Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Why the work done by system not stored as potential energy? Now choose a spring mass system now work done by external agent in slowly moving from equilibrium position is stored as potential energy but where is work done by spring force gone.For genralization work done on system is stored as potential energy but where is work done by system gone during the process?
| To add to the previous answer. The energy stored is known as Elastic Potential Energy to the best of my knowledge. Thus, the energy is in fact stored as a form of potential energy - since the position of the mass in the mass-spring system determines the kinds of energy it possesses.
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Do any quantum interpretations not include nonlocality? Are there any interpretations of quantum theory that have a mechanism such that there is no need to invoke nonlocality?
| I think so. In my work http://link.springer.com/content/pdf/10.1140%2Fepjc%2Fs10052-013-2371-4.pdf (Eur. Phys. Journ. C, (2013) 73:2371) I consider scalar electrodynamics (Klein-Gordon-Maxwell electrodynamics) and spinor electrodynamics (Dirac-Maxwell electrodynamics), show that the matter field can be algebraically eliminated in a certain gauge, the resulting equations describe independent evolution of electromagnetic field and can be embedded in quantum field theories (see references there to other people's results that I used).
To understand how we can do without nonlocality, we need to understand why nonlocality seems inevitable. People say that it is a consequence of the Bell theorem. The latter contains two statements: 1) local hidden-variable theories satisfy some inequalities; 2) these inequalities can be violated in quantum theory. However, the proof of the second statement uses both unitary evolution of quantum theory and the projection postulate, which are, strictly speaking, mutually contradictory (for example, it is shown in https://arxiv.org/abs/1107.2138 (Phys. Rep. 525 (2013) 1-166) that the projection postulate is only an approximation). On the other hand, experimental demonstrations of violations of the Bell inequalities have some loopholes. The "loophole-free" articles of 2015 assume that measurements have definite outcomes at a certain time point, but this assumption is, strictly speaking, not compatible with unitary evolution. .
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Unorthodox way of solving Einstein field equations Usually when we solve field equations, we start with a stress energy tensor and then solve for the Einstein tensor and then eventually the metric. What if we specify a desired geometry first? That is, write down a metric and then solve for the resulting stress energy tensor?
| You can certainly do this, and indeed it is regularly done. For example Alcubierre designed his FTL drive by starting with the metric he wanted and calculating the required stress-energy tensor. It is a straightforward calculation - it is somewhat tedious to do by hand but Mathematica would do the calculation in a few seconds.
The problem is that the resulting stress-energy tensor will almost always contain contributions from exotic matter, as indeed the Alcubierre stress-energy tensor does, and that means it won't be physically meaningful. The chances of solving the Einstein equation by guessing geometries and ending up with a physically meaningful stress-energy tensor are vanishingly small.
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What is it that drives the electrons from one plate to another through a battery in a capacitor? The situation is as follows:
You have a parallel plate capacitor, disconnected from a battery on both ends.
We know the battery tries to maintain a constant potential difference between its two ends. The capacitor is uncharged. When then capacitor is connected to a cell, why do electrons start flowing? What is the need to create a field between the plates?
EDIT: What is the need to maintain a potential difference between the plates?
| Note what you say:
"We know the battery tries to maintain a constant potential difference between its two end"
The wires connecting the battery ends to the capacitor plates, by being conductive extend the function of the two ends
So the question refers to what makes a battery maintain a constant potential at its ends, which is answered in this link.
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Does semi-classical gravity obey the equivalence principle? Question
I was recently wondering about semi-classical gravity :
$$ G_{\mu \nu} = \frac{8 \pi G}{c^4} \langle \hat T_{\mu \nu} \rangle_\psi$$
Does this obey the equivalence principle?
My intuition
Let's say I am in a lift and I want to measure the standard deviation in the acceleration. Then with Heisenberg's equation of motion I can do so ,however , the left hand side $ G_{\mu \nu}$ is a classical object and will predict there is none whereas using the Hamiltonian in quantum mechanics $\hat T_{00}$ will only have zero standard deviation if it is in a acceleration eigenstate. But this view would only work in quantum mechanics and I'm not sure how to argue it for fields. Does this kind of argument still hold in QFT?
| So it seems like your question and intuition are asking two separate questions. Your intuition is asking about the way measurements are made in the classical world and in the quantum world. Semi-classical theories are known to be incomplete and so some parts of the physical process won't actually stand up, ie matching standard deviations.
That being said, we know GR is the most accurate field theory of gravity and most theories of gravity retain the equivalence principle in some form or another. In your question the equivalence principle is buried deep within the $G_{\mu\nu}$ and it's derivation. So with gravity, you will likely always use the equivalence principle it will just be disguised within the mathematics of the problem. But it is always there.
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If entropy decreases for cold systems, isn't the heat death of the universe a state of low entropy? Entropy is a consequence of heat. The heat death of the universe results in an approach to absolute zero temperature. Does this mean the end of the universe is low entropy?
|
If entropy decreases for cold systems,
There is some misunderstanding in this statement, using the verb "decreases".
from the three laws of thermodynamics entropy either remains constant on increases.
If one compares a cold system, to the same system at higher temperatures, the entropy is higher in the hotter system , but in order to get a hot system to a cold state a larger combined system is needed in order to obey the thermodynamic law 2:
Second law of thermodynamics: In a natural thermodynamic process, the sum of the entropies of the interacting thermodynamic systems increases. Equivalently, perpetual motion machines of the second kind (machines that spontaneously convert thermal energy into mechanical work) are impossible.
Entropy as a variable is connected to temperature through differential equations, not linear ones, so it does not necessarity mean "low temperature"= " low entropy" as you assume
It can be demonstrated that all definitions of entropy are equivalent mathematically.
You ask
isn't the heat death of the universe a state of low entropy?
The useful definition of entropy for cosmological purposes , comes from statistical mechanics
Specifically, entropy is a logarithmic measure of the number of states with significant probability of being occupied
You ask:
Entropy is a consequence of heat. The heat death of the universe results in an approach to absolute zero temperature. Does this mean the end of the universe is low entropy?
Consequence and linear dependence are two different concepts. Heat generates microstates and thus increases entropy, but it is not the only variable that can increase entropy.
So it is possible to have a system of particles in a very large volume with very small kinetic energy ( which is directly connected to temperature) to a huge number of states occupied by these particles, so that the entropy is large but the temperature close to zero.
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How can electric fields be used to detect cracks in metals? My physics teacher brought this up in a lecture and I am not exactly sure what he is saying.
| Cracks and other flaws can be found in metal parts by scanning the surface of the part with a small electromagnetic coil, which induces a current flow in the metal part as if the coil and the part were two coils coupled together in a transformer. the presence of a flaw like a crack in the part perturbs the flow of induced current and in turn upsets the current flow in the coil, which can be detected with a sensitive electronic circuit which thereby signals the test operator that a flaw is present at that point. This is called an eddy current test.
If the metal part is ferromagnetic, surface cracks can be detected by temporarily magnetizing the part with a very strong external magnetic field and then rinsing the surface with a special fluid which contains a suspension of extremely finely-powdered iron in which the particles have been coated with a fluorescent dye. In the vicinity of a crack, the field lines in the magnetized part get bunched up and protrude slightly from the crack, attracting and entrapping the iron particles there. The test operator then rinses the excess fluid off the part and shines a UV light at it, which causes the dyed particles to visibly glow and trace out the crack. This test is called magnetic particle inspection (Magnaflux).
Cracks can also be detected in a flat metal part by attaching a piece of photographic film to it and then applying a brief high voltage pulse to the part. Electrons are ejected preferentially from the sharp edges of the crack and ionize the air nearby, causing it to glow faintly and expose the film. Once the film is developed and printed onto photographic paper, the cracks can be seen in it as bright lines. This is called corona discharge imaging.
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Why cannot we determine our own velocity? I know it would violate the principle of relativity. But have there been serious experimental attempts to contradict that principle?
If I am in a moving train without windows and totally sound proof, is there absolutely no way for me to determine that I am moving?
Can this be proved as a mathematical theorem given some other laws/axioms of physics?
| The principal of relativity essentially states the laws of physics are the same in all inertial frames.
As I understand it Galileo conducted (or it has been claimed he conducted) an experiment in which he dropped a rock from the top of the mast of a moving ship to show it would land at the base of the mast and not behind it.(How he was able to make sure the ship was moving at constant velocity is another matter).
As far as the mathematics go you have the Galilean transformations and Lorentz modifications already discussed below.
I suppose to absolutely prove the principle you would have to conduct experiments that demonstrate all currently known laws of physics are the same on the train when it’s moving at constant speed on a straight track as when it is at rest with respect to the track. Or at least conduct enough experiments to be convinced.
Hope this helps.
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Intuition of 2D Velocity Vectors I understand 2D vectors in terms of displacement. For example, Bob moves 3m to the east and 4m to the north, the total change in position (displacement) equals 5m to the north east.
But I don't understand how 2D velocity vectors can be broken down into their horizontal and vertical components.
| Orthogonal axes make right angles, so you can use the Pythagorean theorem to get the hypotenuse. If you multiply the two shorter sides of a right angle by some unit transformation, or divide by a unit of time, the hypotenuse is identically scaled because of the Pythagorean theorem. It is Pythagoras's theorem that gives the meaning to orthogonal vector addition -- which holds regardless of units.
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Why don't we use Ampere's law to find the magnetic field due to a wire of finite length at its perpendicular bisector? I know that finite length doesn't have symmetry and thus it's hard to apply maths here but take the case of magnetic field of a wire of finite length at a distance $r$ from axis of the wire exactly at perpendicular bisector of wire.
The magnetic field is symmetric here, but still, Ampere's circuital law doesn't apply here. Why is it so?
| How is the current in the finite length of wire to be generated?
*
*Using current carrying wires connected to the wire under consideration?
Then these current carrying
connecting wires will also produce a magnetic field which needs to be
included in the line integral $\int \vec B \cdot d \vec s$.
*Using charges stored at each end of the wire?
Then there is a time varying electric field and the $\frac {\partial}{\partial t}\int\vec E\cdot d\vec A$ term has to be included.
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General idea behind simplifying cube resistors I don't know if "cube resistors" is the right way of putting it, but I am wondering if somebody could walk me through the general procedure for solving problems in which you are asked to find the resistance from A to B through a cube with a resistor of some R on each line that makes up the cube.
| If you are trying to find the resistance between two points, you suppose that a pd is placed between these points. You then look for symmetries that show certain points to be at the same potential. When you find such points, you imagine them to be linked by wires of negligible resistance, knowing that there will be no current in these wires, so their presence won't affect the functioning of the circuit. But linking the points at the same potential will usually reveal a much simpler equivalent circuit.
For example, suppose you are trying to find the resistance between two points at opposite ends of a body diagonal, AG, of a cube with equal resistances along each edge. [Two opposite faces of the cube are ABCD and EFGH with the other four edges of the cube being AE, BF, CG and DH.]
A is linked by single resistors to B, E and D. The ways that current can get from these points to G, involve identical resistor combinations [e.g. current from B can go to G via C or F, and you should check the routes to G from E and D.] Thus B, E and D must be at equivalent points in identical potential dividers, and are all at the same potential. If we link B, E and D by wires of zero resistance, there will be no current in these wires. The same goes for G's 'neighbours', points F, C and H.
So imagine these wires to be in place. You'll then see that the 'resistor cube' with 'terminals' at A and G is equivalent to 3 resistors in parallel, in series with 6 in parallel and another 3 in parallel.
You can find the resistance between A and E, A and F and possibly other pairs of points by using variations of this method. Redrawing the cube as a two-dimensional circuit may (or may not) help.
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If the pressure inside and outside a balloon balance, then why does air leave when it pops? Sorry for the primitive question but when we inflate a rubber balloon and tie the end, its volume increases until its inner pressure equals atmospheric pressure.
But after that equality is obtained why does the air goes out when we pop the balloon? If there is pressure equality what causes the air flow?
| Blowing into a balloon is harder than just blowing into the air, because it takes higher air pressure to stretch the rubber. once the balloon is tied the stretched rubber continues to squeeze the air inside, so inner air pressure stays higher than outer air pressure. Untie the balloon and the stretched rubber will squeeze the air out until it shrinks to its normal un-stretched size. Sticking the inflated balloon with a needle will create a flaw in the stretched rubber causing it to split open and release the inner air pressure very fast, pop.
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How do I use a sunstone? How does it work? I have acquired two pieces of Icelandic Spar (optical calcite) with surface and internal fractures that should be serviceable as sunstones.
I have observed the double refraction and cancelling double refraction, a rainbow effect, and varying intensity of stuff. I stood myself in the shadow and looked through the stones (one at a time) at the sky. I observed several effects including the second image rotating around the primary, but everything seems to track with the rotation of the stone rather than the position of the sun.
I tried but failed to observe polarization cancellation from opposed polarizing plates. I was expecting this to be the answer but could not observe this effect.
| Half an answer. I got it working but don't understand it yet:
I spent another half an hour outside trying this way and that to get something out of my sunstones. I got a double-banded rainbow effect when looking through either stone that moved with the stone (definitely not Haidinger's brush) and rotated with the stone, but its brightness depended on the orientation of the stone with respect to the sun. If I turn the stone I find a place where one of the two sets of bands disappears. Once so turned, the top and bottom edges of the stone point at the sun. This works only if the stone is backlit by the sky, and critically, if the stone is not directly illuminated by the sun.
So now it's a pure optics question. I don't understand the rainbow effect; with all the edges forming parallelograms, I wasn't expecting a prism. The brightness has got to be polarization cancelling but it's cancelling something I didn't expect to see at all.
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Radiation of a charged particle Take a uncharged particle. It hits a resting charged particle. Will the charged particle radiate then?
| If one speaks about a neutron and a proton (or also an electron), in a collision an exchange of impulses occurs between them, minus a loss of energy.
In the sentence above there are some remarkable details:
*
*It doesn't matter whether the charge or the neutral particle moves in relation to the observer, only the relative motion between them is relevant.
*The second law of thermodynamics was based on this phenomenon; any energy exchange is accompanied by an energy loss in the form of electromagnetic radiation. The sum of impulses after collision is less that sum before collision.
*It does not matter at all, is the particle charged or not; even for two neutrons an energy loss occurs.
Will the charged particle emit then?
It is more the case that the the energy exchange is accompanied by an energy loss.
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Why are gold nanoparticles red? While watching the latest computerphile video (https://www.youtube.com/watch?v=FGiBHsUkVzU) I came across a solution of nanoparticles of gold
,
but to my surprise, they appeared as red, shouldn't gold be golden color?
| The phenomenon is called Localized Surface Plasmon Resonance (LSPR).
In a nutshell: a plasmon is a charge (quasi) particle oscillation. In the case of light reflecting at an interface between a dielectric (e.g., like your liquid) & a metal,the SPR is excited in the metal and confined to the surface.
Gold nanoparticles suspended colloidally (like yours) are so small, that part of the surface fulfill the conditions for SPR regardless of the light direction. The size of the particle strongly influences the color.
The stained glass window in the Notre-Dame cathedral in Paris uses this effect for some colors.
If you‘re really interested check out this review paper: doi: 10.1088/1361-648X/aa60f3.
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First-order correction to energy in perturbed harmonic oscillator I know, from the perturbation theory, that, if I have the hamiltonian
$$ \hat H = \hat H_0 + \lambda \hat W$$
where $\hat H_0$ is the unperturbed hamiltonian of which I know its eigenvectors and eigenvalues, and $W$ is the perturbation. Then the energy of the perturbed hamiltonian, corrected to the first order, is given by
$$E_n\approx\varepsilon_0+\lambda \varepsilon_1 \tag{1}$$
where $\varepsilon_0$ is the nth eigenvalue of $H_0$ (i.e. $H_0|\varphi_n\rangle =\varepsilon_0 |\varphi_n\rangle $) and $\varepsilon _1=\langle \varphi_n|\hat W|\varphi_n\rangle$.
My question comes from a specific problem in which the hamiltonian given to me is a peculiar "anharmonic" oscillator
$$\hat H = \frac{p^2}{2m}+\frac{m\omega^2}{2}x^2+\alpha x + \beta p^2$$
In this case I don't have just one parameter $\lambda$ but two ($\alpha$ and $\beta$). What should the expressión (1) be here?
Thanks
PS:In addittion I would like to know how to solve this problem in exact way, or at least, if this is possible. I suppose it could work with a change of variables.
| You can solve this Hamiltonian exactly. We can write
$$
\hat H = \left(\frac{1}{2m}+\beta\right)p^2+\frac{m\omega^2}{2}\left(x+\frac{\alpha}{m\omega^2}\right)^2-\frac{\alpha^2}{2 m \omega^2}
$$
Define $\bar{m}$ to satisfy $\frac{1}{2\bar m}=\frac{1}{2m}+\beta$, define $\bar\omega$ to satisfy $m\omega^2=\bar m\bar \omega^2$, and define $\bar{x}=x+\frac{\alpha}{m\omega^2}$. Then the Hamiltonian becomes
$$
\hat H = \frac{p^2}{2\bar m}+\frac{\bar m\bar\omega^2}{2}\bar{x}^2-\frac{\alpha^2}{2 m \omega^2}
$$
which is just a SHO Hamiltonian plus a constant energy offset. So, you should be able to find the energy levels for this Hamiltonian, and write them out in terms of $m,\omega, \alpha, \beta$. Expanding the result to first order in $\alpha,\beta$ will immediately answer your first question.
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Conceptual question regarding transistor biasing I know that in a transistor, the base-emitter (B-E) junction has to be forward biased while collector-base (C-B) junction needs to be in reverse bias.
Question: What would happen if we alter this biasing?
Case 1: Both B-E and C-B junctions are reverse biased.
Guess: Nothing will happen as electrons are being pulled away from the base region leading to no flow of current.
Case 2: C-B junction is forward biased and B-E junction is reverse biased
Guess: Transistor can be operated in normal fashion. But I know that collector is purposefully designed to be larger for heat dissipation. In our case, as emitter (which is not as large as collector) serves the purpose of a collector. It will cause heating in our circuit which can damage the transistor.
Case 3: Both B-E and C-B junctions are forward biased.
Guess: For this one, I suppose one out of two things should happen (I am not sure what).
Possibility 1: As both junctions are forward biased, electrons accumulate at the thinly occupied thinly doped base region. This causes overheating thus damaging the transistor.
Possibility 2: As both junctions are forward biased, electrons accumulate at the thinly occupied thinly doped base region. This constitutes to rise in base current. Thus transistor conducts like two independent separate diodes.
Are these above guesses right or wrong? Detailed explanation would be helpful
| Case 1: Correct, nothing will happen. Only leakage currents will go through.
Case 2: Main disadvantage is that there is much less current amplification and lower breakdown voltage in this mode of operation.
In some cases amplification could be <1, which would make it rarely useful configuration. Heat dissipation is also valid point, but in transistors that I've seen under microscope - areas of collector and emitter are more or less similar, not 10x different.
Main difference between collector and emitter is doping profile which lead to this amplification difference, as well as breakdown voltage difference. In reverse breakdown voltage could be as small as 5V.
Case 3: Transistor conducts like two independent separate diodes. Will it burn or not will depend on power dissipation. Base typically has high resistance (as it is thin), so these diodes would be quite terrible.
I.e. base resistance will severely limit current. If you increase voltage until you get significant current and power dissipation - it will surely burn.
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Setting spinors and $SU(2)$ representations on the same patch I am sorry for the naivety of this question, I am a mathematician and I am trying to put together different ideas.
I am trying to understand the vocabulary of physics, in particular, I want to know:
We know that the isometry group of Minknowski metric is the Lorentz group $O(3,1)$. Furthermore, it acts on $\mathbb R^{3,1}$ by isometries. The isotropy group of this action is precisely $SU(2)$, so it is interested to understand further representations of $SU(2)$.
$1)$ What does mean the notation $(1/2,0)\oplus (0,1/2)?$
$2)$ How can I express an spinor from the $1/2$-representation of $SU(2)$?
| Reference: Srednicki, Quantum Field Theory ch. 33
The Lie algebra of the Lorentz group $SO(3,1)$ can be described in terms of 3 spacial rotations ($J_i$) and 3 velocity boosts ($K_i$):
$$ [J_i,J_j] = i \epsilon_{ijk} J_k $$
$$ [J_i,K_j] = i \epsilon_{ijk} K_k $$
$$ [K_i,K_j] = -i \epsilon_{ijk} J_k $$
To make the $SU(2) \otimes SU(2)$ structure of the (complexified) algebra explicit we need to change to a different set of generators:
$$ N_i = \frac{1}{2} (J_i - iK_i) $$
$$ N_i^\dagger = \frac{1}{2} (J_i + iK_i) $$
This gives:
$$ [N_i,N_j] = i \epsilon_{ijk} N_k $$
$$ [N_i^\dagger,N_j^\dagger] = i \epsilon_{ijk} N_k^\dagger $$
$$ [N_i,N_j^\dagger] = 0 $$
So given that the irredicible representations of $SU(2)$ can be listed by a single index $j= 0, \frac{1}{2} , 1, \frac{3}{2} ...$ , irreducible representations of $SU(2) \otimes SU(2)$ clearly needs two such indices $(j_1,j_2)$ where $j_1$ and $j_2$ are the eigenvalues of $N_3$ and $N_3^\dagger$ above.
Important representations in physics are:
$(0,\frac{1}{2}) $ = Weyl Spinor
$(\frac{1}{2},\frac{1}{2}) $ = Vector
$(0,\frac{1}{2}) \oplus (\frac{1}{2},0) $ = Dirac Spinor
For each of these representations in turn we can calculate the corresponding SU(2) representations of the isotropy group (physicists usually call this the little group).
Physically the little group (isotropy group) corresponds to the subgroup of the Lorentz group which leave the momentum vector unchanged. This (at least for a massive particle) is the group of spatial rotations i.e. the $J_i$ above.
So we need to write $J_i$ in terms of the $N_i$ and $N_i^\dagger$ :
$$ J_i = N_i + N_i^\dagger $$
All physicists then recognise this as the process of "adding two angular momentum" (mathematically it is the process of finding the irreducible representations of the tensor product of two representations):
$(j_1,j_2)$ decomposes into the direct sum of representations $|j_1-j_2|, |j_1-j_2| + 1 , ... j_1 + j_2 $
Applying this we get:
$(0,\frac{1}{2}) : j = \frac{1}{2}$ so this is a Spin $\frac{1}{2}$ particle.
$(\frac{1}{2},\frac{1}{2}) : j = 0,1$ so this contains a Spin 1 (vector) particle.
$(0,\frac{1}{2}) \oplus (\frac{1}{2},0) : j = \frac{1}{2} $ in both parts of the sum, so this again is a Spin $\frac{1}{2}$ particle.
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If the mass of the Earth is decreasing by sending debris in space, does its angular momentum also decrease? We are sending huge amount of debris into space from earth, and also very heavy satellites and rockets, then the mass of earth must be decreasing over time.
If the mass will decrease, then gravitational attraction between earth and sun must decrease, and its angular momentum must decrease, which would result in greater orbit of earth and earth being cooler over the time, but instead the temperature of earth is increasing, why is it so?
| The mass of material sent into orbit may seem huge to a naïve observer, but it is nothing at all compared to the mass of the Earth and wouldn't result in any detectable change in its orbit or rotation. Meanwhile, the mass of the Earth is actually increasing as it picks up dust, meteorites and micro-meteorites on its passage around the sun. A certain amount of the upper atmosphere is gradually leaking into space, but that is more than offset by the space debris which arrives. The majority of orbiting satellites will eventually fall back to Earth.
| {
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Why isn't my calculation that we should be able to see the sun well beyond the observable universe valid? I recently read an interesting article that states that a human being can perceive a flash of as few as 5 or so photons, and the human eye itself can perceive even a single photon. The brain will filter this out, however.
I wanted to calculate how far away you'd have to be standing from our sun for not a single one of its photons to be hitting your pupil over a given second.
The first thing I did was assume that the sun emits $10^{45}$ photons per second, because, well, that's the only number I could find through internet research.
The next step is to assume that the average angle between photons emitted from the sun is pretty much the same, and is equal to $3.6 × 10^{-43}$ degrees.
The next step is to assume that the average human pupil diameter is 0.005 meters, and then draw a triangle like so:
The length of the white line through the center of the triangle equals the distance at which two photons from the sun would be further apart than your pupil is wide, meaning not even one photon should hit your eye.
I broke the triangle into two pieces and solved for the white line by using the law of sines, and my final result is ridiculous.
$3.97887×10^{41} $ meters is the length of the white line. For reference, that's over $10^{14}$ times the diameter of the observable universe.
My conclusion says that no matter how far you get from the sun within our observable universe, not only should some of the photons be hitting your pupil, but it should be more than enough for you to visually perceive.
But if I was right, I'd probably see a lot more stars from very far away every night when I looked up at the sky. Why is my calculation inconsistent with what I see?
| If the sun emits $N=10^{45} $ photons per second, an observer at distance $R$ will receive $N/4\pi R^2$ photons per second per unit area. If human eye has radius about $r=10^{-3} m$, it will receive 1 photon per second if
$$\frac{N\pi r^2}{4\pi R^2} =1, $$
$$R \sim \sqrt{N} r \sim 10^{20} m\sim 1 kpc. $$
This is less than the size of the universe.
| {
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Gravitational field (metric tensor) and speed of light between two massive plates Suppose I have two massive plates of size $l\times h\times w$ mounted parallel to each other with a distance of $d$ and with a mass density of $\rho$. I send a light beam in the middle between them along the length $l$ and in parallel to them.
How long does it take the light beam, in coordinate time, to pass through the empty space between the plates?
I assume the speed between the plates is basically determined by the gravitational field between the plates and how this then determines the speed of light.
Maybe the better question is: what is the metric tensor between the plates and how does it determine the speed of light?
Given the highly regular setup, I would hope that at least for larger $l$ and $w$ there is a nearly homogeneous (read, constant in a plane or all over) metric tensor "near the middle" of the setup.
| The speed of light is always the same in vacuum (which I assume is the case for in between the plates here). Even if the plates were enormously massive, as long as the light ray travels in the middle of the gap there would be no bending and in ANY case no change to the speed of light, that is one of the main ingredients of SR and GR.
| {
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Temperature of vacuum chamber on Earth Does the temperature of a vacuum chamber drop if left for 24 hours or more, since some in space in quite cold somewhere and quite hot how does vacuum become cold near to absolute zero.
| As pointed out by the comments, vacuum itself does not have temperature, since temperature is defined in terms of the kinetic energy of the particles. However, if you put an object (let's say an idealized blackbody) inside the vacuum, its temperature is not simply zero. The temperature will be actually related to the relation of incoming and outgoing radiation. This is the case since electromagnetic radiation can indeed travel through vacuum. In a situation where there is no appreciable source of radiation such as a star, and the object is left by itself inside the vacuum, its temperature will decrease until it reaches about 3 K, the temperature of cosmic background radiation. At such temperature, the object will reach equilibrium.
| {
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How a discrete $Z_2$ symmetry removes flavour changing neutral current from Two Higgs Doublet Model? By applying a discrete $Z_2$ symmetry to the theory of Two Higgs Doublet Model it is ensured that fermions of one type couples to only one doublet. But how FCNC is removed by doing so? Because if all the leptons for example couple to the 1st doublet then there is a chance of coupling of different flavours of leptons to the same neutral mediator. How will it differentiate among different flavours?
| It is not true that any $\mathbb{Z}_2$ symmetry eliminates the flavor-changing neutral currents (FCNCs). A more accurate statement is that some $\mathbb{Z}_2$ symmetries eliminate the tree-level FCNCs. By assigning the quarks and leptons appropriate $\mathbb{Z}_2$ charges, one can avoid that two Higgs doublets couple both to $u$- and $d$-type quarks. The following table, which I found in this link, surveys the possibilities:
Scheme (i) is like the standard model plus a Higgs that does not couple to the fermions at all, so it is fine. The dangerous cross terms are also avoided in schemes (ii) and (iii) but not (iv).
As for "How will it differentiate among different flavours?": the simple traditional models don't, they only distinguish the $u$- and $d$-type quarks and leptons in their $\mathbb{Z}_2$ charges. Settings (i)-(iii) work fine, i.e. do not require additional ad hoc assumptions. On the other hand, models in which the couplings are flavor-dependent are usually less simple.
| {
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Explanation of covalent bond from physics point of view? We can explain ionic bond as the force between charged particles due to Coulomb electrostatic law. This got me wondering how is then covalent bond explained purely in terms of physics?
| Eventually, every kind of bond in condensed matter is reducible to electrostatics. Covalent bond, ionic bond, van der Waals forces, metallic interactions and even more sophisticated interactions like the phonon-mediated electron-electron attraction in a superconductor, are all reducible to purely electrostatic interactions between electrons and nuclei.
What ensures the observed variety of behaviors is the fact that at least one component of the matter does not behave like classical particles but QM is required, in addition to the introduction of effective interactions between a reduced number of degrees of freedom.
| {
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Why is the drag force proportional to $v^2$ and defined with a factor of $1/2$? $$Drag = \frac{1}{2}C_d \rho Av^2$$
I understand that the strength of the drag depends on the density of the fluid the body passes through, the reference area of the body, the drag coefficient, and the velocity of the object.
I don't, however, understand the 1/2 and the $v^2$ in the equation.
| In short, the squared speed $v^2$ appears in the equation because when moving faster, you increase both
*
*how much momentum ($p=mv$) that is transferred to the air molecules (they must be moved away faster) and
*how much air that must be moved away (because you sweep through more air per second).
Increasing the speed means increasing both of these factors that both make it tougher to fall. Thus, speed appears "twice", so to say.
The half $\frac 12$ that also appears in the equation, is - as others also point out - due to the drag coefficient $C_d$ being neatly written as $C_d=\frac{D}{Aq}$, where $q=\frac12\rho v^2$ is the dynamic pressure, an important aerodynamic property.
Sure, you could have included the half in the drag coefficient to simplify the drag formula. But you would simultaneously complicate the relationship $C_d=\frac{D}{Aq}$. You could also ask, why there is a half in $K=\frac12 m v^2$. Why isn't that half just included in the mass $m$? Well, because then many other relationships that include $m$ would become more complicated.
| {
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Recombination of hydrogen Suppose a slow-moving electron and a slow-moving proton are injected into a chamber, such that the two approach each other and are likely to combine to form an atom of hydrogen. How would one calculate the probable sequence of events, including likely emission of photon(s) over time? Will the resulting atom end up in ground state in a predictable sequence?
Note added later: I asked this question under the supposition that it was rather elementary. I was imagining a cold, dark box, sort of like an anechoic chamber, in which a photon might be emitted and never be heard from again, or better yet, a box surrounded by detectors which might record what was emitted and when.
I am still hoping there might be some such simplified approach. But I'm beginning to suspect I might be asking something much harder than I thought. Any comments will be appreciated.
| There are two basic types of recombination (of electrons with protons) into hydrogen atoms:
*
*ternary (three particles), this involves another electron, there is a transfer of energy of the order of the mean kinetic energy from one electron to the other
*radiative, emission of a photon, to conserve momentum (when the hydrogen atom relaxes to ground state), it has two types, SRR (spontaneous photo recombination), and TSRR (two step radiative recombination)
The radiative recombination (RR), the binding of a free electron with a proton(deuteron) accompanied by emission of radiation, plays an important role in astrophysicaland fusion plasmas. The RR is one of the most fundamental processes in atomic physicsand is connected to photo-ionization by the principle of detailed balance. Recombinationof electron to proton with emission of radiation can also take place in the presence of a third particle (electron or photon). The study of RR of electron with proton in the sponta-neous radiative recombination (SRR) channel, in absence of a third particle, dates back to1923 by Kramers [1] within a semiclassical approach.
Please see here:
http://jetp.ac.ru/cgi-bin/dn/e_048_04_0639.pdf
https://www.researchgate.net/publication/252852190_Radiative_recombination_of_cold_electron_with_proton_and_deuteron
| {
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Peskin and Schroeder Section 7.1 Mass Shift I'm slowly reading my way through Peskin and Schroeder. Near the end of section 7.1 they compare the mass shift of the electron from QFT to the classical value, both of which are divergent but in different ways.
The calculation from QFT gives:
$$\delta m = \frac{3\alpha}{4\pi}m_0\log\left(\frac{\Lambda^2}{m^2_0} \right)$$
Which diverges logarithmically as $\Lambda\to\infty$.
Versus the classical expression which diverges linearly, $\alpha\Lambda$.
The bit I don't understand is the argument they use after to explain why the divergence should be logarithmic:
"$\delta m$ must vanish when $m_0=0$. The mass shift must therefore be
proportional to $m_0$, and so by dimensional analysis, it can depend
only logarithmically on $\Lambda$".
(Taken from the paragraph immediately following the $\alpha\Lambda$ result.)
The understand the first part of this statement, its the dimensional analysis I don't get. Why is it logarithmic specifically and not some other dimensionless function of $\Lambda$?
| From a dimensional analysis point of view: Assume you need linearity in $m_0$ (aka. $\delta m \propto m_0$) by the argument given by MadMax above. This means you have $ \delta m \sim m_0 f(\Lambda,m_0)$ for some function $f$. Since the LHS and RHS both must have dimensions of mass, this means $f$ needs to be dimensionless.
Yet it is a function of two quantities which have dimension so you can argue that $f$ needs to be a function of the dimensionless quantity $\frac{m_0}{\Lambda}$. So at this stage you argue that $f(\Lambda,m_0) = F\big(\frac{m_0}{\Lambda} \big)$ for some $F$.
It is true that in principle $F$ could be some really complicated function. However, the since the cutoff of the theory satisfies $m_0 \ll \Lambda$ (i.e. it is much bigger than all the scales in your problem), whatever the function $F$ is, you only need its series for $x\ll 1$.
So at this stage there are only a few options for how $F$ looks like. Either $F(x) \sim x^{n}$ for some $n$ or another option is $F(x) \sim \log(x)$. That's pretty much it (I think).
However, if you take $F(x) \sim x^n$ then you have the mass shift $\delta m \sim m_0 \left( \frac{m_0}{\Lambda} \right)^n$ which is linear in $m_0$ only when $n=0$, so this option is disqualified. You are left to conclude that
$$
\delta m \propto m_0 \log\left( \frac{m_0}{\Lambda} \right)
$$
up to some constants.
| {
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In Feynman lecture Vol1. chapter12 about friction, $F = μN$ will fail because of the excessive heat generated. Why is that? $F = μN$ will fail because of the excessive heat generated if the normal force or the speed of motion gets too big. Why is that?
"$F = μN$, where μ is called the coefficient of friction (Fig. 12-1). Although this coefficient is not exactly constant, the formula is a good empirical rule for judging approximately the amount of force that will be needed in certain practical or engineering circumstances. If the normal force or the speed of motion gets too big, the law fails because of the excessive heat generated. It is important to realize that each of these empirical laws has its limitations, beyond which it does not really work."
| The friction coefficient, $\mu$, varies with temperature, so if excessive heat is generated, the coefficient would change. This doesn't explain why the relationship $F = \mu N$ would fail, however, as the relationship should still hold under these circumstances.
As a general rule of thumb, empirical relationships tend to fail when excess heat is generated for a variety of reasons, and a common reason is material deformation. In the case of friction, the reason is often surface deformations. A common example is a copper pin sliding on a copper plate, in which the heat generated at high speeds would begin to melt the surface, which will cause $F = \mu N$ to fail since it is no longer a simple scenario in which a solid is sliding against another solid.
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Flipping a coin with same initial conditions Today, in my physics class my teacher was talking about how we can never predict the outcome of a coin flip. So I thought:
Will the outcome of a coin flip be the same if we do not change the initial conditions (such as launch angle, force position where force is applied,etc.)?
Intuitively, I feel that the answer would be yes. But is there something related to quantum mechanics that may produce a different answer?
| There are certain ways to find the exact probability of tossing a coin:
First before jumping to this i want to make some things clear over here
This is actually possible because the initial conditions is reproduced to the
entire system but g force can't be done because of distance which is inversely proportional.
*
*Earth is a ellipsoid and the g force is not stable at everywhere.
*You need to know the exact force which will be applied to the object.
*You need to know the position of the coin before we toss it and take a note which one is at top and bottom
So , now according to the first point if you take a rectangle place of 10cm length and 5cm as breadth we can't make sure that the g is same if we then it is easy to calculate.
let it be 5N of force applied to the object.
The way to find is that you need to predict exactly what is the position of the coin in every seconds so that you could actually predict the exact thing.
| {
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In zero gravity! If spacecraft moving in a stable speed or acceleration suddenly stopped would an astronaut continue moving forward (such as a man in a train) ?
| “Zero gravity” is something that doesn't exist. Maybe you mean the situation of freely orbiting spacecraft, in which cosmonauts don't feel any attraction as they are orbiting together with the spacecraft.
Yes, the astronaut will continue to move forward, with all tragic consequences for him.
Note:
In the case of stable or unstable acceleration the astronaut will feel the attraction (in the opposite direction of the acceleration), even if spacecraft don't stop.
| {
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Can measurement of the momentum of a particle can be done without observing its position? Uncertainty principle says that one cannot measure exactly the position and momentum of a particle at same time. As per common understanding when we are measuring momentum of an object it is implicit that we aware of its position. My doubt is in Quantum world, how can we measure momentum of a particle without knowing its position. Are the two momentum and position are mutually exclusive?
| I know a "kind of" answer.
A "velocity selector" AKA "Wein filter" will pass particles with a narrow fixed range of velocities. If we know the species (and therefore the mass) we have measured the momentum of all particles passing the filter without measuring position in the direction of travel (but we have measured position transverse to the direction of travel).
The reason this is interesting is that while the Heisenberg principle limits your precision in measuring both $x$ and $p_x$ at the same time it does not limit your precision in measuring $y$ and $p_x$ at the same time.
| {
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How does a heated lid in a thermal cycler prevent evaporation? A thermal cycler is a chemistry lab device that increases or decreases the temperature of the material inside it. The lid of the device is heated to prevent condensation and evaporation of the mixture inside.
I understand the condensation bit, as a hot lid would keep the gaseous particles gaseous. But I don't understand how a hot lid can prevent evaporation of the mix. Is there a physics law that states liquids cannot evaporate if the temperature of the gas above them is higher than the liquid? Or is it about pressure? Or both?
Edit: Adding some more info since this was migrated to physics. This site provides a decent overview on thermal cyclers. The following paragraph that explains the basics is from there:
The basic idea of a thermal cycler is that it provides a thermally
controlled environment for PCR samples. A thermal cycler usually
contains a heating block with holes or depressions in it that receive
sample tubes (though other types of sample vessels are now possible
also; see below). For the PCR reactions to work properly, the block
must change temperature at specific times, and spend specific
durations of time at specific temperatures. The researcher programs
the temperature cycling information into the thermal cycler either by
computer or via a console on the instrument, or uses a preprogrammed
routine built into the machine.
The main idea is that the machine cycles the temperatures of its contents, the temperatures usually range from around 50 to around 95 C.
| We were discussing about this with a colleague, and we couldn't agree. I propose that it's due to the vapor going near the lid being heated more, therefore developing relatively more pressure than the column immediately under it, and therefore migrating to the lower pressure zone. This brings more vapor/air in its place that gets heated and returns down to the liquid surface. This functionally should stop more vapor from forming, creating an equilibrium that will be reached fairly quickly given the small tubes. Also, the higher vapor pressure due to the heat should reduce the vaporization of the water?
I don't know though, i couldn't find a definitive explanation. What do you think?
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Does electric field vary with cross sectional area in a non-uniform current carrying conductor? Suppose I have a non uniform conductor which is kept in a uniform electric field maintaining a constant potential difference across its ends.Both electric field and current density are a properties of points in space.Since current density is defined as current per cross sectional area, in order to maintain same current through out the conductor the current density changes with cross sectional area. Now according to ohm's law $J=sigma*E$, as current density changes electric field should also changes with cross sectional area. How is this possible if electric field is a property of point in space and we maintain a uniform field across the conductor.How electric field changes with cross sectional area?
|
Now according to Ohm's law $\mathbf J=\sigma\mathbf E$, as current density changes electric field should also change with cross sectional area. How is this possible if electric field is a property of point in space and we maintain a uniform field across the conductor.
We can't maintain uniform electric field across such a conductor, because its resistance is non-uniform.
How electric field changes with cross sectional area?
If you have cross section $A$ depending on distance $d$ from the origin as
$$A = A(d)$$
the resistance will depend on distance too:
$$\textrm dR(d) = \frac 1 \sigma \frac{\textrm d \ell}{A(d)}$$
where $\textrm d\ell$ is a linear element of the conductor and $\textrm dR(d)$ is its resistance.
Since
$$V = IR$$
on a linear element $\textrm d\ell$ there will be a drop of potential $\textrm d\varphi(d)$
$$\textrm d\varphi(d) = I \frac 1 \sigma \frac{\textrm d \ell}{A(d)}$$
Since
$$\int\limits_a^b E\,\textrm d\ell = V$$
we can write for the difference of potentials on an element $\textrm d\ell$
$$E\, \textrm d\ell = \textrm d\varphi(d)$$
Therefore
$$E = I \frac 1 {\sigma {A(d)}}$$
Hopefully this shows the intuition behind the electric field not being uniform.
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Acceleration and motion can be in different direction? I'm not getting what acceleration concept is and how it relates to motion and how motion and acceleration can be in different direction? And what's behind the concept of negative and positive acceleration?
| Try to remember the following things in short which will help you to develop this idea of your own.
(A) Acceleration is in the direction of motion, when you are increasing the speed of the car moving along a straight line.
(B) Acceleration is against the direction of motion, when brakes are applied to the car in motion.
(C )Acceleration is uniform, when an object is under free fall.
(D) Acceleration is non-uniform when a vehicle is driving through a busy city road.
Hope this will help.
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Is the velocity of a string that is being rotated around a central point the same at any point on the string? Say that you have a weight attached to a 2 metre long string, and you are rotating the weight at 5 m/s. Is every point on the string going to be rotating at that same velocity of 5 m/s, or is the velocity of the string going to change according to how far away you are from the centre of rotation?
I'm looking at the linear velocity.
| Depending on the 'type' of velocity.
• Angular velocity will be constant for every particle / unit length of the string i.e. 5/2 revolutions s-1
• Linear velocity will change according to the distance from the point of rotation
V= r. w, where V is the linear velocity, r is the length of the string from the point of rotation and w is the angular velocity
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How is it possible to have different answer using two different equation for same condition?
There are two blocks ${(m_1 ,m_2)}$ of different masses placed on a surface, connected with a spring of spring constant $k$. The coefficient of friction between the blocks and surface is $\mu$. Now we have to find the minimum amount of force applied on $m_1$ in order to shift $m_2$.
The book which I am studying uses work energy theorem — the block $m_1$ is displaced by $x$, and hence spring is stretched by $x$ so we have $${Fx=\frac{1}{2}kx^2+\mu m_1gx}$$ and as we are considering the case when $m_2$ is about to move we have $kx=\mu m _2g$
On solving these two equations we get answer as $${F=\left(\frac{1}{2}(m_2)+(m_1)\right)\mu g}$$
but if we use simply equilibrium condition thinking that just after the force ${F}$ of equilibrium condition equation the block will start moving
so we have $${F-\mu m_1g=kx} $$and as $${kx=\mu m_2g}$$ we get $${F=\mu g(m_1+m_2)}$$
which is different from the work energy equation
how is it possible to have different answer using two method but writing the equations thinking about same condition
| Sorry, a bit overread on my part. For the mass in question which just begins to slide, use balance of forces. For the other mass which is being pulled use energy gained by spring be equal to work doned by pulling force F and frictional force on mass being pulled.
So the answer arrived at in case 1 in original post will be correct. Balance of force on mass being pulled is indeed incorrect as pointed by by others as it is moving.
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What are change of frame and change of coordinates? What's the difference between a change of frame and a change of coordinates?
I feel like both are transformations on the coordinates but change of frame changes also the vectors.
| There's a difference because your coordinates do not always specify your frame. It's true that coordinates always give you one particular choice of a frame, but you can choose to use a different set of basis vectors to represent the vector space at any particular point.
The first example of this you'll probably run into is the difference between the coordinate basis and the orthonormal basis in spherical coordinates. They're based on the exact same coordinate system, but they're totally different frames. If $\mathbf{p}(r, \theta, \phi)$ is the position vector to the point with spherical coordinates $r, \theta, \phi$, then the coordinate basis vectors are defined as
\begin{align}
\mathbf{r} &= \frac{\partial \mathbf{p}} {\partial r} \\
\boldsymbol{\theta} &= \frac{\partial \mathbf{p}} {\partial \theta} \\
\boldsymbol{\phi} &= \frac{\partial \mathbf{p}} {\partial \phi}.
\end{align}
Now, remember that we can express the point in terms of the usual orthonormal Cartesian basis as
\begin{equation}
\mathbf{p} = r\sin\theta\cos\phi\, \hat{\mathbf{x}} + r\sin\theta\sin\phi\, \hat{\mathbf{y}} + r\cos\theta\, \hat{\mathbf{z}},
\end{equation}
which we can differentiate to find
\begin{align}
\mathbf{r} &= \sin\theta\cos\phi\, \hat{\mathbf{x}} + \sin\theta\sin\phi\, \hat{\mathbf{y}} + \cos\theta\, \hat{\mathbf{z}} \\
\boldsymbol{\theta} &= r\cos\theta\cos\phi\, \hat{\mathbf{x}} + r\cos\theta\sin\phi\, \hat{\mathbf{y}} - r\sin\theta\, \hat{\mathbf{z}} \\
\boldsymbol{\phi} &= -r\sin\theta\sin\phi\, \hat{\mathbf{x}} + r\sin\theta\cos\phi\, \hat{\mathbf{y}}.
\end{align}
So already, we see that there are two different bases for the vector space at this point: one that uses the usual $(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}})$ basis, and another that uses $(\mathbf{r}, \boldsymbol{\theta}, \boldsymbol{\phi})$. Any vector can be expanded in either basis, even though they correspond to the same point.
In fact, we can even find another basis based on spherical coordinates. Using the expressions above, it's a simple exercise to see that we have
\begin{align}
\mathbf{r} \cdot \mathbf{r} &= 1 \\
\boldsymbol{\theta} \cdot \boldsymbol{\theta} &= r^2 \\
\boldsymbol{\phi} \cdot \boldsymbol{\phi} &= r^2 \sin^2 \theta.
\end{align}
So this frame is not orthonormal. If you want an orthonormal basis, you need
\begin{align}
\mathbf{e}_r &= \mathbf{r} \\
\mathbf{e}_\theta &= \frac{1}{r}\boldsymbol{\theta} \\
\mathbf{e}_\phi &= \frac{1}{r\sin\theta} \boldsymbol{\phi}.
\end{align}
You'll see and use both of these frames. They are clearly different frames, even though they're based on exactly the same coordinates.
More generally, you're free to define your frame at any point in any way you want (as long as you keep the same number of vectors, and keep them linearly independent). So a change of coordinates implies a change in coordinate frame, but it is not equivalent to a change in frame generally.
| {
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Rotating coordinate frame connection of coordinates and mass Hello I am still confused about rotating coordinate frames and want to ask a question about it.
Is it correct that strictly speaking the mass must be connected with the axis of rotation in the rotating coordinate frame? For example by a rod, or a rotary disk, or whatever. I mean otherwise I do not see how an Euler force or a Coriolis force can act on a particle? (I think I am confused because this connection is usually not showed in figures in the textbooks)
| A rotating coordinate system is a non-inertial coordinate system. For general motion there is translational plus rotational motion. The motion of any object can be evaluated using either an inertial or a non-inertial system. There is no requirement for any fixed relationship between the object and any axes of any coordinate system; however, for some problems this can be the case, such as using a rotating coordinate system for uniform circular motion of a particle, it is convenient to take the particle at a fixed position in the rotating system, and located along one of the rotating axes. In a non-inertial system "fictitious" forces (and torques) are present that must be accounted for in evaluating the motion.
| {
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Thermodynamics: how does the deformation of partially elastic materials produce heat? When a perfectly elastic material is deformed, the energy associated with the strain remains stored in the body as elastic potential energy, called strain energy. Upon the removal of the deforming forces, the body completely regains its original shape, size and configuration with no further loss of heat. However, nothing is perfectly elastic! When a partially elastic material (e.g., a rubber band) is deformed, there is always a remnant deformation even upon the withdrawal of the deforming forces. It exhibits hysteresis and part of the energy converted to heat. Here is a quick reference.
How does this remnant deformation responsible for the heating up of the material?
Edit According to one of the $TdS$ equations, $$dT=\frac{T}{C_V}\Big[dS-\Big(\frac{\partial P}{\partial T}\Big)_V dV\Big],$$ we note that a change in temperature is caused by either a change in volume $V$ or a change in entropy $S$ or both. Can we use this to understand what is going on?
My guess is that for the elastic deformation of a perfectly elastic material, during the process of loading $dT$ is negative because $dV>0$ and during unloading $dT$ is positive because $dV<0$. In the whole cycle, when the system comes to its original state, it does not heat up. But for a partially elastic material, the residual permanent deformation may be responsible for a nonzero $dV$, a nonzero $dS$ or both?
| A partially elastic material (i.e., viscoelastic material) exhibits a combination of elastic and viscous behavior, and it is the viscous part that is responsible for the increase in internal energy (adiabatic case) or the emission of heat (isothermal case).
Imagine a spring and a damper (dashpot) in series. When the combination is stretched, both the spring and damper extend, but the extension of the damper is responsible for dissipation of mechanical energy to either internal energy of heat. So, when the force extending the combination is released, the spring can return to its original length, but the new length of the damper remains locked in. So the overall combination retains some residual extension, and some of the mechanical energy has been dissipated by the damper.
| {
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Different concepts of phase transitions in spin models I am currently revising the lecture notes in which different spin systems
are analyzed, focussing on the occurrence (or absence) of phase transitions.
Different techniques are applied to analyze the different models and - seemingly -
also different notions of phase transitions.
These notions are not mathematically precisely defined and the question is how to
connect them.
I will briefly outline the analysis of three different models and their notions of phase
transitions.
*
*Absence of phase transitions via smoothness of the free energy: 1-D Ising Model.
Here it is shown that the free energy per spin is a smooth (in general not analytic) function of the temperature
and the magnetic field.
*Phase transitions via magnetization: 1-dimensional chain of non-interacting spins in a magnetic field.
Here it is shown that the average magnetization (averaged over all spins, not as in a expected value) converges a.s.
to tanh(hb) in the thermodynamical limit where h is the magnetic field and b is the inverse temperature.
*Phase transition via magnetization: Curie-Weiß-model without magnetic field.
Here it is shown that for high temperatures the average magnetization converges to zero. For low temperatures
however the magnetization is either positive or negative, each with proability 1/2
*Phase transition via existence of Gibbs measure: 2D Ising model without magnetic field. Here it is shown that in the thermodynamical limit there exists a Gibbs measure such that above some critical
temperature the expected direction of every spin is equal and non-zero. This means that the expected average
magnetization must also be non-zero.
I am wondering how to connect these different concept and how these relate to the real-world intuition
we have for phase transitions.
| As for your 1.,2.,3., magnetization is a derivative of free energy with respect to magnetic field, so smoothness of free energy and smoothness of magnetization are closely related.
| {
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How can I understand if an object stay (zero velocity) or moving with constant velocity (zero acceleration) I thought a scenario like; lets say I am looking an object and there is nothing except this object. Is there a way to understand that if this object is stay on its position or if object moving with a constant speed and also I am moving as same constant speed with this object ? (consider there is not any friction etc.)
| Constant velocity and speed have no meaning unless you specify the frame of reference with respect to which it is measured or observed. (The only exception is the speed of light).
If you see an object “moving”, then it is moving with respect to your frame of reference. If you see it as “still” it is still with respect to your frame of reference.
But in the frame of reference of somebody who is with the object in a windowless compartment the object is always at rest and the person has no no way of knowing what you are seeing.
All this assumes constant velocity.
Hope this helps
| {
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What does a coordinate representation of density matrix mean? A coordinate representation of density matrix $\rho$ is defined as
$$ \rho (x, x') \equiv \left<x\right| \rho \left|x'\right> .$$
When $x = x'$, this expresses a probability where a particle is in the state $\left|x\right>$.
Question: what does that mean when $x \neq x'$? Is that related to some probability?
According to Feynman (Statistical Mechanics a set of lectures, (p.72)),
$$ I \equiv \int dx _1 \int dx _2 \cdots \int dx _{n - 1} \left<x\right| \rho \left|x_1\right> \left<x_1\right| \rho \left|x_2\right> \cdots \left<x_{n-1}\right| \rho \left|x'\right>$$
can be interpreted as that "the particle travels from $x'$ to $x$ through a series of intermediate steps, $x_1, x_2, \cdots, x_{n-1}$, which define a path". I don't understand this statement.
| The diagonal entries of the density matrix are called populations and provide information about the probability density of the particles (described by the density matrix), i.e. their probability of "being found" in real space.
This is easily seen from a density matrix $\rho = |\Psi\rangle \langle \Psi|$, and $$\rho(x,x) = \langle x|\rho|x\rangle = \langle x|\Psi\rangle \langle \Psi|x\rangle = | \langle x |\Psi\rangle|^2 = |\psi(x)|^2,$$ which is the usual quantum mechanical probaibility density.
The off diagonal entries of the density matrix are called coherences and provide information about the phase coherence of the system described by $\rho$ between two positions $x$ and $x'$. Is there a fixed phase relationship between $x$ and $x'$, especially as $|x-x'|\rightarrow \infty$? I.e. will constructive & coherent interference occur over a large distance or will it be washed out?
The most famous application of the off diagonal elements of the density matrix is in Off-Diagonal Long-Range Order (ODLRO), which is what manifests in Bose-Einstein Condensates or Superfluids. These are phases where the system breaks a $U(1)$ phase and the wavefunction "picks" a specific phase $\theta$.
The "broken" phase is distinguished from the unbroken phase because it obeys:
$$ \lim_{|x-x'|\rightarrow \infty} \rho(x,x') \rightarrow n_0 \neq 0,$$ i.e. phase coherence is preserved over arbitrarily long distances.
| {
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Is it possible to see individual photons impressioning film? As part of a course in physics teaching, I am developing a small curriculum that will teach (the basics of) quantum mechanics to high school students.
I need a simple way to show the quantization of light. My uni lecturer suggested that low-light photography shows grainy pictures because of individual photons hitting the sensors in discrete points, which shows that they are not diffuse waves.
I have been unable to find anything confirming this on the internet, much less a picture or even better video showing the effect.
I suppose I'm looking for something similar to this video of Tonomura's experiment on the quantization of electrons.
Thank you for your time.
EDIT: I am not actually looking for the equipment. Mere footage of the equipment would be more than adequate. (Just like I won't try to replicate Tonomura in a classroom, but I will show its video)
| Silver halide film doesn't really respond to single photons; it takes four photons, absorbed within a short period of time, to expose a silver halide crystal in the film. An imaging photomultiplier or other photon-counting image sensor would be better, as @Pieter pointed out.
| {
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Does throwing a penny at a train stop the train? If I stand in front of a train and throw a penny at it, the penny will bounce back at me.
For the penny to reverse its direction, at some point its velocity must go to zero. This is the point it hits the train. Two objects in contact have the same velocity, so the train must come to a stop for the penny to change its direction.
I assume I'm getting some principles wrong. Is it because I assumed a perfectly rigid body, when in practice the train actually deforms ever so slightly?
| It looks like there's a hole in your reasoning. If I understand correctly, you're saying this:
*
*At some point, the penny's velocity must be zero.
*At some point, the penny's velocity must be equal to the train's velocity.
*Therefore, at some point, the train's velocity must be zero.
Both 1 and 2 are true (assuming that the collision is exactly head-on). But 3 does not follow, because 1 and 2 happen at different times. Specifically, 1 happens before 2 happens.
The two events both happen during the collision, so they happen within an extremely short interval of each other; but still, they happen at different times.
| {
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Electromagnetic waves according to Maxwell If a variable Electric field creates a variable magnetic field and VICE VERSA (according to Maxwell's equations), then why don't we enter a loop where E vector and B vector keep creating one another until they reach infinite magnitudes?
| Ah, this was actually the great insight of Maxwell. What you are referring to is electromagnetic waves (i.e. light). These waves are just the electric and magnetic field continuously generating each other. Unlike what you may intuit though, looking at the actual mathematical solutions that yield such behavior shows that the magnitude of the fields never actually grow but either stay the same (e.g. plane waves) or shrink (e.g. spherical waves). This can easily be seen by Poynting's theorem which shows that Maxwell's equations conserve energy or, more specifically, the quantity $\frac{\epsilon_0E^2}{2}+\frac{B^2}{2\mu_0}$ is conserved in vacuums.
| {
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Zero mass Kerr metric When mass in Kerr metric is put to zero we have $$ds^{2}=-dt^{2}+\frac{r^{2}+a^{2}\cos^{2}\theta}{r^{2}+a^{2}}dr^{2}+\left(r^{2}+a^{2}\cos^{2}\theta\right)d\theta^{2}+\left(r^{2}+a^{2}\right)\sin^{2}\theta d\phi^{2},$$
where $a$ is a constant. This is a flat metric. What exactly is the coordinate transformation that changes this into the usual Minkowski spacetime metric form $$ds^{2}=-dt^{2}+dx^{2}+dy^{2}+dz^{2}?$$
| The transformation is given in page 15 of this paper: The Kerr spacetime: A brief introduction
BTW there is a 2017 paper that claims that the mass zero Kerr metric is not actually equivalent to Minkowski metric but is a wormhole instead: Zero mass limit of Kerr spacetime is a wormhole
| {
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Finding the quadrature variance of a superposition of squeezed coherent states How do you find the quadrature variance of a state $$\lvert x\rangle =\lvert a,b\rangle +\lvert a,-b\rangle$$ where $\lvert a,b\rangle = D(a) S(b) \lvert 0\rangle$?
$\lvert x\rangle$ is a superposition of squeezed coherent states.
| Starting with the expression for $S(z)$ we have:
$\hat{S(b)} = e^{b\hat{a}^\dagger \hat{a}^\dagger - b^* \hat{a}\hat{a}}$
Through the following BCH formula:
$e^BAe^{-B} = A + [B,A] + \frac{1}{2!} [B,[B,A]] ...$
We can see that, since $ [\hat{a}, \hat{a}^\dagger]=1:$
$\hat{S(b)} \hat{a} \hat{S(b)}^\dagger = \hat{a}cosh(r) + \hat{a}^\dagger e^{i\theta} sinh(r)$
And a simmilar expression for $\hat{a}^\dagger$.
We can also do the same thing for $\hat{D(a)} = e^{a\hat{a}^\dagger - a^* \hat{a}}$ . Here the commutator [B,A] will be a constant and the series will have only two non-zero terms.
$\hat{D(a)} \hat{a} \hat{D(a)}^\dagger = \hat{a} + a$
Then we use the unitarity of these operators to achieve:
$\hat{a} \hat{D(a)} \hat{S(b)} \lvert 0 \rangle = \hat{D(a)} (\hat{a} + a) \hat{S(b)} \lvert 0 \rangle$
$= a\hat{D(a)} \hat{S(b)} \lvert 0 \rangle + e^{i\theta} sinh(r) \hat{D(a)} \hat{S(b)} \hat{a}^\dagger \lvert 0 \rangle $
...
We have expressions for how $\hat{a}$ and $\hat{a}^\dagger$ act on $ \lvert a,b\rangle$:
$$ \hat{a} \lvert a,b\rangle = a\lvert a,b\rangle + e^{i\theta} \sinh(r)D(a) S(b) \hat{a}^\dagger \lvert 0\rangle $$
$$ \hat{a}^\dagger \lvert a,b\rangle = a^* \lvert a,b\rangle + \cosh(r)D(a) S(b) \hat{a}^\dagger \lvert 0\rangle $$
...
Using the definitions of quadrature, and their respective variances:
$$\sigma^2_q = \langle q^2\rangle_x - \langle q\rangle^2_x = \frac{1}{2} \left[\langle x \rvert (\hat{a}^2 + \hat{a}\hat{a}^{\dagger} +\hat{a}^{\dagger} \hat{a} + \hat{a}^{\dagger2}) \lvert x \rangle - \langle x \rvert (\hat{a} + \hat{a}^{\dagger}) \lvert x \rangle ^2\right] $$
| {
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Motion of a charged particle in a constant and uniform magnetic field Assuming the following relationship has been demonstrated
$$r=\frac{m u_{0}\sin \theta_0}{qB\sqrt{1-\left(\dfrac{u_{0}^2}{c^2}\right)}}=\frac{p_0\sin \theta_0}{qB}$$
where $p_0=mu_0/\sqrt{1-\beta^2}$ represents the relativistic momentum of the particle. In the relativistic case, therefore, $\omega$ (relativistic angular velocity $\omega=\frac{qB}{m} \sqrt{1-\left(\frac{u_{0}^2}{c^2}\right)^2}$) is no longer constant but depends on the speed $\bar{u}_{0}$ of the charged particle $q$; in fact the factor $\gamma$ is present.
The step $p$ (different of the momentum $p_0$) is obtained from the product of the parallel component of the initial particle speed $u_{0\parallel}$ and the period $T$:
$$p=(u_{0})_zT=u_{0\parallel}T$$
Making some considerations about the angle $\theta_0$, if $\theta_0=0$ why I obtain a straight line? Don't you have $r=0$?
| When $\theta_0$ is zero, the trajectory is no longer a spiral around the $z$-axis; it is a straight line along the $z$-axis, for which $r=0$ in cylindrical coordinates.
| {
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Varying of momentum for constant kinetic energy How is that If the momentum of the particle is constant with time, its kinetic energy ($E_k$) should also be constant with time a true statement but the converse is false...
When momentum is constant,
\begin{eqnarray}
p&=&mv \\
E_k=(mv^2)/2& =& (mv)^2/2m\\
E_k&=&p^2/2m\\
\end{eqnarray}
therefore when $p $ is constant, $m$ is a constant. So $E_k$ is also constant...
Isn't it same the other way as well.....when $E_k $ is constant, $m$ anyway a constant therefore $p^2$ is a constant which makes $p$ a constant?
But my note says that when $E_k$ is constant with time, $p$ should also be constant with time is false......so how will the momentum ($p$) vary?
| Momentum is a Vector quantity, it is measured with both value and direction. Kinetic Energy is a Scalar quantity, measured only in value, but dependent on momentum.
If momentum is constant this means that we are traveling at a constant speed in a constant direction, our Kinetic Energy is then guaranteed by the equation K=P^2/2m
A constant kinetic energy however, only means that we are traveling at a constant speed; the direction of that speed is not guaranteed. A satellite in a circular orbit, for example, has a constant Kinetic energy but no constant direction, and therefore no constant momentum; while the quantitative value of momentum will be the same everywhere in the orbit, the direction changes, and therefore so does the momentum vector.
| {
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Do objects besides strings, ropes, and rods have tension? Why do we define tension only in strings and ropes and rods and such? Shouldn't every object experience tension force? Like when you pull a paper from opposite sides, it gets taut, and experiences what seems like a state of tension. If every object does experience tension, can you define tension?
| If you are thinking something really fundamental, there's only four acknowledged fundamental force, and the "tension"(of something like a string) was usual treated as a result of electromagnetic interaction(electromagnetic force) from the atoms or particles.
However, tension in sub particle level can also resulted from strong force, like protons inside a nuclei.
Depends, tension is just a more formal wording of the usual "stretch", it's basically a word commonly used by physicists to mention in a way that physicists can easily understand.
| {
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If I stood next to a piece of metal heated to a million degrees, but in a perfect vacuum, would I feel hot? A friend of mine told me that if you were to stand beside plate of metal that is millions of degrees hot, inside a 100% vacuum, you would not feel its heat. Is this true? I understand the reasoning that there is no air, thus no convection, and unless you're touching it, there's no conduction either. I'm more so asking about thermal radiation emitted by it.
| Actually, your friend is probably right but for the wrong reason. That much energy is going to fry you in very short order--and will probably kill the nerves before they can say "hot!"
Remember, energy goes at the 4th power of temperature. 100x the temperature of the sun equals 100 million times the energy. There is no question that's enough to kill you very quickly, the only uncertainty I have here is whether you will perceive anything before that happens.
| {
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Is Lorentz transformation the only way to preserve speed of light? Doesn't this transformation also preserve speed of light?:
$B$ is moving with speed $v$ relative to $A$ in the $+x$ direction. $B$ passes $A$ at $t=0$, and $A$ shines a torch at that moment.
Let $(x', t')$, the coordinates of the light beam in $B$'s frame, be related to the $(x,t)$ coordinates in $A$'s frame by these transformations:
$$x'=x-vt
=ct-vt
=ct\left(1-\frac{v}{c}\right)$$
$$t'=t\left(1-\frac{v}{c}\right)$$
As we can see, this transformation gives $\frac{x'}{t'}=c$. So the speed of light is the same in $B$'s frame. This transformation tweaks the time but lets distance be the same as given by Galilean transformation.
What issues does a transformation like this cause? And what makes 'Lorentz transformation' the right way to go to account for the constancy of speed of light?
| Consider observer $C$, who is traveling at a speed relative $-v$ in the $x$-direction relative to $B$. By your logic, the coordinates $x''$ and $t''$ measured by $C$ should be
$$
x'' = x' - (-v) t' = (x - vt) + v (t - v/c) = x - v^2/c \\
t'' = t'( 1 - (-v)/c) = t ( 1 - v/c)(1 + v/c) = t \left( 1 - \frac{v^2}{c^2} \right)
$$
But $C$'s coordinates should be the same as $A$'s coordinates, i.e., $x'' = x$ and $t'' = t$. This is a contradiction.
The other way to look at it is that the inverse transformation law from $B$ to $A$, according to your equations, is
$$
t = \frac{t'}{1 - v/c} \qquad x = x' + \frac{t'}{1 - v/c}
$$
But this means that the transformation laws to get between reference frames are different between these two frames. This means that the principle of relativity is broken; the transformation laws should have substantially the same form in all reference frames.
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What is a mass moment? I am currently reading through a document Finding Moments of Inertia from MIT, page 4, and I am a little confused as to one of the concepts that they use.
In this document, there is mention of a mass moment. Could someone possibly define this for me please? I can't find anything too clear on the Internet.
Is this synonymous with the first moment of mass?
| To TNTCookie and anyone else who may be looking at this post in need of help
I have found the answer. It lies within the definition of a centre of mass:
$$x_{cm} = \frac{\Sigma_{i=1}^{i=N}m_i x_i}{M},$$
where $M$ is the sum of all masses in a system, and the sum in the numerator is the first moment of mass (mass moment).
If we multiply both sides of our equation by $M$, we get:
$$M\times x_{cm} = \Sigma_{i=1}^{i=N}m_i x_i.$$
This works in accordance with the steps in the MIT document as attached. We are summing individual masses multiplied by individual distances from the centre of mass. In the problem in the document, we have to deal with a lack of mass (which can be considered negative mass), and thus we get:
$$M\times |\vec{OC}| = \big(m_{\mbox{cylinder}} \times 0\big) - \big(m_{\mbox{missing cylinder}} \times \frac{R}{2}\big).$$
*Note that the distance from the centre of mass, to the centre of mass of a normal cylinder must be $0$, and the fraction $\frac{R}{2}$ comes from the problem in the attached MIT document.
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Intuition of Maxwell's Equations Is there an intuitive explanation for Maxwell's equations?
I know they are axioms but is there a logical understanding of why instead of mathematical. Both forms don't explicate the scientific reasoning behind them to me.
I would appreciate a non- or minimally mathematical approach to them.
| The two equations involving the divergence aren't dynamical (they have no time derivatives), and if they're satisfied initially, they're automatically satisfied at all later times. They tell us about the sources and sinks of the fields.
The two equations involving the curl have time-derivative terms and a current term.
The time-derivative terms describe electromagnetic induction. Their signs are opposite, which is what allows negative feedback so we can have oscillating electromagnetic waves.
The current term describes how moving charges create magnetic fields. It has to be there because of Lorentz invariance, i.e., if we had known about electric fields but not about magnetic fields, relativity would have forced us to invent magnetism.
| {
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References and papers to distinguish between the Heisenberg and Ising Model Does anybody have any good papers or references to explain the differences between the Heisenberg model and Ising model?
To the best of my knowledge, I am aware that the Hamiltonians are similar, however the Heisenberg model represents the spins with Pauli operators.
I would like some solid article which I could reference please.
|
Does anybody have any good papers or references to explain the differences between the Heisenberg model and Ising model?
As the comments already suggest, this is more introductory textbook material. Parkinson and Farnell's "An Introduction to Quantum Spin Systems" (Springer Lecture Notes in Physics 816, 2010) might be a good choice for you. It's an overall accessible introduction to quantum spin systems, particularly in one and two dimensions, covering models and common techniques. It also be available as an e-book through many university libraries. (But you should be able to find the distinction between Ising and Heisenberg models in many other books on magnetism or statistical physics too.) From p. 17:
Depending on the types of atom involved and the environment in which they exist the exchange interaction may have different forms. Examples are:
*
*Heisenberg $J\mathbf{S}_1\cdot\mathbf{S}_2$ (as before)
*Ising $JS_1^zS_2^z$
*Anisotropic (a combination of the above) $J[\Delta S_1^z S_2^z + (S_1^xS_2^x + S_1^y S_2^y)]$
*Biquadratic $J\left( \mathbf{S}_1 \cdot \mathbf{S}_2\right)^2$
To the best of my knowledge, I am aware that the Hamiltonians are similar, however the Heisenberg model represents the spins with Pauli operators.
Both models are generally defined in terms of spin operators, as shown above. In the important case of spin-1/2 spins, the spin operator can be represented in terms of Pauli matrices. In fact, there's an equality $\mathbf{S}=\hbar \vec{\sigma}/2$, where $\vec{\sigma}=(\sigma^x, \sigma^y, \sigma^z)$ is the vector of Pauli matrices. Often we just redefine the $J$ to avoid the factor of $\hbar/2$, which is why you might have seen an Heisenberg model written $J\, \vec{\sigma}_1\cdot\vec{\sigma}_2$, but it can make a difference when comparing observables calculated using different notations.
| {
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Queuing by the event horizon I understand that an observer outside the event horizon (EH) of a black hole (BH) will not see anything disappearing from outside the EH - only the effects of the time dilution near the EH.
Assume that a large number of objects are on its way towards the EH and the last objects can observe all the others. It is right to assume that the all – from the last observer’s viewpoint – cross the EH at the same time?
A natural follow up issue is the pattern the observer sees thereafter, i.e. inside the EH.
| General relativity is based on the equivalence principle, and one way of stating the equivalence principle is that in small enough regions of spacetime, the effects of gravity become undetectable for free-falling observers. So if your cloud of observers is small, or if we restrict our attention to a small part of it, then nothing special happens as far as they're concerned. The ones in back continue to see the ones in front. However, if A passes the horizon before B, then any light emitted by A after crossing the horizon will be detected by B after B has crossed the horizon as well.
Personally, I find it extremely difficult to reason about this sort of thing unless I draw a type of diagram called a Penrose diagram. I have a simple, nonmathematical explanation of Penrose diagrams in this book: http://www.lightandmatter.com/poets/ . See section 11.5.
| {
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Anticommutation relation of Grassmann numbers Let $c,c^*$ be the fermion annihilation/creation operators and $\xi,\xi^*$ denote Grasssmann numbers where $$|\xi\rangle = \exp(-\xi a^*)|0\rangle$$ is the coherent state. Then why is it true that
$$
\langle \phi|\xi\rangle \langle\xi |\psi\rangle = \langle-\xi|\psi\rangle \langle \phi|\xi\rangle\tag{1.172}
$$
where $|\phi\rangle,|\psi\rangle$ are states in the fermion Fock space.
I saw this in John W. Negele, Quantum Many-particle Systems, equation 1.172, but it doesn't seem to work for $|\phi\rangle =|0\rangle$ and $|\psi\rangle = |1\rangle$.
| When I read this part recently I got the same question, and I believe it is wrong too. Take a simple example that $\phi$ is a state of definite occupation with $p$ particles and $\psi$ is a state of definite occupation with $q$ particles, then switching the two amplitudes will cause an extra sign of $(-1)^{pq}$, whereas changing $\langle\xi|\psi\rangle$ to $\langle-\xi|\psi\rangle$ will only cause a sign of $(-1)^q$. In general the two signs won't be equal. And your example fits this.
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Application of angular velocity to Euler angles According to a post here Angular Velocity expressed via Euler Angles you can express angular velocity from euler angles. If I choose Y-Z-Y as a rotation sequence the expression becomes.
$\theta_r, \theta_p, \theta_y$ = roll, pitch, yaw
$$
\vec{\omega} = \dot{\theta_r} \hat{y} + R_z(\theta_p)( \left( \dot{\theta_p} \hat{z} + R_y(\theta_y) \left( \dot{\theta_y} \hat{y} \right) \right)
$$
which becomes
according to this
where
which does not make sense.
does it make sense and does it still work in this case?
| Suppose you have a Y-Z-Y scheme with a corresponding sequence of rotation angles $\theta_y$, $\theta_p$ and $\theta_r$.
After the first rotation (yaw), the 3×3 orientation matrix $\mathrm{E}_y$ and angular velocity vector $\vec{\omega}_y$ is
$$\begin{aligned}
\mathrm{E}_y & = \mathrm{rot}(\hat{j}, \theta_y) & \vec{\omega}_y & = \dot{\theta}_y \left(\hat{j}\right)
\end{aligned} \;\tag{1}$$
The above should be self-evident. Now consider the second rotation and the orientation matrix $\mathrm{E}_p$ and angular velocity vector $\vec{\omega}_p$. Since the local axes are rotated by the first rotation we have
$$\begin{aligned}
\mathrm{E}_p & = \mathrm{E}_y \mathrm{rot}(\hat{k}, \theta_p) & \vec{\omega}_p & = \dot{\theta}_y \left( \hat{j} \right) + \dot{\theta}_p \left( \mathrm{E}_y \hat{k} \right)
\end{aligned} \;\tag{2}$$
Finally, with the third rotation we extend this pattern to find the final orientation matrix $\mathrm{E}$ and the final rotation velocity vector $\vec{\omega}$
$$\begin{aligned}
\mathrm{E} & = \mathrm{E}_p \mathrm{rot}(\hat{j}, \theta_r) & \vec{\omega} & = \dot{\theta}_y \left( \hat{j} \right) + \dot{\theta}_p \left( \mathrm{E}_y \hat{k} \right) + \dot{\theta}_r \left( \mathrm{E}_p \hat{j} \right)
\end{aligned} \;\tag{3}$$
The last part is re-written as
$$\begin{aligned}
\mathrm{E} & =\mathrm{rot}(\hat{j}, \theta_y)\mathrm{rot}(\hat{k}, \theta_p) \mathrm{rot}(\hat{j}, \theta_r) & \vec{\omega} & = \dot{\theta}_y \hat{j} + \mathrm{rot}(\hat{j}, \theta_y) \left( \hat{k} \dot{\theta}_p + \mathrm{rot}(\hat{k}, \theta_p) \hat{j} \dot{\theta}_r \right)
\end{aligned} \;\tag{4}$$
This expands out to the following jacobian formulation
$$ \vec{\omega} =
\begin{bmatrix}
0 & \sin(\theta_y) & -\sin(\theta_p)\cos(\theta_y) \\
1 & 0 & \cos(\theta_p) \\
0 & \cos(\theta_y) & \sin(\theta_p) \sin(\theta_y)
\end{bmatrix}
\pmatrix{ \dot{\theta}_y \\ \dot{\theta}_p \\ \dot{\theta}_r }
\;\tag{5}$$
| {
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How is it possible to have four types of generating functions? Since the Hamilton's equations of motion remain unchanged in form under a canonical transformation $(q,p)\to (Q,P)$, the Lagrangians must differ by a total time derivative of a function of $q,t$. In other words, $$L-L'=\frac{dF(q,t)}{dt}$$
$$\Rightarrow (p\dot{q}-H)-(P\dot{Q}-K)=\frac{dF(q,t)}{dt}$$ where $H(q,p,t)$ and $K(Q,P,t)$ are respectively the old and new Hamiltonians. How can there be four types of canonical transformations $$
(i)\ F_1(q,Q,t), \qquad (ii)\ F_2(q,P,t),\qquad (iii)\ F_3(p,Q,t),\qquad (iv)\ F_4(p,P,t)$$ if $F$ were to be a function of $q$ and $t$ only?
| OP's first formula applies to the Lagrangian formalism in configuration space. In the context of canonical transformations (CT), it must be replaced with the corresponding formula in Hamiltonian phase space, i.e. the generating function $F(q,p,t)$ is also allowed to depend on momenta $p$. Since the new phase space variables $(Q,P)$ are supposed to depend on the old phase space variables $(q,p)$ and time $t$, we may assume that the generating function $F(q,p,Q,P,t)$ depends on both new and old variables. The four types of CT are special cases.
| {
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Is there 100% pure white? Is it possible to have an object 100% pure white without sky blue or sun color tinting the pure whiteness of the photons reflecting/deflecting off an object? Are there any lights that can produce pure white photons (RGB)? And can we see that the color is white or is our eyes going to trick us into thinking the white is a different color? Can scientist produce what can be seen macroscopically as "pure white".
| You observe white if your RGB cones produce the same, strong signal. If the signal is not strong it will be seen as grey or black.
A white photon should have a sufficiently wide intrinsic bandwidth. Perhaps a pulsed optical laser with a very short pulse length fits the bill. If the pulse length is of the order of a single wavelength the band width covers R, G and B.
"The lasing medium in some dye lasers and vibronic solid-state lasers produces optical gain over a wide bandwidth, making a laser possible which can thus generate pulses of light as short as a few femtoseconds." https://en.m.wikipedia.org/wiki/Pulsed_laser
| {
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Newtonian Gravitational Waves I was thinking about the classical equations for gravity. I got stuck on two equations:
$$\vec{\nabla}.\vec{g}= 0$$
and
$$\vec{\nabla} \times \vec{g}= 0$$
The first equation is Gauss law of gravitation in vacuum and the second equation comes from the fact that gravitational forces are conservative. If I take the curl of the second equation what I end up with is
$$\nabla^2 \vec{g} =0$$
Whose solution in two dimensions can be written as
$$\vec{g}=\vec{A}\sin(kx)(e^{ky}+e^{-ky})$$
My question is: can we interpret this as a wave equation, where the wave is propagating with infinite speed?
I know that gravitational waves are predicted by GR and they travel at the speed of light but I wanted to know can the above equation be interpreted as a classical gravitational wave equation?
Edit: A general wave equation can be written as
$$\nabla^2y-\frac{1}{v²}\frac{\partial^2 y}{\partial t^2}=0$$
when the velocity tends to infinity the second term becomes 0. So Laplace's equation is a wave equation with infinite speed. But there is no physical difference between wave moving infinitely fast or wave not moving at all. This is what I mean when I say a wave with infinite speed.
| I would say no, since the solution that you found is stationary, it cannot be propagating. For example, in one dimension, a propagating wave needs to have the functional form $\Psi = \Psi (x-ct)$, where $c$ is the (finite) velocity of the wave. In general, a propagating wave in the vacuum constitutes the solution of the equation $\Box \Psi =0$, where $\Box=\nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}$ is the D'alembertian operator in a flat spacetime and we didn't consider the presence of sources in the right-hand side. Actually, the equation that you found for $\vec{g}$ is called Laplace's equation, which also arises when considering stationary and vacuum solutions in Electromagnetism.
| {
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Why is more than one ripple created when a rock is thrown onto the still surface of a pond? I have tried making an analogy with a simple pendulum: if you give it a push it will probably make several oscillations. But why isn't there just one ripple that would carry the energy of the several ripples?
| A way to look at it is this: Every wave consists of a large number of wave components of various pure wavelengths. A very long continuous train of identical sinusoidal waves almost entirely consists of a single wavelength component.
An impulse like that of a rock hitting the water produces an initial disturbance that is highly localized. If we mathematically decompose that disturbance into a set of single-wavelength waves, we find that it is actually composed of a continuous spectrum of waves having a wide range of wavelengths, with each wavelength component itself comprising an infinite train of sinusoidal waves.
That may seem odd, but it works because at the moment the initial disturbance is created, all those waves are on top of each other and they cancel out except at the location of the disturbance where their crests are all precisely aligned. At that location, they add up to form the shape of the initial disturbance.
The speed of a surface wave in water depends on its wavelength: longer wavelength waves travel faster than shorter wavelength waves. So, the wave components from the initial disturbance sort themselves out as they travel away from the center: the longer wavelength components rapidly get ahead of the shorter wavelength components, and the farther they go the more sorted out they become and the more of each wavelength component's wave train becomes visible.
Two other factors are important. First, short wavelength surface waves are attenuated much faster than long wavelength surface waves, so short waves don't travel very far. Second, the size of the initial disturbance determines the spectrum and magnitude of the wave spectrum. Toss a tiny pebble into the water, and the dominant wavelength of the surface waves it produces is roughly the size of the pebble's diameter. Toss a meter-wide boulder in the water, and the dominant produced waves will be on the order of a meter wavelength. If a large slab of rock falls into the ocean, it will produce a tsunami whose dominant surface waves are the size of the slab.
| {
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Flat yet finite Universe I would like to ask you this question:
Let's assume that the universe is perfectly flat and big bang happened as described by our theories some 14 billion years ago. Shouldn´t, therefore, the radius of the universe still be finite, even if the geometry of the universe is flat? Even if inflation happened and even if the universe itself could expand faster than light without violating relativity, both speeds were finite, shouldn´t, therefore, the universe be also finite?
I mean the universe couldn´t start finite time ago, expanded with finite speed and yet be infinite, that is a contradiction, so what gives?
| While I have never studied relativity deeply enough to properly answer this question, I still would like to say that these type of global questions always remind me of Arthur C. Clarke:
Two possibilities exist: either we are alone in the Universe or we are not. Both are equally terrifying.
I like to extrapolate this quote as:
Two possibilities exist: either there's something beyond life, or there's not. Both are equally terrifying.
And finally:
Two possibilities exist: either the universe if infinite (in time and space), or it is finite. Both are equally weird! Maybe terrifying?
@Ben Crowell above said it is a philosophical question. Indeed, the true global answer might be metaphysical. But even if the equations or experiments were to give us an answer... thing about it. Is any of the answers less disturbing? What does it mean to have an infinite universe, one that never ends and we can always move further and further? Infinite is weird! On the other hand, if there's a boundary, what lies beyond it? A periodic case seem less contradictory, but weird anyway.
My point is, any possible answer, raises more questions.
| {
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What is the difference between position, displacement, and distance traveled? Suppose the question is somewhat like this:
If $v=8-4t$ and the position at time $t= 0\ \rm s$ is $2\ \rm m$, find the distance traveled, displacement, and final position at $t=3\ \rm s$
Since $\text dx/\text dt=v=8-4t$, then $\text dx=(8-4t)\text dt$. After integrating we find $x(t)-2=8t-2t^2$, and substituting the value of $t=3\ \rm s$ we get $x(3)=8\ \rm m$.
Is the answer that I found displacement, position or distance?
It can't be distance. I am sure of this. But is it position or displacement?
|
What is the difference between position, displacement, and distance
traveled?
Succinctly:
(1) the position of an object is a vector with tail at the origin of the coordinate system and head at the location of the object.
(2) the displacement of an object is the vector difference of the current position vector of the object and the position vector of the object at an earlier time. That is, the tail of this displacement vector is at the earlier position and the head is at the current position. The length of this vector is, generally, the shortest distance between the earlier and current positions.
(3) the distance traveled from the earlier position to the current position is not a vector but is, rather, a path length. There are an infinity of paths that an object may take from the earlier position to the current position and, in general, each has a different associated length.
| {
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Circuit - solve Kirchhoff's laws with determinant? Here is an exercise from my textbook.
]1
At first I solve it by using Kirchhoff's laws directly and using complex impendance:
$$U_{in}=\frac{1}{Cs}(i_1-i_2)+i_1R_1$$
$$0=R_2i_2+\frac{1}{Cs}i_2+\frac{1}{Cs}(i_2-i_1)$$
By solving above equations, both $I_1$ and $I_2$ become function of $U_{in}$. Then I sub them into $U_o=i_1R_1+\frac{1}{Cs}i_2$ to get the relationship between $U_o$ and $U_{in}$. This method works and lead to the answer provided in the book.
However I found a solution looks quite different in solution manual with determinant:
Let,
$$\Delta=\begin{vmatrix}
G_2+Cs & -Cs & -G_2\\
-Cs & G_1 + 2Cs & -Cs \\
-G_2 & -Cs & Cs + G_2
\end{vmatrix}$$
then,
$$V_j=\dfrac {\Delta_{1j}} \Delta I_1~~~~~~or~~~~~~
\dfrac {V_3}{V_1}=\dfrac{\Delta_{13}I_1/\Delta}{\Delta_{11}I_1/\Delta}$$
Therefore the transfer function is
$$\dfrac {V_3}{V_1}=\dfrac{\Delta_{13}}{\Delta_{11}}=\dfrac{\begin{vmatrix}-Cs & 2Cs+G_1\\-G_2 & -Cs\end{vmatrix}}
{\begin{vmatrix}2Cs+G_1 & -Cs\\-Cs & Cs+G_2\end{vmatrix}}$$
$$=\dfrac{C^2 R_1 R_2 s^2+ 2CR_1 s+1}{C^2 R_1 R_2 s^2+(2R_1+R_2)Cs+1}$$
So far I am quite confused on what does the first determinant stands for and could this determinant be written down directly without any draft? (If so, I think it will be faster and easier to solve problems like this and I really want to learn about it.) Also, why do we need to use the Minor of that determinant?
| I have added some theory before the recipe.
$s= \ j \omega$ where $\omega$ ids the angular frequency and $G$ is the conductance of a resistor of resistance $R$ with $G= \dfrac 1R$
There are three Kirchhoff's current law equations which can be set up for each of the nodes 1, 2 and 3.
For node 1 $(V_1-V_2)sC +(V_3-V_1) G_2 + I_1=0$ where $I_1$ is the current entering the network at node 1 from $+V_{\rm in}$.
Rearranging this equation gives $I_1=(G_2+sC) V_1-sCV_2-G_2V_3$
Repeating the process at nodes 2 and 3 gives $I_2 = -sCV_1+(G_1+2sC)V_2-sCV_3$ and $I_3= -G_2V_1 -sCV_2+(sC+G_2)V_3$
So the admittance matrix for this network is $\mathbf Y=\begin{bmatrix}
G2+sC & -sC & -G_2 \\
-sC & G_1 + 2sC & -sC \\
-G_2 & -sC & sC+G_2\\
\end{bmatrix}$
and the matrix equation which links the currents $\mathbf I=\begin{bmatrix}
I_1 \\
I_2 \\
I_3 \\
\end{bmatrix}$ and the voltages $\mathbf V=\begin{bmatrix}
V_1 \\
V_2 \\
V_3 \\
\end{bmatrix}$ is $\mathbf I = \mathbf Y \,\mathbf V$
Here is the (cookery book) recipe for producing the admittance matrix.
You are dealing with admittances and four nodes which I have labelled 0, 1, 2 and 3 and three nodal voltages $V_1,\,V_2$ and $V_3$ which are relative to the reference node 0 which is at a potential of zero.
For element 11 you see that there are two circuit components, resistor $R_2$ and left-hand capacitor $C$, connected to node 1.
You add the two admittances to give $G_2 + sC$ and this is the value of element 11 of matrix $\Delta$.
For element 22 you see that there are three circuit components connected to node 2 and you add their admittances which produces $G_1+2sC$.
Finally element 33 have a value of $G_2+sC$.
Now you need to look at components which are common to two nodes to find elements $Y_{\rm ij}$ with $i\ne j$.
The left hand capacitor is common to node 1 and node 2 so the entry for element 12 and 21 is minus the admittance of the component which is $-sC$ in this case.
Elements 23 and 32 are also equal to $-sC$.
Elements 13 and 31 are equal to $-G_2$.
You have thus populated all the elements of the matrix $\mathbf Y$.
To finish off the problem Cramer's rule is used.
@PageDavid has provide a reference, Nodal Admittance Matrix.
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Could you extract energy from a rotational white hole? I understand it is possible to do it in a black hole, and its also possible to reach the event horizon of a black hole and increase its size.
But in a white hole, no wave (electromagnetic or gravitational) can reach the white hole's event horizon. So, if you apply the penrose process to a white hole, would you be affecting the speed of rotation near its event horizon?
| A white hole is the time-reverse of a black hole. So, a "white hole penrose process" necessarily is the time-reverse of the ordinary penrose process, which would be a particle absorbing a wave or particle coming from the event horizon, leaving the white hole "spun up".
If this sounds weird, it's because it is weird, in the same way that the advanced solutions of Maxwell's equations are weird.
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Does stacking two magnets together increase the magnetic strength? Just wondering if adding two (or more) identical magnets together increases their magnetic strength (in Tesla) as I am doing a physics write-up on the Lorentz force (Fleming's left-hand rule) as I tried stacking magnets together but found that the force produced is still the same. Wondering if I'm doing anything wrong.
| Yes, stacking magnets does increase the field, but not always by a lot---it depends on the physical arrangement.
For your first test, make a small spacer and put two magnets one on top of the other on either side of this spacer. The field between the two magnets will be strong, approximately twice that of either magnet on its own.
Next put two magnets next to each other, with N poles in the same direction. Now you have roughly the same magnetic field at the end of either magnet as you would get using a single magnet, but notice that this field is now available over a wider region.
Finally, with one magnet stacked on top of another, look at the field at either end of the pair. Now the field is a little bigger than that of a single magnet, but not much (unless the magnets are wide and flat). The reason is that the field from each magnet on its own is reducing in strength quite quickly as you move away from that magnet,
so when they are stacked this way you are not adding two fields at their full strength, but rather the full strength of one is added to a field at one magnet's length away from the other, and this second contribution is smaller.
I won't tell you all the answers here but you might like to learn a bit more about how quickly the field falls in strength as you move away from any given magnet. The field is approximately (but not precisely) that of a dipole. You could make a crude measurement of the field strength by using a small test magnet on the end of a non-magnetic spring, and using Hooke's law for the force from the spring.
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If inflation is an exponential expansion, is the exponent known? Inflation is often called exponential expansion.
I can't seem to find what that exponent is. Is it 2, or 20, or completely unknown?
| In cosmic inflation, the Friedmann scale factor $a(t)$ of the universe doesn’t grow as some power of the time, such as $a\sim t^2$ or $a\sim t^{20}$. These are not exponential expansions at all. Exponential expansion means that time appears in the exponent: $a\sim e^{t/\tau}$ or $a\sim 2^{t/\tau}$ or $a\sim 10^{t/\tau}$ or whatever you prefer to use as the base to be exponentiated. Physicists normally use $e$, the base of the natural logarithms, so let’s go with $e^{t/\tau}$. Using a different base just changes $\tau$ a bit.
In inflation, the characteristic time $\tau$ for the scale factor to increase by a factor of $e$ (which is about 2.7) is fantastically small: typically around a trillionth of a trillionth of a trillionth of a second! That’s $10^{-36}$ seconds. In various inflation models it might be bigger or smaller by a few orders of magnitude, but it is always fantastically small.
So in just $10^{-36}$ seconds everything in the universe got 2.7 times further apart. In another $10^{-36}$ seconds it grew by another factor of 2.7. In another $10^{-36}$ seconds it grew by another factor of 2.7. After 60 $e$-foldings the scale factor is 100,000,000,000,000,000,000,000,000 times larger, and this took just, say, 60 trillionths of a trillionth of a trillionth of a second.
It is truly one of the most astounding ideas in the history of physics.
| {
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Is the acceleration of car greater when hitting the accelerator, or the brakes? Is acceleration of car greater than when pedal the pushed to the floor or when break pedal is pushed hard? I do understand that the signs would change in either but I am more considered about the magnitude....
| The breaking time and distance of a car are affected by many variables, such as road conditions, the condition of the brakes and the type and condition of the tires. Also, of course, the reaction time of the driver and the vehicle speed. The accelerations of cars varies widely.
In order to compare gas pedal acceleration to brake pedal deceleration, you would need reliable data (generated by an independent source) for the same car under the same conditions. But here are some general observations of mine.
I read the typical braking distance of a car going at 60 mph (88 fps) once the brake pedal is pushed is about 180 feet. If I did my math correctly, that works out to a stopping time of about 2 seconds. That does not include the "thinking" distance of 60 feet (according to the source) which would of course increase the stopping time. The average deceleration of the vehicle would then be 88 ft/$s^2$
The 0-60 mph acceleration times for vehicles varies widely. For high performance cars I read they are typically less than 6 seconds, but then those cars typically have shorter stopping distances as well. In the case of exotic sports cars the the 0-60 times can be between 2.5 and 3 seconds. For them the acceleration and braking times (and thus magnitude of acceleration and deceleration) might be more comparable.
But for typical non performance cars (family sedans or SUVs) 0-60 times are greater than 6 seconds. So for them it would appear that 60-0 braking time would be less than 0-60 acceleration time, meaning the braking deceleration would be greater than the car acceleration, than would be the case for the exotic sports car.
Update:
It has been suggested that my answer left open the possibility that acceleration may be greater than braking. That was not my intent. I simply wanted to point out all the variables involved and the observation that the difference between braking and acceleration (e.g., the ratio of acceleration to braking) is generally less for a high performance car than for a typical family car. But braking still generally wins out.
| {
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Minus sign in perturbative expansion via Green's function (Schwartz QFT)? In Schwartz's QFT textbook Section 3.5, the Lagrangian for the graviton $$\mathcal{L}=-\frac{1}{2}h\Box h+\frac{1}{3}\lambda h^3+Jh$$ with EOM $\Box h-\lambda h^2-J=0$ is perturbatively expanded in $h$ to yield:
$$h(x)=\int d^4y\delta^4(x-y)h(y)=-\int d^4y[\Box_y\Pi(x,y)]h(y)=-\int d^4y\Pi(x,y)\Box_yh(y)$$
with the note from Schwartz "where we have integrated by parts in the last step."
The first minus sign I understand with the definition $\Box_x\Pi(x,y)=-\delta^4(x-y)$, but I don't understand why there's not a second minus sign between the last and second-to-last terms.
My impression was that in QFT, we treat terms at the boundary of spacetime as null so that $$\int_U A\partial_\mu B=\int_{\partial U} AB-\int_U B\partial_\mu A=-\int_U B\partial_\mu A.$$
However in this case, wouldn't that mean that
$$-\int d^4y[\Box_y\Pi(x,y)]h(y)=-\int d^4y(\Box_y[\Pi(x,y)h(y)]-\Pi(x,y)\Box_y[h(y)])=\int d^4y\Pi(x,y)\Box_yh(y)~?$$
The last term is definitely supposed to have a minus sign, as Schwartz cites the same equation again with minus sign in (3.84). What am I missing here?
| Figured this out as I was typing the question...
The d'Alembertian $\Box_y=\partial_y\partial_y$, so to 'swap' it with an adjacent element, it requires two uses of integration by parts.
That is, $\int_U A\Box B=\int_U A\partial_\mu\partial_\mu B=\int_{\partial U} A\partial_\mu B-\int_U\partial_\mu A\partial_\mu B=-\int_{\partial U}\partial_\mu (AB)-(-\int_U\partial_\mu\partial_\mu(A)B)=\int_UB\Box A$
The last two terms in the expansion of $h(x)$ then have the same sign.
| {
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Momentum replacement in the axial anomaly calculation in dimensional regularisation (‘t Hooft prescription) I have been studying the axial anomaly and everywhere I see the calculation of the triangle loop using dimensional regularisation (see for example pages 661-664 of section 19.2 of Peskin). In the ‘t Hooft prescription for the $\gamma^5$ they divide the Lorentz space into the usual 4 dimensional one and the rest of dimensions (inside the integration), so the loop momentum can be written as (eq. 19.53)
\begin{equation}
l=l_\parallel+l_\perp
\end{equation}
“ Where the first term has nonzero components in dimensions 0,1, 2, 3 and the second term has nonzero components in the other d—4 ($-2\epsilon$) dimensions.”
Then, we arrive to this integral
\begin{equation}
\int \frac{d^dl}{(2\pi)^d}\frac{l_\perp^2}{(l^2-\Delta)^3}
\end{equation}
and what I do not understand is the following replacement for it (eq 19.57)
\begin{equation}
l_\perp^2\to\frac{d-4}{d}l^2
\end{equation}
“under the symmetrical integration”.
I do understand when we do similar things for an even integral for which any odd term is zero so
\begin{equation}
l_\mu l_\nu\to\frac{1}{d}l^2g_{\mu\nu}
\end{equation}
But in the case with the $l_\perp$ I do not see a proper derivation, could someone help me with that? Thanks in advance.
| As the integral has rotational invariance in $d$ dimension, each $\ell$-component should yield the same value. There are $d-4$ non-zero components in $\ell_\perp$ and $d$ non-zero components in $\ell$, we should thus have
\begin{equation}
\frac{1}{d-4}\int \frac{d^4 \ell}{(2\pi)^4} \frac{{\ell}_\perp^2}{(\ell^2-\Delta)^3} = \frac{1}{d}\int \frac{d^4 \ell}{(2\pi)^4} \frac{{\ell}^2}{(\ell^2-\Delta)^3}
\end{equation}
Therefore
\begin{equation}
\mathcal{I} = \frac{d-4}{d} \int \frac{d^4 \ell}{(2\pi)^4} \frac{{\ell}^2}{(\ell^2-\Delta)^3}
\end{equation}
| {
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Infinite parallel plates have the same electric field between no matter the distance? I saw this in a lecture about gausses law in application to infinite charged planes:
How is it possible that the electric field above the top plane and below the bottom plane is always zero, given that the effect of each plane near its outside surface contributes to the electric field much more than the other plate according to Coulombs Law?
Also, how is it possible that the electric field stays constant between the two assuming they are infinite planes no matter the distance between them by the same principle?
(I assume this deals with the effectively infinite contribution in the +i and -i direction of each individual charged particle.)
Can someone give a more detailed explanation of this?
| The electric field from a uniformly charged infinite plane is constant, it doesn't fall off with distance. Coulombs Law is a statement about the force between two charged particles, which is why it works out differently when you talk about planes. You can actually apply Coulombs Law to reach that conclusion:
$$ V(r) = \frac{1}{4\pi\epsilon_0} \int_{plane}\frac{\sigma dA}{r}$$
To understand it intuitively, imagine the plane in front of you. No matter the distance separating you both, the plane looks the same. So there is no reason for the electric field to change.
| {
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Can the flow be irrotational if the viscous forces act on fluid? I tried to answer the question only using the definitions and the Navier-Stokes equation:
$$\rho \frac{Dv}{Dt} = -\nabla P +\rho g -\mu[\nabla \times(\nabla \times v)] $$
In my opinion if the vorticity is zero, then the fluid is irrotational, regardless of presence of the viscous forces, thus $\mu$ can have a non-zero value which implies the existence of viskeuze forces, while the $\nabla \times v = 0$.
| I'm not sure your momentum equation is correct. For the x-momentum, we should have:
$$\rho\frac{Du}{Dt}=-\frac{\partial \rho}{\partial x}+\rho g_x+\frac{\partial \tau_{xx}}{\partial x}+\frac{\partial \tau_{yx}}{\partial y}+\frac{\partial \tau_{zx}}{\partial z}$$
For Newtonian fluids,
$$\tau_{xx}=-\frac{2}{3}\mu\nabla \cdot \textbf{V}+2\mu\frac{\partial u}{\partial x}$$
$$\tau_{yx}=\mu(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y})$$
$$\tau_{zx}=\mu(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x})$$
If you assume irrotational flow and require the curl to vanish, then the only term left that is related to viscosity is proportional to $\nabla^2u$. Shear stress is gone and you would be right, mathematically.
But this is a highly a-physical assumption. Viscosity would result in shear-stress; so irrotationality assumption with a viscous flow isn't a good one.
| {
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Coulomb's law and the continuous charge model in an infinite charged plane The $E$-field of an infinite plane with a uniform charge density $\sigma$ has a constant magnitude equaling $2\pi\sigma$.
Now, this magnitude assumes a continuous charge model throughout the plane.
In reality, the plane is made up of discrete charges, which obey Coulomb's law.
Does the continuous model accurately describe what happens just on top of the plane?
| No. The continuous model is an approximation which breaks down when you get too close to the discrete charges.
| {
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Huygens-principle Huygens principle states that every point on the wave front acts as a source. If it is true, then why can't a single source (let's say a bulb) illuminate a whole big room? Why is it dark after some distance from the bulb? According to him it should continue to infinite.
Where am I wrong?
| Every point on a spherical wave front acts as a source of a new spherical wavelet. The tangent surface to all of the wavelets becomes the new wavefront. And this process is repeated using the new wavefront to advance (propagate) the wave.
Note that the wavelets are spherical so their amplitude is reduced inversely to their radius as they propagate (as 1/ct or 1/r). This causes the propagating wavefront's amplitude to be reduced in the same way.
So the spherical wave amplitude becomes lower as it propagates.
Note that if the original wave front was a plane wave there would be no reduction in amplitude as it propagates--the tangent surface is a plane--for why, see www.researchgate.net/publication/316994209 the link is given below. Essentially in a planar wave there is a one to one correspondence of between points of the successive planar tangent surfaces, with spherical surfaces there is not. The correspondence is between successive spherical elemental areas which grow in size as the radius increases.
See for Huygens' Principle in general:
http://farside.ph.utexas.edu/teaching/302l/lectures/node150.html
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/huygen.html
https://www.physlink.com/education/askexperts/ae471.cfm
See my paper for a mathematical derivation:
https://www.researchgate.net/publication/316994209
or the update update
https://www.researchgate.net/publication/340085346
| {
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Can force be applied without accelerating? When I push against a wall, I am applying force on the wall and the wall applies an equal force against mine therefore the wall doesn't move and neither does my hand. But isn't acceleration required to apply force? My hand is not accelerating when I am applyin the force. Still let's assume that the muscle fibres are accelerating, but how is the wall accelerating to apply an opposite force. So are the atoms accelerating somehow?
| Acceleration happens when the net force on an object is not zero.
You can apply as much force as you like to an object and it won't accelerate if something else is applying an equal and opposite force.
| {
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Calculating mean velocity: using time or velocity? Background
I have been running tests which involve timing an object moving a certain (fixed) distance, $s$.
Each test has been repeated 3 times, and the 3 times ($t{_{1}}, t{_{2}}, t{_{3}}$) for the object to travel $s$ m are recorded.
However, when calculating average velocity across $s$ for the 3 repeats, $\overline{v}$, I have a problem.
Problem
I can see two ways to calculate $\overline{v}$:
Method 1. Calculate the average time, $\overline{t} = \frac{t_{1}+t_{2}+t_{3}}{3}$, across the trials, then divide the distance by it: $\overline{v} = \frac{s}{\overline{t}}$
$$\overline{v}_{1} = \frac{3s}{t_{1} + t_{2} + t_{3}}$$
Method 2. Calculate the velocity for each trial, $v_{n} = \frac{s}{t_{n}}$, then calculate the average of these velocities:
$$\overline{v}_{2} = \frac{v_{1}+v_{2}+v_{3}}{3}$$
While both seem to follow valid logic, the two methods are evidently algebraically different, and produce different answers (often very similar, but occasionally different enough to be concerning).
Question
What is the physical difference between the two methods?
Which method is the best to use in my circumstance?
Edit
Interestingly, both methods have been supported in answers and comments, and I think this stems from a loose definition of 'Average Velocity', both on my part and intrinsically to the phrase.
CR Drost's answer beautifully addresses this with an explanation of distance- and time-averaged velocity as well as other types.
I have accepted Mindless' answer as it succinctly explains where to use each type
| The average velocity of 3 trials IS the average velocity, you should not be getting different answers if you are doing everything correctly. Are you carrying out your multiplications and divisions to enough decimal points? Leaving out slight remainders could change the answers slightly. Test each equation with whole numbers that will yield whole numbers for simplicity, you should get the same answers.
| {
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Does disentangling A and B always imply entangling A and B with an environment? Does disentanglement of a bipartite entanglement between systems A and B entail actual breaking of this bipartite entanglement or is it rather the beginning of a tripartite entanglement between A, B, and Environment?
I've only found examples of the latter in the literature (for both asymptotic disentanglement and entanglement sudden death). In these cases, the relationship between A and B didn't really end; it just became more complex and more difficult to describe fully because experimenters usually lacked the part of the information encoded in the environment after decoherence.
Are there any examples of mechanisms that simply break the bipartite entanglement between A and B without entangling them with anything else?
| Yes, one can break entanglement without getting entangled with something else.
Consider e.g. two spins interacting via the Heisenberg interaction $\vec S_1\cdot \vec S_2$, which are initially in a state $\lvert\uparrow,\downarrow\rangle$. As time evolves, the system first gets entangled, and subsequently gets disentangled again, without any other system being involved. It is thus perfectly possible to disentangle two entangled qubits without involving any other system.
| {
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Understanding Fermi gas number density According to Wikipedia, the number density of a 3D Fermi gas is given by:
$$\dfrac{N}{V}=\dfrac{1}{3\pi^2} \left( \dfrac{2 E_F m}{\hbar^2} \right)^{3/2}$$
This means that if the mass of the particles increases, the number density will also increase: why is it so?
| If the mass of the particles increases and the Fermi energy stays the same, then the number (density) of the particles must decrease. But the energy levels of the particles depend on their masses as well.
For example, suppose we have a set of $N$ electrons in a cubical box and $N$ protons in another cubical box of the same size. The Fermi energy is defined as the energy of the highest occupied state (more or less.) The most energetic occupied state in each case will have the same values of $\vec{k}$; or to put it another way, the highest occupied wavefunction will be the same in each case. But since the energy of a particle in a box is $k^2 \hbar^2/2 m$, the "most energetic electron" will have a higher energy than the "most energetic proton". Thus, the Fermi energy of the electrons will be higher.
Conversely, if we want to make it so the Fermi energy of the proton gas was the same as the Fermi energy of the electron gas, we would have to add more protons to the box. This would cause the number density of the proton box to go up. When we got to the point where the two Fermi energies were equal to each other, we'd find that there were a lot more protons in the box, and the protons would have a higher number density.
Honestly, it's probably best to think of the Fermi energy as being a function of the density, rather than the other way around.
| {
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How to prove that a $d$-dimensional Hilbert space can only have $d^2$ equiangular vectors (i.e. that a SIC is a maximal collection of that kind)? It is an open question if every $d$-dimensional Hilbert space contains a collection of $d^2$ states, such that for every two of them the squared absolute value of the scalar product is equal to $\frac{1}{d+1}$, i.e. if a SIC-POVM exists for every dimension.
However, I don't understand why $d^2$ is the ultimate upper bound for the size of such a collection. The papers I've seen so far (for example, this review article) just mention that it can be proven, but don't go into detail.
| I found one answer in this paper. Let $S$ be the $d \times N$ matrix whose columns are the SIC vectors, and let $G = S^\dagger S$ be their Gram matrix. The rank of $G$ is equal to $d$ unless the vectors live in a proper subspace of $\mathbb{C}^d$. Now, consider the matrix $G \cdot G^T$ where the dot means the entrywise product of matrices (aka Hadamard product). This matrix has ones on the main diagonal and identical terms $|\langle \psi_i | \psi_j \rangle|^2 < 1$ in all off-diagonal elements. The rank of this matrix is $N$.
The crucial trick is that the rank of the Hadamard product is less or equal than product of ranks of the matrices:
$$
N = \operatorname{rank}(G \cdot G^T) \leq \operatorname{rank}(G) \operatorname{rank}(G^T) = d^2.
$$
| {
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Magnet position for maximum voltage in an ac generator (Lenz's Law)
Animated gif credit
The picture above shows, that the voltage of a phase is greatest, when the magnet alligns with the coil of the phase. Why is that?
To my knowledge of lenz's law the voltage and induced current should be 0 when the magnet aligns with a coil, since there is no opposing emf generated in the coil.
| The smaller is the gap between coil's core and the magnet, the greater is the flux change through the coil from the same magnet displacement. Thus the EMF is the highest at this phase (Faraday's law).
Now, if you add a load to the generator (in the most extreme case, short the winding of the coil), then the induced current in the coil will produce a counter-flux that will resist the magnet's motion. This is what Lenz' law is trying to tell you.
If, on the other hand, you leave the ends of the winding open, there will be no current, and no force resisting the magnet's movement. The apparent confusion in your question is that you must have a changing current in the winding, not just the EMF, to produce the opposing magnetic field.
You can also look at the whole thing from the conservation of energy standpoint. You have an ideal generator. If you give the rotor a rotating push, nothing is going to resist its rotation as long as the generator is unloaded. As soon as you connect a light bulb to the generator, the bulb will light up, at the same time eating up the kinetic energy of the rotor and slowing it down. Thus the Lenz' law can be qualitatively understood as a facet of the energy conservation law.
| {
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Why do all fields in a QFT transform like *irreducible* representations of some group? Emphasis is on the irreducible. I get what's special about them. But is there some principle that I'm missing, that says it can only be irreducible representations? Or is it just 'more beautiful' and usually the first thing people tried?
Whenever I'm reading about some GUT ($SU(5)$, $SO(10)$, you name it) people usually consider some irreducible rep as a candidate field. Also, the SM Lagrangian is constructed in this way. (Here, experimental evidence of course suggests it.)
| Gell-Mann's totalitarian principle provides one possible answer. If a physical system is invariant under a symmetry group $G$ then everything not forbidden by $G$-symmetry is compulsory! This means that interaction terms that treat irreducible parts of a reducible field representation differently are allowed and generically expected. This in turn means that we will instead reclassify/perceive any reducible field in terms of their irreducible constituents.
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How to find the possible measurements of a quantum system? A wave function of an infinite square well is given as
$$ \psi(x) = \frac{1}{\sqrt{5a}}\sin\left(\frac{\pi x}{a}\right) +\frac{2}{\sqrt{5a}}\sin\left(\frac{3\pi x}{a}\right),\quad x\in[0,a] $$
How do I find the possible results of the measurement of the system's energy and the corresponding probabilities?
| In order to find the energies and corresponding probabilities, it's best to rewrite the expression in terms of eigenfunctions. The eigenfunctions of the infinite square well between $x=0$ and $x=a$ are
$$ \phi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right). $$
Using these, your wave function is
$$ \psi(x) = \frac{1}{\sqrt{10}}\phi_1(x) + \sqrt{\frac{2}{5}}\phi_3(x). \tag{1}$$
The coefficients in front of the eigenfunctions are the probability amplitudes. The absolute square of those yields the probability.
So the probability of measuring $E_1$ is $p_1=1/10$ and the probability of measuring $E_3$ is $p_3=2/5$.
...but wait, they don't add up to one. Instead, $p_1+p_3=1/2$. This indicates that our wave function was not normalized. So let's do it again, but first we normalize Eq. (1):
$$ \psi(x) = \frac{1}{\sqrt{5}}\phi_1(x) + \frac{2}{\sqrt 5}\phi_3(x) $$
Now we can read off the amplitudes again as $a_1 = 1/\sqrt 5$ and $a_3=2/\sqrt 5$. Squaring them yields the probabilities: $p_1=1/5=20\%$ to get $E_1$ and $p_3=4/5=80\%$ to get $E_3$.
And you probably know that the energies are
$$ E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}. $$
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If an earthquake can destroy buildings why it cant kill us according to physics? Most earthquakes with magnitude 5.5 and higher can damage or destroy buildings. However, according to my knowledge and experience, I have never seen someone dying from an earthquake itself. Rather, they die from an associated tsunami, damaged buildings, etc.
This seems counter-intuitive, since you need much more force to destroy a building or damage it than to break the human femur or cause similar damage to other species.
| In addition to the other answers provided, resonance cannot be ignored (Resonance is where the object is vibrating at its natural frequency and as a result the vibrations are amplified). To put simply, people and buildings have different resonance frequencies and the lack of being anchored to the ground (for people) mitigates the damage done in the way.
The resonance frequencies of buildings play a large role in the destruction caused by earthquakes. Contrary to common misconceptions, the tallest buildings aren’t always the most damaged during an earthquake.
Check out this great demo for a visual representation of how resonance frequencies amplify building destruction.
(Begin at the 1:25 mark for best experience).
BOSS Demo
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Unable to understand why diffracting an unpolarised source of monochromatic light through a single slit causes the emrgent wave to coherent So in Young's Double slit experiment the source of light that passes through the double slit must be monochromatic and a coherent source (from my undertanding, this is to get a uniform pattern projected on a screen/wall/whatever, correct me if I am wrong). Young without having access to lasers used a light filter to filter light for it be of one wavelength and then exposed it to a single slit for a coherent source. From what I understand, a coherent source of light is a wave which has a constant phase/path difference, meaning that at any point on the wave there is the same phase/path difference as any other point on the same wave. However, I can't figure out why diffracting a monochromatic source of unpolarised light would cause the resultant wave to be coherent, as I am unable to see where phase/path difference are relevant.
Any help is appreciated.
Thanks!
| The single slit is in order to generate a point source in two dimensions.
A point source, by construction will allow only specific wavelengths to its size to go through and spread radially. This means that the wavefront can be described mathematically with sinusoidal functions with fixed phases. For details look at this link.
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Two-photon interference inside Mach-Zehnder interferometer Imagine there's a strong laser beam, not just an attenuated stream of single photons, entering a balanced Mach-Zehnder interferometer.
One-photon picture: Each photon interferes with itself on the second beam splitter. As a result, all photons leave out of the same output.
More realistic picture: If the intensity of the laser is strong enough, a lot of photons will enter the interferometer within the coherence time and will be therefore indistinguishable (at least by their arrival time).
How does this fact change the quantum mechanical description of the state? Shouldn’t there be a two-photon interference at the second beam splitter in this case? How do you describe the evolution of the state throughout the interferometer mathematically: two-photon interference of photons that are themselves in the same superposition of two paths?
| It depends on the quantum state that entered the interferometer. In general the sate is described by a coherent state. To see what happens quantum mechanically one can apply the unitary operator for a beam splitter on the coherent state (with a tensor product with a vacuum state, because we assume nothing enters in the other port). What one would find is a tensor product of two coherent states in the respective path inside the interferometer, each with a slightly reduced amplitude. Inside the interferometer the two path may be slightly different, leading to a relative phase shift, which is easily represented in therms of the complex parameters of the two coherent states. Finally we send them through another beam splitter. At the two output ports one now obtain two coherent states whose amplitudes are respectively the sum and difference of the two input coherent states. The sum and difference are responsible for the interference we observe at the respective output ports.
In conclusion, we do not see quantum interference. Only optical interference, even when we do the calculation in quantum mechanics.
| {
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On proving that charge is linearly proportional to potential for a conductor In Mr. Purcell's Electricity and Magnetism, page 103, it is stated,
An isolated conductor carrying a charge $Q$ has a certain potential $\phi _{0}$, with zero potential at infinity. $Q$ is proportional to $\phi _{0}$. The constant of proportionality depends only on the size and shape of the conductor. We call this factor the capacitance of that conductor and
denote it by C. $$Q=C \phi _{0}$$
I understand that for a given charge $Q_{0}$ and its corresponding potential $\phi_{0}$ we could define a $C_{0}$ as a function of the shape and size of the conductor such that $Q _{0} =C_{0}\phi_{0}$.
When we change the charge to $Q_{1}$, the potential will become $\phi_{1}$. How can we prove that it is the same constant $C_{0}$ that will link $Q_{1}$ and $\phi _{1}$ ? In other words is charge being linearly proportional to potential an experimental result or can we prove it?
If one argues that it is the same constant because it depends only on the shape and size of the conductor, then they must also prove that this constant does satisfy $$Q=C _{0} \phi$$
for every given charge and its corresponding potential.
| Here's a non-finished, non-rigorous attempt:
Let us say we have a conductor having a surface $S$ at a potential $\phi_{0}$ and having a charge $Q_{0}$.
Supposing all the charges are located on the surface, the potential can be defined at any point in space as:$$\phi (\textbf{x}) =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert {\textbf{x}-\textbf{x}'\lvert\lvert }} da'$$
$\phi (\textbf{x})$ is constant inside and on the surface of the conductor. Choosing an arbitrary $\textbf {x}_{0}$ within or on the surface of the conductor, we find:
$$\phi (\textbf{x}_{0})=\phi_{0} =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert {\textbf{x}_{0}-\textbf{x}'\lvert\lvert }} da'$$
where $da'$ is a surface element of $S$.
Since $\vert\lvert \textbf{x}_{0}-\textbf{x}'\lvert\lvert $ is a continuous function of $\textbf{x}'$, we can use the intermediate value theorem for integrals to rewrite the expression above as:$$\phi_{0} =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert {\textbf{x}_{0}-\textbf{x}'\lvert\lvert }} da'=C \int _{S} \sigma (\textbf{x}')da'=C \times Q_{0}$$
Where $C$ is a constant.
The flaw: We must prove that $C$ doesn't depend on the choice of $\textbf{x}_{0}$.
It'd be fantastic if someone could formalize this a little bit more.
| {
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Does the equivalence principle imply that light must move slower when moving away from a massive object? Thought experiment: Elevator going up at an extreme acceleration, pulse of light bouncing up, and down between mirrors on the floor, and the ceiling. Won't it take light longer to travel from the floor to the ceiling, than from the ceiling to the floor? If so,then based on the Equivalence Principle, doesn't this mean that light will move slower from floor to ceiling in an Equivalent gravitational field?
| Light is ALWAYS traveling in the same speed of $c$ in ALL reference frames. This was confirmed by various experiments such as Michelson & Morley experiment and the others.
The only thing that changes - is the light frequency,- if light looses energy somehow then it's frequency is red-shifted, but speed is the same $c$. Of course if photon is traveling in vacuum. If photon is traveling in medium with refractive index of $n > 1$, then phase velocity of light is $v < c$. The only reasonable explanation if light speed is the same everywhere - is that time flow changes and is dependent on reference frame (this was solved by Einstein). I suggest you first to read about special relativity, because Einstein has developed it first. Then study general relativity because it is much much more complex that comes after.
| {
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Is electron capture partially responsible for matter decay? Sorry for the very likely ill-posed question. I have lately started reading a lot about quantum physics and the nature of electrons and I read about a phenomenon known as "electron capture" which can happen with a finite probability when an electron "finds" itself within the nucleus volume.
If an electron capture converts a proton and an electron into a neutron plus the emission of an electron neutrino, does this mean that all matter can "decay" with time?
I hope my question is clear enough for someone more expert than me to answer it.
Thank you again for your time!
| The answer to the title question is "sometimes". Some atomic nuclei decay via electron capture, in which case it is at least partially responsible for decay.
The question asks about something else, whether all matter can decay via electron capture. This time the answer is no. Electron capture (like other forms of nuclear decay) can only happen when it's energetically favorable, and for most nuclei this isn't the case.
You can calculate the energy change for a theoretical decay reaction such as
$^{26}_{13}Al + e^{−} \rightarrow ^{26}_{12}Mg$
by summing up the binding energy of the components on the left and comparing them to the nuclei on the right. If it turns out the process needs energy, as opposed to releases it, then it won't happen naturally.
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Does metal thickness change with rise in temperature? When a metal pipe or metal cylinder is heated uniformly in furness, inner and and outer diameter of cylinder increase and its length also increase but does thickness of the cylinder change with change in temperature?
| Yes. I find the best way to think about it is that, when a metal object is heated, all of its dimensions increase as it expands (assuming its not being constrained externally somehow, has isotropic material properties, etc.). So, hole diameters increase and wall thicknesses will increase.
| {
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Factor of 3 in Photon Diffusion coefficient From definition of Diffusion coefficient:
$$D = c/3(\mu_a+\mu_s),$$
where $c$ is the speed of light front, $\mu_a$ is absorption coefficient and $\mu_s$ is scattering coefficient.
I wonder where does factor of $1/3$ comes from? I assume it is coming from dimensionality, but I didn't find arguments proving that.
UPD:
In this question I refer to diffusion approximation of Radiation Transfer Equation also known as Photon Diffusion Equation.
| The factor of three enters from the flux equation during the derivation. Not a full answer, but this was too large for a comment.
See here equation 5.
| {
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What's the problem with Euclidean geometry for astronomical phenomena? This passage from John Pierce, An Introduction to Information Theory:
"also note that while Euclidean geometry is a mathematical theory
which serves surveyors and navigators admirably in their practical
concerns, there is reason to believe that Euclidean geometry is not
quite accurate in describing astronomical phenomena"
got me wondering. What makes Euclidean geometry inaccurate for this purpose?
The book is neither about geometry, nor about astronomy, so this issue remains unexplained.
| On the scale of most astronomical phenomena, general relativity (GR) is the relevant theory.
There are many aspects in which this theory is incompatible with Euclidean geometry. An illustrative and often used analogy is that Euclidean geometry is already inaccurate for a being confined to the surface of a sphere - if you draw a triangle on a sphere, its interior angles do not in general sum up to 180°. What makes GR a bit strange still is that it posits that it is not space alone that participates in such curved geometry, but spacetime.
That is, general relativity does not assume that space(time) is flat, and it even intermingles space and time so that different observers that move relative to each other will neither agree on whether two arbitrary events are synchronous nor whether they happen at the same place.
The specific physical effects this has are too varied to discuss them here at length and have already been extensively discussed on this site, see e.g. How can time dilation be symmetric?, What is the proper way to explain the twin paradox? and many other questions in the general-relativity tag.
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Is electric potential always continuous? In electromagnetism, we say that any conservative electric field $\vec{E}(\vec{r})$ is associated to a scalar potential $V(\vec{r})$ such that $\vec{E}(\vec{r}) = -\nabla V(\vec{r})$. If the electric field is continuous, the respective electric potential must be differentiable because, if not, its gradient could not be calculated everywhere.
There are some cases, though, in which the electric field is discontinuous, leading to a non-differentiable electric potential. The latter is, however, still continuous.
Why is this? Why is it that even when the electric field is discontinuous, the electric potential is not? Must the electric potential always be continuous everywhere? A mathematical approach (i.e. not just a qualitative insight) is what I'm looking for.
| No. For example, the potential of a point charge is discontinuous at the location of the point charge, where the potential becomes infinite.
Since all charges in nature seem to be point charges (elementary particles such as electrons and quarks), electric potential always has discontinuities somewhere. When we work with continuous charge distributions, we are simply using an approximation that averages over lots of point charges and smears out the discontinuities in their charge density, potential, field, field energy density, etc.
| {
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(Why) Is there only one Lyapunov exponent? Lyapunov exponents describe how two (infinitesimally) close initial conditions behave (exponentially) in the long run. If a system is chaotic, the largest Lyapunov exponent is positive. However, as far as I understand it, if we have say, three ODEs depending on time, each state variable has one Lyapunov exponent. But how?
If we consider
$$
\lambda = \lim_{t \to \infty} \lim_{\delta \mathbf{Z}_0 \to 0}
\frac{1}{t} \ln\frac{| \delta\mathbf{Z}(t)|}{|\delta \mathbf{Z}_0|},
$$
in my understanding, this would always tend to zero as $t\to\infty$ if we consider a chaotic attractor?!
What am I getting wrong?
Thank you!
|
each state variable has one Lyapunov exponent. But how?
The equation you give is for the maximal Lyapunov exponent, while what you're referring to here is Lyapunov spectrum.
You can find how to calculate it in almost any text on chaos theory - Wikipedia as usual could be a good start for getting the overall picture. This previous answer gives some pointers and references for the calculation, and this answer tries to say something about their meaning.
this would always tend to zero as t→∞ if we consider a chaotic attractor?!
Nope. The factor $1/t$ goes linearly to zero, yes, but $\delta\mathbf{Z}(t)$ diverges exponentially for chaotic orbits, meaning that the factor $\ln | \delta\mathbf{Z}(t)|$ should diverge linearly and the whole thing converges, as also explained in this answer for iterated systems.
Also notice the limit $\delta\mathbf{Z}(0) \to 0$, which means that the values that the equations return are local quantities for a specific orbit - so the divergence is local, that's how you can have finite exponents also for spatially confined orbits.
| {
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Energy-momentum Tensor for a Real Scalar Field Lagrangian I'm currently working through Schwartz's QFT book, and I'm trying to find the energy-momentum tensor for the following Lagrangian:
$$ L = -\frac{1}2\phi(\Box+m^2)\phi. $$
Am I correct in thinking that,
$$ \frac{\partial L}{\partial(\partial_{\mu}\phi)} = 0$$
owing to the fact that there are no first derivatives of $\phi $ in the Lagrangian?
This seems to disagree with a solution set that I found for the book, hence why I'm asking here, just to be sure.
The other option I see is that
$$ \frac{\partial L}{\partial(\partial_{\mu}\phi)} = -\frac{1}{2}\phi\partial^{\nu}\delta_{\mu\nu}$$
but I'm not sure why this would be true and the other not.
| Your operator $\Box$ can be expressed via $\partial^\mu \partial_\mu$, so that you
can write the Lagranian density $\mathcal{L}$ as
$$\mathcal{L} = \frac{1}{2} \partial_\mu \Phi \partial^\mu \Phi - \frac{1}{2} m^2 \Phi^2$$
Therefore, the answer to your question is
$$\frac{\partial \mathcal{L}}{\partial(\partial_\mu \Phi)} = \partial_\mu \Phi$$
| {
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Linear momentum of a system remains conserved, but with respect to which frame of reference? I have studied that linear momentum of a system remains conserved. But i can't figure out with which reference of frame it is conserved. Is it conserved with respect to system reference frame or in a reference frame attached to one of body of a system or both?
| It is conserved in any inertial reference frame.
A frame attached to one of the bodies, or even the system reference frame, are not necessarily inertial - non-inertial frames are sometimes easier to work with than inertial frames.
| {
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Equivalence Principle holding in Special Relativity? (let alone QFT) Motivation
I am pretty confused of why people are hopeful to find a version of the equivalence principle ("the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system") within QFT. I personally am of the position QM and the equivalence principle are compatible under proper formulation (and I am aware that this is controversial grounds).
Question
Within Special Relativity the definition of acceleration is limiting (to say the least). In light of this I was curious if there existed a version of the equivalence principle within Special Relativity? Personally if someone asked me this I would have just used GR and tried to find this approximation but please do not post this answer (see motivation).
| Yes, I think you can use the equivalence principle within the realm of special relativity if, at least, the observer(s) is (are) inertial. In short, a person located at the center of a rotating disc, for a small compartment located a distance $r$ away from the center which orbits the observer, measures an anti-gravity (gravity outward the center). His measurements, using the equivalence principle, can be equal to those measured by a Schwarzschild observer for a small compartment located a distance $r$ away from a massive planet if the angular velocity of the disc as well as the mass of the planet are chosen in a way that the centrifugal acceleration in the orbiting compartment equals the gravitational acceleration in the compartment at rest with respect to the planet.
Applying the equivalence principle to special relativity, Newton's law of gravitation can be relativistically revised as a post-Newtonian theory.
P.S. In this regard, I published an article four years ago in a somehow offbeat journal. The article is accessible via this link. For a better edition, I refer you to the first chapter of my book.
| {
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Derivation of Conservation of Energy from Newton's Second Law Given Newtons's Second Law: $$ \frac {d}{dt} (m \boldsymbol{\dot r}) = \mathbf F $$
How is it possible to derive the conservation of energy equation with a constant mass?
That is how can you derive $ \mathbf F = - \nabla V(\mathbf r) $ where $V(\mathbf r)$ is shown to be the potential energy when the force is conservative?
Attempted Proof:
Let $KE = T = \frac {1}{2} m\boldsymbol{\dot r} \cdot \boldsymbol{\dot r} $
or
$$\frac{dT}{dt} = \frac{1}{2}m[\boldsymbol{\dot r} \cdot \frac {d \boldsymbol{\dot r}}{dt} + \boldsymbol{\dot r} \cdot \frac {d \boldsymbol{\dot r}}{dt}] = m \boldsymbol {\dot r} \cdot \frac {d \boldsymbol{\dot r}}{dt}$$
and $\nabla V = \frac {\partial V} {\partial{\mathbf r}} $
Also, a conservative force says $\frac{dE}{dt} = 0$
Newton's Second Law could also be written as: $$m\boldsymbol{\dot r} \cdot \frac{d\boldsymbol{\dot r}}{dt} = \mathbf F \cdot \boldsymbol{\dot r}$$
My question is how is $ \mathbf F = - \nabla V(\mathbf r) $ introduced to Newton's second law properly and then integrated (? maybe) to obtain the energy?
Because I can easily prove that $ \mathbf F = - \nabla V(\mathbf r) $ if $E = T + V(\mathbf r)$ but I am trying to conclude that $V(\mathbf r)$ is the potential energy, not assume it
| Here is another approach to the question.
We know that work done by the force, $dW = \vec{F}\cdot d\vec{r}$.
Now, if the force is conservative, by definition it means that work done is independent of the path taken and only depends on the end state. Hence $dW = \vec{F}\cdot d\vec{r}$ should be exact, i.e. writable as $V(\vec r_f) - V(\vec r_o) $.
This translates to $\vec{F}\cdot d\vec{r} $ being equal to total derivative of some function $V(\vec r)$ that depends on position only.
Hence, $$\vec{F}\cdot d\vec{r} = dV(\vec r)$$
$$\vec{F} = \frac{dV(r)}{d\vec{r}} = \nabla{V(\vec r)} $$
Now, we see that work done by the conservative force $F$ is expressed by $V(r)$, which we define as the potential energy.
| {
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Why do things cool down? What I've heard from books and other materials is that heat is nothing but the sum of the movement of molecules. So, as you all know, one common myth breaker was "Unlike in movies, you don't get frozen right away when you get thrown into space".
But the thing that bugs me is that things in the universe eventually cool down. How is that possible when there are no other things around to which the molecules can transfer their heat?
| You exchange heat with the objects around you.
If the objects around you are hotter than you, you'll heat up.
If the objects around you are cooler than you (neglecting the heat you're generating due to metabolic processes), you'll cool off.
In space, the objects around you (mostly interstellar medium) is cooler than you so you radiate more heat away from you into them than they radiate toward you.
If you were thrown out into space, but very near a star, you might receive more heat from the star more than you could radiate away into space, and you would heat up rather than cooling down.
But the thing which bugs me is that things in the Universe, eventually cool off, and how is that possible, when there's no other things around, to which the molecules transfer their heat?
There are three main heat transfer mechanisms.
conduction is transfer by direct contact between two bodies, or through a body with a temperature gradient across it.
convection is transfer by the flow of a fluid (liquid or gas).
radiation is transfer by the exchange of electromagnetic radiation.
Heat transfer by radiation doesn't require any physical contact between two bodies or any material medium surrounding a body. Radiation is the main heat transfer mechanism for a body floating in space.
| {
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What happens to net acceleration in a non-inertial reference frame with cenripetal force? In a non-inertial reference frame, the centripetal force is balanced by a centrifugal force. Therefore, shouldn't the net acceleration be zero? Yet, you still feel an acceleration outwards? Why is that?
| What you feel is the centrifugal force, not an acceleration. Acceleration is rate of change of velocity. If you are not moving (or starting to move) in the rotating frame, your acceleration is zero in that frame.
| {
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