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How is it that the Earth's atmosphere is not “blown away”? The Earth moves at a high rate of speed around the Sun, and the solar system is moving quickly around the Milky Way. How is it that the Earth's atmosphere is not “blown away”?
| Your intuition about objects facing a head-wind when traveling quickly only works near the surface of the Earth in Earth's atmosphere. In the atmosphere air molecules must be pushed out of the way.
In space though, there isn't anything to do the blowing. There is no interstellar medium / fluid that could drag / push or otherwise effect the Earth's atmosphere simply because we're moving quickly.
The Earth does lose atmosphere due to several reasons though. First, there is a "solar wind" from the Sun which are high energy charged particles (protons mostly) that strike the upper atmosphere and impart so much energy to molecules in the atmosphere that they're able to escape. There are other means too which are listed in the Wikipedia article on atmospheric escape.
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A Musical Pathway Using a small number of sound emitters, could you create a room where certain nodes emitted particular tones, but no meaningful sound was heard anywhere else.
So, for example, by walking down a certain path, you could hear the tones for "Mary Had a Little Lamb." Is there a generalized algorithm to make particular paths for particular tone sets?
| Lower frequencies tend to dissipate in all directions, while higher frequencies tend to be "directed". (For example, you can place your subwoofer anywhere in the room, as the sound waves will propagate in all directions, while your other speakers are more "directed" because they reproduce higher frequencies).
See this Wikipedia article on sound localization for more info.
If you want directional sound, you have a couple of options:
*
*Control the direct (non-reflected) sound using directional speakers (Example: Maestro directional speakers.
*Control the reverberant (reflected) sound by using room shape, and specifically placed absorptive and reflective surfaces.
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Quantum Mechanics, Uncertainty Principle-- help understanding notes There is a section of my notes which I do not understand, hopefully someone here will be able to explain this to me. The notes read (after introducing the uncertainty operator):
If the state $\chi_A$ is an eigenstate of $\hat O_A$ then the uncertainty is zero and we measure it with probability 1. However, if $\hat O_B$ is another observable which does not commute with $\hat O_A$, then the uncertainty in any simultaneous measurement of the two observables will be infinite.
I understand the first sentence, but I can't see how to justify/prove the second one. Can someone tell me how the second sentence is justified, please?
| Assuming we have already proved the uncertainty principle(which can be found here), we know:
$$\sigma_A \sigma_B \geq \sqrt{\Big(\frac{1}{2}\langle\{\hat{O}_A,\hat{O}_B\}\rangle - \langle \hat{O}_A \rangle\langle \hat{O}_B\rangle\Big)^{2}+ \Big(\frac{1}{2i}\langle[\hat{O}_A,\hat{O}_B]\rangle\Big)^{2}}=C$$
Where C is a constant.
Since the state we are looking at is an eigenstate of $\hat{O}_A$, we know $\sigma_A=0$; also since $\hat{O}_A$ and $\hat{O}_B$ do not commute, the right hand side($C$) is greater than zero. Ergo:
$$\sigma_B > \frac{C}{\sigma_A}\rightarrow \infty$$
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Can open, unsafe nuclear fusion reaction burn the atmosphere? I happened to hear people saying that the nuclear fusion bomb tests could set the atmosphere on fire. I have some serious doubts about that - but I have no facts.
Nuclear fusion reaction requires $15*10^{6}$ kelvins to start. If we produce such temperature in "open air" would the atmosphere become a fuel for further fusion? Shouldn't the whole thing just be torn apart by its terrible pressure?
| Around the 60s, a treaty was signed to ban development of nuclear fusion devices with yield greater than about 50 MT (don't remember exact number), in order to prevent fusion of atmospheric hydrogen, thus the uncontrolled multiplication of the device explosive yield. That was before the Threshold Test Ban Treaty was signed in 1974 and entered into force in 1990.
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Momentum of particle in a box Take a unit box, the energy eigenfunctions are $\sin(n\pi x)$ (ignoring normalization constant) inside the box and 0 outside. I have read that there is no momentum operator for a particle in a box, since $\frac{\hbar}{i}\frac{d}{dx}\sin(n\pi x)=\frac{\hbar}{i}n\pi\cos(n\pi x)$ and this isn't 0 at the end points. Nonetheless, we can write $\sin(n\pi x)=\frac{e^{in\pi x}-e^{-in\pi x}}{2i}$, which seems to imply that there are two possible values of momentum: $n\pi$ and $-n\pi$, each with 50% probability.. Is this wrong? If you measured one of these momenta and the wavefunction collapsed to one of the eigenstates then it wouldn't solve the boundary conditions. So, what values of momentum could you obtain if you measured the momentum of a particle in a box?
Edit: I know that you can't measure the momentum of a particle exactly, but normally after a measurement of momentum, or such a continuous observable, the wavefunction collapses to a continuous superposition of momentum eigenstates corresponding to the precision of your measurement. But in this case since the wavefunction seems to just be a superposition of two momentum eigenstates, the wavefunction must have to collapse to one of them exactly, or so it seems.
| I think this is a great question. http://arxiv.org/abs/quant-ph/0103153 This article explains why we shouldn't enforce the boundary conditions that we do (the wavefunction goes to 0 at the boundaries) and instead should use the condition that the wavefunction is equal at both end points. The justification is partly for mathematical reasons, but partly because that condition is too strong physically; the wavefunction isn't measurable. On the other hand the probability of finding the particle between a and b is measurable. We just want to make sure that if a=0 and b approaches 0 that the probability approach 0 continuously. This is achievable even if the wavefunction is discontinuous.
Once we enforce the weaker condition, certain functions that are exponentials inside the box and zero outside are allowed (the ones with the same wavelengths as the energy eigenstates) and these are in fact the momentum eigenvalues. So exactly as you said, if you measure the momentum, the particle will collapse into one of those states.
| {
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What happens when a photon hits a beamsplitter? Yesterday I read that we can affect the path and the 'form' (particle or wave) of a photon after the fact (Wheeler's delayed choice experiment). Part of what is puzzling me is the beam-splitter. Are the individual photons actually being split into two new photons of lesser energy?
This question implies that you cannot split a photon but it seems that beam splitters do exactly that.
| The crucial word is "beam", in "beam splitter". Beam means an ensemble, in contrast to "photon" which is an individual particle.
A light beam is an ensemble of photons and if it is of a single frequency $\nu$, all photons have energy $E= h*\nu$. A light beam can be split in a beam spliter, i.e. the ensemble of photons can be split into two streams of photons: the intensity of the beam goes down, but the individual photons still have frequency $h*\nu$.
Now one can think of impinging photons one by one on a beam splitter. A photon is described by a wavefunction which when squared will give the probability of finding the photon in a particular (x,y,z). It will go either where one stream went or the other according to the probabilities, but it will be seen as a whole photon of energy $E=h*\nu$.
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Discovery of $E=hf$? How was the equation $E=hf$ discovered?
Was the proportional expression between energy and frequency of light $E\propto f$ discovered only by experiment? Or is there some logical(theoretical) senses affected?
| Discrete spectral lines (frequencies) of atoms were known since long ago. So an emitted wave has a quite certain frequency. If an absorber has many "resonators", it is the resonance resonator who likely will absorb the wave entirely. But this exchange says nothing about the relationship of the wave energy and the wave frequency. Classically it is the amplitude square integrated over the space that gives the wave energy. So the Plank relationship is alien to classical notions. It was hard to accept and only experiments helped prove its universality.
The Plank law was first a fit bridging two experimental asymptotics. But it worked so well for different $T$ and with a unique $h$ that Plank started to "derive" this law and introduced those quanta. The original fit is one of those rare cases in Physics when the exact formula is extremely simple and can be guessed.
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Human power on treadmill On an elliptical treadmill a regular person can easily burn 1000 calories in one hour (treadmill reports calories burnt). This translates into:
$$(1\times 10^3\mathrm{cal/hr}\times 4.2\times10^3\mathrm{J/cal})/3.6\times 10^3\mathrm{s/hr} \approx 1.2 \; \mathrm{kW} \approx 1.5 \; \mathrm{hp}$$
On the other hand, Wikipedia says "A trained cyclist can produce about 400 watts of mechanical power for an hour or more..."
Is the problem that the treadmill gives wrong numbers? Or it is true that running using legs and arms - on elliptical machine or cross-country skiing which seems to be similar - a human can produce a lot more mechanical power than cycling? I thought the maximum power is set by the cardiovascular system, so it would be the same, running or cycling.
| The trainers in the gym measure total energy used, based on an estimate of how much oxygen you use for certain levels of workout.
They also notoriously over estimate - if you are at the gym to lose weight you will use the machine that promises 1000cal/hour rather than the one that claims 500. And so the gym will buy the 1000 cal machines.
http://www.ncsf.org/enew/articles/articles-ellipticaltrainersoverestimatecaloricexpenditure.aspx
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SUSY (Supersymmetric) Quantum Mechanics I have seen some books, e.g. by Fred Cooper (Supersymmetry in Quantum Mechanics), define:
$A = \frac{\hbar}{\sqrt{2m}} \frac{d}{dx} + W(x)$,
$A^\dagger = \frac{-\hbar}{\sqrt{2m}} \frac{d}{dx} + W(x)$,
$V_1(x) = W^2(x) - \frac{\hbar}{\sqrt{2m}} W'(x)$,
$V_2(x) = W^2(x) + \frac{\hbar}{\sqrt{2m}} W'(x)$.
In other places, I have seen it defined:
$A = \frac{\hbar}{\sqrt{2m}} \frac{d}{dx} + W'(x)$,
$A^\dagger = \frac{-\hbar}{\sqrt{2m}} \frac{d}{dx} + W'(x)$
$V_1(x) = W'(x)^2 - \frac{\hbar}{\sqrt{2m}} W''(x)$,
$V_2(x) = W'(x)^2 + \frac{\hbar}{\sqrt{2m}} W''(x)$.
It seems like these two definitions will give you completely different Hamiltonians and partner potentials. Could someone comment on if these two conventions are equivalent, and if so, how?
| The second convention only differs from the first one by using the symbol $W'$ for what is called $W$ in the first convention. There is a one-to-one correspondence between (reasonable enough) functions and their derivatives so the translation between the two conventions is completely trivial.
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Particle density operator in second quantization form The particle-density operator is given by
$n(\mathbf{x})=\sum_{\alpha}\delta^{(3)}(\mathbf{x}-\mathbf{x}_{\alpha})$, then how to derive its representation in terms of creation and annihilation operators $n(\mathbf{x})=\psi^{\dagger}(\mathbf{x})\psi(\mathbf{x})$?
| $\newcommand{\bx}{\mathbf{x}}
\newcommand{\psih}{\hat{\psi}}
$This is worked out on p. 20 of Fetter and Walecka. I'll add just a little extra detail here.
Your particle density operator $n(\bx)=\sum_\alpha\delta(\bx-\bx_\alpha)$ is in first-quantized form and $\hat{n}(\bx)=\psih^\dagger(\bx)\psih(\bx)$ is second-quantized.
When a general one-body operator is written in first-quantized form $J=\sum_\alpha J(\bx_\alpha)$, the second-quantized form is:
\begin{align}
\hat{J} &\equiv \sum_{rs}\langle r|J|s\rangle \hat{c}_r^\dagger \hat{c}_s\\
&= \sum_{rs} \int d\bx' \psi^*_r(\bx') J(\bx') \psi_s(\bx') \hat{c}_r^\dagger \hat{c}_s\\
&= \int d\bx' \psih^\dagger(\bx') J(\bx') \psih(\bx')
\end{align}
The last equality follows from $\psih(\bx)\equiv \sum_s \psi_s(\bx) \hat{c}_s$.
In the case of the particle density operator:
\begin{align}
\hat{n}(\bx) &= \int d\bx' \psih^\dagger(\bx') \delta(\bx-\bx') \psih(\bx')\\
&= \psih^\dagger(\bx) \psih(\bx)
\end{align}
Please let me know if you'd like more detail on any of these steps. For somewhat relevant discussion on the relation between first and second-quantized operators, see my answer here: Second quantization. In that answer I only considered the two body interaction operator in detail. I could do the same for (simpler) one body operators if it'd be of help.
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On Einstein notation with multiple indices On Einstein notation with multiple indices: For example, consider the expression:
$$a^{ij} b_{ij}.$$
Does the notation signify,
$$a^{00} b_{00} + a^{01} b_{01} + a^{02} b_{02} + ... $$
i.e. you sum over every combination of the indices? Or do you sum over the indices at the same time, i.e. they take on the same values:
$$ a^{00} b_{00} + a^{11} b_{11} +... ?$$
| Well, you do it one at a time:
$$a^{ij}b_{ij} = \sum_{j}a^{ij}b_{ij} = a^{i0}b_{i0}+a^{i1}b_{i1}+(\dots). $$
Then you expand on the other index
$$a^{ij}b_{ij} = a^{i0}b_{i0}+a^{i1}b_{i1}+\dots = (a^{00}b_{00}+b^{10}b_{10}+\dots)+(a^{01}b_{01}+a^{11}b_{11}+\dots)+(\dots).$$
If you write $a^{jj}b_{jj}$, then you will obtain the second sum you wrote, i.e.,
$$a^{jj}b_{jj}=a^{00}b_{00}+a^{11}b_{11}+(\dots).$$
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Is length/distance a vector? I have heard that area is a vector quantity in 3 dimensions, e.g. this Phys.SE post, what about the length/distance? Since area is the product of two lengths, does this mean that length is also a vector quantity, and why?
| Length and distance are not vector quantities (they are scalar quantities), but position and displacement are vector quantities (at least according to common terminological conventions). Here is how all of these are defined. Note that I am restricting the discussion here to vectors in three-dimensional Euclidean space $\mathbb R^3$.
Every point in three-dimensional space can be specified by a triple of real numbers $\mathbf x = (x,y,z)$ given its coordinates with respect to three axes. This triple is called the position of the point and is clearly a vector.
The length of any vector, such as a position vector, is defined as
$$
|\mathbf x| = \sqrt{x^2 + y^2 + z^2}
$$
Notice that, by definition, length is a positive real number. Given two points $\mathbf x_1=(x_1, y_1, z_1)$ and $\mathbf x_2=(x_2, y_2, z_2)$, the displacement vector pointing from point $1$ to point $2$ is defined as
$$
\mathbf x_{21} = \mathbf x_2 - \mathbf x_1 = (x_2-x_1, y_2-y_1, z_2-z_1)
$$
The length of the displacement vector is called the distance between the two points and is therefore given by
$$
d(\mathbf x_1, \mathbf x_2) = |\mathbf x_{21}| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
$$
Note. I have heard some using the terms distance and displacement interchangeably, or using the term displacement for what I have called distance, and using distance to refer to the total length of a path along which an object travels.
Having said all of this, there is in fact a product that allows one to construct area vectors given two position vectors. It's called the cross product. If you take the cross product
$$
\mathbf x_1\times\mathbf x_2
$$
of two position vectors, then you get a vector whose length is the area of the parallelogram spanned by these vectors, and whose direction is perpendicular to this parallelogram.
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Can deterministic world view be denied by anything other than quantum mechanics If we ignored quantum mechanics and looked at the world with a deterministic Newtonian view. Does not that mean that there is no randomness and that if all the information of the state of the universe during the big bang is accessible one can predict the state of the universe at any period of time and predict that I am writing this question right now.
Of course something like that denies the free will but I am asking if there is any thing other than quantum mechanics that denies the deterministic world view.
| Determinism is not denied by anything (QM included). The so-called no-go theorems against deterministic hidden-variable theories are logically fallacious. They start with the assumption that determinism is false (using fancy names like "free-will assumption" or "no-conspiracy assumption") and conclude, guess-what, that QM is incompatible with classical deterministic realism.
I have yet to see a sound argument against determinism.
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Is a proton collision (collisions like in the LHC) visible to the human eye? I was curious if a proton collision is visible to the human eye.
(This might sound like a really basic question and forgive me if it is. I am very inexperienced in Physics and just wanted an answer to my curiosity)
| These collisions don't produce significant amount of light in the visible range, so the easy answer is "no".
They also take place in a vacuum, inside a beampipe which is itself buried in a detector apparatus that is ten meters plus on a side and packed full of stuff with no room for a human.
That said, there are several ways in which a high energy ionizing particle could---in principle---make light in the visible range. In particular a electromagnetic shower impinging on the eye itself might produce enough Cerenkov light to pass the simple filters the brain imposes on retinal output and be consciously registered as a blue flash. (Each flash would represent a small fraction of the--considerable!--ionizing radiation dose which you would be exposed to in the course of this stunt.)
If we assume a interaction region without a detector and a physicist dumb enough to be in the hall while the beam is on (having carefully bypassed both hardware and administrative interlocks designed to prevent this) who then put his head near the interaction region beam pipe, you might be able to say "I saw one!" every once in a while.
But you won't catch me trying it: I plan to die of something other than radiation poisoning or cancer.
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Why do current-carrying wires heat up? Obviously wires heat up too, but why do they heat up? And for the same reason, why do we get electrical burns?
| All wires which have electric current flowing through them have electrons moving through the wire. The reason for why wires heat up when a current flows through them is that a battery converts chemical energy into electric potential energy. This electric potential energy is given to the electrons, and since the electrons try to minimize their electric potential energy, the electrons convert this electric potential energy into kinetic energy. Due to the wires having electrical resistance, which means that they resist the motion of electrons, the electrons bump into atoms on the outside of the wire, and some of their kinetic energy is given to the atoms as thermal energy. This thermal energy causes the wire to heat up. Electrical burns occur when you hold on to a wire, and due to heat transfer, some of this thermal energy is transferred to you. When the wires get hot enough due to a constant input of heat, there will be more heat to transfer to you if you touch the wire, and all of this excess heat can cause burns and fires.
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Dissolving oxygen into water I was wondering how one would actually calculate how much oxygen would dissolve into water given the necessary initial conditions, and what those initial conditions would need to be. I assume they would be pressure, and initial concentration, but I really don't know where I would go from there. Clearly air and water have different concentrations of gases and liquids, despite having been in contact for thousands of years. And once in water, is oxygen still considered gaseous? I assume it is, but why is it called gaseous-what quality of it deems it a gas despite being surrounded by liquid?
| Well you wouldn't "calculate" it so much as measure it. You'd have some water, change the O2 partial pressure above it and measure how much dissolves. Then you'd have a chart where you can calculate the chemical potential of O2 in water vs. partial pressure, Henry's law coefficient, ect.
You could try to do a simulation. For example a Monte Carlo simulation. The problem here is that you'd need a very accurate and very cheap model of the energy. And then you'd have to do a proper job with the simulation itself.
Finally, there are some things that you can deduce analytically. Like, that the O2 will leave the solution with increasing temperature, or that a hydrophobic liquid (i.e. perfluorinated alkanes) might dissolve more O2.
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Why does the Sun feel hotter through a window? I have this big window in my room that the Sun shines through every morning. When I wake up I usually notice that the Sunlight coming through my window feels hot. Much hotter than it normally does when you're standing in it outside. I know if the window were a magnifying glass that it would feel hotter because it is focusing the Sun's rays, but I'm pretty sure that my window doesn't focus the rays, otherwise things outside would appear distorted.
So my question is, why does Sunlight always feel hotter when it shines on you through a window than when it shines on you outside? I thought it might simply be a matter of convection, but anecdotal evidence would seem to say it still feels hotter even if you had a fan blowing on you. Am I just crazy?
| This is due to the greenhouse effect (that how a normal greenhouse works).
Glass has a low thermal conductivity, but is also transparent. So the sun light enters though the transparent window, turns in to heat, and then cannot leave outside. Trapped, the heat accumulates at the air close to the windows, and that is what you feel.
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Why do metal objects in microwaves spark? I heard that electrons accumulate at points on metals, and this clearly explains the arcing phenomenon, but how does a microwave make an electron imbalance on the fork?
|
Why do metal objects in microwaves spark
Those sparks are due to dielectric breakdown of air.
Microwaves are absorbed by metal and majority of it is reflected off with same phase, just like a mirror.
However, because of this, there can be a large concentration of electric fields in some parts, it causes the massive potential difference that in turn lead to dielectric breakdown of air and the sparks you see.
I think this link explains it quite well and seems to be reliable: https://engineering.mit.edu/engage/ask-an-engineer/why-cant-we-put-metal-objects-in-a-microwave/
The oscillation of the microwaves can produce a concentrated electric
field at corners or an edge of a metallic object, ionizing the
surrounding air “so you can hear it popping away,” says Ross. You
might also see sparking, which “is a little like lightning,” she adds.
This kind of microwave sound and light show isn’t limited to metal.
Ross sometimes puts on a demonstration for her kids: She cuts up hot
dogs, creating sharp edges, and “watches the electric sparks jumping
between them.”
As for the second part of your question:
I heard that electrons accumulate at points on metals, and this
clearly explains the arcing phenomenon, but how does a microwave make
an electron imbalance on the fork?
The electric field is strongest on sharp edges. Why?
Well that has been asked and answered here: Why is electric field strong at sharp edges?
So yes, the concentrated oscillating electric field and resulting high electric field seems very plausible between the tines of the fork. That, in turn, causes the sparks.
Another related point worth noting is that the actual movement of electrons is considerably slower than build up of strong electric field. The electric field causes the dielectric breakdown.
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Radiation: Inverse square law Gamma radiation follows the inverse square law, I understand this as "double the distance, quarter the intensity"
So if you have a gamma source, at the source (distance = 0), the Intensity is $I_0$, and say at distance = 1, the Intensity is $\frac{I_0}{2}$ (You can't work this out just from the fact it follows the inverse square law right? You'd need the constant?)
So at distance = 2, while the intensity be a quarter of the original intensity so $\frac{I_0}{4}$ or a quarter of the intensity at the distance(1) that was doubled, so $\frac{I_0}{8}$?
I ask because I think this graph, which shows intensity of gamma radiation vs distance according the inverse square law, is wrong?
(also I don't see how it gets from 3x to $\frac{I_0}{8}$ because $3^2=9$)
| This isn't the intensity as a function of distance from a point source in open space. It's intensity as a function of penetration through shielding. They're defining $x$ as the amount of material that a gamma ray has a probability of 1/2 of penetrating. Independent probabilities multiply, so the probability of penetrating three such thicknesses is $(1/2)^3$. The label on the horizontal axis probably means "thickness in units of meters."
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Ball jumping from water Few days ago I played with ball(filled with air) in swimming pool. I observed interesting phenomenon.
When I released a ball from 3 meters depth the ball barely jumped above the water surface but when I released it from 50 cm depth it shoot out of the water like nothing.
I observed when released from 3 meter depth the ball goes up in zig-zag trajectory but from 50 cm depth is goes in straight line.
I would be very interested in calculating optimal depth from which the ball would jump the highest above the water surface. And as well I would like to calculate trajectory of the ball under water.
It is obvious that simple drag formula won't help here.
I guess that the zig-zag patterns is happening because there might be something like Karman vortex street behind the ball.
So have anyone idea how to calculate this? Or can you point me to the right literature?
Edit: I forgot one observation I made. It seamed to me that the ball when released from 3m depth was rotating when it hit a surface and that might prevent the jump.
| At 50cm 'depth' the ball was not actually submerged, because the water did not have sufficient time to refill the 'crater' created by the ball entering the water. The ball can be said to be 'floating' on the base of the crater. As the water returns at very high speed to refill the crater the bottom of the crater with the ball 'floating' on it is effectively uplifted through the air in the crater (not water). Once the crater is refilled the ball is now at the surface level of the water; having traveled at high velocity from its original resting place at 50 cm depth. It now leaves the surface of the water with the momentum gained by its high velocity rise through the crater and flies into the air..
[email protected]
| {
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Nuclear Binding energy The nuclear binding energy, is the energy that is needed to seperate the nucleons in a nucleus. And binding energy is also defined as the energy given out when a nucleus forms from nucleons.
Also the larger the nucleus is, the more energy is required to break it apart, so why doesn't that mean that larger nuclei are more stable? I mean Uranium has a lower binding energy per nucleon than Iron, but there are many many more nucleons in Uranium that Iron so the total binding energy is going to be much greater.
Basically I don't understand why whether an element gives out energy by fusion or fission (why the lighter element provide energy by fusion not fission and vice versa for heavy elements) depends on binding energy per nucleon and not "total" binding energy
| Your basic nuclear reaction conserves the number of nucleons present.1
That is important, because at a bit less than 1 GeV each the mass of the nucleons dominates the total energy of all these states.
So the only place available to get or lose energy in a reaction is by
*
*Changing the flavor of nucleons. Every neutron converted to a proton gets you a neutrino and some gammas (once the positron has captured and annihilated).
*By changing the total binding energy. Notice that getting one nucleon more bound doesn't help if another one gets less bound by a large amount. The decision to express this in terms of the average binding energy is completely arbitrary because for $N$ total nucleons (which doesn't change, remember?) $E_\text{total} = E_\text{AVG} * N$.
Another questions addresses what is so special about iron that it has the highest binding energy per nucleon.
1 This is more or less required by Baryon number conservation in the Standard Model, and we will ignore the need for Baryon non-conservation in most beyond the Standard Model candidate theories.
| {
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Why do stars flicker? Why do stars flicker and planets don't? At least this is what I've read online and seen on the night sky. I've heard that it has to do something with the fact that stars emit light and planets reflect it. But I don't get it, isn't this light, just "light"? What happens to the reflected light that it doesn't flicker anymore?
I was thinking that it has to do something with Earth's atmosphere, different temperatures or something (if this has any role at all).
| This is just a sidenote to Nijankowski's nice answer: This twinkling of stars caused by atmospheric turbulence was a major problem for the earlier reflecting telescopes when astronomers tried to look deep into the sky.
Placing the telescope over mountains solved only a part of the problem. A good solution was brought up in two ways in the 1990s. First, sending telescopes to space - Hubble Space Telescope was carried to orbit where it took astonishingly sharp images like the Hubble Deep Field due to the absence of atmosphere's interaction. Then arrived the serious flaw in its mirror. Repairing the telescope (by orbiting in space) was a very big problem as many equipments have to be replaced by service missions.
The second solution was adaptive optics. In telescopes like the Large Binocular Telescope, a secondary mirror was placed which is readily deformable and the shape is modified according to the incoming light source by a number of hydraulic pistons behind, thus correcting for some amount of atmospheric distortion.
| {
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Is this a correct description of bonding in a metal? I am reading the paper "Twenty five years of Finnis-Sinclair potentials" by Graeme Ackland, Adrian Sutton, and Vasek Vitek, Philosophical Magazine 2009, 89, 3111-3116. It is a review-type article describing the notion that pair potentials are inadequate for describing bonding in metals. But it turns out that if the attractive interaction is "described by the square root of a sum of pair interactions to neighbors, centered on each atom in the solid," metallic bonding can be described reasonably well. The authors point out:
The novelty is entirely in the square root. The square root captures
the dependence of atomic interactions on the local density: as the
number of neighbors of an atom decreases, the strength of the
remaining bonds increases. This immediately predicts an inward
relaxation at metallic free surfaces with a tensile surface stress,
both of which are widely observed but not predicted by models in which
the cohesive energy is just a sum of pair potentials.
My question is, is it true for metals that "as the number of neighbors of an atom decreases, the strength of the remaining bonds increases"? Is there any way that I can understand this, either in a qualitative or quantitative way? Also, when it says "the strength of the remaining bonds increases," is this referring to the energy per bond?
| I'm not sure if it's correct. However assuming that it is, keep in mind atoms in a metal are positively charged relative to the "electron sea". That being the case, intuition says if you have more atoms near eachother in a metal, you have larger positive charge density, which diminishes the electrostatic attraction (potential due to separation of charges) felt per atom against the electron sea.
| {
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No well-defined frequency for a wave packet? There are similar questions to mine on this site, but not quite what I am asking (I think). The de Broglie relations for energy and momentum
$$ \lambda = \frac{h}{p},
\\
\nu = E/h .$$
equate a specific frequency and wavelength to a particle, yet we know that a wave packet
is a linear combination of an infinite range of frequencies and wavelengths. How is it that we (or nature) choose one frequency and wavelength out of the range? Does this have to do with the collapse of the wave packet when measured? And if so, is the resulting measured frequency a random outcome? Similarly, when an electron jumps from one energy level to another in an atom, it emits a photon of frequency
$$ \
\nu = \Delta E/h .$$
Since the photon is not a pure sinusoidal wave, how can a single frequency be ascribed to the photon?
| This is a very good question.
As Ben Crowell says, the photon life-time is finite and so its energy is spread by a range of frequencies, i.e. it's not a delta-function over a specific frequency.
The photon's energy can be obtained integrating over all the spectrum and might be the one corresponding to the peak, E= h•f where f is the frequency where there is the peak.
Anyway, in this case should we assign the frequency "f" to this photon and expect that when it's absorbed it will transfer h•f eV to the absorber electorn?
Another question is how the wave packet is in the space since it has to respect Maxwell equations and therefore you cannot place a square window with N periods and left it happily... Gaussian beams may be good candidates...
I am sorry if I cannot be more accurate about this very interesting matter and I also apologize by my English.
Best regards,
Sergio
| {
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Frame of reference of the photon? In the frame of photon does time stop in the meaning that past future and present all happen together?
If we have something with multiple outcomes which is realized viewed from such frame? Are all happening together or just one is possible?
How the communication between two such frame s work meaning is there time delay for the information as $c$ is limited? If there is time delay does it mean that time does not stop?
My question does not concern matter at that speed rather how it looks viewed from the photon reference.
Thanks Alfred! I think I understand it now.
| I think you are asking about how a photon experiences the passage of time? There is no right time. Photons are not ordinary things moving through space. So from the point of view of the photon this time is not moving at all but the point of view of the photon is that it's place in space is changing but no time is passing.
Change - ie. motion - in time and space actually happens in four dimensions in which no point in time or place in space can be preferred. We could create any ( or many) agreeable coordinate system in such space with four indices - they can be anything at all but with them change of any of the four indices for position and time can be described by the difference in these arbitrary agreed numbers. But this "Minkowski space" and the equations that relate motions in space and time allows the one special thing. That one "thing" is that light moves at same speed always, fastest possible. So light photons were they able to experience, could only experience change of space. They cannot experience time having changed position in space in the least possible time.
I have avoided any math based on your question. If you sought a more rigorous treatment sorry, this works for me.
| {
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Is my boss wrong about our mechanical advantage from our pulley system? I work on a drilling rig as a roughneck and we had a lecture today (at the office) about mechanical advantage in pulley systems. Now, I know that my boss is well educated in oil drilling, but my instincts tell me that he may have this one wrong.
A drilling rig works sort of like a crane in that it has a tall structure supporting a pulley system. There is a large winch permanently installed on the base platform and then it goes over the top of the structure (the crown of the derrick) and down through a floating sheave--this has a few wraps to give us more mechanical advantage. I am including pictures to help describe the situation.
Here the picture shows the floating sheave (the blocks) which we use to do most all of our operations. Most importantly, we use it to pick up our string of pipe that is in the ground.
As seen in this picture, the blocks hold the weight of the string of pipe. Now he told us that if the pipe get stuck in the hole (maybe it snags something or the hole caves in), that we lose all of our mechanical advantage. He said that is why the weight indicator will shoot up and go back down after it is freed. He said that because when the pipe is snagged in the hole then we are not dealing with a free floating sheave anymore and that is what is required to have a mechanical advantage.
I disagree with this because even if it is not free, there is still a mechanical advantage such that (say the normal mechanical advantage is 6 to 1) our pulling force is multiplied by 6. I would like somebody to confirm this for me.
First picture taken from www.worldoils.com on June 21, 2013 Second picture taken from www.PaysonPetro.com on June 21, 2013
| As a crane operator since 1972 I work a lot with pulleys blocks and you are a 100% right, looks like you are rig with 6 parts lines which multiplies by six the line pull of the winch.
| {
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Neutrinos: how can they carry information about universe? I know that neutrinos are particles with a very small mass and no electric charge. They infrequently interact with matter and so they can give us information about the "old" universe. But how can they do it?
| In the "old" universe there is interaction of matter by the the 4 fundamental forces. Neutrinos occur in the weak nuclear interactions. By studying the neutrinos one can study these weak interactions. They might not interact much, but they sure do carry energy.
So by a application of the conservation of energy one could determine the loss to neutrinos and try to link these to the interactions that they might think that occur. For example the nuclear fusion in the sun that converts hydrogen into deterium:
$p+p\rightarrow^2_1H+e^++\nu_e+0.42MeV$
gives an electron-neutrino. These neutrino's don't interact and the amount of energy of the neutrino is hence determined by the conservation of energy.
One could construct a model for the "old universe" and determine the energy of the neutrinos, as the missing energy. This could give some extra information I believe.
| {
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If the multiverse theory is true, can there be a Universe where there are different laws of physics? This is probably a very difficult question. But my question is essentially this, if there are other Universes can different laws of physics exist in those Universes and if so, can't there be a Universe where the laws of physics are so different that the multiverse theory has to come out as false?
|
If the multiverse theory is true, can there be a universe with different laws of physics.
Yes, of course. This follows trivially from the way that you phrased the question ie as a syllogism.
However, the 'if' is a very big if. The multiverse hypothesis is highly speculative, not very well understood, and has absolutely no evidence that substantiates it. It is essentially a figment of the physicists imagination and has been hyped endlessly. There are many more interesting things in physics than the multiverse hypothesis.
Philosophically speaking, it makes no sense to think that there can be differing physical laws; simply because one can then ask for the law that regulates these differing laws.
So the answer is: No, there isn't
| {
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Why can't we obtain a Hamiltonian by substituting? This question may sound a bit dumb. Why can't we obtain the Hamiltonian of a system simply by finding $\dot{q}$ in terms of $p$ and then evaluating the Lagrangian with $\dot{q} = \dot{q}(p)$? Wouldn't we obtain then a Lagrangian expressed in terms of $t$, $q$ and $p$?
Why do we need to use
$$H(t, q, p) = p\dot{q} - L(t, q, \dot{q})?$$
Or is it that whatever the Lagrangian is the method of finding $\dot{q}=\dot{q}(t,q,p)$. Will give us that equation for $H$?
| While I appreciate that Wouter's response shows that you can not indeed just perform the aforementioned substitution, it really doesn't answer the question (i.e. why doesn't the substitution work) in my opinion.
The underlying confusion probably comes from the fact that in mathematics if substitution $\dot{q}(p)$ is inserted in $f\left(t,q,\dot{q}\right)$ you will certainly end up with a function $f\left(t,q,\dot{q}\right)=g\left(t,q,p\right)$.
The real answer to the question then, is that Lagrangian and Hamiltonian formalisms are just two different paradigms, and no matter what you put in the Lagrangian instead of $\dot{q}$ you will always end up with another Lagrangian (but for a different problem). In other words the $L$ in front of $\left(t,q,\dot{q}\right)$ has different implications than the $H$, such as stationary action, $n$ second order equations for $n$ coordinates (where $n$ is the number of the degree of freedoms) vs. $2n$ first order equations for $2n$ coordinates, etc. This is unlike the case in mathematics where $f$ carries no other assumption different to $g$.
| {
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Is it possible to accelerate a mass indefinitely using gravitational field? As a particle's velocity increases, its mass increases(gamma times). Therefore, if a particle is in a gravitational field, the gravitational force it experiences must also increase(gamma times). The net acceleration of the particle, i.e.(gamma*force)/(gamma*mass) should therefore remain constant for small distance traversed by the particle and increase as the particle travels closer and closer to the object generating the gravitational field(due to inverse square relation). Hence. is it possible to accelerate a particle indefinitely using gravitational field?
| (The below answer from five years ago is wrong, I will write a corrected answer soon.)
No. It is not possible. The misconception you have is because you are thinking of the gravitational field as being constant. However, this is not true. Ignoring air resistance for the time being, consider a projectile launched off the earth. Now, if it is launched at the escape velocity from the earth, then it will take an infinite amount of time to reach 0 speed, and by then it will be infinitely far away.
Now, consider the opposite situation. If the projectile is dropped from infinitely far away, it will take an infinite time to fall, but when it falls, its velocity would be the escape velocity, i.e. it would be exactly
$$v=\sqrt{\frac{2GM}{r}}$$
Now, we know that this escape velocity cannot be greater than the speed of light. It is the speed of light only when $$r=r_s=\frac{2GM}{c_0^2}$$
Now, consider the situation where the object had some initial velocity towards the planet (not due to the gravity) when it was infinitely far away? Naive Newtonian Mechanics (NNM) would tell you that it would be $v=v_0+\sqrt{\frac{2GM}{r}}\mbox{ could be greater than } c_0$. However, Special Relativity (SR) is required at such velocities. It tells you that the actual final velocity would be:
$$v=\frac{v_0+\sqrt{\frac{2GM}{r}}}{\sqrt{1+\frac{v_0\sqrt{\frac{2GM}{r}}}{c_0^2}}}=c_0\frac{v_0+\sqrt{\frac{2GM}{r}}}{\sqrt{c_0^2+{v_0\sqrt{\frac{2GM}{r}}}}}\leq c_0$$
As required.
| {
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Doubt over Kelvin-Planck statement of thermodynamics' second law I have a doubt over the Kelvin and Planck's statement of thermodynamics' second law, in particular applied to a cycle. Let's take a Carnot cycle as an example, and let's call the first two transformations (the isotherm and the adiabatic) done. Now, isn't it obvious that the machine has to give up heat to go back to the initial state? Isn't it something that follows from the fact that the cycle has to be continuous?
| in simple words we can expain kp statement as it is impossible to completely convert heat into work in a cycle(*).
while in case of a single process say a isothermal expansion as dt=0 therfore ineternal energy which is a function of temp so dE = 0 neglecting k.e and p.e so from 1st law we get dq = dw .
in isothermal expansion we are able to get work but only till the pressure of the gas do not becomes equal to the atmospheric pressure in which we can convert heat into work but not infinitely.
hence the kp statement is that we cannot completely convert heat into work in a cycle of infinitely.....
| {
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Why do some air-conditioned stores blast you with jets of air as you enter? I went to a grocery store on a hot day that was very well air-conditioned, and I noticed as I went through the open entrance that there seemed to be a very powerful downward air current right at the doorway. After crossing the invisible threshold, the temperature immediately dropped a good 15 degrees or so.
How does this process work?
| This is called an air curtain or air door and it actually keeps flying insects from being able to enter the store. It also helps trap the colder air inside.
Edit: see the link for how it works.
| {
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Operator on Function of Momentum (QM) I have exactly 0 clue on how to start this problem, but I would be forever grateful for a hint in the right direction.
Given the operators $\hat x=x$ and $\hat p=-i\hbar \frac{d}{dx}$, prove the following relation:
$$
[\hat x, g(\hat p)]=i\hbar\frac{dg}{d\hat p}.
$$
| Like Prahar had said, the problem reduces fairly simply in momentum-space.
We note that, in such space: $\hat x = i\hbar\frac{\partial}{\partial p}$ and $\hat p=p$, thus, using some auxiliary function $f$:
$$
[\hat x,\hat g(\hat p)]f=i\hbar\frac{\partial (\hat gf)}{\partial p}-i\hbar\, \hat g\frac{\partial f}{\partial p}=i\hbar\frac{\partial \hat g}{\partial p}f
$$
By applying the product rule and reducing, this yields the correct result.
| {
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A water pipe from sea level to beyond the atmosphere If a pipe extended from just above the ocean floor to outside the atmosphere, would water be sucked up it by the vacuum beyond the atmosphere? If a hole was made in the pipe, above sea level, how would that affect the flow of water? Would it stop it completely?
| From what I understand, vacuums such as space don't actually "suck" or "pull" at all.
When a box of compressed air is introduced to a vacuum, the air molecules are actually being pushed by the other, higher-pressure air molecules into the low-pressure region of space. This means that, with regards to your question, there is a limit to how high the water can rise. This is determined by the force of air pushing down on the water to force it up the tube, balanced with the force of gravity pulling the water back down.
http://en.wikipedia.org/wiki/Suction
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Are Lagrangians and Hamiltonians used by Engineers? Analytical Mechanics (Lagrangian and Hamiltonian) are useful in Physics (e.g. in Quantum Mechanics) but are they also used in application, by engineers? For example, are they used in designing bridges or buildings?
| In civil engineering they use it for structures, and strength of materials in the elastic realm. It goes by the name of the enegy method. Google books might give an indication.
Some authors are Beer and the mechanical engineer Stephen Timoshenko.
This is for some what "static" indeterminant structures. So, there is no time element. But, I am sure it could be used dynamicaly for seismic analysis. The structures have virtual loads or displacements applied to them and you get the complimentry diplacement (for load) or load (for displacement) as a result.
It is an elegant use. It can also be applied to warping of beams.
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A rope attaches the Moon to the Earth. What happens? Consider the Earth (mass $M$, radius $R$, rotating about its own axis at $\Omega$) and the moon (mass $m$, radius $r$, with axial rotation equal to $\omega_m$), whose centre of masses are $d$ apart. They rotate around their barycentre at $\omega_e$ and $\omega_m$ radians/second respectively (I believe $\omega_m=\omega_e$ to conserve momentum).
We now tether a point on the Earth's surface to a point on the Moon's surface using a light, inextensible (and very strong) piece of string. Assume the Earth and the Moon are rigid and indestructible.
What happens?
It would be nice to find a solution of the general type, but for a first analysis it may be easier to assume that $M \gg m$, so that the Earth's rotation around the barycentre is negligible and the barycentre $\approx$ the Earth's centre of mass.
Another (perhaps reasonable) assumption is that all rotation takes place in one plane and that the moon's orbit is circular, and that both bodies have uniform density.
Here are some of my thoughts based solely on intuition, feel free to ignore in answering the question:
*
*The moon is immediately pulled in (as $\Omega_e>\omega_m$) and sticks to the Earth like a barnacle. I think this may occur if $\Omega_e \gg \omega_m$ and $M\gg m$.
*If $\omega_m$ is large enough, the moon as before will initially be pulled in, then be thrown outwards again as $\omega_m(t)$ increases.
*If $m\approx M$, the Earth and Moon will form a dumbbell-type system (i.e. the string wouldn't have had much of an effect, as that is the case anyway in 'binary' systems where $M=m$).
| The string breaks. Though on an elementary level, it appears as though the orbit of the moon about its barycentre is circular, its actually elliptical. So the distance between the centres of the earth and the moon is not constant, so the string breaks under the tremendous tension developed in it.
| {
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How is energy extracted from fusion? I understand that combining deuterium and tritium will form helium and a neutron. There are three methods to do this (1) tokamak (2) lasers and (3) cold fusion. I would like to know after helium is formed. How is that energy extracted from tokamak and stored?
| Another, non mainstream thought is the use of Helium-3 instead Helium-2 of as a fuel:
3He + 3He → 2 1H + 4He + 12.9MeV
or:
2H + 3He → 1H + 4He + 18.4MeV
Here the reaction products are all charged, which means that they could work directly on an electrostatic field, thus transferring their kinetic energy directly to a current. Depending on the reactor's design, there would also be bremstrahlung from the decelerating charges - this could be absorbed by water and maybe used as heat in co-generation.
The main technological point of these reactions is that there are no neutrons. There is some concern that conventional fusion may be "dirty", even as dirty as current fission power, because the reactor case will need to be replaced often owing to the huge damage it sustains from absorbing so many neutrons (indeed, as discussed in Anna V's answer, the neutron is the main carrier of kinetic energy, so it would be the case that would be directly converting this kinetic energy to the useable heat). These neutrons cause transmutations in the case, which means that there could be substantial radioactive waste problem.
There is active fusion research underway into the idea of helium-3 fusion. Indeed I understand that Harrison Schmidt is even seeking investment for moon mining operations to win helium-3 as a "clean" fuel. Clearly, once commericial interests and the need to convince investors are afoot, one has to look at touted "advantages" of this approach with due skepticism, especially when touted by a conflicted party, even when they do show up at "physics" conferences to spruik their idea! Nonetheless, I find this interesting as it does show that mainstream fusion would not be without its waste problems.
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Thermal radiation in the Unruh Effect The following formula has been given in 't Hooft's black holes notes ($|\Omega \rangle$ is the vacuum state of Minkowski space, O is a operator):
$$\langle \Omega| O|\Omega \rangle = \sum_{n \ge 0} \langle n | O | n \rangle e^{-2 \pi n \omega}(1-e^{-2 \pi \omega})=Tr(O \rho_{\Omega})$$
How does this mean that the radiation is thermal and follows Plancks black body law?
*
*Here, I read that $\langle O \rangle = \frac{1}{Z}\sum_n e^{-\beta E_n}\langle n |O|n\rangle$. How is this sum the same as that in the above expression?
*How is the density matrix related to the law of black body radiation? How can I derive Planck's law from the expectation value of O in the first expression?
| First, let's see the density matrix of thermal mixture (in some cases is the Planck's law):
$$
\rho_{Thermal}=\frac{1}{Z}\sum_{n=0}^{N}e^{-\beta E_n}|n\rangle \langle n|
$$
Now we note that the expectation value $\langle \Omega|O|\Omega \rangle$ is equal to
$$
\frac{1}{Z}\sum_{n=0}^{N}e^{-\beta E_n}\langle n|O|n \rangle ,
$$
with $Z=(1-e^{-2 \pi \omega})^{-1}$ and $\beta E_n=2 \pi n \omega$ (this expression give us the $\beta$).
The operator $O$ in $|n\rangle 's$ basis is simple a matrix $O_{nm}=\langle n|O| m\rangle$. Now is easy to see that $\langle \Omega| O|\Omega \rangle = Tr(O \rho_{Thermal})=$
$$
\sum_{n=0}^{N}O_{n}\frac{e^{-\beta E_n}}{Z}.
$$.
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How is time measured inside the bubble created by an Alcubierre drive? I am a layman. I am aware that the Alcubierre Drive has not yet been proven to be possible, but there is something about the concept itself that I am confused about. If there is no movement within the bubble, how is time measured inside the bubble? If there is no velocity, what is time measured against?
| The passage of time inside the bubble will be the same as outside the bubble; the passengers inside the bubble will be able to transfer between points separated by vast distances as if they were effectively moving faster than the speed of light but locally this is not the case since it is impossible to move faster than the speed of light locally. The bubble itself will have regions of incredible tidal forces and time dilation however the positive and negative curvatures will in a way cancel out around at the bubbles surface making this possible.
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Rate of twinkling of stars According to "Why do stars flicker?", the twinkling of stars (stellar scintillation) is caused by the thick layers of turbulent air in the Earth's atmosphere. While the explanation is convincing, it would be better to have some quantitative justification.
Since the twinkling of stars is easily observed, it is reasonable to estimate the twinkling rate as ~ $1\textrm{s}^{-1}$. How can one obtain a similar estimate from the underlying theory?
| The underlying theory would be atmospheric turbulence with the size of turbulent elements ranging from several hundred meters to millimeters. Therefore, the twinkling rate will not be a single frequency but a continuous spectrum of frequencies ranging from 0.001 Hz to 1000 Hz. Of course, the amplitude of the lower frequencies will depend on the turbulent state of the atmosphere. The higher frequencies however, belong to the inertial subrange of atmospheric turbulence in which energy is transfered from larger to smaller eddies. This range is independent of any large scale motion.
To answer the question, the visible twinkling rate is the superposition of all larger motions in the atmosphere between the star and the observer.
| {
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Is sound relative? Is sound relative?
For example, if I and my friend are having a ride at 1000 mph and I shout towards him (speed of sound 700 mph). What would happen? Will the speed of the sound relative to the ground be 1700 mph? Or will it be something else? Why?
| The speed of sound is a property of the medium(air). So speed of sound is fixed with respect to the medium. As the medium is at rest with respect to the ground(assuming no wind), the speed of sound w.r.t ground will stay 700 mph.
If there is wind, you vectorialy add velocity of the wind and velocity of sound relative to air to get the velocity of sound w.r.t ground.
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How does rolling resistance of rail wheel depend on diameter? Freight train is more efficient than truck due to lower rolling resistance. And I wonder which one has lower rolling resistance<>, small diameter or larger one or it doesn't not depend on diameter at all? Both are steel wheel on steel rail.
| If you look at the Wikipeda article on rolling resistance that @fibonatic pointed out, you can find an equation for the coefficient of rolling friction:
$C_{rr}=\sqrt {\frac z d}$
There is another equation after it that is for steel on steel, but that also shows that C is proportional to $d^{-\frac 1 2}$. So from this, you can easily see that a smaller diameter will have a higher C, and thus a higher resistance. A freight train will have a larger diameter than a truck, meaning less resistance, thus being more efficient.
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What are the forces acting between two air bubbles in water? The exact question is
Two air bubbles in water
*
*attract each other
*repel each other
*do not exert any force on each other
*may attract or repel depending upon the distance between them.
The chapter is about gravitation.
The given answer is
A lighter body inside a denser medium behaves like negative mass as far as gravitational force is considered. Two air bubbles i.e. two negative masses will attract each other.
What is negative mass in this context and how can it be applied to such macroscopic objects? How would it result in attraction?
My reasoning is:
Consider the bubble A in the above image.
The air particles forming the bubble A would be attracted more to the left(away from B) as there are more dense particles towards that side-the air particles making up bubble B are less dense than the medium and they will attract the air from bubble A to a smaller extent than if the volume of bubble B was filled with the medium.
A similar case would apply to B due to lesser density of particles forming A and the bubbles would be (indirectly) repelled.
So what is happening in this case?
| I think I see the argument:
One should go to the line joining A and B. And draw two vertical tangent on the right for A and on the left for B. There is more water volume to attract the water in the region on the left of A, so there will be a force towards B, and vice versa.
Anyway bubbles in liquids are not a simple matter like your problem states.
see : http://www.tandfonline.com/doi/pdf/10.1080/18811248.2001.9715037
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Buoyancy in muddy water Consider the buoyancy force in water with very small but macroscopic particles in it. Such particles (suspension) will very slowly drift downwards and will eventually settle on the bottom. If one did not know that the particles are present there then for calculating the buoyancy force, $F = \rho V g$, one would just use the average density of water with suspended particles in it, which is larger than the density of pure water. Would this be a correct calculation?
Suppose we do an experiment with a cylindrical vessel filled with water and a fully submerged float in it, attached to the bottom with a cord, and then we drop some amount of very fine powder into the water. The powder will form a cloud that will slowly drift downward. What would be the observable effect (if any) on the tension in the cord, once the cloud of particles fully covers the float? Assume that the cloud transverse size is large enough to fully cover the cross-section of the vessel. It is assumed that the dust particles don't stick to the surface of the float.
| I would think the tension force will increase, as the average liquid density has increased. Note, for example, that the water pressure on the bottom will increase more than it would be if the same volume of water is added. The particles move downwards with a constant speed, so their full weight acts on the water (until they are on the bottom).
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Is there an EMF in a conductor moving at constant speed across the uniform magnetic field If a conductor - a long rod - moves at constant speed across the "lines" of a uniform magnetic field, is there an EMF within this conductor? Or, if a conducting rod rotates at uniform rate, pivoted in the middle or at one of its ends in a uniform magnetic field perpendicular to the plane of rotation, is there an EMF generated within the conductor?
| If the setup you have in mind is like the image below, there will be a voltage across the conductor.
This is due to the fact that there no closed path for a current.
The mobile electrons in the conductor "feel" a magnetic force towards the b end of the rod.
The electrons "bunch" up at that end resulting in an electric field that points from a to b.
Assuming the velocity is constant, the force on the electrons due to the electric field cancels the magnetic force.
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What causes a black-body radiation curve to be continuous? The ideal black-body radiation curve (unlike the quantized emission seen from atomic spectra), is continuous over all frequencies. Many objects approximate ideal blackbodies and have radiation curves very similar in shape and continuity to that of an ideal black-body (often minus some emission and absorption lines from the atoms in an object, such as radiation curves seen from stars). I am wondering what exactly gives rise to a basically continuous black-body radiation curve in real objects? Since atomic energy states are quantized, it seems real life black-body curves would have some degree of measurable quantization to them (or perhaps the degree of quantization is so small the radiation curves look continuous).
|
perhaps the degree of quantization is so small the radiation curves look continuous
Yes, this is the reason. The correspondence principle says that quantum mechanics has to become classical in the appropriate limit. One way to obtain an appropriate limit is with large numbers of particles. As you increase the number of particles in a material many-body system, you get more and more ways of putting together combinations of states for your material object. The density of states of the object grows very quickly (roughly exponentially) with the number of particles. Therefore the number of possible transitions between states also grows very rapidly.
The number of particles in a tungsten lightbulb filament is something like Avogadro's number. The exponential of Avogadro's number is really, really big.
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What is the invariant associated with the symmetry of boosts? Noether's Theorem states that if a Lagrangian is symmetric for a certain transformation, this leads to an invariant: Symmetry of translation gives momentum conservation, Symmetry of time gives Energy conservation etc.
The Galilean principle stating that all reference frames that move with constant speed relative to each other are equivalent is also a symmetry principle: Setting up a physical system that is identical to the original except for a constant velocity (boost) added will have the same behaviour.
Shouldn't there be an invariant associated with this symmetry? If yes, what is that invariant?
| The conserved quantity corresponding to boost symmetry is
$$
\int d^3 x (P_0 x_i - P_i t)
$$
which is the relativistic analogue of $x_{CM} - v_{CM} t$, the position of the center of mass at $t=0$. It is quite a useless conserved quantity, and that is why people don't talk about it.
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Why does positronium decay into 2 photons more often than into 3 photons? I cannot find the answer to the above question.
I know that para-positronium is created with a probability of $25\%$ and decays into 2 photons, while ortho-positronium is created with a probability of $75\%$ and decays into 3 photons.
I also know that ortho-positronium has a way longer life time than para-positronium. This, in my understanding, should not affect the number of decays per time, but just means that the ortho-positronium will decay LATER into three photons. But in the end there should be $75\%$ 3-photon-decays and $25\%$ 2-photon-decays. But in reality 2-photon-decay happens about 300 times more often than 3-photon-decay.
What information am I missing?
Thank you!
| Jezstarski is mostly correct,
The para-positronium (p-PS) state ends up being the main mode of annihilation of positronium (PS). Positrons can annihilate in at least eight different ways but once ortho-positronium (o-PS) forms in a void/vacuum, it has additional time to undergo another mode of annihilation.
P-PS annihilates in under 125 picoseconds.
O-PS annihilates at or less than 142 nanoseconds (vacuum).
A consequence of this long lived lifetime is "pick-off" (the correct term) annihilation where an opposite spin electron from some surrounding material will annihilate in para-orientation before the ortho-bound electron can collapse and annihilate with the positron.
Source: Member of the Positron Science Lab at UMKC.
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Difference between a quantum process and a thermal process? I was reading an article online pertaining to quantum mechanics and I stumbled across these few sentences.
A look at the corresponding energy regimes shows (Beck and Eccles 1992) that quantum processes are distinguishable from thermal processes for energies higher than $10^{-2}$ eV (at room temperature). Assuming a typical length scale for biological microsites of the order of several nanometers, an effective mass below 10 electron masses is sufficient to ensure that quantum processes prevail over thermal processes.
I would like to know what they mean when they say "is sufficient to ensure that quantum processes prevail over thermal processes". Or possibly just what the difference is between a quantum process and a thermal process.
Original Text (Section 4.4 Beck and Eccles: Quantum Mechanics at the Synaptic Cleft)
http://plato.stanford.edu/entries/qt-consciousness/#4
| The quantum uncertainty in position of particles (micro sites) of mass $m$ moving or vibrating at thermal speeds characteristic for a temperature $T$, is given by the thermal De Broglie wavelength $\sqrt{\frac{2 \pi \hbar^2}{m \ kT}}$. If this uncertainty in particle (micro site) position is larger than or comparable to its size, the inter-particle (inter site) interactions must be described quantum mechanically.
For the example quoted, $m$ is 10 electron masses and $T$ is 300 K (room temperature), it follows that the De Broglie wavelength is a few nanometer. Hence, micro sites weighing 10 electron masses can be treated classically (using Newton's laws rather than quantum mechanics) provided they are significantly larger than a few nanometers. If these micro sites are of size a few nanometer (or smaller) quantum uncertainty kicks in an the full quantum physics machinery needs to be brought in.
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Any physical example of an "explosive" differential equation $ y' = ky^2$? I was told that in physics (and in chemistry as well) there are processes that may be described by a differential equation of the form $$ y' = ky^2. $$ That is, the variation of a variable depends from the number of pairs of the elements.
I understand the mathematical meaning of that equation, but I cannot conceive of any physical process which leads to this. Maybe some nuclear reaction could, but I am at a loss. Any hints?
| This is perhaps stretching the definition of "physical" but Metcalfe's law says that the value of a telecommunications network is proportional to the square of the number of users. A growing network might reasonably have a rate of growth proportional to its "value", which would lead the the differential equation you describe, so some networks might grow hyperbolically. See https://en.wikipedia.org/wiki/Hyperbolic_growth#Applications for more examples.
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Are orbits reversible in general relativity? It seems if I reverse velocities then things begin orbiting backwards, at least in classical mechanics.
From here:
Every orbit and trajectory outside atmospheres is in principle reversible, i.e., in the space-time function the time is reversed. The velocities are reversed and the accelerations are the same, including those due to rocket bursts. Thus if a rocket burst is in the direction of the velocity, in the reversed case it is opposite to the velocity. Of course in the case of rocket bursts there is no full reversal of events, both ways the same delta-v is used and the same mass ratio applies.
What's up when I put relativistic effects into the mix?
So for example I watch a super light test particle orbiting a black hole in a highly precessing flower shaped orbit. Then I put a bouncy wall into it's path that's at rest from my viewpoint when the particle hits it, so the particle bounces back reversing it's velocity from my viewpoint.
Would it begin running its orbits backwards?
The actual reason I'm asking this, because I want to know whether I can use backwards ray-tracing to render a black hole.
| For the simple case of a black hole in a locally flat spacetime(so that it does not accelerate) you don't have to worry about losing precision due to relativistic effects, because Schwarzschild and Kerr black holes emit exactly zero gravitational radiation(because they are both cylindrically symmetric), that is: the metric does not change over time so you can reverse the photons and get arrive at the same path.
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What are the advantages in using 2 identical capacitors? What advantage might there be in using two identical capacitors in parallel connected in series with another identical parallel pair, rather than using a single capacitor?
| The resultant capacitance of capacitors in parallel is the summation of the capacitances:
$$C_{eq} = \Sigma C$$
While, the inverse of the resultant capacitance of capacitors in series is the summation of their inverses:
$$\frac{1}{C_{eq}} = \Sigma\frac{1}{C}$$
Connecting capacitors in parallel will therefore help in increasing the capacitance of the circuit.
If each capacitor has a capacitance of $C$, your circuit will give a resultant capacitance of:
$$\frac{1}{C_{eq}} = \frac{1}{C+C} + \frac{1}{C+C}$$
$$= \frac{2}{2C}$$
$$= \frac{1}{C}$$
$$\therefore C_{eq} = C$$
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What is the sign of the work done on the system and by the system? What is the sign of the work done on the system and by the system?
My chemistry book says when work is done on the system, it is positive. When work is done by the system, it is negative.
My physics book says the opposite. It says that when work is done on the system, it is negative. When work is done by the system, it is positive.
Why do they differ?
| Taking it in a different sense: chemistry is taken the approach that you want to create a new configuration inside the reaction vessel (and input energy like heat or such) and the physics course book is taking about letting the system doing work (eg. burning what is in the reaction vessel).
In both ways entropy will be at least equal or positive.
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Overtaking with non-constant acceleration I have tried to solve this problem by adding the sum of the displacements during acceleration, constant velocity and deceleration, but it does not work out.
Question:
A car accelerates from rest to $20~\text{m/s}$ in $12$ seconds ($a =5/3~\text{ms}^{-2}$), it travels at $20~\text{m/s}$ for $40$ seconds, then retardation occurs from $20~\text{m/s}$ to rest in $8$ seconds ($a = -2.5~\text{ms}^{-2}$). As the car accelerates an RC car, moving parallel to the car, is moving at $14~\text{m/s}$. When will overtaking occur and what will the distance be? The RC car passes the car just as it starts to accelerate.
I can do this without a problem if acceleration is a constant. Is there a differential equation I can use to compute this as that is my better area or must I stick with the SUVAT equations?
Again, if I could be pointed in the right direction that way I can learn.
| Plotting the displacement time graph helps a lot in these kind of problems.
From this graph we can say the RC is ahead of the car initially. The car catches up and overtakes the RC at t=20 seconds.
Note:
You do not need graph plotting tools to plot these graphs. Basically, you don't need to plot these graphs to accurately too.
Rough back-of-the-envelope calculations can easily determine the displacement of the car and RC at various time intervals. Then you can use the fact that an accelerated S-T graph would be quadratic, hence a parabola, again giving you approximate plots.
This is enough to determine that the car will overtake the RC during its uniform motion. And then it becomes pretty easy to use SUVAT!
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A Rolling Quarter A U.S quarter is rolling on the floor without slipping in such a way that it describes a circular path of radius $R=4 \text{cm}$. The plane of the coin is tilted at an angle of $\theta=45^{∘}$ with respect to the horizontal plane. Find the coin's period $T$ in seconds, that is, the time it takes for the coin to go around the circle of radius $R$. The radius of a U.S quarter is $r=1.2 \text{cm}$.
| This is a well known problem. So, I will try to solve it in the general case. For a somewhat more detailed answer, please look at Chapter 9 of David Morin's Introduction to Classical Mechanics book. Also, since this looks like a homework, whenever you think you can do the rest of the problem yourself, stop reading and actually do it yourself!
Let's first go to the CM frame. As it can be seen in the figure below, I will illustrate the principal axis by $x_1$(into the page), $x_2$ and $x_3$.
Let $\Omega$ be the angular velocity of the attaching point of the coin with the floor around the center. This means $T=\frac{2\pi}{\Omega}$.
Using the non-slipping condition, one will find out that the total angular momentum is:
$$\vec{\omega}=\Omega \hat{z}- \frac{R}{r}\Omega \hat{x}_3$$
Now writing $\hat{z}$ in the principal coordinates as $\sin{\theta}\hat{x}_2+\cos{\theta}\hat{x}_3$ we will have:
$$\vec{\omega} = \Omega \sin{\theta} \hat{x}_2 - \Omega \left(\frac{R}{r}-\cos{\theta} \right)\hat{x}_3$$
The principal moments are:
$$I_1=I_2=\frac{mr^2}{4} \ \text{ and } \; I_3=\frac{mr^2}{2}$$
so the angular momentum will be:
$$\vec{L}= \frac{mr^2}{2}\left(\frac12\Omega \sin{\theta} \hat{x}_2 - \Omega \left(\frac{R}{r}-\cos{\theta} \right)\hat{x}_3\right)$$
Only the horizontal component of $\vec{L}$ is changing:
$$|\frac{d \vec{L}}{dt}|=\Omega L_{\perp}=\frac{1}{4}mr\Omega^2\sin\theta\left( 2R-r\cos\theta \right).$$
Now we have to calculate the torque relative to CM, as well. The torque comes from the forces at the contact point, the horizontal force is $m(R-r\cos\theta)\Omega^2$ and the vertical force will be simply $mg$.
$$|\vec \tau|= mg(r \cos θ) − m(R − r \cos θ)Ω^2 (r \sin θ)$$
Using $|\vec\tau|=|\frac{d\vec L}{dt}|$, we can find:
$$\Omega=2\sqrt{\frac{g}{6 R \tan\theta - 5 r \sin{\theta}}}$$
$$ \Rightarrow T= \pi \sqrt{\frac{6 R \tan\theta - 5 r \sin{\theta}}{g}} \approx 0.446 \text{s}$$
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Can spheres leaking charge be assumed to be in equilibrium? I am struggling with the following problem (Irodov 3.3):
Two small equally charged spheres, each of mass $m$, are suspended from the same point by silk threads of length $l$. The distance between the spheres $x \ll l$. Find the rate $\frac{dq}{dt}$ with which the charge leaks off each sphere if their approach velocity varies as $v = \frac{a}{\sqrt{x}}$, where $a$ is a constant.
This is embarrassingly simple; we make an approximation for $x \ll l$ and get
$$
\frac{1}{4 \pi \epsilon_0} \frac{q^2}{x^2} - \frac{mgx}{2l} = m \ddot{x}.
$$
We can get $\ddot{x}$ from our relation for $v$, so we can solve for $q$ and then find $\frac{dq}{dt}$.
However, in general, $\frac{dq}{dt}$ will depend on $x$ and hence on $t$. The answer in the back of the book and other solutions around the web have $\frac{dq}{dt}$ a constant.
You can get this by assuming that at each moment the spheres are in equilibrium, so that you have $\ddot{x} = 0$ in the equation of motion above.
Does the problem tacitly imply we should assume equilibrium and hence $\frac{dq}{dt}$ is constant, or am I missing something entirely? I.e. why is the assumption of equilibrium justified? I understand reasoning like "the process happens very gradually, so the acceleration is small compared to other quantities in the problem," but I don't understand how that is justified by the problem itself, where we are simply given that the spheres are small (so we can represent them as points) and $x \ll l$ (which we have used to approximate the gravity term in the equation of motion).
| I think the answer is sonething that you have overlooked, a (. ) AKA FULL STOP.
You state that the web results say the answe is dt/dq a this is a constant because (a) is A constant.
The question you ask is
"
Does the problem tacitly imply we should assume equilibrium and hence dt/dq is constant, or am I missing something entirely?"
I reckon you've overlooked the fact its (a) NOT (dt/dq) thats the constant.
Have I answered your question?
Sorry for the technical lomotations of my keyboard.
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Statistical analysis of data in Physics Analysis of data is integral in bridging the gap between theory and experiment. How much do the results of the analysis depend upon the choice between Bayesian and frequentist methods?
For instance, consider experiments in particle physics. This paper by Louis Lyons states that particle physicists use a hybrid approach to analysis (frequentist hypothesis testing and Bayesian parameter estimation). If a purely Bayesian/Frequentist approach were applied, would there be a difference in the results of the analysis? What implication would any difference have on the interpretation of data?
| If there are enough data and the prior is not completely unreasonable, the frequentist and the Bayesian approach give essentially the same answer. This is related to the central limit theorem.
If data are fairly scarce, the two approaches may differ a lot. In this case the Bayesian approach is far preferable but only if the prior reflects true prior knowledge and not just prejudice. (With a drastically wrong prior and limited data, the Bayesian approach tends to reaffirm the prejudice, and hence will be far worse than the frequentist result.)
To see this, take a prior and assume that just one observation. It is clear that the Bayesian outcome is just a small change of the prior. So if the prior was appropriate (reflected true knowledge), the result is an improvement, while if the prior was bogus (just prejudice), the outcome is as bad. If the number of observations is large, their contribution dominates the outcome, and the result is essentially prior-independent, and is easily seen to agree with the frequentist (maximum likelihood) result.
All this is completely independent of physics. But in statistical mechanics we have a good enough theory that enables us to choose appropriate priors. This is the (only) reason why the maximum entropy principle works there.
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Sliding force less or equal? Why is the force required to slide a magnet off a steel plate A LOT less than the force required to directly pull it off?
The force required to pull the magnet can be: 20lb
While the force required to slide the magnet can be: 1lb more/less.
Why is that?
| After much studying, and experimentation.
I conclude that indeed, the force required to "slide" the magnet's is less than the force required to "pull" them directly off.
| {
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How to prove that we are living in a 3+1D world? Is there any scientific experiment that can lead us to conclude we live in 3 spatial dimensions without the premise of the conception of limited dimensions?
Thank you all who helped in the improvement of this question (which was not clear at first).
EDIT:
I know that this can be a little philosophical, but it is also a scientific question.
Let's consider the scenario where the mankind was not ever able to see.
Let's also consider that this limitation could be surpassed thus not limiting us to reach a scientific and technological knowledge "similar" to what we have today.
Would this civilization of blind people reach the conclusion that they are living in a 3D spatial world?
Is the sense of touch enough to reach that conclusion? Is there any scientific experiment that can lead us to that conclusion without the premise of the conception of limited dimensions?
Would it be easier, harder, or just different to reach a conclusion predicted by the M-Theory? (please do not focus only on this last question)
| Verify any inverse-square law process, like gas diffusion or classical forces.
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The relation between Gauss's law and Coulomb law and why is it important that the electric field decrease proportionally to $\frac{1}{r^{2}}$? My question relates to the third MIT's video lecture about Electricity
and Magnetism, specifically from $21:18-22:00$ : http://youtu.be/XaaP1bWFjDA?t=21m18s
I have watched the development of Gauss's law, but I still don't quite
understand the link between Gauss's law and Coulomb law: How does
Gauss's law change if Coulomb law would of been a different one.
I also don't understand why is it so important for Gauss's law that
the electric field decrease proportionally to $\frac{1}{r^{2}}$ ?
For example, what would of happened if the electric field decrease
proportionally to $\frac{1}{r}$ , or $\frac{1}{r^{3}}$ ?
| Gauss's law states that the ratio of charge and the dielectric constant is given by a (two-dimensional) surface integral over the electric field:
$$\int E\cdot dA=\frac{Q}{\epsilon_0},$$
where I have omitted vector notation for simplicity.
It can be linked to Coulomb's law by assuming spherical symmetry of the electric field and performing the integration.
The other way around, you can start by assuming that Coulomb's law,
$$E=\frac{1}{4\pi}\frac{Q}{r^2},$$
holds, and take the divergence on both sides of the equation. This leads to the differential form of Gauss's law:
$$\nabla\cdot E=\frac{\rho}{\epsilon_0},$$
where $\rho$ is the charge density. In order to reproduce its integral form, just integrate both sides of the equation. Detailed calculations can be found on the wikipedia page on Gauss's law. If you really want to understand how it all comes together (which I assume to be true, otherwise you wouldn't ask here), I would recommend reproducing the calculation on your own. Then you can also try to see what happens if you assume other forms of Coulomb's law.
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Projectile Motion Question involving a ball and a ramp inclined at an angle The question is to finde the initial horizontal velocity of the ball at end of the ramp, where it is released.
I know how to do this using gravitational potential energy and kinetic energy ($v=\sqrt{2gh}$), assuming all potential energy is converted into kinetic energy but the question is asking me to find the error in the experiment.
It gives me the values on $y$ as the ramp is lifted up (it is lifted up about 10 cm each time) and the corresponding values of $x^2$. I drew a graph of $x^2$ vs $y$ and found the gradient.
How can I use this to find the experimental initial horizontal velocity?
This sketch shows what the experiment looks like:
| There is an 'error' in your reasoning. The velocity of the ball at the end of the ramp is not given by $v=\sqrt{2gh}$. This assumes that the ball slides down the ramp. Presumably it rolls and gains rotational as well as translational kinetic energy. If the ball has uniform density and rolls without slipping then the velocity after rolling down an incline is $v=\sqrt{\frac{10}{7}gh}$, which is smaller. If the ramp is quite steep there could be a mixture of rolling and sliding, so the launch velocity will be somewhere in between the two values.
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How to block neutrons What is a good way to block neutrons and what is the mechanism that allows this? It's my understanding that polyethylene is somewhat effective. Why?
| Being bulk neutral neutrons participate only weakly in electromagnetic interactions which is the dominate interaction for charged particles.
Instead neutron scattering can be thought of as primarily a contact interaction with the nuclei of atoms in the way. Light atoms (and hydrogen in particular)
*
*have a larger cross-sectional area per nucleon than heavy ones
*take up more of the energy of the interaction in recoil than heavy ones
making them much more effective at reducing the kinetic energy of non-thermal neutrons per unit areal mass density.
Historically waxes, water and plastics have been the neutron shielding materials of choice, though concrete or rammed earth are cheap and not too bad.
Once down to thermal energies neutrons get as much kinetic energy as they lose on average and you just have to wait for them to decay or capture.
Doping your absorber with boron, chlorine or even gadolinium will help to capture the thermalized neutrons faster. PVC gets you the chlorine for free in your plastic, and boron can be added easily to concrete or to a number of plastics.
It should not be overlooked that it takes a lot of space to slow, thermalize and capture neutrons (that contact interaction thing means they go through more material before interacting than charged particles); especially if you need to get them all. They are notorious for penetrating large quantities of shielding, and distance is one of your best friends when it comes to neutron shielding.
| {
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How to derive the metric for a 2-sphere I have a question in Polchinski's string theory vol I p 167. It is said
For example,
$$ds^2= \frac{ 4 r^2 dz d \bar{z} }{(1+ z \bar{z})^2} = \frac{ 4 r^2 du d\bar{u}}{ (1+ z \bar{z})^2} \tag{6.1.3}$$
describes a sphere of radius $r$ and curvature $R=2/r^2$.
Why Eq. (6.1.3) describes a sphere of radius $r$ and curvature $R=2/r^2$? How to derive it?
[I tried to use $z=\sigma_1 + i \sigma_2$ and $\bar{z} = \sigma_1 - i \sigma_2$ coordinates. By $dz d\bar{z} = 2 d \sigma_1 d \sigma_2$, I got
$$ ds^2= \frac{ 8 r^2 d \sigma_1 d \sigma_2 }{ (1+ \sigma_1^2 + \sigma_2^2)^2} $$
By any means I have imagined, it does not look like a sphere.]
| Writing :
$x_1 = \sin \theta \cos \phi$,
$x_2 = \sin \theta \sin \phi$,
$x_3 = \cos \theta $
The unit radius $2$-sphere metrics is $ds^2=(d\theta^2 + \sin^2 \theta ~d\phi^2)$
We are going to use the stereographic projection : $ \large z = \frac {x_1+ix_2}{1-x_3}$
This gives : $z = cotg(\theta/2) ~e^{i \phi}$
So, $$ dz = \frac{1}{2}(\frac{-1}{\sin^2 (\theta/2)})~e^{i \phi} ~d\theta + i~cotg(\theta/2) ~e^{i \phi}~ d\phi$$
$$ d \bar z = \frac{1}{2}(\frac{-1}{\sin^2 (\theta/2)})~e^{-i \phi} ~d\theta - i~cotg(\theta/2) ~e^{-i \phi}~ d\phi$$
We have :
$$1 + z \bar z = 1 + cotg^2(\theta/2) = \frac{1}{sin^2(\theta/2)}$$
$$dz d \bar z = \frac{1}{4} \frac{1}{sin^4(\theta/2)} ~d\theta^2 + cotg^2(\theta/2)~~d\phi^2$$
Finally,
$$\frac{4 ~dz ~d \bar z}{(1 + z \bar z)^2} = d \theta^2 + sin^2\theta ~~d\phi^2$$
We then multiply by $r^2$ to get the metrics of a sphere of radius $r$
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What happens when the black hole at a galactic core eats the galaxy? I'm making several assumptions, not sure if any are correct:
*
*there is a black hole at the center of a galaxy
*the black hole is eating the galaxy
Eventually the galaxy will be gone, right?
Has this been observed? Do we know what happens afterwards?
Posting here since astronomy got merged into physics
| A black hole is "just" a massive object. Interesting things happen when close to the black hole, because the high gravity makes all the Einsteinian effects more apparent, including the "horizon" and the trapping of light, and so on. But from afar, this is "just" a massive object, which other objects handle like any other, i.e. by orbiting it. If you replace a massive star with a black hole of the same mass, then planets orbiting it just keep on orbiting it with the same orbits.
Stars in a galaxy have no more reason to "fall" towards the central black hole and be eaten by it than the Earth would have to fall towards the Sun and collide with it.
As for long term behaviour, black holes are supposed (theoretically) to evaporate over time.
| {
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What is the percentage of useful energy do we get from matter-antimatter annihilation? This is a theoretical question since we haven't made enough antimatter to try it in reality of course. But I am asking about the physics part in this.
Also, by "useful energy" I mean the energy we are able to use either as a heating source for something like a nuclear reactor, or as energy for an explosion like nuclear explosions.
If I am not mistaken, a large part of the energy we get from the annihilation is in the form of neutrinos, which we for some reason can't consider them useful energy.
So now, if we subtract the energy of the neutrinos, is it safe to consider the rest as useful energy as I explained ?
Please try to be as simple as possible because I don't speak English very well.
| I found a interesting paper on the use of antimatter for rocket propulsion from NASA that addresses this subject. Your question has a pretty complicated answer.
I think you will find more formation than you wanted reading that article. In the report it says about an electron-positron collision:
His results indicate that the neutrinos carry off ~22% of the
available pion energy (~1248 MeV) whereas the muons retain ~78%. The
unstable muon, having an average energy of ~300 MeV, also decays (in
~6.2μs) into an electron, or positron, and two neutrinos as shown in
Table 2. The energy appears to be about equally distributed among the three
particles with the neutrinos carrying off ~2/3 of the available
energy. Ultimately, the electrons and positrons can also annihilate
yielding additional energy in the form of two 0.511- MeV gamma rays.
The neutrinos are considered to be massless and move at essentially
the speed of light. They are extremely penetrating and rarely interact
with matter. Under vacuum conditions the various muon and electron
neutrino particle–antiparticle pairs carry off ~50% of the available
annihilation energy following a p–p reaction.
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Why does light diffract only through slits? We can see diffraction of light if we allow light to pass through a slit, but why doesn't diffraction occur if we obstruct light using some other object, say a block? Why are shadows formed? Why doesn't light diffract around the obstruction as it does around the slit?
| Diffraction can also occurr around abstruction . Suppose u put ur finger infront of light w.front such that ur fingers shadow will appear on the screen placed behind ur finger . When light w.front strikes ur finger then the light ray at the upper and lower extremes of obstruction will bend and enter in to shadow region generated by the obstruction and in this case light will not spread rather it will converg into shadow region and form argo spot
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Localization of wave or particles
Nonlinear field theories contain a large number of localized solutions.
I have found this text in a article. What I don't understand is "what is localized?". Is it refer defining position of a particle or a wave?
Can someone give me an elaboration with example?
| I would guess that the article is referring to solitons. I'm not sure if every non-linear system gives soliton solutions, but many do. The Wikipedia article I've linked gives lots of examples of classical solitons, but I'm not sure to what extent (if at all) they're important in the Standard Model. Perhaps one of the QFT specialists hereabouts could comment.
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Why is there a factor of $4\pi$ in certain force equations? I mean to ask why there is $4\pi$ present in force equations governing electricity? Though all objects in universe are not spherical and circular, the constant of proportionality in both equations contain $4\pi$. Why?
| The main reason is that it makes the calculation easier and the results look nicer. For example, suppose the field (or force) is given by
$$ E = \frac{1}{4\pi} f(\mathbf{r})$$
for some function $f(\mathbf{r})$.
If the system process rotational symmetry, after sum over the density, you will get a factor of $2\pi$. If the system process spherical symmetry, you will get a factor of $4\pi$. In both case, the resulting equation contains no $\pi$ factor. It is particular useful in EM as we usually consider a density distribution process some symmetry.
So, why there is no $\pi$ in the Newtonian gravity $F=GMm/r^2$ ?
Because there is no such need. You cannot craft a planet like macroscopic object into a infinitely long rod, a disk shape, or some funny shape. Gravity only have significant effect when it accumulates a large mass, at the same time, it become a sphere by its own gravity. So, basically, it is already a good point like particle when we look at it some radius away. The extra $\pi$ won't make any calculation easier.
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Degrees of freedom of the graviton versus classical degrees of freedom I have a puzzle I can not even understand.
A graviton is generally understood in $D$ dimensions as a field with some independent components or degrees of freedom (DOF), from a traceless symmetric tensor minus constraints, we get:
*
*A massless graviton has $D(D-3)/2$ d.o.f. in $D$-dimensional spacetime.
*A massive graviton has $D(D-1)/2-1$ d.o.f. in $D$-dimensional spacetime.
Issue: In classical gravity, given by General Relativity, we have a metric (a symmetric tensor) and the Einstein Field Equations(EFE) provide its dynamics. The metric has 10 independent components, and EFE provide 10 equations. Bianchi identities reduce the number of independent components by 4. Hence, we have 6 independent components.
However, for $D=4$, we get
*
*2 independent components.
*5 independent components.
Is the mismatch between "independent" components of gravitational degrees of freedom (graviton components) one of the reasons why General Relativity can not be understood as a quantum theory for the graviton?
Of course, a massive graviton is a different thing that GR but even a naive counting of graviton d.o.f. is not compatible with GR and it should, should't it? At least from the perturbative approach. Where did I make the mistake?
| I used to count dofs in terms of spinor representation. There are two spinor representation of SO(3,1) (1/2,0) and (0,1/2) denoted by dot and no-dot indices.
Vector (spin 1) representation in spinor indices is (1/2,0) X (0,1/2) = (1/2,1/2) which has 4 dofs. If the theory is massless, the theory is gauge invariant, there is one gauge dof, which can be eliminated by the gauge fixing condition. One dof is time-like one, which is not physical. So 4-1-1 =2 physical dofs remain.
For symmetric tensor we have (1/2,1/2)X (1/2,1/2) = (1,0)+ (0,1)+ (1,1)+(0,0) The irreducible symmetric representation is just 9-dimensional. The general coordinate transformation eliminate 4 dofs (which essentially is 4 space-time coordinates). So the massive spin 2 has 9-4=5 dofs. The massless spin 2 theory will be gauge invariant, where we can eliminate 1 dof. Two dofs are time-like which do not propagate. So only 5-1-2 dofs remain.
We can extend the argument to arbitrary dimension.
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Path of least resistance vs. short circuit Some sources on the web claim that "electricity follows the path of least resistance" is not true, e.g. this physics SE question. However, in every explanation of "short circuits", the author says that current flows through the short because it's following the path of least resistance. How do I reconcile these two facts?
| The statement electricity flows through the path of least resistance means that electricity flows more into the path with less resistance, because it is proportional to the inverse of resistance, obeying:
$$I_1=I\frac{R_2}{R_1+R_2}$$
$$I_2=I\frac{R_1}{R_1+R_2}$$
From these relations you can find out that in two cases all of the current flows through the 1st resistor:
*
*$R_1=0$
*$R_2\gg R_1$ $($mathematically $R_2\to\infty$ $)$
The important fact is that what matters here is the ratio of the resistances:
$$I_1=I\frac{R_2}{R_1+R_2}\to I_1=I\frac{1}{\frac{R_1}{R_2}+1}$$
If you plot $I_1\over I$ from the above relation (w.r.t $\frac{R_1}{R_2}$ ) you can see this better:
This is the meaning of the quoted statement.
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Melting and freezing point Why is the melting and freezing point of a substance are always the same? This was quoted in my textbook but they didn't give a reason for this being so.
| It's the same temperature because it's the only temperature at which the liquid phase and the solid phase may co-exist – which is a symmetric description of the temperature.
When we add heat to this mixture of "ice" and liquid, it will keep the temperature at the same point but the percentage of "ice" will be decreasing, and only when all the "ice" is gone, the temperature will stop rising. In the same way, if we remove heat from the mixture, the temperature will be constant for a while as more liquid turns into "ice", and only when the whole body is frozen, the temperature starts to drop.
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Does alternating current (AC) require a complete circuit? This popular question about "whether an AC circuit with one end grounded to Earth and the other end grounded to Mars would work (ignoring resistance/inductance of the wire)" was recently asked on the Electronics SE.
(Picture edited from the one in the above link)
Though I respect the AC/DC experts there, I think (with the exception of the top answer) they are all wrong.
My issue is that they all assume that AC requires a complete circuit in order to function. However, my understanding is that a complete circuit is necessary for DC, but not AC. My intuitive understanding is that AC is similar to two gas-filled rooms with a pump between them - the pump couldn't indefinitely pump gas from one room to another without a complete circuit (DC), but it could pump the gas back and forth indefinitely (AC). In the latter case, not having a complete circuit just offers more resistance to the pump (with smaller rooms causing a larger resistance).
Is my understanding correct - can AC circuits really function without a complete loop?
More importantly, what are the equations that govern this?
If larger isolated conductors really offer less AC-resistance than smaller AC conductors, how is this resistance computed/quantified? Would its "cause" be considered inductance, or something else?
|
Does AC current require a complete circuit?
Can AC circuits really function without a complete loop?
I think that neither AC nor DC theoretically needs a complete loop. (ie, without reusing the electrons/charge carriers that flow in the circuit)
And you don't even need to think about capacitors for that. A capacitor in a DC circuit still has an electric field between the plates (in case of a parallel plate capacitor): the same electric field responsible for current in the circuit, and that, to me constitutes a closed circuit.
Of the top of my head, I think the following illustrations would suffice to back what I am saying:
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Red Giant branch and Asymptotic Giant branch What's the difference between the RGB and the AGB? I can't seem to find an clear distinction anywhere.
Thanks.
| Red giants and asymptotic giants have some close similarities, and one actually evolves into the other. Both have an extended envelope of relatively cool, non-burning material (mostly $\rm{H}$, $\rm{He}$). They also each have a core of dense, non-burning material; in the case of the red giant this is mostly $\rm{He}$, while for the asymptotic giant it's $\rm{C}$ and $\rm{O}$.
The burning shell in the red giant is $\rm{H}$. For stars of the right mass, the conditions (density, temperature) in the core will periodically be sufficient to ignite the $\rm{He}$ causing a "core flash".
Red giant structure:
After the red giant branch of stellar evolution there is a brief period where the $\rm{He}$ core burns called the horizontal branch. Once the He core is exhausted (it's been converted to $\rm{C}$ and $\rm{O}$), the star starts on the asymptotic giant branch. This branch has two parts, the early asymptotic giant branch (E-AGB) and the thermal-pulse asymptotic giant branch (TP-AGB).
E-AGB structure:
Stars on the E-AGB are like red giants, but in addition to a $\rm{H}$ burning shell there is a $\rm{He}$ burning shell (the energy output is dominated by the He burning shell). In the TP-AGB, the $\rm{H}$ shell picks up again and dominates the energy output, but periodically as the $\rm{He}$ produced by $\rm{H}$ burning is accreted onto the $\rm{He}$ shell, "helium shell flashes" occur, analogously to helium core flashes in red giants.
Source/Reference: Carroll & Ostlie "An Introduction to Modern Astrophysics: 2nd Edition" (Pearson)
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When is temperature not a measure of the average kinetic energy of the particles in a substance? I had always thought that temperature of a substance was a measure of the average kinetic energy of the particles in that substance:
$E_k = (3/2) k_bT $
where $E_k$ is the average kinetic energy of a molecule, $k_b$ Boltzmann's constant, and $T$ the temperature. (I'm not sure of the 3/2 coefficient.) Then I heard from several folks that this is a simplistic notion, not strictly true, but they didn't explained what they thought was flawed with this idea. I'd like to know what (if anything) is objectionable about this idea? Is it that the system must be macroscopically at rest? Is it that it ignores the quantum mechanically required motion of particles that persists at low temperatures? When is it not valid?
| There are a number of ways of defining temperature, for example using:
$$ {\partial S \over \partial E} = {1\over T} $$
This definition is the basis of negative temperatures. This is a case where the temperature is not a measure of average kinetic energy.
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Unpolarized light vs. randomly rotating polarized light? I am confused with physical picture about unpolarized light.
Is unpolarized light very fast rotating polarized light? or co-existing state of two orthogonal polarization? (or something else?)
If there is a linear polarizer which rotates very very fast and randomly (the polarizer in imagine), the output light is same to unpolarized light? I don't think so but I am not sure.
--
or, instead of linear polarizer, a Faraday rotator with magnetic field whose amplitude is randomly chnaged can be considered, I think.
| Unpolarized light can be thought of as a superposition of wave trains of a finite duration of order $0<\tau<\infty$, each of which has a certain pure polarization, which may be elliptical, with a random direction. The phases of the pulses and their start and end times are also random.
What this means in practice is that any unpolarized light source has a coherence time $\tau$. If you look at the polarization with higher temporal resolution than this, you will see a pure polarization (per spectral component! If the light source is not monochromatic the picture is more complicated). If you measure with a lower resolution, the randomly rotating polarization will average out and you will observe no polarization effects.
To put things in scale, the coherence length ($=c\tau$) of sunlight is about $0.6\,\mu\text m$ (doi). In practice this means that any polarization-dependent interferometry must involve path differences shorter than that, or you will be seeing the (lack of) interference between two different pulse trains with random relative polarizations and phases.
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What are magnetic field lines? Does a magnetic field have concentrations of magnetic force lines as seen when putting iron filings over top a bar magnet or are these imaginary? I.e. are they just an artifact of the iron being a 'conductor' of the magnetic field lines making them look like they are concentrated along these path lines but are really continuum of strength around the bar magnet when there are no filings are present?
Also I learned that field lines do not cross, yet there are magnet configurations who's forces are explained as the magnetic force vectors are indeed crossing and are additive such as a Halbach Magnet Array. So what is actually happening here?
| Field lines are a visualization tool for the magnetic field (and for other fields, too). They are a way of representing the field (which is a physical---if intangible---thing) in a drawing. They are not unique in this: there are other ways of representing vector fields, but they have a long history.
Now, magnetic fields are vector fields subject to superposition, which means that the total field can be given by adding up the contributions for multiple fields. If you visualize each on the contributions separately then the field lines from one may cross the field lines from another, but it remains true that
*
*No single contribution will be represented with lines that cross
*The net (total) field will not be represented with lines that cross
| {
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Frequency of small oscillation of particle under gravity constrained to move in curve $y=ax^4$ How to find the frequency of small oscillation of a particle under gravity that moves along curve $y = a x^4$ where $y$ is vertical height and $(a>0)$ is constant?
I tried comparing $V(x) = \frac 1 2 V''(0) x^2 + \mathscr O(x^3) = \frac 1 2 kx^2$ (assuming $V(0)$ is ground state and $V'(0)$(slope) remains is horizontal at extrememum, but unfortunately $V''(0) = 0$. I am pretty much clueless. Thanks for your help.
ADDED:: Letting $m=1$ the Lagrangian is $L = \frac 1 2 (\dot x^2 + \dot y^2) - gax^4 = \frac 1 2 (\dot x^2 + (4 ax^3 \dot x)^2)-gax^4 = \frac 1 2 \dot x^2(1 + 16a^2x^6)-gax^4$
The above Lagrangian gives the equation of motion as
$$\ddot x(1+16a^2x^6)+\dot x^2 96 a^2x^5 + 4 gax^3 = 0$$
Since we are considering the system a small oscillation whose potential is of order 4, $\mathscr{O}(x^{k>4})$ can be ignored which reduces into $\ddot x = -4agx^3$.
To solve this,
$$\frac 1 2 \frac{d}{dt}(\dot x^2) = -\frac{d}{dt}(agx^4)$$
which gives $\dot x = \sqrt{k - 2agx^4}$, Assuming the system begins from $t=0$ at $x=x_0$ with $\dot x = 0$, $k = 2agx_0$, which turns the integral into
$$\int_0^{T/4} dt= \int_0^{x_0} \frac 1 {\sqrt{2agx_0^4-2agx^4}} dx=\frac{1}{\sqrt{2ag}x_0}\int_0^1\frac{1}{\sqrt{1-y^4}}dy$$
Which gives $$T = 2 \sqrt 2 \sqrt{\frac{\pi}{ag}}\cdot \frac{\Gamma(5/4)}{\Gamma(3/4)}\cdot \frac 1 {x_0}$$
which is a dubious result. Please someone verify it. Any other methods are welcome.
| You got the correct answer on your own, but I'd like to point out a useful treatment for this type of questions, from Landau and Lifschitz, Course of theoretical physics.
It will get you only the result up to a numerical constant, but it's
beautiful and absolutely general.
I quickly recall the definition of an Euler homogeneous function: $f(x)$ is said to be homogeneous of degree $p$ if $f(\alpha x) = \alpha^p f(x)$.
Absolute generality
Let $L = K - U$ be a lagrangian. Usually, the kinetic energy is quadratic in the velocity, so that it is homogeneous of degree 2 in space and -2 in time. Or, more clearly, if you "stretch" the time by a factor of $\alpha$ and the space by a factor of $\beta$ the kinetic energy will transform as
$$
K^\prime =\frac{\beta^2}{\alpha^2}K.
$$
Now, if the potential energy is homogeneous of degree $p$ in space you will get similarly $U^\prime = \beta^p U$, so that the lagrangian becomes
$$
L^\prime = \frac{\beta^2}{\alpha^2}K - \beta^p U = \frac{\beta^2}{\alpha^2}(K - \beta^{p - 2}\alpha^2 U).
$$
That is equivalent to the original lagrangian whenver $\beta^{p - 2}\alpha^2 = 1$, or
$$
\alpha = \beta^{1 - p/2}.
$$
When two lagrangians are equivalent, any solution of one is
also a solution of the other. So, in this large class of problems when you
stretch the distances by a factor of $\beta$, you have to multiply the time
by a factor of $\beta^{1 - p/2}$. If you have a period of time $T$ for a
solution, then it must be proportional to the distance with the above dependence $x^{1-p/2}$.
Applications
The current question
You derived the approximate lagrangian with a quadratic kinetic energy and a
potential proportional to the fourth power of the distance from the origin.
Hence $p = 4$ in the above formula, and you get straight away $T \propto
1/x$. With dimensional analysis you can actually get
$$T \propto \dfrac{1}{\sqrt{ag}x}.$$
Isochronous oscillations
For a potential proportional to the square of the distance (small
oscillations of a pendulum, spring), $p = 2$, and $T$ doesn't depend
on the amplitude of the oscillations.
Kepler's third law
Gravity has a potential with $p = -1$. Hence $T \propto x^{3/2}$.
| {
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Why does boiling water in the microwave make a cup of tea go weird? When I boil water in the kettle, it makes a nice cup of tea. Sometimes I need to use a microwave because a kettle isn't available. I boil the water in the mug and it looks pretty normal, but when I drop in the teabag the water froths up and looks foamy. I don't see what the chemical difference is here, so I assume it must be some physical difference. I have noticed this with multiple types of tea and multiple microwaves, the results being consistent so it's not just a weird microwave or something like that.
What is the reaction here and how/why does it occur?
Here is a photo of the 'fizzy' looking tea just after dunking in the teabag.
| A common problem with microwave is that you lack control of the
temperature of the water.
Second problem is that the water is heated from the top and the sides of the mug mostly.
Result is, that the content of the mug is not really boiling hot everywhere,
altogether (after mixing) it is well below 100 °C
A side effect from this is, that the water is not dearated
like it is when You boil it on a hot plate.
So: You have some water close to 100 C at the surface
and colder (containing air) water below.
When You dip in the bag, all that mixes, the air bubbles out.
| {
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Could we see a lower dimension within our universe? My question refers to the fact, that, say if our universe were to be sitting on a 4 (spatial) dimensional plane, that we cannot see, then within our universe could there be a point mass in which 1 or 2 (spatial) dimensions could exist. We may not be able to see them, this could be a result of our world only consisting of 3 Dimensions. Although the computer screen, for example, is 2Dimensional, it sits on a 3 Dimensional object. Could a 2Dimensional object or world, exist in our 3D plane without having a 3D object that it is displayed upon, I have illustrated it below to build a clearer picture...
I hope this makes sense!
| Based on the string theory tag in your question, you're probably asking for something like a brane world. In such a scenario, standard model interactions, which are mediated by open strings, and are thus bound to the branes. According to some research (e.g.), it is possible to ensure general relativity also behaves normally on the brane despite the higher dimensionality of spacetime.
| {
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What is the difference between phase difference and path difference? The path difference is the difference between the distances travelled by two waves meeting at a point. Given the path difference, how does one calculate the phase difference?
| Path difference and Phase difference are very similar things.
Let me show this in a bit more intuitive way!
In our world, we have "good" waves and "bad" waves (atleast for calculations)
For example :
*
*A good wave is y=sinx plotted on a graph having crests and troughs at $\frac{n\pi}{2}$ alternatively.
*
*A bad wave is basically what I like to call a "light wave". I mean light waves do look like sine waves (amplitude=1, wavelength=$2\pi$) but they have different amplitudes and wavelengths.
What happens to a bad wave?
Well, we really love to work with $\pi$ rather than with plain numbers like $632nm$. You will get to know why.Just read till the end.
Suppose a physicist is given two coherent sources of light of wavelength $600nm$ (lets say, for performing Y.D.S.E).
Now, what physicists secretly do is that they do a mathematical transformation of the light wave (to make a new wave) in such a way that the wavelength of the light wave is made equal to $2\pi$ units
Now the physicist does all his calculations on the new transformed wave. Suppose he transformed the light waves of the two sources in the similar way and finds that the difference between two nearest crests of the two waves is $\pi$ (a.k.a phase difference)
What the physicist found is called as phase difference of two waves. But as you know, the physicist is doing all his calculations on a transformed wave. Meaning, that the real life phase difference of the wave will be very different than the calculated value $\pi$.
So our task is to get the "real life phase difference". We use pretty basic unitary methods to find the relation between the phase difference of our transformed wave and the real wave.
Since $600nm$ (wavelength of real light) in real light corresponds to $2\pi$ in transformed wave, therefore ,
$\Delta x$ phase difference in the real waves is equal to $\frac{2\pi\phi}{\lambda}$ where $\phi$ is the phase difference of transformed waves.
Now, the moment of truth. This $\Delta x$ that you calculated is the phase difference of real waves. This is also known by the term "path difference"
Therefore path difference is nothing but phase difference of the transformed waves. Phase difference is just an aid for ease of calculations. The real thing that matters to get correct results is the path difference.
| {
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Does light loses its energy when it passes through denser medium? I know it does not because it emerges out of denser medium at 300,000 KM per second, but according to $E=mc^2$ and given that speed of light decreases inside denser medium with refractive index greater than 1, does not it suggest that energy of light inside denser medium is less?
| A light wave consists of many photons that give its amplitude. While the frequency of the light stays the same the number of photons or waves amplitude may change if there’s absorption.
Every real substance absorbs and scatters even the transparent. But statistically the frequency is same.
The way you asked the question implies that you ask if the frequency can change. If you consider scattering I.e. the Compton effect yes it is possible to change even the frequency and this is the quantum nature of light. Here we are considering single events.
| {
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Are dark energy and zero-point energy the same thing? According to Quantum Mechanics is it possible that the famous "dark energy" and "zero-point energy" are the same thing that drives the accelerated expansion of the universe or maybe related to each other?
| Nobody knows. There are multiple explanations for dark energy that haven't been eliminated. One of the explanations that hasn't been eliminated is zero point energy:
http://arxiv.org/abs/1205.3365.
Another possible explanation is that the alleged expansion is actually a result of neglecting the effects of inhomogeneities on averaging:
http://arxiv.org/abs/1311.3787.
| {
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Controlling Neutrinos for Communication Neutrinos travel straight through earth at the speed of light. Therefore, it seems that they could be great for intercontinental communication. Of course, I assume a lot still needs to be learned about detecting, producing and controlling neutrinos before they can be used for the practical purpose of communication.
My question: In principal, could neutrinos be manipulated similarly to radio waves for the purpose of communication? I mean, modulation, filtering, etc. ?
| Radio waves normally transmit information by amplitude or frequency modulation. This assumes there is a carrier wave that can be modulated, and as twistor59 says in his comment, creating a carrier wave using neutrinos would be very difficult.
However many radio and TV stations are already streamed digitally, and in principle neutrinos could be used for this. I say in principle because, as you say in your question, neutrinos interact too weakly for this to be possible in the near future.
| {
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What does imaginary number maps to physically? I am taking undergraduate quantum mechanics currently, and the concept of an imaginary number had always troubled me. I always feel that complex numbers are more of a mathematical convenience, but apparently this is not true, it has occurred in way too many of my classes, Circuits, Control Theory and now Quantum Mechanics, and it seems that I always understand the math, but fail to grasp the concept in terms of its physical mapping. Hence my question, what does imaginary number maps to physically?
Any help would be much appreciated
| @Prathyush's answer is completely correct and provides nice reference, but I just wanted to add that behind his discussion about phases and path integrals is the concept of unitary and conservation of probability, which is a physical concept, and the OP asked about what imaginary numbers "map to physically". Thinking in the Heisenberg picture one can write every operator as a real diagonal matrix and a set of mappings (rotations) between these operators, but in order for everything to conserve probability it is absolutely essential that these rotations contain imaginary numbers in general as any unitary operator may be written as $\exp{i \hat{H}}$, where $\hat{H}$ is a Hermitian operator (with real eigenvalues). It is similar for another very special rotation in Hilbert space, which is time evolution, without which there would be no physics as such. This has a deep connection to the relation between Lie algebras and Lie groups.
Without complex numbers we would have no unitarity and no conservation of probability or would be led to a trivial quantum theory which is not dynamical!
| {
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Expansion of the universe and strain From cosmological models that involve expansion of the universe, can we not say that there are ever increasing tidal forces felt by solid bodies?
If so, the material in solid bodies like metal blocks, glass rods, skeletal systems will tend to separate causing a strain on the body which counteracts the expansion
If the tidal forces increase with time, will there not come a point in time, when the stress limit is reached and our bodies begin to shatter under intense tidal forces (similar to 'spaghettification')?
If the above is true, using present cosmological data, has anyone calculated how long it will take before our bones shatter?
| This is really just a footnote to Luboš's answer, but for completeness (and because it's fun :-) we should note that the equation of state of dark energy has not been determined and it remains possible that the ratio between the dark energy pressure and its energy density is less than or equal to -1. If so, this is known as phantom energy, and it causes an ever increasing acceleration leading eventually to a Big Rip and the destruction of, well, everything!
In this scenario there is indeed a growing stress on anything with a finite size, and our bones will indeed shatter. Fortunately this seems a remote possibility.
| {
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Which way do black hole jets spin? The centers of black holes and quasars often have jets coming out the two poles of an accretion disk, say north and south. Is it known if the two jets spin in the same direction or opposite directions to each other?
| Looking at Earth's spin down upon the north pole, it is counter-clockwise. Looking at Earth's spin down upon the south pole, it is clockwise. The sense of helicity depends upon the observer's position. (If the spin is relativistically propagating, the observer only has one POV. Beta-rays are chiral left-handed until they are slowed.)
Looking down in one direction all the way through, there is a problem with angular momentum disk vs. jets if the mirror-direction jets are not spinning in the same sense while traveling in opposite directions. Choose your nomenclature to define the relative directions of spin.
This is not a frivolous problem! Chemical optical rotation is "coming toward," viewed through a polarimeter. Geometric rotation is "going away," whether clockwise rotation drives the screw into wood (if yes, right-handed). Optically left-handed quartz is observed to be crystallographic space group P3(1)21, which is a right-handed 3(1) screw axis (and 3(2) left-handed). It gets worse! P3(1)21 quartz and P3(2)21 berlinite are both levorotatory, as optical rotation is not sourced by physical atomic mass distribution, J. Appl. Crystallogr. 19, 108 (1986).
When you say "same direction of rotation," define it very carefully.
| {
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Background radiation: radon vs potassium 40 In doing a little research into natural background radiation, I came upon a table from the National Council on Radiation Protection and Measurement (NCRP). It shows that inhaled radon gas is by far the largest contributor to average annual dose equivalent from background radiation, for people living in the United States. [1] That table lists "other internally deposited radionuclides" as contributing only a small fraction of what inhaled radon gas contributes. Among those "other internally deposited radionuclides" is potassium-40.
According to another source, the typical activity of potassium-40 in the human body is about 0.1uCi. [2] Typical activity of radon in outside air is said to be 0.4pCi/L, while for indoor air the average is 1.3pCi/L. [3] Given that the typical volume of air the average adult inhales is 0.5L, at any given time an adult has about 0.4pCi of radon in their lungs. [5] Note that the stated activity levels for radon are about a million times lower than those for potassium-40 (0.4 pCi versus 0.1 uCi).
I also learned that radon typically decays via alpha decay. [4] Potassium-40, on the other hand, mostly decays via beta decay. [2] I understand that alpha radiation is given more weight than beta radiation, generally by a factor of 20 or so, when estimating the biological effects of radiation. [6] However, a factor of 20 falls far short of the 10 million or so times higher effect stated for radon versus all other internally deposited radionuclides in the NCRP table. The table implies that the one million times greater activity of potassium-40 results in only about a tenth (or less) the equivalent dose. That seems way beyond what could be accounted for with just a weighting factor.
Why doesn't potassium-40 contribute much more to the equivalent dose? What am I missing?
| In addition to the deposition location when EnergyNumbers mentions in the comment the types of radiation and the length of the decay chain are an issue.
Radon sits at the top of a long sequence of decays many of which are alpha emitting (quality factor $\approx 10$--$20$) including Po-210 (5.34 MeV alpha, yikes!). Also the radon has a non-zero fission branch (quality factor $\approx 20$+). Basically per deposited MeV the radon is doing a lot more damage.
Then the alphas do their damage in a highly localized place and the K-40 (being a beta emitter) spread the damage around.
| {
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Feynman's subscript notation Consider this vector calculus identity:
$$
\mathbf{A} \times \left( \nabla \times \mathbf{B} \right) = \nabla_\mathbf{B} \left( \mathbf{A \cdot B} \right) - \left( \mathbf{A} \cdot \nabla \right) \mathbf{B}
$$
According to Wikipedia, the notation $\nabla_\mathbf{B}$ means that the subscripted gradient operates on only the factor $\mathbf{B}$. Can somebody explain the term $\nabla_\mathbf{B} \left( \mathbf{A \cdot B} \right)$ in detail, give a concrete example, or an expression in components because I do not understand it at all. I encountered this identity in electromagnetism.
| The notation $\vec \nabla_B$ means simply that the derivative are applying only on the vector $\vec B$.
That is :
$$(\vec \nabla_B)_i (\vec A\cdot \vec B) = \vec A\cdot\frac{\partial \vec B}{\partial x^i}\tag{1}$$
| {
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Proof of conservation of energy? How is it proved to be always true? It's a fundamental principle in Physics based on all of our currents observations of multiple systems in the universe. Is it always true to all systems? Because we haven't tested or observed them all.
Would it be possible to discover/create a system that could lead to a different result?
How are we 100% sure that energy is always conserved?
Finally, why did we conclude it's always conserved? What if a system keeps doing work over and over and over with time?
| A simple solution can be like this-
Firstly, we find the rate of change of kinetic energy.
The process is as follows.
$$\frac d{dt}\left(\frac 12 mv^2\right)=\textbf{F}\cdot\textbf{v}$$
Now imagine this case of a constant force acts on the body which is equal to $-mg$ (minus indicates downward direction). So, the RHS of the equation is $-mgv$. Now the velocity is the rate of change of vertical position of the body. So the RHS of the equation becomes $ -mg\cdot\displaystyle\frac{dh}{dt}$. Now this is the rate of change of $-mgh$. Therefore we see that the rate of change of kinetic energy $=$ negative rate of change of potential energy. So,these two always add up to give the same sum and we can say the energy (sum of kinetic and potential energy) is always constant.
| {
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What is the Momentum Operator? I know the equation for the momentum operator, but what exactly is the momentum operator? It's bizarre to me that taking the derivative of the wave function, which is an operator, should return something that isn't just a function. Specifically, my confusion is that first momentum is written as being multiplied by position, but then, after the derivative, it's acting on a function. I feel like there is some formalism missing. How does the operator go from being multiplied to acting on something?
| The momentum operator, like other operators in quantum mechanics, acts on a given wave function (state). Multiplication of operators in linear algebra is the same as them "acting on" a mathematical object. Originally quantum mechanics was called 'matrix mechanics'; so when you study linear algebra you're really studying quantum mechanics too.
Momentum is a Hermetian operator, and it's eigenvalues correspond to the possible values the momentum can take on in a given measurement.
When the momentum operator $ \mathbf{A} $ "acts on" a given state $ \langle a | $ ("state" here is equivalent to "eigenvector"), the state has a corresponding eigenvalue $ a $. We would write this in an equation as
$$
\mathbf{A} \langle a | = a \langle a |
$$
This means that in a measurement of the observable $\mathbf{A}$ of a system in a state $ \langle a | $ you would return a value of $a$ within limits of uncertainty.
| {
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If gravity doesn't exist,what are the implications? I just heard about new theories proposed by Erik Verlinde about the fact that Gravity doesn't exist..or at least it's not a foundamental force. My question is : if this is true what are the implications on current models like string theory , eternal inflation ecc ecc ? what can change in the understanding of our universe ?
| I take the question being asked to be:-
"If gravity doesn't exist, what are the implications ?"
Ergo, anything to do with Erik Verlinde is irrelevant to the question.
Gravity is one of the two infinite range forces; the other being the Coulomb force between electric charges.
Unlike electric charges, eg proton and electron, if they meet, effectively cancel as far as external fields are concerned. Like electric charges mutually repel, and can only be compressed together by Coulombic forces, outside of them; which in turn requires more outer charges.
Earnshaw's theorem tells us that no stable configuration of electric charges exists, so large amounts of matter cannot be compressed to high density by any static Coulomb field.
Gravity, is the only long range force that pulls instead of pushes (between like objects).
So gravity sucks. Without gravity there would be no stars; and no ground or apples to fall on it.
Doesn't matter a jot why gravity does or does not exist; without it there would be nothing that we would recognize.
| {
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Are there two types of D-term and two types of F-term in SUSY? I've noticed that one can obtain D-terms either by integrating a vector superfield (the vector multiplet) over superspace or by integrating a Kahler potential over superspace. In both cases we get functions of the D-auxiliary field (albeit different functions) that contribute to the Lagrangian density. A subsequent integration over space gives the action contribution.
Similary we can get an F-term either by integrating a chiral superfield over superspace or by integrating a superpotential over superspace. These just lead to different functions of the F-auxiliary field for the Lagrangian density contribution. Again, integration over space then gives the action contribution.
So it seems to me that we have two different types of D-term and two different types of F-term depending on whether we choose to write our lagrangian density in terms of superfields or in terms of Kahler/Superpotentials.
Is it correct to say that it is strictly one or the other? In other words we couldn't have a Lagrangian density containing a vector superfield as well as a Kahler potential, thus resulting in two separate D-terms when these are integrated over superspace.
Is there some relation between the superfield approach and the potential approach that I'm missing? Why use the two seperate approaches at all?
| The D-term is the last term in the Taylor expansion of a vector superfield over fermionic coordinates, $D \theta^1\theta^2\bar\theta^1\bar\theta^2$. Similarly, the F-term is the "middle" term $F\theta^1\theta^2$ which only contains the unbarred fermionic variables. Chiral superfields only depend on these coordinates (half of the superspace) so they're the last terms, too.
Chiral superfields have F-terms in them. Vector superfields have D-term in them. Kähler parts of the action are composite vector superfields so they also have D-terms. Similarly, the superpotential part of the action is a composite chiral superfield so it has F-terms.
The D-terms and F-terms that define the potentials and dynamics are those obtained from the Kähler potential and the superpotential in the Lagrangian. The D-terms and F-terms talked about when particular fields are discussed are the D-terms and F-terms of these particular fields. There can't ever be any confusion. One never sums or has to distinguish "two types of F-terms" or "two types of D–terms". There are as many types of D-terms or F-terms as the number of elementary or composite vector superfields or chiral superfields one may invent.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why should nature of light(or any quantum object) depend on observation? We know that, in the double slit experiment, observation changes the behavior of a quantum object, that it behaves like a particle when observed and a wave when not observed. But why should its nature depend on observation? What if we didn't exist and hence no observation...? The nature of the quantum objects must remain same, right? Why is it based on observation? Am I trying to understand the wave-particle duality in a wrong way?
| This question is still an open question in science. We know that a particle when it is not observed is described by a wave function, that wave function evolves in time. The wave function is described as a superposition of possible states of the particle (called eigen states)
If a measurement was made, the wave function collapses into one of those eigen states which means it becomes well/defined particle with position and momentum.
The collapse of the wave function is still not understood. No body knows what happens exactly. For more information see:
http://en.wikipedia.org/wiki/Wave_function_collapse
"The nature of the quantum objects must remain same, right? " Intuition says so but nature has a different opinion, as far as we understand it.
Hopefully that helped!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/77738",
"timestamp": "2023-03-29T00:00:00",
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Does the black hole only increase? I don't understand this:
The more mass a Black Hole has, the more gravity it create! The more gravity it has, the more mass it can get!
With this lines, someday, all the universe will be a black hole.
Is this correct?
| That conclusion does not (logically) follow from those two premises.
It's true that if a block hole gains mass, then it's gravitational pull on matter will be stronger. If the gravitational pull is stronger, then matter doesn't have to come as close to the black hole in order to be pulled in. It does not mean, however, than everything will be pulled in.
Someone will hopefully. speak about the expansion of the universe and argue it that way.
| {
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"timestamp": "2023-03-29T00:00:00",
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Surface gravity of Kerr black hole I'm going through Kerr metric, and following the 'Relativist's toolkit' derivation of the surface gravity, I've come to a part that I don't understand.
Firstly, the metric is given by
$$\mathrm{d}s^2=\left(\frac{\Sigma}{\rho^2}\sin^2\theta\omega^2-\frac{\rho^2\Delta}{\Sigma}\right)\mathrm{d}t^2-2\frac{\Sigma}{\rho^2}\sin^2\theta\omega \mathrm{d}\phi \mathrm{d}t+\frac{\Sigma}{\rho^2}\sin^2\theta \mathrm{d}\phi^2+\frac{\rho^2}{\Delta}\mathrm{d}r^2+\rho^2 \mathrm{d}\theta^2$$
With
$$\rho^2=r^2+a^2\cos^2\theta,\quad \Delta=r^2-2Mr+a^2,$$
$$\Sigma=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=\frac{2Mar}{\Sigma}$$
The Killing vector that is null at the event horizon is
$$\chi^\mu=\partial_t+\Omega_H\partial_\phi$$
where $\Omega_H$ is angular velocity at the horizon.
Now I got the same norm of the Killing vector
$$\chi^\mu\chi_\mu=g_{\mu\nu}\chi^\mu\chi^\nu=\frac{\Sigma}{\rho^2}\sin^2\theta(\Omega_H-\omega)^2-\frac{\rho^2\Delta}{\Sigma}$$
And now I should use this equation
$$\nabla_\nu(-\chi^\mu\chi_\mu)=2\kappa\chi_\nu$$
And I need to look at the horizon. Now, on the horizon $\omega=\Omega_H$ so my first term in the norm is zero, but, on the horizon $\Delta=0$ too, so how are they deriving that side, and how did they get
$$\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2}{\Sigma}\nabla_\nu\Delta$$
if the $\Delta=0$ on the horizon? Since $\rho$ and $\Sigma$ both depend on $r$, and even if I evaluate them at $r_+=M+\sqrt{M^2-a^2}$ they don't cancel each other.
How do they get to the end result of $\kappa$?
| You are right that the $(\Omega_H-\omega)^2$ term doesn't contribute. This is because it is the square of something that vanishes at the horizon: when you take the derivative, there remains a vanishing factor. As for the other term, since $\Delta$ vanishes at the horizon, this term vanishes except when the derivative hits $\Delta$. This yields the last formula you wrote.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is the receiver of this solar tower cream-colored? I wonder, why the receiver of this solar tower is not black? Wouldn't a black receiver be more efficient? If not, why is it not white then, but is an intermediate color?
It seems that the receiver of other solar towers, such as Solar One is indeed black:
Does it mean that the color is not important at all?
| In the picture you posted the collector is on the other side of the tower so you can't see it. Have a look at this PDF for detailed pictures of the tower - the collector is shown on page 40, and it is indeed black.
| {
"language": "en",
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Why doesn't Bernoulli's Principle apply to Current and Resistors in a circuit? Bernoulli's principle makes sense when you apply it to fluids. If you decrease the diameter of a pipe then the velocity of the fluid increases because it needs to keep the same rate of fluid moving through the pipe.
So my question is:
If Voltage == Diameter of the pipe
and
Current == Rate of which the fluid is moving
Why do resistors work?
Shouldn't the resistor only actually work within itself but then return the current to it's actual rate once you have passed it?
Or have I taken the analogy of wires being like pipes of water to far?
| True but in my experience traffic is usually jammed up before going through a tunnel but when you get to the other side the traffic magically disappears....same amount of cars same amount of roads space.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there an easy way to get water at roughly 70°C in our kitchen? Some green tea requires to pour water at 70°C. I have no thermal sensor or kettle with adjustable temperature with me.
Do you know a way to get water at roughly 70°C like “boil water and wait for x minutes” or “mix x part of boiling water with 1-x part of fresh water” ?
| "Wait for $x$ minutes" is difficult because the cooling rate will depend on a lot of non-universal details (room temperature, the surface of the water, the material the pan is made of and its thickness).
However, there are two temperatures which are easy to attain in your kitchen. $100\,^\circ\mathrm C$ (don't need to explain this) and $0\,^\circ\mathrm C$ - for this, you can let ice cubes sit on the counter until they melt. Stir/mix the water/ice mixture and as long as they are some ice cubes left, you're guaranteed that it is at $0\,^\circ\mathrm C$.
Now the heat capacity of water stays almost constant if you go from $0\,^\circ\mathrm C$ to $100\,^\circ\mathrm C$, see for example this table - fifth column:
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
Informally speaking, this means that you need the same amount of energy to get a kilogram of water from $2\,^\circ\mathrm C$ to $3\,^\circ\mathrm C$ as you need to get it from $92\,^\circ\mathrm C$ to $93\,^\circ\mathrm C$.
Because of this property, you can mix $30\,\%$ of the $0\,^\circ\mathrm C$-water and $70\,\%$ of the $100\,^\circ\mathrm C$-water and you'll end up with a liquid that is at $70\,^\circ\mathrm C$, up to a small error.
For liquids that have a heat capacity that varies strongly with temperature, you can do the same thing but the calculation is more difficult.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/78183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Adiabatic filling of a container Suppose a thermally insulated container is filled with atmospheric air until the pressure reaches 5000 psi. This could represent the filling of a diving cylinder, before thermal dissipation becomes significant.
Initially, then, some mass of air has atmospheric temperature $T_1$ and $p_1$. When forced in to the tank adiabatically, we expect its final temperature to be
\begin{align}
T_2 &= T_1\left(\frac{p_2}{p_1}\right)^{1-1/\gamma} \\
&= (300\,\mathrm{K})\left(\frac{5000\,\mathrm{psi}}{14.5\,\mathrm{psi}}\right)^{1-1/1.4} \\
&\approx 1590\,\mathrm{K}
\end{align}
This final temperature seems far too hot. A diving cylinder would likely melt at this temperature.
More likely, though, something is wrong with my reasoning. Is this not the correct use of the temperature-pressure relations, assuming an ideal gas? If not, what other information is needed to predict the final temperature of the container?
| The practical thing that you are missing here is that the air compressor is cooled constantly so that it doesn't melt during operation (can't be adiabatic). So the upper limit on the temperature is driven by the operating temperature of the compressor. Then the other posters comments about heat transfer out of the tank apply.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Period $T$ of oscillation with cubic force function How would I find the period of an oscillator with the following force equation?
$$F(x)=-cx^3$$
I've already found the potential energy equation by integrating over distance:
$$U(x)={cx^4 \over 4}.$$
Now I have to find a function for the period (in terms of $A$, the amplitude, $m$, and $c$), but I'm stuck on how to approach the problem. I can set up a differential equation:
$$m{d^2x(t) \over dt^2}=-cx^3,$$
$$d^2x(t)=-{cx^3 \over m}dt^2.$$
But I am not sure how to solve this. Wolfram Alpha gives a particularly nasty solution involving the hypergeometric function, so I don't think the solution involves differential equations. But I don't have any other leads.
How would I find the period $T$ of this oscillator?
| Since
$$\frac1 2mv^2+U(x)=U(A)$$
We have
$$dt=\frac{dx}v=\frac{dx}{\sqrt{2(U(A)-U(x))/m}}=\frac{dx}{\sqrt{c(A^4-x^4)/(2m)}}$$
Then
$$\frac T4=\int_0^{\frac T4}dt=\int_0^A\frac{dx}{\sqrt{\frac{c}{2m}(A^4-x^4)}}$$
Thus
$$T=4\int_0^A\frac{dx}{\sqrt{\frac{c}{2m}(A^4-x^4)}}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/78415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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