Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Fluid velocity in a vertical pipe Consider a pipe with length $L$ and uniform radius $A$ is held vertically. According to the continuity equation, the velocity of water going into the pipe seems to be the same as the velocity of water coming out. But according to Bernoulli's equation:
$$P_{atm}+\frac{1}{2}\rho v_1^2+\rho gL=P_{atm}+\frac{1}{2}\rho v_2^2$$
$$v_2=\sqrt{v_1^2+2gL}$$
Which means that the e water would come out faster, which makes much sence. What is wrong with my equations?
| Your version of Bernoulli's equation only applies to steady flow -ie. to motion in which the velocity at any point is indepenedent of time. In the case of your pipe open at both ends the fluid will accelerate. In that case you must use the time-dependent version of Bernoulli. For the irrotational flow of in incompressible fluid (so that we can write ${\bf v}=\nabla \phi$) Bernoulli says that
$$
\frac 12 |{\bf v}|^2 + \frac{P}{\rho} +\frac{\partial \phi}{\partial t}= const.
$$
You are omitting the $\partial \phi/\partial t$.
| {
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Calculating reflection of light when the refractive index changes continuously Suppose you have 2 materials, one with refractive index $n_1$ and the other with refractive index $n_2$, and a plane-wave coming from the first material hits the interface with an incident angle of $0^\circ$.
Fresnel tells us that the reflected power will be
$$r=\frac{n_1-n_2}{n_1+n_2} \Rightarrow R=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2$$
Now, if you have a set of materials each with thickness $d_i$ and refractive index $n_i$, you can use the simple formula above and interference to calculate the net reflection and net transmission (for example, one can multiply the matrices associated with the transmission and reflection of each interface and the matrices propagating in each material).
I'm having trouble with a similar but different problem. I have this optical fiber with refractive index $n_0$, and at some point along the fiber the refractive index changes periodically and continuously:
$$n(x)=n_0+\epsilon\sin\left(\frac{2\pi}{d}x\right), \; \epsilon\ll1$$
where the wavelength $\lambda$ is not negligible compared to the period $d$.
After $N$ periods, the refractive index returns to $n_0$.
The question is: How do I calculate the net reflection and transmission of such grating? The refractive index varies continuously, not in discrete steps for which I can use Fresnel's equations.
Thanks a lot!
| I will outline two approaches. Both involve numerical solution.
Method 1. Use the transmission matrix concept, but for a very large number of small steps through the material. For each small step $\delta z$ find the matrix based on the local change of index, and multiply it onto the matrix you have so far.
Method 2. Solution of wave equation. You have
$$
\frac{\partial^2 E}{\partial t^2} - \frac{c^2}{n^2}
\frac{\partial^2 E}{\partial z^2} = 0
$$
(I think that's ok even when $n$ is a function of $z$, but to be honest I am not completely sure and you would need to look into this before proceeding. Born and Wolf is a good resource.) Assuming this differential equation is the right one (or replacing it with another one which is right), then all you need to do is use a standard numerical partial differential equation solver. Matlab has them, so does Python, and many other languages.
I wrote this answer just off the top of my head. I would not trust it until I had done some further working out, and I would probably want to set up both methods and thus use them to check each other.
| {
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On the Heisenberg uncertainty relation Are there fundamental limits on the accuracy for measuring both position $q$ at time $t$ and momentum $p$ at time $t+\Delta t$, with tiny $\Delta t$?
If yes, why?
If no, why can't one then measure (in principle) both $q$ and $p$ arbitrarily well at the same time $t$ (which is not allowed by Heisenberg's uncertainty relation), by taking $\Delta t$ sufficiently small and noting that any measurement takes time?
| Measuring arbitrarily well means observing eigenstates of the observables, so the answer depends on whether the state can evolve from a position eigenstate to a momentum eigenstate in time $\Delta t$.
| {
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Are vectors in integral infinitesimals? There are many equations that deal with the integral, like Gauss' Law and Coloumb's Law. For example, to find the electric field from a continuous distribution requires integrating $d\overrightarrow E = k \frac {dQ}{r^2}\hat r$.
In these situations, it is common to see things like $dQ$ is an infinitesimal of the total charge. However, I've read posts like this where the first answer states that $dy$ and $dx$ are not infinitesimals because that brings in many problems, with the answerer specifically citing the Archimedian property. So is technically correct to use the word infinitesimal for $dQ$ for the equation given or for others, or is it just a concept to use in the integral?
| In physics, we think of them as ratios. They work like ratios algebraically in that you can divide them and cancel them out.
The Archemedian property thing is not really relevant because we are not saying they are real numbers, just that "$d\vec{E}$ is a symbol for an arbitrarily small change in $\vec{E}$, and $\frac{d\vec{E}(t)}{dt}$ is the symbol for the slope (really tangent vector) of the curve which describes how $\vec{E}$ behaves as $t$ varies."
| {
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Is metric a classical or a quantum field in General Relativity? I am currently reading the article of A Castro about $AdS_3/CFT_2$. I have a confusion in reconciling several definitions. It appears that I've understood them imprecisely or may be wrong. These definitions are:
Einstein gravity, Quantum gravity and General Relativity.
Here are several quotes which I want to understand:
Semi-classical regime of pure general relativity...
Does it mean that there can be quantum regime of general relativity?
I have found that in semi-classical gravity we consider matter fields to be quantum while gravity field(metric) to be classical one. "Pure" means that we consider only gravity field without matter fields
Strongly coupled gravity theory where AdS radius is of Plank scale...
Einstein gravity with cosmological constant of Plank scale...
Gravity in strongly coupled regime, where quantum effects are of order one...
As I believe now, in order for all these quotes to hold simultaneously I have to state that Einstein gravity, Quantum gravity and General Relativity are all synonyms. I also must assume that strong coupling equals quantum regime of gravity theory.
I was under impression that General Relativity (the same thing as Einstein gravity) is classical regime of Quantum gravity. Hence GR doesn't consider metric to be quantum field. Yet this doesn't seem to hold simultaneously with aforementioned quotes.
| General Relativity is a classical theory (specifically it is not a quantum theory: people sometimes use 'classical' to mean something like 'newtonian', which it clearly is not). The metric in GR is therefore a classical field.
We do not have a working quantum theory of gravity: if we did, and if it was curvature-based, then the metric would presumably be a quantum field. That theory would not be GR although its classical limit might be GR.
| {
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The 'right' way to express the line element $ds^2$ Taking Minkowski space as an example, the line element is commonly expressed as,
$$ds^2 = dt^2 - dx^2-dy^2-dz^2.$$
However, when expressing the line element in terms of an orthonormal basis $e^a$ it may be expressed as,
$$ds^2 = e^t \otimes e^t - e^x \otimes e^x - e^y \otimes e^y - e^z \otimes e^z.$$
In the case of Minkowski space, we of course have $e^t = dt$ and so forth. This then makes me question whether in the first line element $dt$ is somehow different from the $e^t = dt$ since there is no tensor product symbol. As I understand it, $e^t$ is a differential form and so lives in a section of the cotangent bundle of the underlying manifold.
What's the correct way to express the line element?
| I'm not entirely sure the question but with regards to defining a line element $(ds)^2$ in terms of basis vectors we have the relation $g_{ij}=\mathbf{e}_i \cdot \mathbf{e}_j$ so the line element is given by
$$ (ds)^2 = g_{ij} dx^i dx^j = (\mathbf{e}_i \cdot \mathbf{e}_j )dx^i dx^j$$
| {
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Why is there less information in a written book (which makes me see things) than in a random string of letters and spaces? If I read a book I can see with my inner eye (or ear) see the plot developing. With a random string of letters, with random spacings, this isn't the case. Then why contains the former less information than the latter?
| The English language contains about 170 000 words. Some words occur more frequently than others. So if we assign the number 0 to the spaces between words (which doesn't reduce the information) the number 1 to the most frequently used words, the number 2 to the second most frequently used ones, etc. until we arrive at the less frequently used word to which we assign the number 170 000, we can replace the string of words by a string of decimal digits (with a space between them) varying from 0 to 170 000, which obviously contains lesser characters (decimal digits) than the number of alphabetic characters in the original text. For simplicity, I omitted the grammar (syntax) of the words, which gives us a bigger number of words (I am, you are, he/she is, etc.)
Of course, you need a "dictionary", in which you can see what number to assign to each word, and because of that, the text will not be easy to read. But the number of characters is reduced, and so is the information.
The meaning of the words (for example the associations I have in my mind when reading the word "woman") is not information inherent in the written text, but (literally) contextual information, which doesn't count as an objective measure.
| {
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What are the de Sitter Killing vectors? I'm trying to find the Killing vectors for de Sitter space using global coordinates, i.e. the spherical foliation. In the end, I want to know how to perform a boost on the 1+1 and 3+1 de Sitter manifolds so that I can make two points lie at the same spatial position (the spatial origin). I know how to perform a Lorentz transformation in Minkowski, but not for curved spacetimes.
Two things I've tried are re-deriving the Lorentz transformations for a curved space using the steps on the Wikipedia page for Lorentz transformations, as well as performing a regular Lorentz transformation on the hyperboloid embedded in a higher dimension.
Any hints?
| To assist with your problem, I will describe the general approach, when we don't necessarily know what we're looking for. As you are aware, finding the Killing vectors $X^\mu$ requires solving,
$$\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0$$
which is an over-determined system of differential equations. As you know, these Killing vectors are precisely those for which $\mathcal L_X g = 0$ and are the infinitesimal generators of isometries of the manifold being described. If you split them into a linearly independent set, $\{\xi_i\}$ you can work out the Lie algebra they generate and sometimes may be able to identify what they are physically.
Alternatively, one can explicitly find the finite form of the transformations they generate by solving the equation for the integral curves of the fields, that is,
$$\frac{d x^\mu}{d\lambda} = X^\mu(x)$$
where $x^\mu(\lambda)$ parametrises the integral curve, that is, you can think of it as an embedding function and $X^\mu(x)$ is the Killing field, seen as a function of the components $x^\mu$.
Choosing the right coordinates often makes the calculation simpler; it is certainly desirable to have the metric be diagonal, and well-behaved at as many points as possible.
| {
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Can a mirror reflect only one color while allowing all other visible light pass through? Is it possible to construct a mirror such that it only reflects certain wavelengths of visible light while allowing other wavelengths of visible light to pass through? I was reading about microwave ovens and how the doors are designed to block microwaves from exiting the oven and was wondering if the same principle can be applied to visible light.
Thanks!
| There are mirrors which reflect only particular color wavelengths and not the others. There are mirror which can reflect particular symmetry and not the other. By using proper coating and proper atomic symmetry elements we can develop different kinds of mirrors. In Kurukshetra, India In science panaroma, both these kinds of mirrors are available, Where people look at the alphabets of different colors and there are only few colors which gets reflections others dont. There is this mirror which can reflect bisymmetric alphabets like T or I but not G or K etc.
| {
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Is short circuit technically the same as overloading? Taking the simplest circuit: battery and resistors.
If I connect lots of resistors in parallel, wouldn't that increase the current to an extent that it would be technically be very similar to shorting the circuit?
| Yes. The equivalent resistance for $n$ equal resistors of value $R$ connected in parallel is $R(n)=\frac{R}{n}$. As $n \to \infty$ then $R(n) \to 0$, provided that $R$ is finite.
| {
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Gravitational time dilation and neutron stars How great is the gravitational time dilation close to a neutron star? How would the effects of gravitational time dilation compare with the event horizon of a black hole?
| As it says in the Wikipedia article referred to by Kyle, the relevant formula applicable to clocks situated in the gravitational field of a spherically symmetric object is that an observer at infinity sees a clock in the gravitational field slowed by a factor $(1 - r_s/r)^{-1/2}$, where $r_s$ is the Schwarzschild radius.
For a non-rotating black hole, the Schwarzschild radius marks the event horizon and at $r = r_s = 2GM/c^2$, and the time dilation becomes infinite.
Neutron stars have radii that are at least $\sim 1.5$ times their Schwarzschild radii, and more likely closer to 2 times as big. Thus typical time dilation factors at their surfaces are 1.7 to 1.4.
A clock falling towards a neutron star will appear to run slower as it falls inwards until it reaches these maximum dilation factors just as it hits the neutron star surface. Unlike for a black hole, an external observer will see objects hitting the neutron star surface in a finite time. This phenomenon is observed in X-ray bursters.
| {
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Isn't D'Alembert's wave equation enough to see that Galilean transformations are wrong? The D'Alembert equation for mechanical waves was written in 1750:
$$\frac{\partial^2u}{\partial x^2}=\dfrac{1}{v^2}\dfrac{\partial^2u}{\partial t^2}$$
(in 1D, $v$ being the propagation speed of the wave)
It is not invariant under a Galilean transformation.
Why was nobody shocked about this at the time? Why did we have to wait more than a hundred years (Maxwell's equations) to discover that Galilean transformations are wrong? Couldn't we see that they are wrong already by looking at the D'Alembert equation for mechanical waves?
Am I missing something?
| There's no problem with the non-invariance of D'Alembert equation for mechanical waves, if I understand what you mean, because mechanical waves do have a preferred inertial frame, an "aether".
For example, a sound wave in a fluid satisfies the wave equation with speed: $$c^2=\left(\frac{\partial p}{\partial \rho}\right)_s$$ in the rest frame of the fluid.
The point is that Maxwell's equations are supposed to be valid in every inertial frame of reference. Since, in vacuum, they lead to the wave equation, the wave equation must be valid in every inertial frame, that's the problem.
| {
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How is it possible for other animals to have better night vision than humans, who can detect individual photons? According to the Wikipedia article on night vision,
Many animals have better night vision than humans do, the result of one or more differences in the morphology and anatomy of their eyes. These include having a larger eyeball, a larger lens, a larger optical aperture (the pupils may expand to the physical limit of the eyelids), more rods than cones (or rods exclusively) in the retina, and a tapetum lucidum.
But a recent study has shown that the human eye is capable of detecting individual photons of visible light. It seems to me that this should be the highest physically possible sensitivity to light, since QED requires excitations of the E&M field to be quantized into integer numbers of photons.
How is it possible for animals to have better night vision than humans, if humans can detect individual light quanta? Is it just that while the human eye can sometimes detect individual photons, other animals' eyes can do so more often?
| That research shows that humans can detect single photons, not that we're particularly good at it.
Averaging across subjects’ responses and ratings from a total of 30,767 trials, 2,420 single-photon events passed post-selection and we found the averaged probability of correct response to be 0.516±0.010 (P=0.0545; Fig. 2a), suggesting that subjects could detect a single photon with a probability above chance. (emphasis mine)
This study showed that we could do better than random chance, but not that we could do substantially better than random chance.
Based on the efficiency of the signal arm and the visual system, we estimate that in ∼6% of all post-selected events an actual light-induced signal was generated (Methods section).
| {
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Maximally entangled quantum state - not in hilbert space? I am reading a paper about Gaussian quantum states and the mathematical formalism used to describe them. At one point the authors say
"An important example of a Gaussian state is the maximally entangled state $\Phi$. In their endnotes, they then note that
"Although this maximally entangled state does not belong
to the Hilbert space, it can always be considered as a limit
of a proper pure state."
Why does the maximally entangled state not belong to the Hilbert space? Is it in some sense ill-defined as a pure state?
| The maximally entangled state in the Gaussian framework is the state that one gets when taking the limit of infinite energy/photon number of a particular class of state (see equation (4) of the paper you linked). From equation equation (2) it is clear that such a state is not bounded, and thus is not in $\mathcal{H}=L^2(\mathbb{R}^n)$.
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Synchronized Clocks in Inertial frame Will the synchronized clocks placed in an inertial frame remain synchronized forever?
| Yes.
Assuming, of course, they are mathematical clocks and this is a thought experiment in flat spacetime. Reality introduces too many unquantifiable variables.
They will drift apart under the influence of any acceleration in the direction of their displacement from one another.
If they are held a fixed distance apart in their own reference frame, the time difference between them will add up the accumulated acceleration they experience or 'feel' in their own reference frame.
| {
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Should Beta Minus decay put an upper limit on the mass of a neutrino? Beta minus decay emits an electron with a range of energies. Within the nucleus, the following is happening: $n\rightarrow p+e^-+\bar{v}_e$. For this reaction to be possible, by lepton number conservation, the neutrino must be present. Since this neutrino accounts for the range of electron energies, can this not be used to be a constraint on the mass of the neutrino?
For the maximum electron energy, the neutrino will have no kinetic energy, and only its mass energy, $mc^2$. So, how come this principle has not been utilised in putting limits on neutrino mass-energies; there must be a problem somewhere?
| Your idea is correct. However, tyical energy releases in a beta decay are of the order of MeV, while neutrino masses are in the range of an eV or so. Current experiments are not sensitive enough to give a definite value of the neutrino mass, but produce upper limits. The best data so far, combining the Minz and Troitsk experiments (using Tritium beta decay), leads to an upper limit of $2~\text{eV}$ (see the particle data book, http://pdg.lbl.gov/2016/listings/rpp2016-list-neutrino-prop.pdf, page 4).
KATRIN (https://www.katrin.kit.edu/) is a new experiment, currently being assembled and supposed to start measuring this year, that aims to improve the sensitivity by a factor of ten, down to $0.2~\text{eV}$. (Note that from neutrino oscillations, one neutrino mass must be at least $0.04~\text{eV}$.)
| {
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Fourier transforming the wave equation twice The wave equation
$$\nabla^2 u(r,t)-\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(r,t)=0$$
can be Fourier transformed with respect to time, using $\frac{\partial}{\partial t}=i\omega$, to obtain the Helmholtz equation:
$$\nabla^2 u(r,\omega)+\frac{\omega^2}{c^2}u(r,\omega)=0\,.$$
What I don't understand is what happens if we Fourier transform is subsequently with respect to space:
$$-k^2 u(k,\omega)+\frac{\omega^2}{c^2}u(k,\omega)=0\,,$$
since we now get
$$\left(-k^2+\frac{\omega^2}{c^2}\right)u(k,\omega)=0\,.$$
If this should hold for all $k$ and all $\omega$, the solution is the zero function, right?
| When solving PDE we usually allow solutions to be distributions (aka generalised functions). I'll spare you the details; you'll have to read a book on PDE by a mathematician to understand that there is a deep connection between PDE and distributions. Long story short, the distributional solution of
$$
(k^2-\omega^2)u(k,\omega)=0
$$
is
$$
u(k,\omega)=f(k,\omega)\delta(k^2-\omega^2)
$$
for an arbitrary function $f$. This function satisfies $(k^2-\omega^2)u=0$ but $u\neq 0$. (Note that I set $c=1$ to simplify the notation).
You can find some more details about this approach in this answer of mine, where I solve $(\partial^2+m^2)\phi=0$. Note that in that post, $\partial^2=\nabla^2-\partial_t^2$; to get the standard wave equation, you can take $m=0$.
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Clashing definition of rays in Weinberg and Sakurai and Born interpretation without normalizability In Sakurai's Modern Quantum Mechanics, it is stated that
One of the physics postulates is that $|\alpha\rangle$ and $c|\alpha\rangle$, with $c\neq 0$, represent the same physical state. In other words, only the "direction" in vector space is of significance.
Apart from the complex number $c\neq 0$, shouldn't we also require $|c|=1$? Shouldn't the normalizability of physical states force $|c|=1$?
As far as I know, physical states are represented by rays which contains $|\psi\rangle$ and $e^{i\theta}|\psi\rangle$, $\forall\theta\in\mathbb{R}$. This is how Weinberg defines a ray, in his QFT text (Vol. 1, section 2.1). It says that
A ray is a set of normalized vectors (i.e., $(\psi,\psi)=1$) with $\psi$ and $\psi^\prime$ belonging to the same ray if $\psi^\prime=\xi\psi$, where $\xi$ is an arbitrary complex number with $|\xi|=1$
*
*So which definition of a ray is the correct one? Weinberg's or Sakurai's? I read this post. Does it mean Weinberg's definition is not the general one?
*More displeasingly, if the criterion of $|c|=1$ (or $|\zeta|=1$, in Weinberg's notation) is relaxed, how does the Born's probability interpretation work? For example, how would one interpret a state $|\psi\rangle$ whose norm is $\langle\psi|\psi\rangle>1$ or $<1$. Shouldn't the probability amplitude for a particle to be in a state $|\psi\rangle$ when it is known to be in a state $|\psi\rangle$ must be unity? This is also not addressed here.
| The quantum state associated to a Hilbert space vector $\psi $ is isomorphic to the orthogonal projector on the subspace spanned by $\psi $. Since $\psi $ and $c\psi $ span the same subspace for any $c\neq 0$, the state associated to either one is the same.
| {
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Charging by friction Why does charging by friction charges an insulating material even if an insulator does not allow flow of electrons between an object to another? Is it because of the TriboElectric effect but doesn't it violate the law with the material being an insulator?
Does being a conductor or an insulator affect charging by friction?
| Let's consider the following case: an object made of ebonite and some cat fur. Electron's from cat fur will move to the ebonite.
The friction energy $E$ is transferred to the electrons on the outside layer which are in this way freed from their bonds.
Since the material is an insulator they have no where else to go except on the other object that caused the friction, which gets charged negatively (they can't stay on the cat fur either because this happens during the friction process while energy is transferred to the electrons on the fur).
On the other hand if both materials are conductors while the objects are in touch there is going to be a current flow back from where they came creating heat in the process due to $P=RI^2$ - for example train tracks and train wheel when the train is suddenly stopping - you will see sparks and if you touch the wheel it's going to be hot.
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Why does a wave not interfere with its secondary wavelets? Huygens principal says that every point wavefront is the source of a secondary wavelet. If this is true, why do those wavelets not interfere with the main wave? Shouldn't waves looks like a circle of interfering circles?
| Huygens principle is taught at the high school level because it was an important historical concept that broke down the old particle only theory of light and showed that a wave model would explain the observed interference pattern. It worked well until in the early to mid 1900s quantum concepts about light as a wave function more fully explained the pattern, not as interference but as viable pathways or modes for the wave function which are limited through a slit. In the modern understanding there is no interference.
An advanced understanding is really that a photon is a wave function, and a wave function is a predetermined path for the photon to take , I believe it requires a starting electron and a finishing point also an electron. Photons never superimpose but for example the electron in the nerve cell in the eye will not see 2 photons if they are out of phase with each other (180 deg), these photon will get absorbed as heat in deeper tissues.
| {
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What constitute the secondary particles beam when a high energy proton beam hits a target material? Basically I want to know what particles emerge along with high energy photons(not sure about it) as the second beam when a specific high energy proton beam is incidented upon a target material such as iridium or gold. Can the composition of secondary beam be calculated theoretically?
Does the ratio of photon energy and particles energy of the secondary beam hold any specific value for a given energy of primary beam?
| At the level of complexity suggested by this question, the only reasonable answer is "it depends." Lots of types of particles are produced when hard protons interact with matter, and a clever experimentalist can build an extraction system to focus on any number of them.
The 800 MeV proton accelerator at Los Alamos was built in the 1970s to produce secondary beams of mesons, and for decades was called the "Los Alamos Meson Physics Facility," LAMPF. That program ended and a new target station was built to produce cold neutrons for condensed-matter research. Same accelerator, but now the "Los Alamos Neutron Science Center."
When I was doing a nuclear physics experiment using a neutron beam at LANSCE, we discovered that our beamline had a lot of hard photons that we didn't really want. To characterize our detector's neutron response versus its photon response, we purified the beam by putting stuff in its way. Blocking the beam with a piece of lithium-loaded plastic absorbed most of the neutrons and gave us a pure photon beam; blocking the beam with a wall of lead bricks absorbed most of the photons and gave us the pure neutron beam that we had in mind when we designed the experiment. But when, as a grad student, I described this process to my high-energy colleagues, they would ask me questions about the kaon component of the beam. Their questions made it clear they were imagining a completely different system for extracting the secondary particles.
| {
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Why does a force not do any work if it's perpendicular to the motion? I have a book that says the Moon's orbit is [in this context assumed to be] circular. The Earth does no work on the moon. The gravitational force is perpendicular to the motion. Why is there no work done if support force is perpendicular to the motion?
| I think you are right @Shrodingers cat. It would be better to clarify the answer this way.
Work done (w) = F . d
= F d Cos θ
At 90 degrees, θ = 90 and Cos 90 = 0,
W = F x d x Cos 90
= F x d x 0 = 0 Joule.
So, when the force is applied perpendicularly to the surface, the work done will be zero.
| {
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Why are there no stable isotopes with an atomic mass of 5 or 8? One of the things I've encountered in my travels is the mass-5 roadblock. Rod Nave writes about it on his excellent educational hyperphysics website:
The helium-4 nucleus or alpha particle with a mass of 4 is particularly stable. But there are no is stable isotope with a mass of 5. Helium-5 isn't stable, nor is lithium-5. They decay almost immediately. Lithium-6 however is stable, but it has a lower binding energy than helium-4, which sounds relevant. Something else that sounds relevant is that lithium-7 is stable too but lithium-8 is not, and nor is beryllium-8, or boron-8. The $64,000 question is why?
Why are there no stable isotopes with an atomic mass of 5 or 8?
| The extreme stability of He-4.
Look at the decay modes of the the eights and they produce two alphas and if it is necessary convert a neutron/proton to a proton/neutron with an appropriate beta decay.
The production of two alphas is energetically favourable.
Li-6 and Li-7 lack nucleons to form two alphas.
Li-5 kicks out a proton and He-5 kicks out a neutron to form He-4.
More about 5 in LubošMotl's answer.
| {
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What is the significance of the phase constant in the Simple Harmonic Motion equation? The displacement of a particle performing simple harmonic motion is given by $x = A \sin(\omega t + \phi)$ , where $A$ is the amplitude, $\omega$ is the frequency, $t$ is the time, and $\phi$ is the phase constant. What is the significance of $\phi$. How is it used? Please explain the meaning of the phase constant
| All the phase angle does is to give you a facility to decide on the displacement of the particle undergoing shm at time $t=0$ or any other time.
With your phase angle of $\phi$, assuming it to be positive, the graphs of $x_1 = a \sin (\omega t)$ (grey) and $x_2= A \sin (\omega t + \phi)$ (red) are shown below.
In this case the motion $x_2$ is in advance of the motion of $x_1$ by a time $t$ (shown in the diagram) or a phase angle of $\phi= \dfrac t T 2 \pi$ where $T$ is the period of the motion and equal to $\dfrac {2 \pi}{\omega}$.
So everything that particle with displacement $x_2$ does the particle with displacement $x_1$ does a time $t$ later.
| {
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Relative velocity with relativistic speeds in 3D I have a question regarding the relative velocities of two objects in $3$D space. If objects $A$ and $B$'s velocities are being observed by stationary observer $C$, what is the velocity vector $A$ will see $B$ with or vice versa. This question also assumes $A$ and $B$ are also going at relativistic speeds.
$$\mathbf{v}_A=(a,b,c)$$
$$\mathbf{v}_B=(d,e,f)$$
If we assume all the vector components are being multiplied by $c$ (the speed of light) and can't be greater than $1$, what would $B$'s velocity vector be from $A$'s perspective?
| Let $C$ be rest frame $S$ (or lab frame, earth frame, etc) and $A$ be the moving frame $S'$.
Transformation of velocity
Velocity of $B$ observed by $A$ is
$$\mathbf{v}'=
\frac{\mathbf{u}+
\left(
\dfrac{\gamma_{v}-1}{v^2}\mathbf{u}\cdot \mathbf{v}-\gamma_{v}
\right)\mathbf{v}}
{\gamma_{v}
\left(
1-\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}
\right)}$$
where $\gamma_{v}=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$, $\mathbf{u}=(d,e,f)$ is the velocity of $B$ observed by $C$ and $\mathbf{v}=(a,b,c)$ is the velocity of $A$ observed by $C$
Please refer to the derivation in pp. 33-4 of the following notes.
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Calculating work to move a particle I have this question : There is a negatively charged spherical shell. Within this shell is a positive charge q which is at the center of the shell. How much work is required to move the charge from the middle to a position where the particle is right next to the inside wall of the shell.
The answer is 0 but I don't know how to get that. I understand work is the change in potential energy and in this case that would just be $ {kqq\over r^2} $ so wouldn't work just be the potential energy at the center minus the potential energy after the change in position?
| The electric field due to the negatively charged shell is zero inside due to symmetry.
There is no field inside ,so force on the particle will be zero and since force is zero you don't need to push the particle against electric field to bring the particle from the middle to a position where the particle is right next to the inside wall of the shell.
So work done on the particle must be zero.
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Leading/Lagging terminology for sinusoidal waves Is there a method to identify which wave is leading and which wave is lagging from their equations?
For example, if the two waves are $A\sin(\omega t+\pi/6)$ and $A\sin(\omega t+2\pi)=A\sin(\omega t)$ (by trigonometric relations), is the second wave leading in phase or lagging in phase compared to the first? Or we can say both is true?
What is the convention to be followed when we say "wave A leads wave B..." ?
| If you start from $y_1= A\sin(\omega t)$ and compare it with $y_2= A\sin(\omega t+ \phi)$ you find that time $t=0$ motion $1$ has a displacement of $y_1=0$ and motion $2$ has a displacement of $y_1=\sin \phi$.
You will see that whatever motion $2$ does motion $1$ does a little later in time so motion $2$ leads motion $1$ by phase angle $\phi$.
So the more positive $\phi$ represents the motion which is leading.
However it is also true the motion $1$ lags motion $2$ by phase angle $\phi$.
The way you have written the two motions your second motion leads your first motion by $2\pi-\frac \pi 6$.
However if your two motions had been given as $A\sin(\omega t + \frac \pi 6)$ and $A\sin(\omega t)$ and no other information was given you would say that the first motion leads second motion by phase angle $\frac \pi 6$.
| {
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A question on Collison of macroscopic particles
Hello,
In the above question I could solve for average elastic force by taking velocity with respect to wall and finding change in momentum of the ball after that divided change jn momentum by time interval. Answer comes out to be option b.
But as it is written in the question that collision is elastic, therefore, Kinetic energy before the collision should be equal to kinetic energy after the collision and option d should also be correct. But the correct answers according to the book are b and c.
Please Explain why d is incorrect and c is correct.
| Alright this is what I got:
sorry for being so slow :(
| {
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Understanding a step in the derivation of the length contraction effect in special relativity I've been trying to figure out how the equation for length contraction is derived in my textbook (Krane, Modern Physics 3e) since a few of the final steps are omitted. The equation in question is:
$$
L = L_0/\gamma = L_0 \sqrt{1-u^2/c^2}
$$
Where $u$ is the velocity of the object. I managed to derive myself the previous equations
$$
\Delta t = \Delta t_0\sqrt{1-u^2/c^2}
$$
and
$$
\Delta t = \frac{2L_0}{c} \frac{1}{1-u^2/c^2}
$$
after which my book states:
Setting the two equations above equation to each other and solving, we obtain:
$$
\Delta t = \Delta t_0\sqrt{1-u^2/c^2} = \frac{2L_0}{c} \frac{1}{1-u^2/c^2} \rightarrow L = L_0 \sqrt{1-u^2/c^2}
$$
without any appearance of $L$ in the above equations. I have a hunch that it might be implicitly defined as $u\Delta t$, but I'm not certain. What is the missing step here in getting that last equation from the two above?
| It isn't clear which bit is foxing you, so let's go through the argument:
This is viewed from the Earth frame i.e. the frame in which the light clock is moving. In this frame the length iof the clock is $L$. In the outward trip the light moves a distance $L+ut_1$ in a time $t_1$, and since light moves at the speed of light we get:
$$ L+ut_1 = ct_1 $$
Likewise for the return trip the light moves a distance $L-ut_2$ in a time $t_2$ so:
$$ L-ut_1 = ct_2 $$
The total time is therefore:
$$ t = t_1 + t_2 = \frac{2L}{c}\frac{1}{1 - u^2/c^2} \tag{1} $$
Now switch to the rest frame of the clock. In this frame the length of the clock is $L_0$. In this frame the light simply moves a distance $2L_0$ in a time $t_0$, and again light moves at the speed of light so:
$$ t_0 = \frac{2L}{c} $$
Now the book uses the previously derived result that:
$$ t = \frac{t_0}{\sqrt{1-u^2/c^2}} = \frac{2L_0/c}{\sqrt{1-u^2/c^2}} \tag{2} $$
This is just the usual equation for time dilation. The time $t$ is the same time in equations (1) and (2), so we just set them equal to get:
$$ \frac{2L}{c}\frac{1}{1 - u^2/c^2} = \frac{2L_0/c}{\sqrt{1-u^2/c^2}} $$
And this rearranges to the final result:
$$ L = L_0\sqrt{1-u^2/c^2} $$
But I have to say that this is a dreadful derivation of the Lorentz contraction because it give you no insight into what actually happens in special relativity. The contraction is not really a contraction it is a rotation in spacetime. Have a look at "Reality" of length contraction in SR to get an idea of what is actually going on.
You might also be interested to look at How do I derive the Lorentz contraction from the invariant interval? to see how the Lorentz contraction is related to the symmetry that underlies special relativity.
| {
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Would it be possible to walk up a wall under right circumstances? For example, you are in a box that is connected a distance $R$ from a clockwise spinning centre. If I understand correctly, the spinning box is a result of the resulting centrifugal force $F_{centrifugal}$ = $\frac{mv^2}{R}$. The person then would be pushed against the wall opposite to the direction of $F_{centrifugal}$
See this picture in top view:
Where distance $R$, the direction of $F_{centrifugal}$ and v are indicated. The person in the box is looking in the opposite direction of the center and perpendicular to $v$ (speed).
If $R$ is small enough or $v$ large enough, then at a certain combination of $R$ and $v$, $F_{centrifugal}$ would be larger than $F_z$ = $mg$ (gravity).
Question
If $F_{centrifugal}$ > $F_z$ would it be possible to walk up the wall that is in front of him?
(The forces $F_{centrifugal}$ and $F_z$ are perpendicular on each other which mean they do not counteract each other. Similarly for example in a bus with the vertical gravity force and the horizontal acceleration of a person due to the acceleration of the bus.)
| In simple words: Depends on your velocity.
Because if we look from the frame of reference of the box we find the the centrifugal force would provide the normal reaction (assuming cofficient of friction to be constant) and hence the friction force to balance gravitational force acting on the man.
u*(mv^2/r) ≥ m*g(this is the minimum condition for the man to stand on wall. If you want to be extra safe then you should solve the above equation with strict inequality to find v.) Once this condition is achieved the man can move on the wall. But remember while walking he must not remove both his legs at the same time because the frictional force is a contact force.It would only act when at least one the man's leg is in contact with the wall.
Where u is coefficient of friction between man's shoes and the wall
m is the mass of the man
v is the speed with which the box is rotating
| {
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Short question about moment force I have a question about moment Forces. Let $\mathbf{e_1}$, $\mathbf{e_2}$ be the unit vectors defining a Cartesion coordinate system $Oxy$.
Let $\mathbf{F}$ be the force applied at point $A$.We have:
$$\mathbf{F} = F_x \ \mathbf{e_1} + F_y \ \mathbf{e_2}$$
where
$$\begin{cases} F_x &= a\\F_y &=-b \end{cases}$$
The moment of force $\mathbf{F}$ about point $O$ at point $A$ is, by definition:
$$\mathbf{\mathcal{M}_{/O}}\left(\mathbf{F}(A)\right) = \mathbf{OA} \times \mathbf{F}$$
Therefore the magnitude $M$ of this moment force is
$$M = \left(\begin{array}{c} x \\y\end{array}\right) \times
\left(\begin{array}{c} F_x \\F_y\end{array}\right)=
\left(\begin{array}{c} x \\y\end{array}\right) \times
\left(\begin{array}{c} a \\-b\end{array}\right) =
-(xb+ya)$$
However, we can also have
$$M = \left(\begin{array}{c} y \\x\end{array}\right) \times
\left(\begin{array}{c} F_y \\F_x\end{array}\right)=
\left(\begin{array}{c} y \\x\end{array}\right) \times
\left(\begin{array}{c} -b \\a\end{array}\right) =
xb+ya$$
How to define which one to use? What is the convention used for the 1st and 2nd equations? It should depend on the convention used for a positive moment but I can't figure out how it's done.
Edit: Added my intuitive answer
I'll post my intuition just below but ... this is not really solid as it is only intuitive. I'd still like a solid proof.
*
*Convention used: Moments are positive when rotation is clockwise (opposite of the geometrical convention)
*For a positive moment, as rotation is clockwise, The vector along the rotation axis must be pointing outward (away) (defined by $\mathbf{e_3}$)
*Therefore, $\mathbf{u_r} = \mathbf{e_2} \times \mathbf{e_1} = - (\mathbf{e_1} \times \mathbf{e_2})$ . Then, coordinates of $\mathbf{F}$ and $\mathbf{OA}$ are defined by $(\mathbf{e_2},\mathbf{e_1})$ and not $(\mathbf{e_1},\mathbf{e_2})$
*Equation 2 for $M=xb+ya$ is correct for this convention (opposite to the geometrical/mathematical one)
Is this correct? How to demonstrate it?
Thanks!
| There is a strict convention for cross product in three space. Your plane $Oe_1e_2$ is viewed as sitting inside the three space $Oe_1e_2e_3$ with orthonormal basis vectors $e_1, e_2, e_3$ and you have the cross product between two vectors $OA = x \, e_1 + y \, e_2$ and $F= a \, e_1 - b \, e_2$. Then the cross product is linear
$$OA \times F = ( x \, e_1 + y \, e_2) \times (a \, e_1 - b \, e_2) = $$ $$= (x\,e_1) \times (a \, e_1) - (x\,e_1) \times (b \, e_2) + (y\,e_2) \times (a \, e_1) - (y\,e_2) \times (b \, e_2)= $$ $$= x a \, (e_1 \times e_1) - x b\,(e_1 \times e_2) + y a\,(e_2\times e_1) - y b\,(e_2 \times e_2)$$ However $e_1 \times e_1 = e_2 \times e_2 = 0$ and $e_1 \times e_2 = e_3$ while $e_2 \times e_1 = - e_3$ so finally
$$OA \times F = - x b\,e_3 - y a\, e_3 = -(xb + ya) \, e_3$$ This is the mathematical convention. In your convention, you have ordered your basis differently: $O e_2 e_1 e_3$ and you get
$$OA \times F = x b\,e_3 + y a\, e_3 = (xb + ya) \, e_3$$
| {
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Glasses underwater So I figured the refraction of the cornea is based on the index of air and the vitrous humor to make a perfect image. Underwater this is messed with because water has the same index as the eye. Hence you can't see clearly underwater without making an air pocket with scuba goggles for example.
Would it be possible to compensate this by wearing regular glasses with a diopter that is equivalent to the difference in index between air and the eye, normally?
| Yes, it’s possible to make glasses that allow you to focus underwater. You could use a high-index material with a concave lens, or a low-index material (such as air) with a convex lens.
| {
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Why doesn't current decrease in series combination? I know that the question is quite stupid but I want to get an insight of this case. consider 3 resistors connected in series with a battery, after the current passes through resistor 1 it loses some of its energy, the kinetic energy of the charge carriers will definitely also decrease and so does the drift velocity then why doesn't the current decrease? Its quite confusing.
| You are right - but not for long...
The situation you describe is unstable. This is not steady current. More electrons enter the resistor than what leave it. It will be "filled up" quickly. There will be a pile-up, a queue, of incoming electrons waiting to enter.
When electrons accumulate at a point like that, the electric field increases there. This will repel new arriving electrons more and more. Less electrons then show up.
The current before the resistor is reduced because the current after is smaller.
In the end it equalizes, so that just as much enter as what leaves per second. And the current is then constant and equal on either side; it is steady.
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Radial motions moment of inertia I'm looking for some references on a specific moment of inertia, for radial motions of a spherical body. In my calculations, I got this integral :
\begin{equation}\tag{1}
\bar{I} = \int r^2 \, dm = \int_{\mathcal{V}} \rho(r) \, r^2 \, d^3 x,
\end{equation}
where $\rho$ is the matter density and $r^2 = x^2 + y^2 + z^2$ defines the usual radial coordinate (the coordinates origin is located at the center of the spherical body). For an uniform mass distribution, this integral is easy to do :
\begin{equation}\tag{2}
\bar{I} = \frac{3}{5} \; M R^2.
\end{equation}
Please, don't confuse this with the well known moment of inertia of the sphere, around some rotation axis. This is about radial motions, and not rotation.
I never saw this in any books on mechanics.
Notice that expression (1) above is also half the trace of the inertia tensor :
\begin{equation}\tag{3}
I_{ij} = \int_{\mathcal{V}} (r^2 \, \delta_{ij} - x_i \, x_j) \, \rho \; d^3 x,
\end{equation}
Then we have this :
\begin{equation}\tag{4}
\bar{I} \equiv \frac{1}{2} \; I_{kk}.
\end{equation}
I'm not sure the "radial inertia moment" defined by (1) (if it have a proper interpretation) is getting the proper factor.
Any thoughts on this ?
| The moment of inertia is the measure of resistance to angular acceleration about an axis. Unless I'm mistaken, what you're after is the modulus of elasticity $E$ (or Poisson's ratio $\nu$) of the object. That dictates the response to radial motion given a uniform pressure field acting on the surface of the sphere.
| {
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Why the generator of $U(1)$ is the particle number operator $N$? When $U(1)$ symmetry is broken, the particle number is not conserved any more. What is the relation between u1 and particle number. Why the generator of $U(1)$ is the particle number operator $N$?
| For a single-particle state, a $U(1)$ rotation operation is simply multiplying a phase factor of the form $e^{-i\theta}$. The same operation on an $N$-particle state multiplies $e^{-i\theta}$ for each particle and hence $e^{-iN\theta}$ on the quantum state in question.
It is easy to see that the operator acting on the Fock space with the above property is $e^{-i\hat{N}\theta}$, where $\hat{N}$ is the number operator. Thus, the generator of $U(1)$ rotation is the number operator.
| {
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Finding Gravitational Force b/w 2 People I just need some clarification. The question is:
What is the gravitational force between two 100 kg people standing 1 m apart, and how does this compare with the gravitational force of either of them relative to Earth?
I'm guessing the formula to use here is Newton's Law of Gravitation. So, I'd use $M_1, M_2$ = 100 kg, and use $1$m as $r$, right? And to find one of the people's force relative to the Earth, what value do I use for $r$? Isn't the distance between the Earth and any person almost negligible? Thank you.
| $$F = \frac{G m_1 m_2}{ r^2}$$
where $r$ is the distance between the center of gravity of the individuals.
$$
F = \frac{6.67\cdot10^{-11} \text N \cdot { \text m^2/ \text k\text g^2}\times 100~\text k \text g \times 100~\text k\text g}{1~\text m^2}=6.67\cdot10^{-7} \text N
$$
The gravitation force of one of them with the earth would be about 1000 N.
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} |
If magnetic fields are created by moving charges, how do magnets work? This is probably is stupid question but I think it must clear up some misconception I have. Magnets, presumably have magnetic fields. But where are the moving charges? Don't we need a current?
| Most of the magnetism in a magnet comes from the magnetic dipole moment of the electrons orbiting its atoms. If we think of the electron as a small spinning charged sphere, then it is the spinning chrage that makes the current. Unfortunately this is a not a good model because the surface of the electron (assuming its "classical radius'') will be moving faster than the speed of light. For this reason Pauli originally did not believe in "spin" --- even though he gets most of the credit for the idea.
| {
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Is the Moon in a "Freefall" Around the Earth?
The force of gravity keeps our Moon in orbit around Earth. Is it correct to say that the Moon is in “free fall” around Earth? Why or why not?
I think the answer is yes. The moon is falling towards the Earth due to gravity; but, it's also orbiting the Earth as fast as it's falling towards it. This balance between the 2 forces means the moon is essentially "freefalling" towards the Earth. Is my thinking correct? Thanks.
| Earth's moon, Luna, is free falling. More particularly, it is in
orbit around the sun, Sol, and is perturbed in its orbit by a nearby
planet, Earth.
Solar gravity acceleration at Earth/Luna orbital distance ($1.5\times 10^{11}m$):
($\frac{2\pi R_{orbit}}{3.16 \times 10^7})^2/R_{orbit}$ = $0.0059 ms^{-2}$
Earth gravity acceleration at Earth surface: $9.8 ms^{-2}$
Earth gravity acceleration at Luna: $0.0028 ms^{-2}$
It seems more appropriate to refer to Luna as orbiting Sol, than
Earth. Earth is the less important influence.
| {
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Sea quark parton annihilation? Consider the figure below1:
This can be read as follows (please correct me if I am wrong): two particles come in and 'fragment', a parton from each particle $C$ and $D$ annihilate to form the particle $X$. An intuitive guess is that the partons $C$ and $D$ must correspond to valence quarks whist the 'jets' of $A$ and $B$ must contain the remaining valence quarks and the usual gluon and sea quarks. My question is this: Is such a reaction possible but in which $C$ and $D$ represent sea quarks (or even gluons)? Please can you explain either way.
1Image adapted from that given on page 20 of http://www.hep.man.ac.uk/u/hanl/lecture/Lecture1_LHC+TeVatron.PDF
| Yes, this can really happen. One example is a process called Drell-Yan production (at a hadron collider). A Drell-Yan event is the interaction of a quark and an anti-quark, which annihilate to form a photon or Z-boson (which then at some point decays to a pair of leptons).
At a proton-proton collider this can only happen if the anti-quark is a sea-quark (because the proton has only three quarks as valence quarks).
The Feynman graph for this process looks something like this:
| {
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How can we compare the ratio of strenghts fundamental forces? I have read in many books that the ratio of strengths of gravitational force, electromagnetic force, nuclear force is 1:10^36:10^38 (one: 10 raised to thirty six: 10 raised to thirty eight). On what basis are we able to compare this? I mean, there is no connection between any two forces in the above mentioned list of forces. For example, we measure gravitational force by measuring mass, while we measure electrostatic force with charge. So how can we compare those two forces?
Or is it based on SI Units ? If it is, the ratio changes with system of units. Isn't it?
Hope someone could throw some light on this. Thank You.
| A ratio is produced by comparing the gravitational force $F_g=\dfrac{Gm^2}{r^2}$ between two protons or electrons of mass $m$ and separation $r$ with the electrostatic force $F_e=\dfrac{e^2}{4\pi\epsilon_o r^2}$ with the same separation.
$$\dfrac{F_e}{F_g}=\dfrac{e^2}{4\pi \epsilon_o Gm^2}$$
The exact ratio you get depends on the charged particles that you choose.
Try putting in the values and see what you get.
| {
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Why can I assume the force to be constant in this particular interval? If I have force, or any function $f(z)$, I was told that I can assume it to be constant only in the interval $dz$.
However, in this case, I had to calculate the work done by the spring force as a function of $y$
Over here, I assumed the spring force, which is a function of its elongation $x$ ($F = -kx$) to be constant in the interval $dy$ and integrated and this gave me the correct answer
I want to know why the error vanished over here. Shouldn't spring force only be constant in the interval $dx$ and not $dy$?
I also want to know, in general, if I have a function, how to decide whether it is constant in some particular interval/in which cases the error will vanish as I take the limit and integrate.
Note: I do know I can assume a function $f(x)$ to be constant in the interval $[x,x+dx)$ while integrating, but over here I've assumed it to be constant in the interval $dy$. I want to know why I can do that and also if I can assume a force/function to be constant in any infinitesimal interval such as $Rdθ$, $dy\over cosϕ$,$dz$ etc.
| It is, in general, by definition. $dx$ and $dy$ are defined in such a way that the force can be assumed to be constant in such an interval.
That's the general case, the way we use infinitesimals in physical problems mostly. Exceptions to this may occur.
| {
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How does the temperature of an ideal gas exhausting into vacuum vary? Since a gas at a certain pressure exhausting into vacuum has no atmospheric pressure to push against, there shouldn't be any adiabatic cooling taking place.
But looking at the energy conservation:
$TdS=dU+VdP+PdV$
$TdS = 0$ {Adiabatic process}
$PdV = 0$ {No change in volume of container}
Hence: $dU=-VdP$, leading to a lower temperature?
| The gas in the container does work on the elements of gas that are moving towards the orifice and out. As the the gas remaining in the container expands, it does work and thus cools down.
One can use the dX description for any element in the container, if it does not move too violently. Increase of internal energy of the element is heat accepted minus work done (gas expands):
$$
dU = dQ - dW
$$
If the process is slow, gradient of temperature is low and the heat transferred to the element will be negligible in comparison to work it does. The work done by the element is $pdV$, so the internal energy decreases:
$$
dU = -pdV < 0
$$
If the gas is ideal, decrease of energy is sufficient to conclude temperature decreases as well. The gas in the container cools down.
| {
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Calculating the (expected) kinetic energy of an electron in the ground state of a Coulomb potential? I've been struggling with this all week to no avail.
I'm asked to calculate the expectation value of kinetic energy for an electron in the ground state of a Coulomb potential. I know that it ought to be $ 13.6 \, \mathrm{eV}$, but I am having a difficult time arriving there.
In general, the expectation value of, say, $Q$, is
$$\langle Q \rangle = \int \psi^* \hat Q \psi \, \mathrm dV $$
over all space. In the case of kinetic energy, $\hat Q$ would be equal to
$$ \frac{-\hbar^2}{2m} \nabla^2 $$
and, in the case of a ground-state electron, we would have
$$ \psi = \sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}} \exp(- r / a) $$
with $a$ being the Bohr radius.
However, for the life of me, I cannot get this integral to work. For a while, I was continually coming up with either 0 or a non-converging integral, until I stumbled on some piece of information (that I can't find convincing proof of, either in my textbook or on the internet) that the square angular momentum (that is, $(\mathrm d^2/ \mathrm d \theta^2 + \mathrm d^2/ \mathrm d \phi^2) \theta\psi$) is equal to $l(l+1)$ - in my case, 0, since $l = 0$ in the ground state $(1,0,0)$. This simplified things and gave me an integral I could get to converge. However, it seems to converge to
$$ \frac{\hbar^2}{a^2} $$
which not only has the wrong units of (energy time per length)^2 but also has the wrong value.
Please help. This homework problem has taken an embarrassingly long time and a lot of scratch paper to do already.
| The factorization into a radial part plus the angular momentum operator is true, but you don't really need it; instead, you can simply use the Laplacian in spherical coordinates,
$$
\nabla^2
= \frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}
+\frac{1}{r^2\sin(\theta)}\frac{\partial}{\partial \theta}\sin(\theta)\frac{\partial}{\partial \theta}+ \frac{1}{r^2\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2},
$$
and note that the angular derivatives vanish with your spherically-symmetric wavefunction.
From here you just need to calculate the action of the kinetic energy operator on the ground state,
\begin{align}
\hat{Q}\psi
&= \frac{-\hbar^2}{2m}\nabla^2\psi
\\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\nabla^2 \exp(- r / a)
\\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\frac{1}{r^2}\frac{\partial}{\partial r}\left[r^2\frac{\partial}{\partial r} \exp(- r / a)\right]
\\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\frac{1}{r^2}\frac{\partial}{\partial r}\left[r^2\frac{-1}{a} \exp(- r / a)\right]
\\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\frac{1}{r^2}\left[2r\frac{-1}{a} \exp(- r / a)+r^2\frac{1}{a^2} \exp(- r / a)\right]
\\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\left[\frac{-2}{ar} \exp(- r / a)+\frac{1}{a^2} \exp(- r / a)\right],
\end{align}
and then integrate against the wavefunction itself:
\begin{align}
\langle\psi|\hat{Q}|\psi\rangle
&=\int \psi(\mathbf r)^* \, \hat{Q}\psi(\mathbf r)\mathrm d\mathbf r
\\& = \int_0^{\infty} \psi(r)^* \hat{Q}\psi(r) 4\pi r^2\mathrm dr
\\& = \int_0^{\infty} \left(\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\exp(- r / a)\right)
\\& \qquad\times\left(\frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\left[\frac{-2}{ar} \exp(- r / a)+\frac{1}{a^2} \exp(- r / a)\right]\right) 4\pi r^2\mathrm dr
\\& = \frac{-\hbar^2}{2m} \frac{4}{a^3}\int_0^{\infty} \left(\frac{-2r}{a} e^{- 2r / a}+\frac{r^2}{a^2} e^{-2 r / a}\right) \mathrm dr
\\& = \frac{-\hbar^2}{2m} \frac{4}{a^3} \left(\frac{-a}{4} \right)
\\& = +\frac{\hbar^2}{2ma^2}
\\& = 13.6\:\mathrm{eV},
\end{align}
as required.
| {
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Spherical lens ray-tracing simulation is not focusing rays I am trying to make a very basic ray-tracing simulation of lens.
*
*A single ray origin trigger light photons of a supposed single frequency in all directions. (I do not consider chromatic aberration).
*Some lens emulated as a couple of spherical superficies with refraction index.
After detecting an intersection between a ray and the closest lens, I get the normal at this point to the lens and apply the following refraction formula wikipedia:
$$\frac{sin(\theta_1)}{sin(\theta_2)} = \frac{n_2}{n_1}$$
The formula is applied twice: a first time when entering the lens with (e.g. index $2.0$) and a second time when exiting the lens with the inverse index (e.g: $0.5$).
I checked the results numerically, but the rays do not focus:
As you may observe, the focus "point" is closer for rays crossing the outer part of the lens, while is further for rays crossing the lens in the middle.
Some suppositions:
*
*The formula is too simplistic for achieving a correct focus point
*Lens are not exactly spherical
*I do some mistake somewhere
**Why are those rays not converging to the same focus point? **
Example of calculation:
*
*Ray: Origin=$(400,300)$ Direction=$(0,-1) $
*Lens: Circle center=$(400-50\sqrt{2}, 100)$ Radius=$100$
*Intersection: Position=$(400, 170.7107..)$ Normal=$(\sqrt{2}, \sqrt{2})$
*Refracted ray inside the lens: Origin=$(400,170.7107...)$ Direction=$(-0.41144, -0.91144)$ $\theta_1=-45$ $\theta_2=-20.7048$ $\frac{n_1}{n_2}=2$
*Intersection2: Position=$(323.027, 0.19627)$ Normal=$(0.062622, 0.99804)$
*Refracted ray outside the lens: Origin$(323.027, 0.19627)$, Direction=$(-0.75, -0.66144)$ $\theta_1=20.7048$ $\theta_2=45$ $\frac{n_1}{n_2}=0.5$
| You have found the defect of a perfect spherical lens called spherical aberration which manifests itself when dealing with rays which are far from the principal axis.
It is precisely the fact that the lens is spherical which causes this lens defect.
The simple lens formula is only an approximation for rays which are near to the principal axis.
| {
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Is Lcm really equal to Icm x w? Let's suppose a body is rotating about some axis passing through the centre of mass (cm) with a angular velocity $\vec \omega$.
The angular momentum about the centre of mass axis is given by $L = I_{cm} ||\vec \omega||$, with $I_{cm}$ being moment of inertia about the axis.
Now, does it, in any way, mean that angular momentum about the centre of mass (see, I stress the cm point here) is indeed $I_{cm} ||\vec \omega||$?
| The angular momentum of a rigid body rotating about some point is given by $\vec L = \mathrm I_\text{cm}\, \vec \omega$, where $\mathrm I_\text{cm}$ is the moment of inertia tensor about the center of mass. It's neither a vector nor a scalar. There are some special cases where the moment of inertia can be treated as if it were a scalar. Introductory physics students are only given problems where these special cases apply.
Those special cases do not apply in the case of a precessing top or a tumbling textbook. The problem with pretending $\mathrm I_\text{cm}$ is a scalar is that it misses some interesting physics. The tensorial nature of the inertia tensor means that angular velocity and angular momentum are not necessarily parallel to one another.
| {
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Visual interpretation, on the Bloch sphere, when Hadamard gate is applied twice It's known that the Hadamard operation is just a rotation of the sphere about the $\hat{y}$ axis by 90 degrees, followed by a rotation about the $\hat{x}$ axis by 180 degrees.
On the other hand, $H^{2}=I$, where $H$ is the unitary matrix corresponding to the Hadamard gate and $I$ is the identity matrix.
If we do the rotation corresponding to the Hadamard matrix twice, then based on $H^{2}=I$, we would come out to the original situation, right? But, somehow, I can not see that. Could someone shed some light on this problem?
| It can be nice to represent things geometrically. We can represent qubits using vectors, as I'm sure you know. So let's start with a qubit in the $|0\rangle$ state, or in the state $\begin{bmatrix}1\\0\end{bmatrix}$. We can graph this as
(That blue there is the vector representing the qubit.) So, now that we have our initial state, let's apply the Hadamard gate to it. The Hadamard gate is represented by $\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\1&-1\end{bmatrix}$. When we multiply the vector $\begin{bmatrix}1\\0\end{bmatrix}$ and the above matrix, we get the Hadamard gate applied to the zero state, or $\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{bmatrix}$. Graphing that, we get
Now, let's apply the Hadamard gate again. We have our vector $\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{bmatrix}$ and the matrix representing the Hadamard gate, which we can rewrite as $\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{bmatrix}$ for simplicity. Now to multiply the two, of course, we do $\frac{1}{\sqrt{2}}\cdot\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{bmatrix}+\frac{1}{\sqrt{2}}\cdot\begin{bmatrix}\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{bmatrix}$. $\sqrt{2}^2$ is of course $2$, so we get $\begin{bmatrix}\frac{1}{2}\\\frac{1}{2}\end{bmatrix}+\begin{bmatrix}\frac{1}{2}\\-\frac{1}{2}\end{bmatrix}$. This comes out to $\begin{bmatrix}1\\0\end{bmatrix}$, and the graph at the beginning. We're back where we started!
Hopefully this helps; if it doesn't, let me know and I'll try to think up another way to picture it.
| {
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Eigenvalues of fermionic field operators Consider the fermionic field operators $\psi_a(x), \psi^{\dagger}_b(y)$ with the canonical anti-commutation relations $$\{\psi_a(x),\psi_b(y)\} = 0 $$ and $$\{\psi^{\dagger}_b(t,\vec{x}),\psi_a(t,\vec{y})\} = \delta_{ab} \delta(\vec{x}-\vec{y}).$$
What can we say about their eigenvalues? Are they real or Grassmann-numbers?
I'm a bit confused about this at first I would guess they are Grassmann-numbers since $$\psi_a(x)\psi_a(x) = -\psi_a(x)\psi_a(x) = 0$$ but I'm not sure if this conclusion is true.
| *
*An eigenvalue $\lambda$ of an operator $\hat{A}$ (with definite Grassmann-parity $|\hat{A}|$) is a complex supernumber of the same Grassmann-parity. See also this Phys.SE post and links therein.
*Note however that an annihilation operator $\hat{a}$ and a creation operator $\hat{a}^{\dagger}$ of definite Grassmann-parity $|\hat{a}|$ do not supercommute
$$ [\hat{a}, \hat{a}^{\dagger}]~:=~\hat{a}\hat{a}^{\dagger}-(-1)^{|\hat{a}|}\hat{a}^{\dagger} \hat{a}~=~ \hbar~\hat{\bf 1}~\neq~0, $$
and are therefore not supernormal, and hence not diagonalizable, cf. e.g. this Phys.SE post. OP's fermionic fields are field-theoretic versions of Grassmann-odd annihilation & creation operators.
| {
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Can you power a car by using air pressure In my fluid mechanics class, I've learned that a fluid traveling in a pipe will generate a force when the exit area is smaller then the entrance area. Suppose a pipe is attached to a car that will use the kinetic energy of the air passing over the vehicle. Assuming that the car is moving at a constant velocity and the pipe is straight and the exit area is $\frac{1}{4}$ the size as the frontal area, is it feasible for a moving vehicle to generate enough force from the air to reduce energy use?
| I just learned about air pressure, and to my knowledge you can't really generate enough force form just air pressure to power a full-sized SUV going at a constant velocity. However, if you did create some sort of high and low pressure system using the engine and exhaust pipe, you could make the car move, but the system would have to be on a huge scale to generate enough power to make a vehicle go forward.
| {
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"timestamp": "2023-03-29T00:00:00",
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Universe expansion / contraction ambiguity I've read everywhere that the universe is expanding, and accelerating the expansion. But it is our single point observations of the universe enough to resolve this result?
What gives us 100% conclusion that the universe is expanding, and not contracting, aren't this equivalent?
Imagine there is massive black hole (orders of magnitude bigger than all the mass measured in the observable universe), which contains all the remaining mass of the big bang that did not manage to escape its own gravity.
This black hole is outside our observable cone of spacetime.
Every object in the universe is falling towards it.
*
*Object A (close to it) is being attracted a lot
*We are further away from it, but we still get attraction
*Object B (far away from it) would be still falling but slowly.
From our point of view, both objects are accelerating away from us, and their acceleration is not constant but increases over time (as we all get closer to the gravity source).
This effect may even be accentuated by the long distances and the time the gravitational waves take to travel.
| You are right that objects arrayed radially with respect to a black hole will move apart from each other. This would mean they would appear red shifted according to any observer looking along this radial direction. However, along the plane perpendicular to that radial direction objects would appear blue shifted. They would be moving towards the observer. If we lived in a black hole universe then galaxies would appear red shifted or blue shifted in an anisotropic fashion. This is not what is actually observed in the universe.
| {
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Finding the vector potential of a spinning spherical shell with uniform surface charge? I have problem solving the following magneto-static problem. I would greatly appreciate help and guidance.
This is how the problem is stated:
A spherical shell of radius $R$, carrying a uniform surface charge $\sigma$, is set spinning at a angular velocity $\mathbf{ω}$, see figure 5.45. Calculate the vector potential by solving the corresponding partial differential
$\nabla^2 \mathbf{A} = \mu_0\mathbf{J}$.
I have attempted to solve the problem by first writing down what the surface current is, using spherical coordinate system.
$$
\mathbf{J}(\mathbf{r^\prime}) = \sigma \mathbf{v} = \sigma \mathbf{ω} \times \mathbf{ρ} = \sigma\omega r^\prime \sin\theta\:\delta(r^\prime - R)\mathbf{\hat{e}_{\phi}}
$$
where $\mathbf{ρ}$ is the distance from the z-axis, if the z-axis is taken to be the axis of rotation and $\theta$ is the angle measured from the z-axis. From this we conclude that $\mathbf{A}$ only depends on the $A_\phi$ component since the surface current only depends on the phi component, and thus we have $\nabla^2 A_\phi = \mu_0 J_\phi$. And we're left with solving a "simple" poission equation. Our problem clearly has azimuthal symmetry. We then proceed to solve Laplace's equation $\nabla^2 A_{\phi}(r,\theta)=0$. We solve this partial differential equation with help of separation of variables. In the end we find that the general solution is a linear combination of separable solutions.
$$
A_{\phi}(r,\theta)=\sum_{l=0}^\infty \Big(A_lr^l + \frac{B_l}{r^{l+1}}\Big)P_l(\cos\theta)
$$
For the interior region where $(r\leq R), B_l=0$ otherwise the potential would blow up at the origin.
$$
A_{in}(r,\theta)=\sum_{l=0}^\infty A_lr^lP_l(\cos\theta)
$$
In the exterior region $(r\geq R)$ we get that $A_l=0$, because they don't go to zero at infinity.
$$
A_{out}(r,\theta)=\sum_{l=0}^\infty \frac{B_l}{r^{l+1}}P_l(\cos\theta)
$$
We need now to join these to functions together at the boundary $r=R$. The vector potential should be continous at the boundary. So we have the following boundary conditions
$$
\label{eq:one}
A_{out}=A_{in}
$$
$$
\Big( \frac{\partial A_{out}}{\partial r}-\frac{\partial A_{in}}{\partial r}\Big)\Big|_{r=R}=-\mu_0J_{\phi}=-\mu_0\sigma\omega R\sin\theta
$$
From the first boundary condition we find that
$$
B_l=A_lR^{2l+1}
$$
Plugging it in the second boundary condition equation we get
$$
\sum_{l=0}^\infty(2l+1)A_lR^{l-1}P_l(\cos\theta)=\mu_0\sigma\omega R\sin\theta
$$
Using fouriers trick by multiplying both sides with $P_{l^\prime}(\cos\theta)\sin\theta$ and integrating. We can determine the coefficients $A_l$ since the Legendre polynomials are orthogonal functions,
$$
\int_0^1 P_{l^\prime}(x)P_l(x)\textrm{d}x = \int_0^\pi P_{l^\prime}(\cos\theta)P_l(\cos\theta)\sin\theta\:\textrm{d}\theta = \frac{2}{2l+1}\delta_{ll^\prime}
$$
$$
A_l= \frac{\mu_0\sigma\omega R}{2R^{l-1}} \int_0^\pi \sin\theta\:P_l(\cos\theta) \sin\theta\:\textrm{d}\theta
$$
This is where I hit a brick wall. I have attempted to solve the problem by expressing $\sin\theta$ in terms of Legendre polynomials and then use the orthogonal condition to determine the coefficients $A_l$. I know that $\sin^2\theta$ can be expressed using Legendre polynomials in the following way.
$$\sin^2\theta=\frac{2}{3}(P_0(\cos\theta)-P_2(\cos\theta))$$
But if I just want to have $\sin\theta$ I need to take the square root of everything and then it just gets really messy. Where did I go wrong in my solution?
| The $\phi$ component of $\nabla^2\mathbf A$ is not $\nabla^2 A_\phi$, but
$$ (\nabla^2 \mathbf A)_\phi = \nabla^2A_\phi - \frac{1}{r^2\sin^2\theta} A_\phi + (\text{terms involving $A_r,A_\theta$}) $$
Thus what we need to solve is not a Laplace equation. The solutions are the associated Legendre polynomials $P_l^1(\cos\theta)$:
$$ A_\phi = \sum_{l=1}^\infty \left( A_l r^l + \frac{B_l}{r^{l+1}} \right)P^1_l(\cos\theta). $$
In particular, $P^1_1(\cos\theta)=-\sin\theta$, which should solve your problem neatly.
| {
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What is the need for angular magnification We've been told that angular magnification is defined as the ratio of the angle subtended by the image to the angle subtended by the object. But why would we need a quantity called angular magnification? Won't the simple magnification formula do?
| Linear magnification is only useable in situations where lenses produce real images such as projection onto a screen. Optical instruments with eyepieces such as microscopes produce virtual images whose linear dimensions cannot be measured. Therefore, with these devices, angular magnification is used.
Read more: https://en.m.wikipedia.org/wiki/Magnification
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Is voltage across parallel branches still equal if there's a break?
I was wondering if there is a break in one branch, will the potential difference across it still be equal to that of the branches parallel to it? I know that the top wire would still have to be at the same potential as the other top wires because the wire is an equipotential, and the same idea applies to the bottom wire, which would give it the same potential difference across the branch. However, if no current can flow across the resistor, where would the potential difference come from? Also, would anything different happen if instead it were a capacitor in the broken branch instead of a resistor (like in an RC circuit when the capacitor is almost completely charged and acts like a break in the circuit)? Thank you!
| The potential across each resistance is the same, that which is established by the battery (it's EMF). This ignores any resistance in the wires. The break in the middle resistor wire in your diagram simply causes that path to act like a resistor with infinite resistance; again, the same as measuring the potential directly across the battery.
The voltage across a single capacitor replacing one of these resistors would also be equal to the voltage across the battery. Again, the EMF of the battery and ignoring any resistance in the wires.
| {
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Conceptually, why is acceleration due to gravity always negative? As the title states, why is acceleration due to gravity always (-). Say you assign "up" as the positive direction. If an projectile is thrown at a 24 degree angle above the horizontal, I get that acceleration due to gravity before the vertex is negative. However, why is it not positive after the vertex? If acceleration due to gravity is negative and we assign downwards as negative, wouldn't that make acceleration positive?
What I think is that acceleration due to gravity is always towards the ground. Even if a projectile is going downwards, and we assign downwards as (-), the acceleration due to gravity is still (-), because the object still accelerating downwards. Despite it going upwards or downwards, the net acceleration of the object is downwards.
| It is about the perspective we are using about the forces. When you put a +Q ball near to another +Q ball you increase the potential energy of the system against the Electrical forces. When you leave the system electrical forces will then push both balls a far from each other. So when we make work against a system we consider it + because we increase the potential energy of the system but when natural forces make work just the opposite way we did we say that it's - because it works "against" what we did. If we were to be electrical forces we would say that we did + work and the others make - work.
On your case it's really the same. We work against gravity to increase the height of a thing and when we stop gravity takes it back.
| {
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Can there be really a theory of everything? If quantum theory becomes compatible with Einstein's theory of gravity, we get the theory of everything. But if it can predict anything in the universe (provided boundary equations are given) wouldn't it predict itself. I mean won't the theory tell us when the theory will be formed? This sounds quite absurd.
| There are three flaws in what you're suggesting :
*
*A theory of everything need not be deterministic. Current mainstream thinking would be to expect a non-deterministic theory of everything.
*Even if a deterministic theory was found, you can't use it to deduce when it will be found until after you have it. And that's ignoring how unrealistically complex such a deduction would be.
*A theory is a model for reality. It simply models the behavior we see around us to some level of accuracy and in some circumstances. A theory of everything is the ideal theory, but there's no reason to assume we can actually develop such a theory or would ever stop refining it as the accuracy we can measure at increases and reveals more detail than we thought was there. A theory of everything could be a dragon's tail that we chase forever.
| {
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Why are the Xi baryons called "cascade" also? Between the Lambda, Sigma, Xi, and Omega baryons, only the Xi family has an alternate name - cascade. This must be a result of the "particles zoo" and my guess is that the particles cascade in the accelerator is related. Why is this family the only one with this "unconventional" name?
| The first observation paper:
Neutral Cascade Hyperon Event
Luis W. Alvarez, Philippe Eberhard, Myron L. Good, William Graziano, Harold K. Ticho, and Stanley G. Wojcicki
Phys. Rev. Lett. 2, 215 – Published 1 March 1959
Has the world Cascade Hyperon in the title
And in this wiki article
Ξ0 and Ξ− are also known as "cascade" hyperons, since they go through a two-step cascading decay into a nucleon.
If the discovery paper did not mention the new particle as Ξ ( I do not have access to the PRL) it is the same reason as the J/psi double name, two different namings, the "cascade" part honoring the first discovery.
Ξ is the next unused letter after Λ , μ being used for muon, and ν used for the neutrino.
| {
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Is weight a scalar or a vector? My professor insists that weight is a scalar. I sent him an email explaining why it's a vector, I even sent him a source from NASA clearly labeling weight as a vector. Every other source also identifies weight as a vector.
I said that weight is a force, with mass times the magnitude of gravitational acceleration as the scalar quantity and a downward direction.
His response, "Weight has no direction, i.e., it is a scalar!!!" My thought process is that since weight is a force, and since force is a vector, weight has to be a vector. This is the basic transitive property of equality.
Am I and all of these other sources wrong about weight being a vector? Is weight sometimes a vector and sometimes a scalar?
After reading thoroughly through his lecture notes, I discovered his reasoning behind his claim:
Similarly to how speed is the scalar quantity (or magnitude) of velocity, weight is the scalar quantity (or magnitude) of the gravitational force a celestial body exerts on mass.
I'm still inclined to think of weight as a vector for convenience and to separate it from everyday language. However, like one of the comments stated, "Definitions serve us."
| I think your professor is mixing up terms. Mass is scalar, weight is a vector. But many people get into the habit of using the terms interchangeably. Also, don't always believe everything someone in a "superior position" tells you. Sometimes they are wrong, so question everything.
| {
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Why do we invent non-physical concepts (like e.g. point particles) to study physical phenomenons? There is nothing exist like point particles in reality then why did we invented the notion of point particles and how does it relate to real world?
| To be able to understand very physical properties and ideas, simpler theories and concepts have to be developed.
Physics on the whole is not perfect, it is a way to model what is seen around the world. With something like a planet, understanding what happens to the mass if it is considered as a point source simplifies a lot of the mathematics and gives a very good approximation. In fact, a lot of very complicated problems cannot be solved if objects are not considered as point particles.
Applying a force for example if the object is not considered a point source would be very difficult. The force would have to be summed over for all the particles in a system which is not feasible. Instead if the object is simplified it is much easier to understand what is going to happen to he object when it is for instance interacting with something else.
Also, especially in fields such as Quantum Mechanics, non-physical concepts are very important (especially at an undergraduate level) to gain very good intuition into the subject area and understand reasons for why things happen.
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Is a tensor which is symmetric in two indices still symmetric after raising/lowering one index? I have had this question for a while. I have yet to find information on this online or use this property in any calculation. I believe myself to have proven that it will still be symmetric but I am somewhat unsure of my proof.
Suppose you have $A^{ij}=A^{ji}$, then,
$$ A^{ij}=A^{i}_{\phantom{i}k}g^{kj}=A^{\phantom{k}i}_{k}g^{kj}=A^{ji}\tag{1}$$
So $$A^{i}_{\phantom{i}k}g^{kj}=A^{\phantom{k}i}_{k}g^{kj}\tag{2}$$
Which implies $$A^{i}_{\phantom{i}k}=A^{\phantom{k}i}_{k}.\tag{3}$$
Is this true?
| TL;DR: No, it is per definition not symmetric, although OP's eq. (3) indeed holds.
In more detail, if
$$S~:=~S^{ij}~e_i \otimes e_j~\in~ T^2V~=~V\otimes V \tag{A}$$
and
$$g~:=~g_{ij}~e^{\ast i} \otimes e^{\ast j}~\in~ T^2V^{\ast}~=~V^{\ast}\otimes V^{\ast}\tag{B}$$
are symmetric tensors
$$S^{ij}~=~S^{ji}\qquad\text{and}\qquad g_{ij}~=~g_{ji},\tag{C}$$
then the mixed tensor
$$ M~:=~ M^i{}_j~e_i \otimes e^{\ast j}~\in~ V\otimes V^{\ast},\tag{D}$$
given by
$$ M^i{}_j~:=~S^{ik}g_{kj},\tag{E} $$
is not symmetric. For starters, the transposed tensor
$$ M^T~:=~ (M^T)_i{}^j ~e^{\ast i} \otimes e_j
~:=~M^j{}_i ~e^{\ast i} \otimes e_j~\in~ V^{\ast}\otimes V\tag{F}$$
lives in a different space, so the potential symmetry condition
$$M~=~M^T \qquad\qquad(\longleftarrow \text{Wrong!}) \tag{G}$$
is formally meaningless. What holds instead is
$$ M^j{}_i~=:~(M^T)_i{}^j~=~g_{in} ~M^n{}_m~ (g^{-1})^{mj}~=:~M_i{}^j. \tag{H}$$
Example: If
$$S^{ij}~=~\begin{pmatrix}0&1\cr1&0\end{pmatrix}
\qquad\text{and}\qquad g_{ij}~=~\begin{pmatrix}1&0\cr0&2\end{pmatrix},\tag{I}$$
then the matrix
$$ M^i{}_j~=~\begin{pmatrix}0&2\cr1&0\end{pmatrix}\tag{J}$$
is not symmetric.
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Quantum mechanics and gender of unborn baby Having a debate with my other half here, I'm currently pregnant and we have not had any scans or tests to determine gender of the baby.
I'm seeing this as similar to a Schroedinger's cat experiment in that my baby is both male and female until it is determined either way. However my partner believes that it is unequivocally fixed because the sperm and egg determined the gender, the fact that it has not yet been observed is irrelevant.
I know that this is all mainly theory but I wanted to get a physicist's opinion. How determined is my baby's gender right now?
| Schroedinger's Cat is a thought experiment and an analogy used to explain superposition and indeterminacy. Instead of a cat you should actually understand a quantum system. This system is under the rules of Quantum Mechanics and in the thought experiment it is in a superposition of two states (live and dead, up and down, whatever, those are just labels). Once you measure the system (open the box) you project the superposed state into one of its components (live or dead).
Now it comes to the question: Is a fetus, or even a zygote, a quantum mechanical system? It is definitely small but yet too big to show effects originated from state superposition. Otherwise one would detect a interference pattern in a double slit experiment made out of cells. This has never been observed. The largest structure showing such a quantum mechanical effect is a structure called bucky-ball which is made out of 60 atoms. On the other hand, a single cell has approximately $10^{14}$ atoms. That is too much to show quantum effects. So the zygote is not in a superposition of states (male and female) and the baby genre's is determined right after the fecundation.
| {
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Why do we not feel weightless at equator but feel in satellite A person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. My question is, why does he/she not feel weightless as a satellite passenger does?
If we compare a geostationary satellite with the earth's equatorial surface then we know they both revolve around the centre of earth with same Angular velocity. So if Normal force is zero in satellite then why not at equator's surface.
| The fact that there is a special orbit called "geostationary" should ring a bell. What is so special about this orbit, corresponding to a specific radius? It is the only radius at which the satellite's angular speed is the same as the angular speed of a point on the surface of the Earth (the point has to be on the Equator).
For all the other orbits the orbital speed will result in different angular speeds. All orbits below this have faster angular speeds. All the orbits above the geostationary have slower angular speeds.
For example, for a satellite orbiting just a little bit above the Equator, the orbital speed is about 8 km/s whereas the points on the Equator move with a little less than 500 m/s. So there is a factor of about 16 between the two speeds which corresponds to a factor of about 256 between accelerations (centripetal acceleration is proportional to speed squared).
This explain why moving with the Earth you don't feel weightless. At the speed of the satellite, the centripetal acceleration is equal to the gravitational acceleration and you feel weightless. If you move with the Earth, your centripetal acceleration is about 250 times less than the one of the satellite corresponding to the very low orbit. The effects is a reduction of gravitational acceleration by a factor of about 260 or about 10/260 or about 0.04 $ m/s^2 $
| {
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Green's function for a driven, damped oscillator There's an example given in chapter 4 (Differential Equations) of Mathematical Tools for Physics by James Nearing:
$$m\ddot{x}+kx=F_{ext}(t) \, .$$
Obviously this is a undamped driven oscillator. The exercise is to solve the equation using the method of Green's functions. After an impulse, the motions follows $A \text{sin}(\omega _0(t-t^{'}))$. The change in momentum is $m \Delta v_x=F\Delta t'$, and $\Delta t$ is sufficiently small such that the mass is subject only to $-kx$. Just after $t=t'$, $v_x=A \omega_0=F \Delta t' / m$ $\therefore A=F \Delta t' / \omega _0$. Then the solution $x(t)$ is:
$$x(t)=\begin{cases}
(F\Delta t' / m \omega_0) \text{sin}(\omega_0(t-t')) & t >t' \\
0 & t\leq t'
\end{cases}$$
The problem I've been given is to solve the equation for a damped, driven oscillator using the same method.
$$m \ddot x+b \dot x+kx=F_{ext}(t)$$
So my idea is that the change in momentum is the same, but afterward the mass is subject not only to $-kx$ but also to $-b \dot x$. So after $t=t'$,
$$v_x=A \omega _0=\frac{\Delta t'(F-b \dot x)}{m}; \therefore A=\frac{\Delta t'(F-b\dot x)}{m \omega _0} \, .$$
Is this reasonable? I just want to make sure I'm on the right track, since this is my first attempt at solving the problem in this way.
| When you have damping, the motion will not be a pure sinusoid anymore. You will have an exponential decay as well: $$x(t>t_0) = A \sin (\omega (t-t_0)) \exp(-\gamma (t-t_0)).$$
Try to compute the values of $\omega$ and $\gamma$, and you can use the Green's function technique after that.
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Why electron spin being anti-parallel to the orbital angular momentum gives lower energy than the electron spin being parallel to the angular Take hydrogen atom as example. The 2P orbital will split due to spin-orbit coupling to the $^{2}P_{3/2}$ and $^{2}P_{1/2}$ terms.
The term $^{2}P_{1/2}$ has anti-parallel orientation of electron spin and orbital angular moment and is lower in energy than the $^{2}P_{3/2}$ which has parallel spin and orbital angular moment.
Why is that and not the other way around?
From electrodynamics we know that magnetic potential energy is minimal when two magnetic dipoles are parallel which confuses me here.
| The magnetic potential energy of two magnetic dipoles is minimal when the magnetic dipoles are anti-parallel. You can easily check this with two bar magnets.This also explains why anti-parallel electron spin and orbital angular momentum (and thus anti-parallel magnetic dipole moment) gives, in general, lower energies than parallel orientation.
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Does the system of a spherical and massless mirror, at rest in our reference frame, with photons moving around in it, has a rest mass? Can we say that if a big number of photons is travelling in a perfect, massless spherical mirror which is at rest in our reference frame, the combined system of the massless mirror and the photons moving in it has a rest mass?
| Yes. Imagine that there are only two photons, one with 4-momentum $(p,p,0,0)$ and the other with $(q,q cos(\theta),q sin(\theta),0)$, the sum of their 4-momentum will be $(p+q,p+q cos(\theta), q sin(\theta), 0)$, and because we take the mirror as massles (I will asume that it doesn't have a 4-momentum asociated with no one more thing than it's movement, like if it wass massive and we trow it's mass to zero), then the mirror+photons inside will have a rest mass $(p+q)^2-(p+q cos(\theta))^2-q^2 sin^2(\theta)=2pq(1-cos(\theta))$. A more detailed analysis will take into account the energy asociated with the potential that forbidds photons traveling outside it, probably.
| {
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What determines the direction of a path on a line integral (vector case)? Line integrals are very important to use in Physics. For example, we calculate work by: W=∫<F,dr>. But I just got confused about something. What determines the direction of motion? The integral limits, or the vector dr?
Well, when we do the internal product of the Force by the path(dr), we are aligning this force on the path's direction(I mean, the shape of the curve) . But if I want a path from the position B to the position A, I determine this inserting on Integral's limits(B inferior limit and A superior limit), or the vector dr would indicate the direction of motion (from B to A) and the integral limits would be (A inferior, B superior)?
| The correct form of an integral over a path takes the form $\int_{\overrightarrow{r_i}}^{\overrightarrow{r_f}}d\overrightarrow{r}$. The direction of motion was determined when you chose the path and the limits were determined when you chose the path.
If you know the velocity, the direction of the path is obvious because $\frac{d\overrightarrow{r}}{dt}=\overrightarrow{v}$.
| {
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Why do we regard $W^\pm$ as physical fields but not $W^1$ and $W^2$? When expanding the electroweak Lagrangian, we obtain terms like
$$
\mathcal L = \frac{g}{2}(\bar e_L (W^1_\mu + iW^2_\mu)\gamma^\mu \nu_L + \bar \nu_L (W^1_\mu -iW^2_\mu )\gamma^\mu e_L)+ \dots
$$
and then redefine the fields as $W^\pm =W^1_\mu \mp iW^2_\mu$.
It seems that we don't want two different gauge fields interacting with the same fermions, but I don't know why is this. Shouldn't any linear combination be on the same footing? Although I guess some eigenstates are phenomenologically more convenient.
| It's not that we don't want two different gauge fields interacting with the same fermions, it's that you can clearly see from this Lagrangian that if an antineutrino interacts with an electron, they can produce a particle given by $W^- =W^1_\mu + iW^2_\mu$. The two gauge bosons that are produced are coherent, so they behave as a single particle. Since there are no terms in the Lagrangian that couple to only $W^1_\mu$ or $W^2_\mu$, there is no process in nature that will produce just one of them, so it is more convenient to redefine the fields in terms of single particles that are actually produced.
| {
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What does it mean if an intertwiner respects a group action? This is more of A maths question however it cropped up when I was researching isospin so I think it belongs here. I am wondering what exactly is meant by 'respects' and how it works in the context of isospin and hilbert spaces. The link to the source I have been using is: http://math.ucr.edu/~huerta/guts/node4.html#sec:isospin
Specifically about half way down, the paragraph has the words 'Intertwining operator' in bold text
| There are some details in [this paper][1], On the page 95 it says"...an orthonormal basis of the
invariant subspace of a tensor product of vector spaces, are usually known as
intertwiner operators or simply intertwiners.
"
[1][Daniele Regoli]-The relation between Geometry and Matter in classical and quantum Gravity and Cosmology (PhD thesis): https://arxiv.org/abs/1104.2910
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How does voltage affect boiling ability of a kettle? I was reading an article about the lack of electric kettles in the USA - the claim at the end states one reason is the lower voltage at the socket. Personally, I think that the main reason is that morning tea is not as big in the US as it is in the UK or Australia, but that's not the point.
Would a 120V kettle really take a "really, really long time" compared to a 240V one? Couldn't it just draw more current?
| The boiling time is a function of power, not voltage. However, to get the same power at a lower voltage requires a proportionally higher current. Yes, a 110 V kettle could draw more current to get to the same power as a 220 V kettle, but there are limits and downsides to drawing more current:
*
*The cord needs to be thicker. More current means more copper crossection area. That costs more, and makes the cable thicker and less flexible.
*Outlets have limited current. The typical "standard" outlet in the US can provide 15 A. 20 A versions are available. For high power appliances, like a electric range or clothes dryer, a special 220 V outlet is used. These are large and klunky, and not something you find in a typical kitchen. As a result, manufacturers don't make kettles with such plugs.
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How Do I Increase The Range Of An Home Made Electromagnet? Please forgive me, I am new to this forum and I am not a physics guy but any help would be appreciated. I would like to know how to I increase the range or reach of my electromagnet. By range, I mean the distance from my magnet to a metal object (paper Clip). Currently, my electromagnet has to touch the paper clips to affect them, but I what it to be able to attract them by just coming close with out having to touch. What materials and techniques would be best to achieve this? Thanks for reading
Jay
| *
*Increase the current
*Increase the number of turns of wire
*Place an iron core inside the coils of wire
*Bend the electromagnet into a horse-shoe shape
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Is the magnetic quantum number dependent on my choice of coordinate system? I'm reading on wikipedia: https://en.wikipedia.org/wiki/Magnetic_quantum_number
that
"The axis used for the polar coordinates in this analysis is chosen arbitrarily. The quantum number $m$ refers to the projection of the angular momentum in this arbitrarily chosen direction"
I don't think I understand this. If m is the projection, to me it seems like I can change the axis, and make the projection, hence m, any number between $0$ and $l$. But I don't think it is possible to change a physical quantity like that by just changing the coordinate system. What is wrong with my understanding?
| If we denote $\hat{L}^2$ as the total angular momentum (squared) and $l_i$ its projection onto the i-axis than:
$$[\hat{L}^2, l_i] = 0 \phantom{text}\text{but}\phantom{text}[l_i,l_j] \neq 0$$
This means that, given some state $\psi$ we cannot measure its total angular momentum and all projections. The "best" thing that we can do is to make a measurement of its total angular momentum $L^2$ and pick out one projection e.g. $l_x$ and call this last quantity m.
What they mean with "The axis used for the polar coordinates in this analysis is chosen arbitrarily" is that we could also have measured $l_y$, $l_z$ or any linear combination of $l_x,l_y,l_z$ and called it m.
Measuring some projection say $l_x$ changes the state of the system such that there is no memory of its $l_y$ and $l_z$ values. This is demonstrated in the Stern Gerlach experiment
The source emits a beam of electrons. We measure its projected spin onto the z axis(which corresponds to the choice $l_z=m$) and eliminate those with $l_z = -1/2$.
Than we try to measure its projected spin onto the x axis(which corresponds to the choice $l_x=m$. We observe both $\pm1/2$ as was to be expected.
In a final stage we remeasure its projection onto the z axis. We observe both $\pm1/2$.
What happens here is the following
$$Z_+ = \frac{1}{\sqrt{2}}(X_+ +X_-)$$
$$X_+ = \frac{1}{\sqrt{2}}(Z_+ -Z_-)$$
so after the first measurement of $Z_+$ we know that $l_x$ can be either + or - 1/2. Once we measure $l_x$ to be +1/2 we lose all previous information about $l_z$ so that it may be $\pm 1/2$ again.
So indeed, making a measurement of $l_x$ changes the value of $l_y$ and this has been proven experimentally as explained above.
And because you define m to be the projection that you just measured you are indeed changing m by performing these measurements.
I hope this helps :)
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Why is potential difference in ECG considered a vector? In the electrocardiogram, there are leads placed on the limbs and an "electrical vector" is calculated as the difference in potentials measured I the two leads. The direction is given by the line joining the position (or representation of) of the two leads.
I don't understand what an electrical vector is because we were taught in school that neither potential (difference) nor current is a vector. Only electric field is. Are we measuring the field? I am a medical student and have just high school understanding of physics. Could someone pls explain what it means?
What I know - that ECG is a representation of the propagating wave front of electric conduction in the heart.
enter link description here
| In an anisotropic material (like the human body) the Ohm's law is written as
$$E_i=\rho_{ij}J_j$$
Where $\rho_{ij}$ is the resistivity tensor, $J_j$ the current density and $E_i$ the electric field.
The voltage between two points is defined as
$$ \Delta V=-\int_{\ell}\mathbf E\cdot d\mathbf l $$
where $\ell$ is the line connecting the two points.
As you correctly note $\Delta V$ is a scalar, while $\mathbf E$ is a vector but when you make a voltage measurement between two points of an extended material (and not between two points of an idealized 1-dimensional wire) the result of your measurement will depend not only on the distance between these points, but also on the exact position. This is because different points are connected by different paths that are characterized by different resistance (because $\rho_{ij}$ is a tensor and not a scalar).
So, my guess is that even if you are measuring a scalar quantity during an ECG, the result is heavily dependent on the vector quantities involved and that justifies the improper way of speaking about eletrical vectors.
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Why doesn't saturation current in the photoelectric effect depend on the frequency of light absorbed by the metal emitter? If current $I$ is given by $I = nAev$, where $n$ is the number of electrons per unit volume, $A$ is the area, $e$ is the charge of an electron and $v$ is the velocity of the electron, it must mean that the current increases with increase in velocity of the electron which increases with the frequency of light incident on the metal emitter. Why doesn't then saturation current increase with increase in frequency?
| In the relation, $I=neV_dA$, the velocity is the drift velocity of the electrons inside the conducting material.
The saturation current depends on the intensity of the light,for a given frequency. More photons correspond to more electrons. One photon knock out one electron and if you increase the number of photons (or increase the intensity) you increase the current.
If you increase the frequency of the photon, you increase the electron's kinetic energy outside the surface. When the increase you increase the frequency you increase stopping potential but the saturation current remains the same for a given intensity.
At zero potential, the value of photoelectric current is higher for the photon whose frequency is greater than the other frequency of other photons. But the saturation current doesn't depend on the frequency. When the Photo-electons collide with the anode plate they lose some of their kinetic energy. More energy implies loss of more energy.
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Probability of finding the oscillator in some position in $[x,x + \mathrm d x]$ Given that the position of a one-dimensional harmonic oscillator is given by
$$x(t) = Acos(\omega t + \phi)$$
where $A,\phi,\omega \geq 0$ are constants of real numbers I'm trying in some sense
find the $p(x)\mathrm dx$ probability of finding the oscillator between position $x$ and $x+\mathrm d x$.
There is a hint that this is the same as calculate $\mathrm dT/T$ where $T$ is the period of oscillation and $\mathrm dT $ is an interval of time within the period.
My attempt: So, I'm trying to calculate using $\mathrm dT/T$ but I do not know how to do it. If we have that $x(t+T) = x(t)$ then we use $\mathrm dx / \mathrm dT = (\mathrm d x/\mathrm dt)(\mathrm dt/\mathrm dT )$ but this do not go much further. I also tryed to look at other derivatives as a trick of considering the constants as variables but for all the constants I got stuck in calculations that didn't go anywhere; for exemple for $\omega$ I just found $\mathrm d T/ T = -\mathrm d \omega/\omega$. Do not give a full answer, give just a hint so that I can conclude the question.
Physical concept involved: This is a statistical mechanics half-problem; the harmonic oscillator is one of the few examples of problems that it is possible to chek the validity of two important elements of the theory: Ergodic Hypothesis and equal-a-priori postulate.
| This can also be solved using the standard microcanonical method given in statistical mechanics textbooks. Start with the Hamiltonian:
$$ H = \frac{p^2}{2m} + \frac{1}{2} k x^2 $$
It gives us an ellipse in phase space as the phase trajectory. Calculate the total number of microstates ($\Sigma_{tot}$) with energy less than or equal to $E$, as the area inside the phase trajectory divided by $h$. This gives us:
$$ \Sigma_{tot} = \frac{\pi a b}{h} = \frac{2 \pi E}{h} \sqrt{ \frac{m}{k} } $$.
Then the total number of microstates ($\Omega_{tot}$) with energy between $E$ and $E + \delta E$ is
$$ \Omega_{tot} = \frac{\partial \Sigma_{tot}}{\partial E} = \frac{2 \pi }{h} \sqrt{ \frac{m}{k} } $$
Now we need the number of microstates with a position between $x$ and $x + dx$, $\Omega$ favourable. $\Sigma$ favourable is the region inside the ellipse positioned at $x$. This has to be divided by $h$. It has a width of $dx$, and the height is twice the height of the ellipse.
$$ \Sigma_{fav} = \frac{1}{h} dx \ 2 \sqrt{2mE-mkx^2} $$.
Then $\Omega_{fav}$ is equal to:
$$ \Omega_{fav} = \frac{\partial \Sigma_{fav}}{\partial E} = \frac{2mdx}{h \sqrt{2mE-mkx^2}} $$
Then the probability of the particle being between $x$ and $x+dx$ is
$$ p(x)dx = \frac{\Omega_{fav}}{\Omega_{tot}} = \frac{\sqrt{k} dx}{\sqrt{2} \pi \sqrt{E-\frac{1}{2}kx^2}} = \frac{dx}{\pi \sqrt{A^2-x^2}}$$
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Hypercharge and Isospin as additive quantum numbers in $SU(3)$ flavour symmetry I am studying the $SU(3)$ flavour symmetry and I'm reading that we use the fact that hypercharge and isospin are additive quantum numbers in order to decompose the tensor products of the fundamental representations $3\otimes3\otimes3$ and $3\otimes{\overline3}$ in the direct sum of the irreducible components.
I don't understand why, from a physical and mathematical points of view, hypercharge and isospin should be additive.
| The (strong) hypercharge operator $Y$ and the (strong) isospin operators $(I_1,I_2,I_3)$ are generators of a $u(1)\oplus su(2)$ Lie subalgebra of the Lie algebra $su(3)$ of flavor symmetry. A Lie algebra is a vector space, and hence has a linear structure. Representations can be decomposed in eigenspaces for $Y$ and $I_3$.
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Musical notes and colors of a rainbow I have wondered that in an octave in piano there are seven primary notes, and also we observe mostly seven primary colors of a rainbow. I know we perceive logarithmically, that means we only care about relative differences.
Is there any relation between $7$ musical notes (in an octave) and $7$ colors of a rainbow?
EDIT: I agree that the primary term for the $7$ notes in an octave is more or less the matter of taste. However, if we take the western musical taste as a guide, we can justify ourselves to use $12$ notes in an octave and place piano keys in the present way. Take a look at here.
| As requested in comments:
There is a connection in the sense that Isaac Newton regarded both musical harmony and optical physics as branches of mathematics (Kepler did the same with harmony and astronomy, and this kind of thing was not original to them), and deliberately chose 7 rainbow colours to match the common Western scale, despite his poor eyesight initially only spotting 5 colours; he later added orange and indigo
Wikipedia's article on the rainbow says
Newton chose to divide the visible spectrum into seven colours out of a belief derived from the beliefs of the ancient Greek sophists, who thought there was a connection between the colours, the musical notes, the known objects in the Solar System, and the days of the week.
and includes a reference to a 2004 article by Niels Hutchison, MUSIC FOR MEASURE, On the 300th Anniversary of Newton's "Opticks"
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How is entropy a state function? Is there only one reversible way to move from one state to another?
If we consider two states $A$ and $B$ on an isotherm and we move from $A$ to $B$ by first reversible isochoric process and then reversible isobaric process. Now the path followed should be reversible since both the processes were reversible. But what about simply following the reversible isothermal process?
According to me both processes should be reversible. Now entropy is the heat added reversibly to move from one state to another divided by the temperature at which it is added. But we know that the heat added to the system is different in both the cases. Then how is entropy a state function?
| There are generally many reversible paths between different states. As an example consider a Carnot cycle. If I start at the beginning of the cycle at the point where the working medium is fully contracted I can, since every step is reversible, get to the point half way through, where the working medium is fully expanded, by going around the cycle in either direction, giving me two different paths between the two states.
Notice however that both these paths have two stages. On one you expand isothermally and then adiabatically, and on the other you apply the same steps in the reverse order.
There are, however, limits on which states can be accessed on certain types of path. In particular you can only move between two states adiabatically if they lie on the same adiabat and can only move between them isothermally if they lie on the same isotherm.
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Difference between note and tone: which one has only one frequency? What is note and tone both in physics and musical terms? Are the two used in different ways in the different fields or are they the same thing. Moreover, which one has only one frequency?
Well this one got me really confused: https://music.stackexchange.com/q/3262/.
| Both note and tone are terms used in music rather than physics. Both refer to the pitch of the sound but in music they include other information also. As far as physics is concerned all 3 terms mean essentially the same thing : a sound of a particular frequency.
In music, notes are usually identified by letter - eg middle C - instead of frequency (261.6 Hz). The term can also refer to duration - eg crotchet, quaver, minim. Tone also refers to the quality or timbre of the sound.
A tone played by a musical instrument - or even an electronic instrument - is rarely a single pure frequency. It usually includes overtones which are multiples of the fundamental frequency. The relative strength of the different overtones compared with the fundamental gives the tone its quality. Quality describes the difference between the same note played by different instruments, also by the same instrument under different conditions.
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Field inside a plate capacitor with a dielectric material Suppose I have two plate electodes with a dielectric material between them with a permitivity of $\epsilon=10$. I now put a voltage V between them. What is the electric field in the dielectric region?
Well since the electrodes are plates we have simply:
$E = V/d$ which is independent of $\epsilon$.
On the other hand, my intuition tells me that this should depend on $\epsilon$. What is wrong with the argumentation above?
| In general, a dielectric will decrease the $\vec E_{in}$-field inside the plates as this dielectric will become polarized.
You are correct in pointing out that $V=E_{in}\times d$. Here,
$\vec E_{in}$ is the net electric field, which is the sum of the external electric field and the polarization, so that the magnitudes are related by $E_{in}=E_{ext}-E_{pol}$.
With a dielectric this net $\vec E_{in}$ would be smaller than without a dielectric, so $V$ would be smaller for a given separation.
If you insist on keeping $V$ fixed, the external electric field $\vec E_{ext}$ can be made larger than if there was no dielectric. This external electric field $\vec E_{ext}$ is determined by the charges present on the plate, i.e. for the same voltage difference you can accumulate greater charge on the cap.
Since $Q=CV$ or $C=Q/V$, if you can increase the charge but still keep $V$ fixed, you have increased the capacitance.
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How does inflation get rid of initial Baryon asymmetry (if any)? Why is it said that even if the baryon asymmetry existed as an initial condition, the asymmetry would have been destroyed by inflation. How does inflation get rid of initial Baryon asymmetry (if any)?
EDIT: For reference see the section 1.1 (last line of second paragraph) of this review.
| There can be a relation with inflation and leptogeneis which also can lead to asymmetric baryogenesis. The inflation will indeed smooth out everything, but there were no baryons then, as @Anna correctly states. Could not be, temperature was too high for baryons, though if there had been they would have been smoothed out by many orders of magnitude.
It's a problem and not really likely. Besides the temperature not being right and there being no baryons then (a sort of not too small problem), there is an extreme fine tuning that would have been required. The smoothing was 50 or more e foldings, whereas the observed asymmetry now is about 1 part of about $10^{7}$ would have required some real fine tuning before inflation, Of many orders of magnitude. That mechanism is deemed to have been not likely to have been the cause of the observed remaining asymmetry.
The Phys Rev review said that if there had been baryon asymmetry at the Big Bang (i.e., primordial) it would have been diluted. There were none till reheating, but the statement is not inaccurate, just misleading.
Inflation does however enter in, and there are models of baryogenesis and leptogenesis that depend on the model for the inflation field, the Inflaton. Thus, depending on the inflation field model (and there are many) there are papers that show an asymmetry from the inflation field that then leads to asymmetric leptogenesis including parity breaking neutrinos) and baryogenesis, all after inflation and during or after reheating. Some of the models show how the baryon asymmetry then comes about. But really nothing with it being diluted by inflation.
See also various other papers like https://arxiv.org/abs/hep-ph/0103229, or Google leptogenesis and inflation.
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On the derivation of the canonical distribution in deriving the canonical distribution $\rho_c$ for a sistem (labeled by 1) of N identical particles in thermal equilibrium with a heat bath (labeled by 2), at one point I get into the Taylor expansion of the entropy of the reservoir . It is justified with the assumption that the dimensions of the latter are way more big that the ones of the system,
$$ E_2 \gg E_1; \quad N_2 \gg N_1; \quad E=E_1 + E_2 $$
So, doing the expansion,
$$ S_2(E_2 = E - E_1) \approx S_2(E_2 = E) -E_1 \Big(\frac{\partial S_2}{\partial E_2}\Big)_{E_2 =E} + (...) =k_B ln[\Gamma(E_2 = E - E_1)]$$
So, as I've already shown that $\rho_c \propto \Gamma_2(E_2 = E - E_1),$ taking the exponential of the expansion I can write
$$ \Gamma(E_2 = E - E_1) \approx exp\Big[\frac{S_2(E_2 = E)}{k_B}\Big]\cdot exp\Big[\frac{E_1}{k_B T}\Big]$$
After this, it's said that $S_2$ is evaluated at the fixed energy of the universe, and so it's a constant term, suggesting that it can be eliminated from the calculation. In fact, when it gets to the point of finding $\rho_c$, it is formulated as
$$ \rho_c = \frac{1}{Z}e^{-\frac{E_1}{k_B T}}$$
where Z is the partition function, that does not contain any term with $S_2$. So I don't get why I shall drop the term with the entropy of the reservoir. One idea is that I can directly eliminate the term $S_2(E_2 = E)$ in the Taylor expansion, or think of it as a sort of renormalization of the entropy, but I'm not sure these ideas are legitimate.
| Here, $\rho_{c}(E_{1})$ is just the probability of the system $1$ being found at the given energy $E_{1}$, and so the constant pre-factor factors out of expression.
$Z$ is just the sum of all the $\Gamma(E_{2})$ as $E_{2}$ is varied over all possible discrete values (or we need an integral if the number of admissible energy values are continuous). Also, $\rho_{c}(E^{*}_{1})=\frac{\Gamma(E^{*}_{2}=E-E^{*}_{1})}{\sum_{E_{2}} \Gamma(E_{2})}$. The $exp[\frac{S_{2}}{k_{B}}]$ factors out of both the numerator and denominator of this expression.
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Relationship between orbital radius, mass, and orbital velocity I have a question regarding the relationship between a body's orbital radius, mass, and orbital velocity.
I understand that there is the equation $V = \sqrt{\frac{GM}{R}}$, where $V$ is the orbital velocity, $G$ is the gravitational constant, and $R$ is the orbital radius. Does the equation imply that the orbital velocity is inversely proportional to the square root of the orbital radius? So, can it be accurately stated that the orbital velocity decreases as the orbital radius increases?
However, this article states that "In general, the speed with which stars orbit the centre of their galaxy is independent of their separation from the centre; indeed, orbital velocity is either constant or increases slightly with distance rather than dropping off as expected."
These observations seem contradictory. Can someone please help clear up the misinterpretation?
| The article states:
To account for this, the mass of the galaxy within the orbit of the
stars must increase linearly with the distance of the stars from the
galaxy’s centre.
So the mass inside the orbit of radius $R$ is $M = M_oR$ where $M_o$ is a constant.
$\Rightarrow V = \sqrt{\dfrac{GM}{R}}= \sqrt{\dfrac{GM_oR}{R}} = \sqrt{GM_o} = \rm constant$
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Incorporation of adiabatic phase into quantum effective action Suppose we have a system (or a subsystem) in the quantum state $|\text{in}\rangle$ and the same system in the state $|\text{out}\rangle$, which differs from $|\text{in}\rangle$ only by a phase:
$$
\tag 1 \langle \text{in}|\text{out}\rangle = e^{i\theta},
$$
Next, suppose $\theta$ is a phase which can be written through an integral over dynamical variables $p,q$:
$$
\theta = \int d\epsilon F(\epsilon), \quad \epsilon = (p, q)
$$
Could we immediately conclude that $(1)$ provides additional summand in quantum action
$$
\Delta \Gamma = \theta?
$$
| You might find the paper "Effective Action for Adiabatic Process" (Prog. Theo. Phys. 74 (3), 439 (1985)).
https://academic.oup.com/ptp/article/74/3/439/1894676/Effective-Action-for-Adiabatic-ProcessDynamical
There in authors discuss about a correction coming from berry phase term (under adiabatic conditions for some internal degrees of freedom like spin for example) to the action of the spatial coordinates, which will certainly affect effective action if calculated for the spatial coordinates.
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Quark explanation of reaction I've got an issue. There's a reaction in cosmic rays which I need to discuss on 'subatomic' level. I do not need calculations or anything, it has to remain very basic. Could you please explain to me how does this reaction happen on quark level?
$$K^{-} + p^{+} \rightarrow K^{0} + K^{+} + \Omega^{-}$$
I am aware that the $strange$ quark never becomes a $down$ quark and that these mesons and baryons can be written as:
$$\overline{u}s + duu \rightarrow d\overline{s} + u\overline{s} + sss$$
What are the main reactions between the quarks and how to discuss it in Layman's terms?
Thank you in advance.
| This is a famous reaction, that led to the discovery of the $\Omega^-$ baryon.
The quark-line diagram is shown below. The important features are:
*
*It's all quarks and gluons, no weak bosons or photons, so it's a purely strong-interaction process.
That means the reaction rate will be relatively large. (Whereas the subsequent decay of the $\Omega^-$ requires a weak-interaction vertex, which is why $\Omega^-$ lives long enough to be observed in a bubble chamber.)
*The diagram is only schematic; it shows the quark rearrangements, and pair creations and annihilations, needed to account for the flavor content of the hadrons. There are many more gluon exchanges, not shown, that bind the quarks into hadrons.
| {
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What is the force that pushes a car upwards when moving on a banked road? On banked roads, if the road is frictionless, we can drive a car in circular motion only at a particular velocity called safe velocity. Now if the velocity is less than the safe velocity, then the car would slide down the road(so that radius is decreased) and if velocity is more than safe velocity then the car would slide up.
What is the force acting on the car that accelerates it up or down with respect to ground or in the inertial frame?
| What makes the car go up and down the slope is the centrifugal "force". I have put the word force in quotes, because it is not a real force, but rather an apparent one.
What causes the centrifugal force is inertia. The car wants to go in a straight line, but it is prevented from doing so by the friction of the tyres and/or the slope of the road. This is the same force as the one you feel when you swing a weight at the end of a string: the weight wants to go in a straight line because of inertia, but the string keeps it going in a circle. The only real force is the one you use to stop the weight from flying away.
If the centrifugal force gets too strong (the car is going too fast around the curve) then the wheels can no longer keep it on its path and it slides outwards, towards the higher end of the bend. If the bend gets steep enough near the top, the car will find a new equilibrium point. Similarly, if you slow down, the equilibrium will be at a less steep part of the bend.
In case you think that it's weird that the centrifugal force is not real, there are other forces that are equally "unreal". One example is the Coriolis force, which creates cyclones and hurricanes. It is caused by the earth rotating, while the air mass wants to follow inertia and go in a straight line. If you're being blown away by a hurricane, remember that you're just being pushed by inertia. But you'll land just as badly at the end ;-)
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How to calculate the vacuum expectation value of a scalar field? Let's consider a complex scalar field $\phi$ with potential $\frac{\lambda }{4!}(|\phi|^2-v^2)^2$ (Goldstone model).
Applying the Schwinger-Dyson equations, I obtain $\langle \phi \rangle=0$. This result is intuitive because if we have an interaction involving even powers of the field, we will never be able to draw a tadpole of the type
But it is known that if we compute the expected value of $\phi$ at the true vacuum instead of the Fock vacuum, we would get $\langle \phi \rangle \propto v$. I don't see how we can represent this result with Feynman diagrams. Is it even possible?
| You have to rewrite the Lagrangian in terms of the field $\varphi = \phi - v$. Then quantize perturbatively around $\varphi = 0$. This will give you (by Schwinger-Dyson eq) $\left< \varphi \right> = 0$, which in turn means that $\left< \phi \right> = v$.
Or are you asking something completely different?
P.S. Exploring the physical vacuum state through the perturbative expansions around $\phi = 0$ is a bad idea. In some cases it might give you something physically meaningful, but generally lots of interesting physics gets lost. This is because perturbation theory is approximate, and each perturbative expansion has an (analogue of) a radius of convergence, which most of the times doesn't allow you to pass between different vacuums without getting something completely nonsensical.
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Why as distance decreases capacitance increases? I understand mathematically that as the distance between the capacitor plates decreases, capacitance increases, but I find there is a small contradiction that I want to understand. If the capacitance equals the charge on the capacitor divided by the voltage, isn't the charge that is accumulated on the capacitor triggered by the force that results from the charge on the other plate, which is constant and equals to $\sigma/2\epsilon_0$?
| By definition, capacitance measures the ability of a system to hold charge. Mathematically, it can be stated as the amount of charge the system can hold per unit potential difference across it.
$$CV = q \tag{1}$$
where $V$ is the potential difference (or potential) of the system, $C$ is the capacitance of the system and $q$ is the charge stored in the system.
An electric field is what causes a change in potential. The relationship between the change in electric potential and the electric field its simplest form can be expressed as follows:
$$V = E.d \tag{2}$$
Consider a parallel plate capacitor. If you keep the amount of charge on the system constant and then reduce the distance between the plates, the potential across the capacitor decreases. As the electric field between the plates depends solely on the surface charge density (assuming that the plates are very large), from equation $(2)$, you can infer that $V$ decreases as $d$ decreases. Therefore, you are holding the same amount of charge for a smaller potential difference. Aha! The capacitor is now able to store more charge per unit potential, therefore its ability to hold charge, that is capacitance, has increased.
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What causes like electric charges to repel and opposite electric charges to attract at the smallest level? When talking about charged particles, the law of charge dictates that two particles with opposite charge will attract each other and two particles with the same charge will repel each other.
However, I have never seen why this works. So, on a fundamental level why does the law of charge work? What causes like electric charges to repel and opposite electric charges to attract at the smallest level?
| Electric charge is fundamental to the structure of matter.
The atomic nucleus contains protons, which attract electrons that occupy different levels of energy, or electron shells around the nucleus. Atoms can become electrically charged ions by gaining or losing electrons from their outer shells, unbalancing electrical charge within the atom.
Benjamin Franklin coined the terms "positive" and "negative" to describe his single-fluid theory of electricity. He described electricity as a fluid that flows from objects with excess electrical fluid (positive) to objects with a deficit of electrical fluid (negative).
By convention, objects likely to lose electrons are called negative, and objects likely to gain electrons are called positive. But also by convention, the flow of electricity is considered to move from positive to negative (thanks to Franklin). Thus, electric current is by convention said to flow opposite to the actual flow of electrons in a conductor.
The source of electrical charge is the attraction between protons and electrons in the atom, and the repulsion of each for its own kind. So far as I know, there is no classical explanation for why such attraction and repulsion exists in protons and electrons. It's a fundamental force in the Universe.
| {
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Have quarks been observed in free states? I had been reading a book in particle physics very recently. The book has probably been written in the late 90's, (and obviously before 2000), and states that quarks, the underlying particles in all baryons and mesons, has not yet been detected in free state.
Is this still true, or have we been able to detect quarks in free states?
| The quark-gluon plasma is a hot dense soup of (mostly) free quarks and gluons. It is created in heavy ion experiments like ALICE, and ATLAS Heavy Ion at the LHC or STAR at RHIC, among others, via two large (typically gold or lead) nuclei smashing together at close to the speed of light. However, this "free" quark state is very short lived, and it almost immediately cools down and re-hadronizes into confined states which are measured in the detectors for these experiments. So, we don't actually "see" free quarks but we can deduce that they must have been. In nature, the only place you might be able to find deconfined quarks is at the center of a neutron star, but that hasn't been observed.
| {
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Why do gases have higher internal energy than solids and liquids, when at the same pressure? I was given the following question:
Three containers filled with $1 kg$ of each: water, ice, and water vapor at the same temperature $T = 0C$.
Now apparently, the vapour has the highest internal energy, the water has the next highest and the ice has the lowest internal energy.
It seems as though it has something to do with the gas having more degrees of freedom and thus a higher entropy?
I'm confused because when I use the formula $U = \frac{3}{2}nRT$ , they should all have the same $U$. Because they are each $1kg$, the $n$ should be the same. They all have the same $R$ and the same $T$. So how can they even have different internal energies?
| I figured out the answer to my own question.
Internal energy is higher when there is more degrees of freedom.
The formula U=3/2nRT only works for monotomic substances.
Every time that more degrees of freedom are added, the 3/2 constant increases.
So in fact, the potential energy formula should yield correct results had I been using the right one.
Essentially, my main error was assuming that the given substances had the potential energy formula U = 3/2nRT, when in reality this was not the case.
However, I'm still unsure why the formulas would be different for solids, liquids and gases.
It would be nice if someone could clear that up.
| {
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Why does the Earth accelerate upward, according to Einstein? I recently watched a video on the YouTube channel PBS Space Time which was called "Is Gravity an Illusion?". In this video, the host explains that Einstein claimed that it is not the apple that accelerates towards Earth but the other way around. In Newton's theory, the apple accelerates down because the Earth's gravity is pulling it down, but in GR, the Earth accelerates up because of––what?
I am familiar with geodesics and spacetime warping in GR, but am new to this concept of "Earth accelerating upward" and do not understand it very well.
| It depends upon the frame of reference.
If one adopts the frame of reference of the apple then the Earth accelerates towards it.
If one adopts the Earth's frame of reference then it is the apple that accelerates towards it.
In GR, there is an equivalence to frame of references and so either scenario is correct.
An object moves up or down relative to another object and so the Earth "moving up" is the result of the apple's frame of reference being adopted.
| {
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The water in the tank are pressurized to 50- 100 PSI? but water cannot be compressed I was looking up about how domestic water heater (DHW) works, and couldn't understand the following:
considering the fact that water (and all other liquids) cannot be compressed (practically), how can they state that the tank is a " heavy metal tank that holds 40 to 60 gallons (151 to 227 liters) of hot water at around 50 to 100 pounds per square inch (PSI), within the pressure range of a typical residential water system".
Do they mean that the water shoots out of the pipe at pressure of 50-100 PSI?
How can one pressurize water (or any other liquid) ? If I put a membrane in a sealed tank that separate the water from the air, and then I compress the air to lets say 50 PSI, then it can be said that the water are pressurized to 50 PSI? ( and that means that if I open a valve in water side of the tank, they'll shoot out at 50 PSI ? can you remind me how to find their velocity? it comes from bernoulli?)
| They mean that if you were measuring the pressure on the walls of the tank (without opening up the tank or anything like that), it would read 50-100 PSI.
You can pressurize a liquid by applying force on it while it is constrained. Since that force cannot go towards compression very well, it just pushes out in every direction, increasing the pressure.
Looking into hydrostatics should be able to give you a lot more information on the subject.
| {
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Why do power lines use high voltage? I have just read that using high voltage results in low current, which limits the energy losses caused by the resistance of the wires.
What I don't understand is why it works this way. Does it have anythnig to do with electromagnetic induction in the wire which resists the current?
| From the formula $P=VI $, $I=\frac {P}{V} $. So, if the voltage is high, current becomes low for same power. Now, $H=I^2RT $, so lower the current, lower is the heat production. Mainly to reduce heat production, the voltage is increased.
| {
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Where does the reaction force act? Does it act on the centre of mass? I know that the reaction force acts perpendicular to the surface in contact with the body. But where exactly (on the body) does it act?
When we deal with the forces on bodies of finite size, we usually refer to the force on the centre of mass. What's our reference when it comes to reaction force? Where does it act?
| A reaction force is basically an electromagnetic force. When we push on a rigid surface, what we are doing is that we are decreasing the distance between two surfaces. This results in increase of the unbalanced electromagnetic forces on the two surfaces since they are not perfectly neutral thus the reaction force acts on these points at the surfaces which are pushed very closely to each other (which appears to be in contact when seen by a naked eye) so if you are analyzing a whole surface then you can say that the reaction force acts on the point of contact between the surfaces.
| {
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Doubt on precession So we are studying rotation of rigid body. Our teacher talked about precession but not in much detail
It got me thinking so if we have axis of axis of rotation then
*
*how many such axis of axis of .......of body can be there
*if the answer is many can't we ultimately represent any motion as sum of such motion
Sorry if it's nonsense as it is coming from just physics enthusiast and not a scholar or something
| Refer to the bible of physics :http://www.feynmanlectures.caltech.edu/I_11.html
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What is intuitive or physical meaning of wave functional and field configuration and field eigenfunction? what is the physical meaning of field configuration in quantum field theory. I have come across such terminologies in Schrodinger field theory and path integral field theory. What is the actual difference between quantum field operator and quantum field functional or wave functional of the given field configuration? Also is there any eigenfunction and eigenstate there for field operator? what is the physical meaning of the eigen of a field operator in quantum field theory? Also can someone give intuition for how functional derivatives defines and explains quantum field theory in the place of usually partial differential operator?
| The field operators $\psi^{\dagger}$ and $\psi$ are creation and annihilation operators for particles in the position eigenstate $\left|x\right>$. A wavefunctional is the analogue of a wavefunction in ordinary QM, it gives you the probability amplitude when applied to an entire classical field configuration, just like an ordinary wavefunction gives you an amplitude when you apply that function to a position of a particle. This book explains the formalism of QFT much better than standard textbooks, it allows you to understand what QFT is including advanced concepts, without getting bogged down in intricate mathematical details.
| {
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What makes the number of neutrons the number of proton similar? In basic chemistry, we are taught that an atom has roughly the same number of neutrons and number of protons, this doesn't seems to hold for larger atoms, but it is always roughly proportional (i.e. you seldom find an atom with 100 protons but 1 neutron, that just does not happen). Why is that?
| This is my first answer, be nice please.
A different approach (previous answers are already very good) that might give you a better intuition:
Back in the day, Weizsäcker proposed a parametrization of the binding energy of nuclei by analogy with a liquid drop such as
$BE(A,Z) = a_VA - a_SA^{2/3} - a_CZ^2A^{-1/3} - a_A(A-2Z)^2A^{-1}$ (not considering the pairing term here)
where the coefficients $a_V,...$ are fitted on nuclear masses tables (e.g. AME2012), Z is the number of protons and A is the total number of nucleons (protons + neutrons).
The different terms in this relation respectively correspond to the bulk energy, the surface energy, the coulomb energy, and the symmetry energy.
Considering $BE(A,Z)>0$, the most stable nuclei for a given baryonic number A corresponds to $(A,Z^{eq})$ with $Z^{eq}$ associated to the higher value of the binding energy. You could "reverse" the problem to find the value of A that maximize the binding energy for a given Z.
For heavy nuclei, such as uranium, there is a significant difference between the number of protons and the number of neutrons (which is higher). Assuming that our previous relation is true, this is because the balance between the repulsive force between protons and the asymmetry tilts in favor of Coulomb.
This is the basics of nuclear phenomenology. There are a lot of microscopic approaches (e.g. DFT, ab initio) that consider the nucleon-nucleon interaction in a different way.
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Why is Maxwell's theory insufficient to explain the photoelectric effect? You can argue that electromagnetic waves from a UV light source travel towards a metallic plate, and by the time they reach where a loose electron is located, they affect it with a electromagnetic field (force), so the energy might be enough to knock off the loose electrons. So why is a photon hypothesis proposed?
| Maxwell Equation do not take into account the quantum nature that comes after introducing the Planks constant. The energy of a photon is Planck constant X frequency. There is no Planck constant in Maxwell equations in this sense they are termed classical. To explain photoelectric effect the quantum nature of light has to be assumed.
| {
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Problem of the ideal in the GNS construction I read this in the wikipedia's article about the GNS construction:
"the degenerate elements, a in A satisfying ρ(a* a)= 0, form a vector subspace I of A. By a C*-algebraic argument, one can show that I is a left ideal of A (known as the left kernel of ρ)."
Is there a full proof online?
A related quetion: As C* algebras are equipped with a * norm
is the set of X such that $\|X^*X \| = 0$ a "natural" ideal of the C* algebra?
| I had the answer in one of my own books. $\|$
I did not find the answer online but in Haag's book: Local Quantum Physics.
He defines a state over a C* algebra $\mathcal{A}$ as a
normalized positive linear complex form.
Positive means that $f(A^*A) \ge 0$
Having defined the set $\mathcal{I}$ of the X with $f(X^*X) = 0$
we want to show that it is an ideal i.e $f((AX)^* AX)) = 0$.
this is possible with a Schwarz like inequality.
Let us expand the non negative
$$f((tX+Z)^*(tX+Z)) =$$
$$t^2 f(X^*X) + tf(X^*Z + Z^*X) +f(Z^*Z) =$$
$$t^2 f(X^*X) + 2t Re f(Z^*X) +f(Z^*Z)$$
As it is always non negative the discrimibant is null or negarive
$$[Re f(Z^*X))^2 \le f(X*X) f(Z*Z)$$
If X belongs to I we have $f(X^*X) = 0$ so
$Re f(Z^*X)$ is null.
Let us take $Z = (AX)^*A$
in this case the positivity of f implies that it is a positive or null number.
This proves that AX belongs to $\mathcal{I}$
| {
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A rod is moving in space and an insect is on it. How many degrees of freedom does the insect have? Is the answer 7? The number of degrees of freedom of a system can be viewed as the minimum number of coordinates required to specify a configuration. Applying this definition, we have:
*
*For a single particle in a plane two coordinates define its location so it has two degrees of freedom.
*A single particle in space requires three coordinates so it has three degrees of freedom.
*Two particles in space have a combined six degrees of freedom.
*If two particles in space are constrained to maintain a constant distance from each other, such as in the case of a diatomic molecule, then the six coordinates must satisfy a single constraint equation defined by the distance formula. This reduces the degrees of freedom of the system to five, because the distance formula can be used to solve for the remaining coordinate once the other five are specified.
| 3 for rod and 1 for insect, total 4
| {
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Do particle and antiparticle annihilate when they meet? Do particle and antiparticle annihilate when they meet? As we know, an electron and a positron will annihilate when they meet. However, many quarks and antiquarks do not annihilate, but coexist as mesons. For example, a neutral pion $\pi^{0}$ is made up of $u, \overline{u}, d, \overline{d}$; a meson $\eta$ is made up of $u, \overline{u}, d, \overline{d}, s, \overline{s}$; a meson $\eta_{c}$ is made up of $c, \overline{c}$; a meson $\phi$ is made up of $s, \overline{s}$; and a meson $\Upsilon$ is made up of $b, \overline{b}$ . Why do these quark-antiquark pairs not annihilate?
| Mesons are unstable and decay , either through the weak decay of a quark in the charged cases, or because quark + antiquark annihilate in the neutral through the strong interaction. The strong interaction allows also for other decays, with no annihilations by the ability of gluons to generate a quark antiquark pair and create new resonances or particles ( example phi to K+K-) if the energy is enough and quantum numbers conservation allow it .
The pi0 can only decay electromagnetically, quark annihilating on antiquark and giving two photons, because there is nothing composed of quarks it can decay to through the strong interaction. Example the phi, which has a branching ratio to s s_bar and can decay to rho pi, i.e. the s annihilates on the s_bar, (it also goes into K+K-). The decay is very fast because it is a strong interaction one.
In the interactions, they form a temporary bound state . Similar to the positronium bound state, but much more complicated because of the strong interaction. Given the appropriate conditions a bound state exists temporarily, but there is a probability of annihilating from an overlap of the wavefunctions. When looking at the Y decays (page 147) there are so many channels open by quantum numbers and energy so some decays carry the b b_bar through, but the dominant involve an annihilation.
BTW within a proton, quark antiquark pair are continuously created and annihilated, for example.
| {
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Conservation of angular momentum in a glass of wine? I've been pondering about something i experienced the other day, hope you smart people might help :)
I was holding a glass of wine, filled half way more or less. And at some point i was swirling the wine, by holding the glass at its neck and making circular clock-wise motions parallel to the table. Sort of like drawing circles with it on the surface of the table.
While I was doing that, I noticed that the glass itself was spinning counter clock-wise on its axis, which is perpendicular to the table.
So my question is,how does this happen? Is it the product of conservation of the angular momentum? By that the glass is responding to the wine spinning inside of it, and spins back trying to keep the total angular momentum at 0? does the friction between the wine and the inner part of the glass makes a difference in the outcome?
I wonder...
By the way, after a few sips, I tried spinning the glass once again. Same results, but not as strong of a reaction regarding the more full glass.
What do you think?
| As you move the full glass, this tosses the contents around the inside (swirling).
Even if you're not attempting to tip the glass, the force of the liquid on the vessel is going to (slightly) tip it. You can imagine that as the liquid is farthest from you, the glass is also slightly tipped away from you. This means the strongest contact with the table is also at the point on the base farthest away from you.
At this point in the motion, you are pushing the glass to the right. But since it is leaning, the frictional forces are not equal over the base. They are strongest at the far edge. This creates a torque that serves to spin the glass to the left (or counterclockwise).
| {
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