). Thus there is a time delay between the capture of the pixels on the left and those on the right; if the axis is not horizontal then this would lead to a vertical displacement between them.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What does E/M field look like when I close a circuit? Suppose that we have a charged capacitor with two pins: $ C_+ $ and $C_-$.
Suppose that we have a long wire with fixed geometry, that is already connected to the pin C+.
Let the distance along the wire be called $s$.
I closed a circuit by connecting another end of the wire to the pin $C_-$ in the moment $t=0$.
What is the general way to find electrical current in wire as a function of $s$, $t$ and geometry ?
Is there a difference between if the first end of wire was connected to the $C_+$ pin, or unconnected at all?
| Recall $$I = n q v_d A \tag{1}$$
Note that the electrical current depends only on the cross sectional area of the wire and not on its length...
However if we manipulate eq.1 we find that
$$I = \frac{nqsA}{t}$$
Since $v_d = s/t$ if we continue further we see that
$$I = \frac{nqV}{t }$$
Where $V$ is the volume of the Wire.
I don't get your second question, please be more clear.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the difference between phase difference and path difference? The path difference is the difference between the distances travelled by two waves meeting at a point. Given the path difference, how does one calculate the phase difference?
| The optical path difference is the length difference $d$ (dimensions of length: $[L]$) in the paths travelled by two different rays from one plane, at which they are typically assumed to have the same phase, to a second location (typically another plane, surface or point).
Light is a wave with a wavelength of $\lambda$, or corresponding angular spatial frequency of $k = 2\pi/\lambda$ (dimensions of inverse length: $[L^{-1}]$). As a wave propagates through space it accumulates phase at a rate of $k$.
Thus to convert optical path difference to optical phase difference we multiply by $k$:
$$
\phi = kd.
$$
Note that when $d=\lambda$ the phase difference is $\phi=2\pi$ as expected.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Why do meteors explode? A report on the Chelyabinsk meteor event earlier this year states
Russian meteor blast injures at least 1,000 people, authorities say
My question is
*
*Why do meteors explode?
*Do all meteors explode?
| Remember that meteoroids are celestial objects. At these scales, it is quite common for relative velocities to be extremely large.
When such a fast-moving object enters the atmosphere of the Earth, the air slows it down due to its viscosity. However, the meteor is moving very, very fast. Unlike most falling objects, it doesn't get a chance to even reach terminal velocity.
With the meteor moving at this speed, there is a lot of heat generated (compression, as well as friction/viscosity). This boils the meteor (many meteors contain ice or dry ice), leading to a buildup in pressure. If the boiling happens too quickly, the meteor will explode.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/76045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Double Slit Experiment: How do scientists ensure that there's only one photon? Many documentaries regarding the double slit experiment state that they only send a single photon through the slit. How is that achieved and can it really be ensured that it is a single photon?
| Quantum dots. nanoscale semiconductor materials that can confine photons in 3 dimensions and release them a measurable time after. Based on material used the decay time is known empirically. frequency is also known. the latter is sufficient to calculate the energy of one photon. The former is then sufficient to calculate the rate of photon re emission from the QD. If the peaks at the detector are further apart than the decay time and each peak is measurable to one photon's worth of energy then you know you have a beam of single photons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/76162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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CMBR temperature over time? How has CMBR temperature dropped as function of time? A graph would be nice, but I'd be happy with times (age of universe) when it cooled enough to not be visible to human eye, became room temperature equivalent, or reached some interesting temperatures regarding matter in the universe.
If there is a nice formula giving the temperature as function of time, that would be great too.
| In terms of the redshift, the background temperature is
$$
T(z) = T_0\left(1+z\right)
$$
where $T_0\sim2.725$ K is today's CMB temperature. For simplicity, one can invoke a uniformly-expanding universe to get the relation between $z$ and $t$ as
$$
1+z\propto\frac{1}{t^{2/3}}
$$
So as $t\to0$, $T(t)\to\infty$ and as $t\infty$, we see $T(t)\to0$.
This $1/t^{2/3}$ relationship can be plotted by Wolfram Alpha.
Note that the relationship between $z$ and $t$ is a bit more complex when considering a non-uniformly expanding universe, but the $T$ & $z$ relationship should still be valid.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Superconformal approach to supergravity In the book (Supergravity - Daniel Z.Freedman & Antoine Van Proeyen - Cambridge), there is (Chapters 16-17) a presentation of pure supergravity or supergravity with matter, from a superconformal approach.
The "simplest" link, is to begin with a superconformal gauge multiplet coupled to a chiral multiplet, then gauge fix which will break the scale and special conformal transformation symmetries, and finally get a pure supergravity (in the same dimensional space-time). Here one speaks about $\mathcal N=1$ supersymmetry in a $D=3+1$ space-time.
I have some questions about this approach.
*
*Is it only a mathematical approach, or it is also a physical approach, that is, is it possible to associate some physical quantities of the $2$ theories in some way?
*Thinking about $\mathrm{AdS}_4$/$\mathrm{CFT}_3$, there is some regime, where supergravity is trustable. In this regime, with the above approach, we have $2$ sides of a triangle, so it may be tempting to look at the 3rd side of the triangle, that is a link between a superconformal theory in $3+1$ dimensions, with a superconformal theory in $2+1$ dimensions, or maybe a step further, that is looking at the superconformal theory in $3+1$ as a "mother" theory, as a united point of view of $\mathrm{AdS}_4$/$\mathrm{CFT}_3$, at least in the supergravity regime. Does all this makes sense ?
| *
*The "simplest" link, is NOT to begin with a Superconformal gauge multiplet coupled to a chiral multiplet, but to couple the Weyl multiplet to a superconformal chiral multiplet.
*It is just a mathematical tool to make your life easier. As a matter of fact, you can take the superconformal action and make a field redefinition to get the Poincare action, thus the superconformal symmetry is nothing but a redundancy to be removed. You can see an example of that for a five dimensional $\mathcal{N}=2$ theory in https://arxiv.org/abs/1107.2825 see equation (3.1).
| {
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"timestamp": "2023-03-29T00:00:00",
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What does imaginary number maps to physically? I am taking undergraduate quantum mechanics currently, and the concept of an imaginary number had always troubled me. I always feel that complex numbers are more of a mathematical convenience, but apparently this is not true, it has occurred in way too many of my classes, Circuits, Control Theory and now Quantum Mechanics, and it seems that I always understand the math, but fail to grasp the concept in terms of its physical mapping. Hence my question, what does imaginary number maps to physically?
Any help would be much appreciated
| Both sides of a complex number are each real numbers. Either the real part or the imaginary part can be used for computing the value of a measurable quantity. And the results are always the same. Then we are left with only 'i', to determine it's Mathematical and then physical meaning. The allegory of 'i' harbors a real Mathematical meaning which when unveiled, discloses the 'reality' of a 'complex' number. In other words, 'complex numbers' are 'real numbers' in disguise.
One can say the above in another way by saying that (-1)^1/2 is completely a real number.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/76595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Intuition behind defining divergence as flux divided by volume? For a continuously differentiable vector field $F$ the divergence theorem can be used to give
$$(\nabla\cdot F)(a) = \lim_{r\to 0} \frac{3}{4\pi r^3}\int_{|x-a|=r} F \cdot n dA$$
This should mean that for $c<3 $
$$\lim_{r\to 0} \frac{3}{4\pi r^c}\int_{|x-a|=r} F \cdot n dA=0$$
In particular for $c=2$ that
$$\lim_{r\to 0} \frac{3}{4\pi r^2}\int_{|x-a|=r} F \cdot n dA=0$$
Although I understand mathematically why divergence is associated with the first equation given above, I don't have a good grasp about why intuitively it is correct to divide by $r^3$ instead of $r^2$. I would have thought that since the flux is a surface integral we should divide by $r^2$ to account for the changing size of the surface area. Can anyone give an intuitive explanation for what I am missing?
| The basic reason for this is that we want the limit to be finite. As $r$ approaches zero, there are two reasons why the flux shrinks.
One is that the flux is proportional to the surface area, which goes like $r^2$. The other is that flux tends to cancel on opposite sides of the volume. If the field is constant, this cancellation is always exact. The change in the field from one side of the volume to the other is proportional to $r$.
Putting these two factors together, the flux shrinks in proportion to $r^3$. To make the limit finite, we need to divide by $r^3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What are the equations of motion of a hole in a soap bubble? Imagine the following situation: I have a thin stationary water film, like a soap bubble, suspended inside a large ring. I throw a small loop of string onto the film and punch a hole inside it. How can I describe the motion of the hole in the water film bounded by the loop of string? The surface tension of the surrounding film will tend to minimize the ratio between the length of the boundary and the circumference, so the hole will disk shaped. Furthermore, if the weight of the string that bounds the hole is smaller than the weight of a water film with the surface given by the inside of two, then the effective mass of the hole will be negative, i.e. if the film is subject to a gravitational field, then the hole will tend to move upwards.
What is the correct way to describe such a system? How can I derive its equations of motion?
| Let the large ring be placed vertically. For simplicity, consider a weightless string, which thickness is equal to the film thickness. Also, assume that the hole is quite far away from the edge of the ring so that we can ignore the surface phenomena of the film. Then the speed of the rising hole can be estimated as follows:
The buoyant force acting on the hole:
$$F_b=\rho gV=\rho g\pi R^2h$$ where $\rho$ is density of water, $g$ is acceleration of gravity, $R$ is radius of the hole and $h$ is the thickness of the film.
Next, we need a formula for the drag on the moving hole . We can use a formula for an infinite cylinder, which moves slowly in a fluid, perpendicular to its axis:
$$F_d=\frac{4\pi\eta v}{\ln\frac{3.7\nu}{Rv}}$$ where $\eta$ is dynamic viscosity and $\nu$ is kinematic viscosity of water.
This is the drag per unit length of the cylinder. The derivation of the formula is given for example in: H.Lamb, Hydrodynamics.
Now, equating the buoyant force and the drag force we get an equation for the rising speed $v$ of the hole:
$$gR^2=\frac{4\nu v}{\ln\frac{3.7\nu}{Rv}}$$ Here we used the formula $\nu=\frac{\eta}{\rho}$. This is a transcendental equation.
To get some estimation we use $\nu=0.01\frac{cm^2}{s}$ at $20^\circ C$ and $g=1000\frac{cm}{s^2}$. Let's introduce a new variable $x=\frac{3.7\nu}{Rv}$. Then the equation can be written as follows:
$$x\ln x=\frac{14.8\nu^2}{gR^3}$$ Now we have seen that $x\approx 1$ holds for all real values for the radius of the hole $R$. That means we get a simple estimation for $v$:
$$v=\frac{3.7\nu}{R}$$ For example a hole with radius $R=1cm$ moves up with speed $v=0.037\frac{cm}{s}\approx 0.4\frac{mm}{s}$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/76861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Quantum states as rays as opposed to vectors I recently read that a quantum state is actually defined by a ray and not a vector. That is it is possible to multiply a state $\psi$ by any complex number $c\in \mathbb{C}$ and you won't be changing the physics in any way. I understand this mathematically, but I don't understand what the physical meaning of such an "equivalent state" would be since the new state need not be normalised if $c$ is not of the form $e^{i\phi}$.
| There is no particularly interesting new physical significance to such a state vector. As you already stated, it represents exactly the same physical state. The only difference is that, on taking the modulus squared, the new state gives an unnormalised probability distribution over possible measurement outcomes. You can easily extract the probability of obtaining a measurement outcome corresponding to the (possibly unnormalised) state $|\phi\rangle$ from an unnormalised state $|\psi\rangle$ by using the Born rule:
$$ \mathrm{Pr}(\phi) = \frac{|\langle \phi | \psi \rangle |^2}{\langle \phi | \phi \rangle \langle \psi | \psi \rangle }. $$
Clearly, using normalised states is just a handy convention that avoids any worries about calculating the denominator above. There is nothing wrong with formulating quantum mechanics without normalising state vectors to unity. Indeed, most people avoid bothering with the overall normalisation factors until they are needed at the very end of the calculation, since they just add unsightly clutter to the mathematics and have no physical significance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/76939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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