Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Is nuclear power desireable in the long term, given the fact that it's an unnatural heat input to our planet? I've been reflecting on whether we want nuclear at all in the long term (compared to renewables like wind, solar, and hydro). There's a certain amount of heat (energy) entering our planet and leaving it. Greenhouse gases reduce the amount leaving, causing the planet to warm up. Nuclear power increases the input because it's energy that would not be released here without us. But the question to ask is what's the significance of the energy input from nuclear power. Say for example that future society becomes fully powered on fusion reactors, the energy input from these reactors would be roughly $10^{22}$ joules/year (approx 20 times 2013 world energy consumption). We can compare to the total energy input from the sun, which is $10^{25}$ joules/year. From these numbers, the input from nuclear would be about 0.1% the total solar input. Is that enough to cause disturbance to the energy balance of our planet and to worsen any global warming symptoms?
| All known other energy sources including solar and wind energy produce extra waste heat into the planet that would otherwise be lost. All human energy needs are adding extra tiny amounts of energy- nuclear is not special.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/378506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 5
} |
Interval Preserving in Minkowski Space The squared line element in any spacetime is given as $$ds^{2}=g_{ab}dx^{a}dx^{b}.$$ The use of tensors helps us to infer that the line element in some other frame would be $$ds'^{2}=g'_{ab}dx'^{a}dx'^{b}$$ where simply $dx'^{a}=\frac{\partial x'^{a}}{\partial x^{b}} dx'^{b}$.
My question is, in special relativity, there is further a condition on the line element that it should $$c^{2}(s-t)^{2}-(x_{1}-y_{1})^{2}-(x_{2}-y_{2})^{2}-(x_{3}-y_{3})^{2}=c^{2}(s'-t')^{2}-(x'_{1}-y'_{1})^{2}-(x'_{2}-y'_{2})^{2}-(x'_{3}-y'_{3})^{2}$$ which gives us the Lorentz transformations. How can we prove this condition using the postulates of special relativity?
Also where and how do we employ the condition that the frames we are transforming to are inertial?
| Here why $\eta_{ab} = \eta_{mn} \Lambda^m_a \Lambda^n_b$ in SR (special relativity).
$ds^2 = \eta_{ab} dx^a dx^b$ (1)
$ds^2 = \eta_{mn} dx^m dx^n$ (2)
but:
$dx^m = \Lambda^m_a dx^a$ Lorentz transformation
so:
$\eta_{mn} dx^m dx^n = \eta_{mn} \Lambda^m_a \Lambda^n_b dx^a dx^b$ (3)
by equating (1) and (2), taking count of (3):
$\eta_{ab}=\eta_{mn}\Lambda_a^m\Lambda_b^n$
Note:
In SR you have linear transformations as Lorentz transformation is linear in the coordinates as a consequence of the principles of special relativity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/378824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why must the wavelength of light be smaller than uncertainty in position? I have read that to measure the position of a particle to an accuracy $\Delta x$, we need light of wavelength $\lambda < \Delta x$. Is it true? Why is it so?
| This may not be the answer expected, however it is not true if you have a light source that is very narrow bandwidth and stable.
You can interpolate between fringes on an interferometer. For example, a commercial laser interferometer using a 633nm source can achieve a 1nm resolution.
Even if you simply bounce light off the object the round trip path changes 2λ for every λ you move the object.
There are additional and alternative techniques that can improve upon that considerably.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/379301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Why does it take time to melt ice when the loss of magnetization of a material at its Curie temperature is immediate? Everything is in the title: Why does it take time to melt ice when the loss of magnetization of a material at its Curie temperature is immediate?
Have you an explanation for this difference?
| In addition to Valerio answer, while liquid water has a higher energy level than ice (which you need to provide for the phase transition. Magnets, on the contrary store some potential energy (You can observe it if you break a magnet parallel with the field lines, parts will turn to make a shorter magnetic circuit, and even more if you reduce it to dust. And so, passing the Currie point will free some energy instead of absorbing it, which can happen instantaneously.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/379440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to fall into a black hole, then come back out? Everybody knows that you can only fall into a black hole and you can never get out. But on second thought, this flies in the face of time reversal invariance, which says that every process that goes forward and also go backwards. Moreover, the standard Schwarzschild black hole is invariant under time reversal because it's a static spacetime.
In Newtonian physics, if a ball falls into a deep well, it can bounce around inside for a long time. But if it ever hits anything that reverses its velocity, it'll just retrace everything it did in reverse, then come right back out of the well. Does this argument also apply for something that's fallen into a black hole?
| From the point of view of an outside observer, the object never reaches the event horizon, so it can come back. Also, it's possible for the black hole to emit Hawking radiation that coalesces into a spaceship emerging from the black hole. The probability of this happening is astronomically small. Thus, this provides a "reversibility" of a spaceship flying into a black hole that is analogous to the sense in which an ice sculpture melting is "reversible"; the reverse process is physically possible, but thermodynamically impossible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/379549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Why is static friction providing centripetal force? This could be tagged as a duplicate, but I couldn't understand why static friction causes centripetal acceleration in a car taking a circular turn which is moving. Mustn't it be kinetic /sliding friction as the object is moving?
| Regarding things that are rolling such as wheels of a car, remember one key thing: kinetic friction is not about moving, but about sliding.
Even though a wheel is moving, it isn't sliding over the surface. There is no kinetic friction going on. Only static friction which holds the contact point still while it is in contact.
Since there is no kinetic friction happening when the car is driving in a circle, only static friction is left to cause the centripetal acceleration.
Now, as @JohnForkosh mentions in a comment, another way to answer your question is that the driving direction and the centripetal (radial) direction are perpendicular and thus completely seperate.
There can easily be sliding (kinetic friction) in one direction but stationarity (static friction) in the other. And this is the case here. Even if the car was sliding in the driving direction, it is still not sliding in the radial direction (it is not moving further away from the circle centre).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/379678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What does Mobius group/transformations have to do with special relativity? The group of Mobius transformations, denoted by ${\rm Mob}(2,\mathbb{C})$, is isomorphic to ${\rm SL}(2,\mathbb{C}))/\mathbb{Z}_2$ which in turn is isomorphic to the Lorentz group ${\rm SO}^+(3,1)$.
This connection, to me, seems very intriguing. After all, Mobius transformation is the most general, one-to-one, conformal map of the Riemann sphere to itself, given by $$w=f(z)=\frac{az+b}{cz+d}\tag{1}$$ where $a,b,c,d$ are arbitrary complex constants satisfying $(ad-bc)=1$. Apparently, (1) has nothing to do with spacetime transformations.
But the aforementioned isomorphism makes me curious whether there is any deep physical consequence(s) related to this isomorphism.
| *
*A future-directed light-ray $$n^{\mu}~=~(1,{\bf p}/|{\bf p}|), \qquad {\bf p}~\in~\mathbb{R}^3\backslash\{\bf 0\}, \tag{1}$$ can be identified with a non-zero future-directed light-like 4-vector $$p^{\mu}~=~(E,{\bf p})\tag{2}$$ if we mod out with the energy $E\equiv |{\bf p}|>0$.
*Therefore the set of future-directed light-rays (through a fiducial point) can be identified with the Riemann sphere $$\mathbb{C}P^1~\cong~S^2~\cong~(\mathbb{R}^3\backslash\{\bf 0\})/\mathbb{R}_+.\tag{3}$$
*The restricted Lorentz group $SO^+(3,1)$ acts transitively on the set of future-directed light-rays, cf. e.g. my Phys.SE answer here. Hence they can be identified with Moebius transformations.
*For a proof of $SO^+(3,1)\cong SL(2,\mathbb{C})/\mathbb{Z}_2$, see e.g. this Phys.SE post.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/379945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Electric potential just outside a spherical shell We know that as we get closer and closer to a point charge, the electric potential approaches infinity.
Since electric potential at the surface of a spherical shell is finite (Gauss law) , so on moving away from the surface it would fall. In other words, it would be finite as well.
But if I think this way: Take an infinitesimal charge element on the surface so that we can take it as a point charge, then potential due to this charge would be similar due to a point charge. So as we approach this element, shouldn't potential go to infinity.
What's wrong with this reasoning? Please help.
| The problem here is symmetry. Use Gauss's law on an infinite sheet of charge. You get a finite (and constant field):
$$ E = \frac{\sigma}{2\epsilon_0} $$.
($\sigma$ is the charge density). It does not matter how close you get, or how far you get: the problem is scale independent, because at any point, you always see the same thing: and infinite sheet of charge occupying $2\pi$ str. Even if your get infinitely close, it's the same--and all the charge matters, even the infinite amount of charge infinitely far away. You can not ignore it.
Now consider your situation: infinitely close to a shell of charge. The field is now:
$$ E=\frac{\sigma}{\epsilon_0}$$
It doubled because there is no field inside the shell of charge, so the inside flux has to come out (doubling the outside flux). Other than that nuance, it looks like the infinite sheet of charge as you get infinitesimally close, and the field does not diverge.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/380080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Does in theory spacetime holds all the data of everything? A Question that is bothering me lately.
In my mind spacetime has been in contact with all existing particles and continues to be in contact.
If mass bends spacetime then in quantum scale no matter how small it is there is got to be bend in spacetime and those bends are like piece of information no?
| There is active research on a correspondence between spacetime and information, but it goes the other way round, with the idea that information may give rise to spacetime via entanglement.
The catch words are "It from Qubit"; see an introduction in this Scientific American article. It is yet another venue from the AdS/CFT correspondence which has been explored for many years now as a pathway towards quantum gravity.
For a critical point of view, also see this blog post.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/380251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Heat produced when dielectric inserted in a capacitor When a capacitor is connected to battery, it stores $\frac{C V^2}{2}$, while battery supplied $CV^2$ energy. Therefore, $\frac{C V^2}{2}$ energy gets lost as heat. When a capacitor is already charged and a dielectric is inserted in this charged capacitor (which is still connected to the battery), will there be any heat produced ?
| Yes, because when you add the dielectric more charge will flow onto the plates of the capacitor.
The energy on the capacitor is initially:
$$E_{c,i} = \frac{1}{2}CV^2 $$
After you add the dielectric:
$$E_{c,f} = \frac{1}{2}C'V^2 $$
So the change in energy is:
$$\Delta E_{c} = \frac{1}{2}V^2 (C'-C)$$
However, for the battery,
$$E_{B,i} = CV^2 $$
$$E_{B,f} = C'V^2 $$
$$\Delta E_{B} = V^2 (C'-C)$$
Finally,
$$\Delta E_B - \Delta E_c = \frac{1}{2}V^2 (C'-C)$$
which is the amount of heat produced.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/380558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can we explain the linear relation between the temperature scales? In the proof of the relation F=(9/5)C+32 (degrees celsius to fahrenheit) we assume that there is a straight-line (linear) relation between the two temperature scales: F=mC+b. Then we need two "points" to find the equation for that line.
How can we explain to a student the very form (i.e. linear) of the relation among these temperature scales (or between kelvin and degrees celcius, rankine and degrees fahrenheit as well as kelvin and rankine)?
| If we demand that our temperature scale respects the zero'th law of thermodynamics and it is an integrating factor to the heat vectorfield, then our temperature scale is unique up to linear rescaling (affine rescaling not even allowed): the elaboration required for such a statement is adequately contained in my answer to this question, so I will not repeat it here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/380892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
Gravity in vector We know that gravity is a force. But what is it's direction? Can it be expressed by vector and how can we do that? This question can also be asked for Coulomb's Law.
| Imagine you have only two masses in the universe $M$ and $m$, then the gravitational force that $m$ feels due to $M$ is indeed a vector that points towards $M$. This is called a central force, and as you point out Coulomb's force also behaves like that.
If you add another mass $M'$ into the picture the problem becomes more complex, in the sense that $m$ will feel now two forces: one pointing towards $M$ (${\bf F}$) and the other one pointing to $M'$ (${\bf F}'$). The resulting force ${\bf F} + {\bf F}'$ is also a vector but not necessarily point in any particular direction
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/381533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
L-band for search radars? I'm making my way through the Linesman-related section's of Gough's Watching the Skies. Linesman was developed to counter the carcinotron jammer.
The main solution was the Type 85, which had 12 frequencies switching each pulse at random.
However, they also deployed a second radar, the Type 84. This was a normal single-frequency pulse radar, but in the L-band rather than the S-band. This was to be used as an early-warning system.
I'm trying to understand why... for one, as a single-frequency system it would be highly vulnerable to jamming, and for another, the longer wavelength would mean lower resolution or much larger antennas.
Gough mentions a single possible reason for this, in passing, that L-band was less susceptible to clutter. I'm not sure why this would be.
Does anyone have any ideas on why the L-band would be better for early warning?
| If the radar’s mission is to search a given solid angle in a given period of time, then there is no real advantage (other than accuracy) to using a higher frequency, because a narrower beam will not be able to dwell as long on any given position. If the design features a phased array antenna, it is less costly to build it from lower-frequency modules, since the number of modules required per unit area scales as the square of frequency. L-band technology offers a nice compromise between cost and accuracy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/381791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
soft $Z_2$ symmetry breaking in the context of the Two-Higgs doublets model Imposing additional symmetries to the Higgs doublets in the Two-Higgs doublets model, like the $Z_2$ symmetry, aims to reduce the number of parameters in the potential. But it is said that the mass term $m^2_{12}$ term breaks it "softly" what is the meaning of softly in this particular case ?
| Actually, imposing $Z_2$ on the two Higgs doublet model (2HDM) is usually a strategy to prevent flavor changing neutral currents, which are very restricted by experiment. Alternatively, it is also a way of impose CP invariance in the model.
With this in mind, $Z_2$ symmetry removes the lagrangian parameters $\lambda_6$, $\lambda_7$ and $m_{12}$ (as they are usually denominated). This is an exact symmetry on the theory. When considering $m_{12}\neq 0$, we are still retaining CP conservation on the theory, but we "softly" break $Z_2$ as to keep the desired physical properties of the model but not oversimplifying it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/382234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Harnessing permanent magnetism? Putting aside any energy generating schemes that would break the laws of thermodynamics, is it possible or is there a motor which generates power using a permanent magnet? So that the energy wouldn’t be coming from nothing but from the atoms in the magnet being misaligned.
| The idea of a "magnetic motor" pops up on the interwebs almost weekly, accompanied often with youtube clips of the device "actually running", conspiracy theories about the idea being suppressed by government agents, and so forth. Too bad it is impossible.
The best disproof I have read about the fundamental idea of building a motor out of permanent magnets with no electrical power input runs like this- and please, I invite the experts here to check the reasoning to make sure I am remembering it correctly:
The field created by any magnet (except a monopole, which I will omit from the discussion) exhibits a property called "divergence". The divergence of the magnetic field happens to be zero, which is another way of expressing the fact that every field line emerging from one pole of the magnet loops around and ends on the other pole of the magnet. The argument goes that if the divergence of the field is zero, then there is no way ever for a magnet to be cleverly smuggled into the proximity of another magnet in such a manner that the two magnets will magically ignore one another- until at some point they wake up and suddenly repel each other, and in so doing insinuate yet another magnet into position and repeat the process, etc., etc. and thereby create a motor that performs work with no power input.
This fact does not stop people from tinkering and dinkering with fistfuls of artfully contrived links, levers, cams and gearwheels with precisely-machined magnets stuck in strategic locations throughout, in the mistaken belief that a "free-energy" or "over-unity" perpetual motion machine is just one experiment away from reality.
It also does not stop fraudsters from duping venture capitalists and even governments into believing that their free energy motor is real and worth spending millions of dollars developing. This took place in Europe not too long ago, with the whole shebang going down to wreckage and ruin, winning prison sentences for the perps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/382341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Original 1925 paper by Einstein on Bose-Einstein Condensation? Does anyone know if it is possible to retrieve the original 1925 paper by Einstein on Bose-Einstein Condensation? Possibly a translation into english, but german would be fine if no translation is available. I have managed to find a translation of Quantentheorie des einatomigen idealen Gases (Quantum Theory of a Monoatomic Ideal Gas) from 1924 where it is referenced as Sitzungsberichte der Preussischen Akademie der Wissenschaften, Berlin, Physikalisch-mathematische Klasse, 1925, p. 3–14, but couldn't find it anywhere.
| I found the following translations :
*
*Quantum Theory of the Ideal Monoatomic Gas (1925).
*The first paper (1924) can be found in translation here Quantum Theory of a Monoatomic Ideal Gas.
As I don't understand German I can't speak to the accuracy of the translations, but they do seem to have been done by someone who understands physics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/382713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the equivalent of piezoelectricity for optics/photonics? Piezoelectricity can be considered as the electromechanical transduction behaviour, where a material can convert mechanical deformation/energy directly to voltage difference / electricity or vice versa. Are there any materials which have the same behaviour but for optics/photonics? For example when they are under pressure emit photons, or deform when they absorb photons. What would this type of materials should be called?
| Photons don't interact the same way with matter as electrons in a crystal lattice. It's the rearrangement of electrons that causes the material to deform. Photon's on the other hand are packets of energy that are said to be mass-less while they travel through space and are converted or release their energy when they collide with matter. see Crookes radiometer. The movement of the vanes only occurs while there is some gas present (partial vacuum). If photons had mass that was causing the movement of the vanes then they (the vanes) would continue to move right through to a deep vacuum. To date I know of no material that emits light under pressure.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/382967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Milky Way: where does it end? The milky way galaxy, as is typical for spirals (and others?), has a relatively flat rotation curve except very near the center:
Eventually, at great enough distances, almost the entire mass of the galaxy will be enclosed and the rotation curve must become nearly Keplerian and fall off (dark matter, if it is important at all at these scales, would make the curve fall off even faster). Do we have any data that shows the curve becoming nearly Keplerian at great distances? Do we even have an idea of what that distance would be?
| Galaxies generally have quite flat rotation curves even at very large distances (much farther away from all the stars forming the galaxy).
At such distances most if the ordinary matter of the galaxy is just a heated gas. The rotation velocity of this gas is measured by observing the Doppler shift of some particular splectral lines.
Astronomers explain these rotation curves by means of a dark matter halo which surrounds the galaxies at large distances.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/383161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Force without magnetic or electric field In a long primary solenoid,if we introduce a secondary one on its outer surface,due to mutual inductance-an emf is generated in the secondary one according to Lenz's law if there is a time varying current in the primary coil.
But,
*
*1.we know that for an infinite solenoid,there is no magnetic field outside it-hence there is no magnetic field in the secondary coil placed over it.All field is inside the primary coil.
*2.There is no electric field as the primary coil is electrically neutral overall-irrespective of current flowing through it.
*3.The electrons in the secondary coil experiences a force which causes the current.
Considering all these together,we can conclude that the electron in secondary coil-which is in a region of no electric or magnetic field still experiences a force.How is it possible?
| According to Faraday's Induction Law, the integral form of the Maxwell-Faraday equation $$ \oint_{C} {\bf E} \cdot d{\bf l} = -\frac{d}{dt} \int_S {\bf B} \cdot d {\bf S}$$ (valid for stationary surface/boundary), you don't need a magnetic field at the location of the outer coil for an electric field to be induced there according to the time derivative of the magnetic flux. Each winding of the outer coil, the integration path on the LHS of eq. (1), encloses an area, the core of the inner coil, where there is a magnetic flux that changes with time. No magnetic field needs to be at the boundary of the enclosed area $S$ of the integral on the RHS of eq. (1). Thus, for an electric field induced in the outer coil, there needs not be any magnetic field produced by the inner coil at the location of the outer coil, as is the case in the described situation of a perfect, infinite inner coil.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/383269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In what sense are quasiholes and quasiparticles "excitations" in Fractional Quantum Hall (FQH) systems? In the theory of Fractional Quantum Hall states, one often sees quasi-holes and quasi-electrons (or quasi-particles) being called "excitations" from the ground state initially given by Laughlin (Jastrow-Laughlin style wavefunctions). Usually when one speaks of excitations, it is to a higher energy level. However from the wavefunctions of the quasi-holes and quasi-particles, it seems to me that we are still in the Lowest Landau Level. My question is: in what sense are these "excitations"?
| The full microscopic Hamiltonian of the Hall effect
$$ H = \sum_j \frac{1}{2m} \left [\frac{\hbar}{i} \nabla_j + \frac{e}{c} A(\mathbf{r}_j) \right ]^2 + \sum_{i<j} \frac{e^2}{|\mathbf{r}_i - \mathbf{r}_j|}$$
In the quantum Hall effect, the dynamics is restricted to the lowest Landau level due to the application of very high magnetic fields.
Restricted to the Lowest Landau level the first part of the Hamiltonian is a multiple of the zero point energy, thus an unimportant constant. Thus the Hamiltonian of the restricted dynamics becomes:
$$H_{LL} = P_{LL} \sum_{i<j} \frac{e^2}{|\mathbf{r}_i - \mathbf{r}_j|} P_{LL}$$
Where $P_{LL}$ is the lowest Landau projector. It is of course very hard to write a closed analytic expression of the projected Hamiltonian.
The quasiparticle states are very good approximations to the excited eigenstates of this interaction Hamiltonian.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/383450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why Hubbard model cannot be solvable exactly? The Hubbard model is a model to describe electrons in a lattice. In general, the Hubbard model Hamiltonian $H$ contains two terms:
*
*The kinetic term:$$T=-t\sum_{\langle ij\rangle\sigma} [c_{i\sigma}^\dagger c_{j\sigma} + h.c.] $$
*The onsite Coulomb interaction term:
$$U=u\sum_{i=1}^N n_{i\uparrow}n_{i\downarrow}$$
So my question is: why can we not diagonalize the Hamiltonian: $H=T+U$? Some books attribute the reason that $T$ doesn't commute with $U$, therefore we need to formulate a perturbation theory. In particular, I want to know whether the space of solution of $T$ has the same dimension compared to the space of solution of $U$ due to $[T,U] \neq 0$?
Edit:
For $T$ operator we have the following eigenequation:
$$T|n\rangle=T_n|n\rangle$$
For $U$ operator we have another one eigenequation:
$$U|\alpha\rangle=U_\alpha|\alpha\rangle$$
Due to $[T,U] \neq 0$, I am wandering whether $|n\rangle$ has the same dimension compared to $|\alpha\rangle$? And why
$$[T+U]|N\rangle \overset{?}{=} H_N|N\rangle.$$
| As for the question of dimensionality: Yes, the two sets of solutions for $T$ and $H$ have the same dimensions (Because they act on the same Hilbert space).
The pedestrian reason why you cannot solve the Hubbard model analytically is that the $U$ term contains quartic interactions ($c^\dagger c^\dagger c c$), and it is only in special cases that these can be diagonalized exactly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/383854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Moment of Inertia of an Equilateral Triangular Plate I was reading about moment of inertia on Wikipedia and thought it was weird that it had common values for shapes like tetrehedron and cuboids but not triangular prisms or triangular plates, so I tried working it out myself. I will post my attempt below, but for some reason I cannot find any source online that confirms or denies my solution. Please let me know if you find anything wrong with it. Thanks.
Q: What is the moment of inertia of an equilateral triangular plate of uniform density $\rho$, mass $M$, side length $L$, rotating about an axis perpendicular to the triangle's plane and passing through its center?
*
*First I modeled an equilateral triangle using three lines with its center of geometry at the origin as follows: $x=\frac{1}{\sqrt{3}}y-\frac{1}{3}L \\
x=\frac{1}{3}L-\frac{1}{\sqrt{3}}y \\
y=-\frac{\sqrt{3}}{6}L$
I used the fact that the circumradius of an equilateral triangle is $\frac{\sqrt3}{3}L$ and that its height is $\frac{\sqrt{3}}{2}L$ .
*Next, using the definition of moment of inertia ($I$) and with the help of Wolfram Alpha, I obtained the following result:
$$I=\int r^2 dm=\rho \int r^2 dA\\
=\rho \int_{-\frac{\sqrt{3}}{6}L}^{\frac{\sqrt{3}}{3}L} \int_{\frac{1}{\sqrt{3}}y-\frac{1}{3}L}^{\frac{1}{3}L-\frac{1}{\sqrt{3}}y} x^2+y^2 dxdy\\
=\frac{\rho}{16 \sqrt{3}}L^4=(\frac{4M}{\sqrt{3} L^2})(\frac{L^4}{16\sqrt{3}})\\
=\frac{1}{12}ML^2$$
| Another solution is to integrate the triangle from an apex to the base using the $\iint{r^2 \text{d}m}$, which becomes $(x^2+y^2)\text{d}x\text{d}y$. Using the limits of $x$ to be $0$ to $h$, and the limits of y to be $-x\tan30°$ and $+x\tan30°$, you get the moment of inertia about an apex to be $0.32075h^4M/AL$, where $h$ is the height of the triangle and $L$ is the area. Using the parallel axis theorem, you can find the moment of inertia about the center by subtracting $Mr^2$, where r is $(2/3)h$. The final result is $ML^2/12$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/384336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Is there a type of magnet that will only attract one type of metal? Is there a type of magnet that will only attract one type of metal. For example if I were to me in a room full of small metal objects and I turned on a 'special' magnet, would it be possible that only one of those metals in that room react?
| All magnets producing the same magnetic field $B$ have the same effect on other objects. Under normal circumstances you will only be able to sort out (attract) ferromagnetic objects with a magnet. Some of these ferromagnetic objects (materials) might be more strongly magnetized and thus attracted than others. With ea really strong magnet you could also attract paramagnetic objects and repel diamagnetic objects. But there is, in general, no way to attract only a specific single metal.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/384707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there evidence that a = dv/dt and a = F/m are always equivalent? If the rate of change in velocity in a particle (of mass m) caused due to a force F is dv/dt, then
F = m dv/dt
It may be argued that this is how we define force. But my question is:
Can there be any kind of force, which is so strange that no matter how we write the formula for the force, we will find that F = m dv/dt fails in at least some cases?
| If $F = \frac{d(mv)}{dt}$ fails in just one example, then you have disproven Newton's 2nd law by falsification. Then you will get the next Nobel price.
The thing is that Newton's 2nd law is not provable. But after infinitely many experiments, no one has ever been able to disprove it with normal size scales and speeds. We therefore trust it to be always true. We can't prove that a pen will fall when we let go - but we are still pretty sure it will, because it always does.
Newton's 2nd law is therefore called a law of nature, and it is one of the most established ones in physics, I would say.
So the answer to your question is without a doubt, although I can't prove it: No.
As the other answers mention, though, note that the mass $m$ is included in the derivative in the complete description of Newton's 2nd law, since it was originally formulated by I. Newton as the change in momentum:
$$F=\frac{dp}{dt}=\frac{d(mv)}{dt}$$
So, we are talking about a change in momentum and not only a change in velocity. And this is what my explanation above coves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/385010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Friction during pure rotation I have already read
*
*Role of friction during pure rolling
*What is the direction of the friction force on a rolling ball?
*Rolling resistance and static friction
*and many others
But I still have the following problem:
*
*On one hand I've been told that if pure rotation takes place then the object goes on forever. This must imply no frictional force in any form acting on the body.
*On the second hand I've been told that during pure rotation(no external forces like wind) on a plane surface friction acts opposite to velocity of center of mass (accompanied by change of normal) and slows it down.
Both seem logical enough:
The point of contact doesn't slip therefore no friction is required; So no retarding force and it must go on.
Things don't go on forever; Friction must be present to slow it down.
So, which hand is correct? Or is there a third hand altogether?
| *
*Pure wheel rotation in outer space: will rotate forever.
*Pure rotation in outer space plus CM linear velocity: will rotate forever and CM will move forever.
*Same as (1) on a frictionless table: rotate forever.
*Same as (2) on a frictionless table: Rotate forever and move forever.
*Behavior on a table with friction depends on initial conditions and friction coefficients. For example, drop a rotating wheel on the table, friction will accelerate the CM and de-accelerate rotation. But push a non-rotating wheel on the table, friction will de-accelerate CM and accelerate rotation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/385115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What would happen if the cathode ray experiment by jj Thompson was done in complete vacuum?
Cathode tube has complete vacuum and potential difference of 10,000. Volts
There is a hole in anode for electrons to pass.
| I can't find the details of the equipment J. J. Thomson used, but I believe it did not use a heated electrode i.e. it did not use thermionic emission as the source of the electrons.
In that case the electrons were generated by an avalanche process in the gas in the tube. Any stray electrons, for example produced by cosmic rays, were accelerated and collided with gas molecules ionising them and creating more free electrons. These electrons in turn collided with more gas molecules and so on, and the result is a lot of free electrons to form the electron beam.
If the tube were completely evacuated there would be no free electrons and hence no beam. I would guess that J. J. Thomson had to experiment to find the best gas pressure for the avalanche process to create a beam.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/385280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Software for simulating gas of hard particles (i.e. polyhedra?) I was wondering if there existed a kind of software I could use to simulate a gas of polyhedra, such as tetrahedra. They would interact through entropic interactions only, i.e. excluded volume.
I'm not an expert in simulation so any kind of suggestion is welcome. This is just for recreational purposes.
| LAMMPS is a good option, as suggested by lemon.
The region command can be combined with the create_atoms command in order to fill a region of space of defined shape with atoms, with the crystal arrangement that you prefer. You have to
The region command allows you to define the following shapes:
*
*cube
*cone
*cylinder
*prism
*sphere
*plane
Plus the union/intersection of different shapes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/385433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof of constructing action-angle coordinates on Hamiltonian system By Liouville-Arnold Theorem, we know we can construct action-angle coordinates such that the Hamiltonian system, when described in these coordinates, will have a form that is integrable by quadratures. I am looking at a proof of the construction of these coordinates, and I am not certain of a certain part. On page 180 of Mathematical Aspects of Celestial Mechanics it says that the Poisson bracket $\{F_i,\varphi_j\}$ is constant on $M_f$, the Lagrangian tori. I tried to prove it but I am not sure of the proof. Here is my attempt:
Choose Darboux coordinates $\{\bf{p},\bf{q}\}$ and by the coordinate representation of the Poisson bracket, we have
$$\{F_i, \varphi_j\} = \sum\limits_{k=1}^n \dfrac{\partial F_i}{\partial p_k} \dfrac{\partial \varphi_j}{\partial q_k} - \dfrac{\partial F_i}{\partial q_k} \dfrac{\partial \varphi_j}{\partial p_k}
= \omega_j\sum\limits_{k=1}^n \dfrac{\partial F_i}{\partial p_k} \bigg(\dfrac{\partial F_1}{\partial p_k}\bigg)^{-1} - \dfrac{\partial F_i}{\partial q_k} \bigg(\dfrac{\partial F_1}{\partial q_k}\bigg)^{-1}
= 2\omega_j, $$
where the second equality is because of the Hamiltonian equations (with $H = F_1$) and the fact that $\dfrac{\partial \varphi_j}{\partial t} = \omega_j$ where the $\varphi_j$ are the angle coordinates that are described by linear flow on $M_f$, and the third equality is because of $\dfrac{\partial F_i}{\partial F_1} = \delta_{i1}$. However, I am not so sure of the third equality because it just feels odd to differentiate the $F_i$'s by the $F_1$'s.
Any form of help will be appreciated as I am doing my thesis now and struggling :/
| It seems to me that they are making the identification $F_i = I_i = y_i$ where the $y_i$ are the ones discussed in Theorem 5.3 (page 174) and, as they mention, they assume the hypotheses of Theorem 5.3 to hold. As a consequence, they say that Hamilton's equations with respect to all $F_i$ have the form $\dot{\phi}_m=\left\{\phi_m,F_i\right\}= \mathrm{constant}$ as well as $\dot{y}_m=\left\{y_m,F_i\right\}= \mathrm{constant}$. In particular, with the choice $y_s=F_s$, we have by hypothesis $0=\left\{F_m,F_i\right\}=\dot{F}_m$. The $F_m$ are then independent constants pariwise in involution. Hence we have $\left\{\phi_m,F_i\right\}= \mathrm{constant}(F_k) \,$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/386062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Energy in simple harmonic motion ─ where is the kinetic energy stored, and where is the potential energy? When a mass connected to a spring is in simple harmonic motion and somewhere between the mean and extreme positions the mass is cut from spring. Then instantaneously after cutting the mass will only have its kinetic energy right? (Or it will have total energy kinetic+potential?)
Or I mean to say that in a system in simple harmonic motion, the kinetic energy is stored in the mass while the potential energy is always stored in spring. Am I correct?
| You are quite correct. Elastic potential energy is always stored in the spring. Then why does the block move when compressed though it does not have any energy?
Let us consider a situation in which a spring is compressed by a (non attached) block. When compressed the spring will acquire elastic potential energy given by
$$ U_i=\frac {1}{2}kx^2 $$
Any system acquiring any energy will always configure to a position such that it has minimal energy. So the spring will move towards its equilibrium position. In this process the block also moves as it is in immediate contact with the spring. Thus gaining kinetic energy.
The block reaches maximum speed when the spring reaches its equilibrium length - that's the point where all the energy stored in the spring is converted to kinetic energy. And the law of conservation of energy is also consequently followed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/386298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why do charged particles deflect one way but not the other in a magnetic field? I am well aware that a charged particle moving in a magnetic field will experience a force perpendicular to that magnetic field. But why is it that positive and negative particles experience a force in opposite directions?
What exactly determines the direction that a given charge will experience a force? I.e. why does a negative particle experience a force in one direction and not the other?
| Electromagnetism is symmetric with respect to parity. That symmetry is broken by the convention we choose to use for defining the magnetic field vector. Aliens on another planet could define magnetic fields to point in the opposite direction compared to our definition. They would then use a left-handed rule $\textbf{F}=-q\textbf{v}\times\textbf{B}$ rather than our right-handed $\textbf{F}=q\textbf{v}\times\textbf{B}$. If you get in radio contact with these aliens and try to get them to tell you whether their definitions are the same as ours or opposite, you can't tell without some external reference point that tells them which hand you consider right.
What exactly determines the direction that a given charge will experience a force? I.e. why does a negative particle experience a force in one direction and not the other?
why is it that positive and negative particles experience a force in opposite directions?
You can express the rules in ways that don't refer to the magnetic field or its arbitrarily defined flippable direction. For example, parallel current-carrying wires attract each other if the currents are in the same direction. Such rules are independent of which charges you define as positive and which way you define the magnetic field.
When expressed in these ways that avoid the arbitrary conventions, these rules follow from special relativity. The classic presentation at the freshman physics level is in the textbook by Purcell.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/386391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Is this a black hole? I came across this metric definition:
$ds^2=-\left(1-\frac{r_s}{r}\right)^2dt^2 + \left(1-\frac{r_s}{r}\right)^{-2}dr^2+r^2d\Omega^2$
I was trying to figure out if it describes a black hole, but I was unable to find the proper definition.
My hints so far:
*
*The time coordinate does not change its sign when r crosses $r_s$
*There are two singularities, one at $r=r_s$ which I guess arises from coordinate singularity and another at $r=0$, which looks like the center of a black hole?
What do you think?
| We need to be careful about terminology because the phrase black hole tends to be used to mean a vacuum solution. The only two vacuum solutions are the Schwarzschild and Kerr metrics, so in everyday use black hole means either the Schwarzschild or Kerr metric.
But if we relax the requirement for the geometry to be a vacuum solution then we can get other forms of black hole. For example if we add a cosmological constant we get the de Sitter-Schwarzschild solution and if we add null dust we get the Vaidya metric. Your geometry is also a non-vacuum geometry.
In principle we can tell it's not a vacuum solution by calculating the Einstein tensor from your metric, because this gives us the stress-energy tensor and that will turn out not to be zero. However the GR Mathematica notebook I have does not include a routine for calculating the Einstein tensor and calculating it by hand would be something of an ordeal. I did calculate the Ricci tensor and it seems fairly obvious from a glance that this is going to give a non-zero Einstein tensor.
But I would guess your question is really whether this geometry has a horizon. In general finding horizons is a surprisingly hard thing to do because we can only find them by considering the entire spacetime. However in a simple, static, geometry like this one we can simply look at the coordinate speed of light and see if it's zero. We do this by noting that for a light ray $ds = 0$. If we choose a radial trajectory, so $d\Omega=0$ and make this substitution we get:
$$ 0=-\left(1-\frac{r_s}{r}\right)^2dt^2 + \left(1-\frac{r_s}{r}\right)^{-2}dr^2 $$
Giving us:
$$ \frac{dr}{dt} = \pm\left(1-\frac{r_s}{r}\right)^2 $$
And since this is zero at $r=r_s$ this is indeed an event horizon.
Courtesy of AccidentalFourierTransform we can calculate the Ricci and Kretschmann scalars. The Ricci scalar turns out to be zero everywhere, which isn't much help, but the Kretschmann scalar is:
$$ K = \frac{8r_s^2 ( 6 r^2 - 12 r r_s + 7 r_s^2)}{r^8} $$
This remains finite at $r=r_s$, but goes to infinity as $r \to 0$. So the former is a coordinate singularity while the latter is a true curvature singularity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/386551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Thermal energy generated by collision observed from two different frames of reference An isolated system is composed of two bodies $A$ and $B$, with masses $m_A$ and $m_B$, $m_A \ne m_B$, which are in route of collision. The relative velocity between them is $v$. The collision is inelastic and I want to calculate how much thermal energy is generated at the collision.
I understand that the thermal energy $\Delta T$ generated equals the variation of the kinetic energy, so that the total energy is conserved. I expected $\Delta T$ to be independent of the reference frame.
However, if I consider a reference frame $S_A$ fixed in body $A$, I get the following change in kinetic energy of the system:
$$ \Delta T_A = \left( \frac{m_A 0^2}{2} + \frac{m_B v^2}{2} \right) - \left( \frac{m_A 0^2}{2} + \frac{m_B 0^2}{2} \right) = \frac{m_B v^2}{2}. $$
And if I consider a reference frame $S_B$ fixed in body $B$, I get a different change in the kinetic energy of the system:
$$ \Delta T_B = \left( \frac{m_A v^2}{2} + \frac{m_B 0^2}{2} \right) - \left( \frac{m_A 0^2}{2} + \frac{m_B 0^2}{2} \right) = \frac{m_A v^2}{2}. $$
Could you please point out the flaw in this reasoning?
| The usual laws of kinematics are valid in inertial frames of reference only. In particular, inertial frames of reference move with constant velocity. So you can't apply the laws of physics as you've learned them if your reference frame is actually fixed to a body that undergoes acceleration. You can have the reference frame start out with the same velocity as one of your bodies before the collision; you can have it end up with the same velocity as one of your bodies after the collision; but you can't have it both ways.
So a correct analysis of your problem could use a frame $F_A$ that starts out with the same velocity as body $A$, or a frame $F_B$ that starts out with the same velocity as body $B$... or literally any inertial frame, though it's probably easiest in either of those two. Using conservation of momentum, and arbitrarily picking the sign of $v$, we have
\begin{align}
F_A: \qquad -m_B\, v &= (m_A+m_B)v_{f, A} &\qquad \Delta T_A &= \frac{m_B v^2}{2} - \frac{(m_A+m_B)v_{f,A}^2}{2} \\
F_B: \qquad \hphantom{-}m_A\, v &= (m_A+m_B)v_{f, B} &\qquad \Delta T_B &= \frac{m_A v^2}{2} - \frac{(m_A+m_B)v_{f,B}^2}{2}.
\end{align}
Solve either of the equations on the left for either $v_f$, plug that into the corresponding equation on the right, and you get
\begin{equation}
\Delta T_A = \Delta T_B = \frac{m_A m_B}{m_A+m_B} \frac{v^2}{2}.
\end{equation}
It's the same for either frame.
Another good frame is the center-of-mass frame, $F_{AB}$, in which the velocity of the final combined object is $0$. Momentum conservation is expressed by
\begin{equation}
m_A v_A + m_B v_B = 0,
\end{equation}
and by the definition of the relative velocity $v$ we have
\begin{equation}
v_A - v_B = v.
\end{equation}
Various rearrangements of these equations let us write
\begin{align}
v_A &= \frac{m_B}{m_A+m_B} v, \\
v_B &= -\frac{m_A}{m_A+m_B} v.
\end{align}
So we get
\begin{equation}
\Delta T_{AB}
=
\left( \frac{m_A v_A^2}{2} + \frac{m_B v_B^2}{2} \right) - \left( \frac{m_A 0^2}{2} + \frac{m_B 0^2}{2} \right)
=
\frac{m_A m_B}{m_A+m_B} \frac{v^2}{2}.
\end{equation}
Again, it's the same answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/386652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Counting massive degrees of freedom after gauge fixing Consider the theory of scalar QED with the Lagrangian
$$\mathcal{L} = - \frac14 F^{\mu\nu} F_{\mu\nu} + (D^\mu \phi)^* (D_\mu \phi) - m^2 \phi^* \phi \tag{1}$$
where $\phi$ is a complex scalar field with mass $m$. Counting the degrees of freedom, we have
*
*two massless real degrees of freedom from $A_\mu$
*two massive real degrees of freedom from $\phi$
Now, even though there's no symmetry breaking going on here we can still choose to go to unitary gauge, i.e. fixing the gauge so that $\phi$ is real. We now have the gauge-fixed Lagrangian
$$\mathcal{L} = - \frac14 F^{\mu\nu} F_{\mu\nu} + (D^\mu \varphi) (D_\mu \varphi) - \frac12 m^2 \varphi^2\tag{2}$$
where $\varphi$ is a canonically normalized real scalar field, and there is no gauge symmetry. Then we have
*
*three real degrees of freedom from $A_\mu$
*one massive real degree of freedom from $\phi$
where I know there are three real degrees of freedom in $A_\mu$, because gauge fixing always removes one and we have no gauge fixing here.
What confuses me is that there must be two massive degrees of freedom, just as there were in the original theory. So that somehow means that one of the degrees of freedom in $A_\mu$ is massive while the other two aren't -- but how can that be? There's no mass term for $A_\mu$ in sight.
| Before we can compare with the second Lagrangian (2),
the first Lagrangian (1) should include a gauge-fixing term ${\cal L}_{\rm gf}$, e.g. ${\cal L}_{\rm gf} = \lambda~ {\rm Im}(\phi),$ where $\lambda$ is a Lagrange multiplier. After integrating out $\lambda$ and ${\rm Im}(\phi)$ the Lagrangian (1) becomes the Lagrangian (2).
Why we need to consider gauge-fixed (rather than un-gauge-fixed) Lagrangians is e.g. discussed in my Phys.SE answer here.
For both Lagrangians, the scalar field effectively induces a mass term for the $A_{\mu}$-field.
$\downarrow$ Table 1: Real DOF of OP's Lagrangians.
$$\begin{array}{ccc}
\text{Lagrangian}&
\text{Off-shell DOF}^1 &
\text{On-shell DOF}^2 \cr
(1)& 2+4-1=5 &2+3-1=4 \cr
(2)& 1+4-0=5 &1+3-0=4
\end{array}$$
$^1$ Off-shell DOF = # (components)- # (gauge transformations).
$^2$ On-shell DOF = # (helicity states)= (Classical DOF)/2, where Classical DOF = #(initial conditions).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/387049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Noether charge for Lagrangian with higher-order derivatives I'm trying to find the Noether charge for the symmetry
$x\rightarrow x+f\left(x\right)$
This transformation should leave the action invariant, so
\begin{align*}
dS&=S\left(x+f\left(x\right),\dots\right)-S\left(x\right)=0\\
&=\int dt\ \mathcal{L}\left(x+f\left(x\right),\dot{x}+\dot{f},\dots,t\right)-\mathcal{L}\left(x,\dot{x},\dots,t\right)
\end{align*}
Using $f\left(x+\epsilon\right)-f\left(x\right)\approx \epsilon \frac{df}{dx}$
\begin{align*}
dS=\int dt\ \frac{\partial \mathcal{L}}{\partial x}f+\frac{\partial \mathcal{L}}{\partial \dot{x}}\dot{f}+\frac{\partial \mathcal{L}}{\partial \ddot{x}}\ddot{f}+\dots
\end{align*}
Writing the second term as a total derivative
\begin{align*}
dS&=\int dt\ \frac{\partial \mathcal{L}}{\partial x}f+\frac{d}{dt}\left[\frac{\partial L}{\partial \dot{x}} f\right]-f\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{x}}\right)+\frac{\partial\mathcal{L}}{\partial \ddot{x}}\ddot{f}+\dots\\
&=\frac{\partial\mathcal{L}}{\partial \dot{x}}f\bigg|_0^T+\int dt\ \frac{\partial \mathcal{L}}{\partial x}f-f\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{x}}\right)+\frac{\partial\mathcal{L}}{\partial \ddot{x}}\ddot{f}+\dots
\end{align*}
For the higher order terms we can do the same
\begin{align*}
\frac{\partial \mathcal{L}}{\partial \ddot{x}}\ddot{f}&=\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \ddot{x}}\dot{f}\right)-\dot{f}\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \ddot{x}}\right)\\
&=\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \ddot{x}}\dot{f}\right)-\boxed{\frac{d}{dt}\left(f\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\ddot{x}}\right)\right)}+f\frac{d^2}{dt^2}\left(\frac{\partial\mathcal{L}}{d\ddot{x}}\right)
\end{align*}
So now the integral becomes
\begin{align*}
dS=\sum_{n=0}^N\frac{\partial\mathcal{L}}{\partial x^{\left(n+1\right)}}f^{(n)}\bigg|_0^T+\int dt\ f\left[\sum_{n=0}^N\left(-1\right)^n\frac{d^n}{dt^n}\left(\frac{\partial \mathcal{L}}{\partial x^{(n)}}\right)\right]-\boxed{\frac{d}{dt}\left(f\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\ddot{x}}\right)\right)+\dots}=0\end{align*}
Where the sum under the integral represents the Euler-Lagrange equations for the unperturbed action. I am kind of expecting the boxed terms to vanish as well, leaving only the first term. Did I do it right, what steps am I missing?
| In this answer, we will just list the result without a proof. For a higher-order action
$$ S[q]~=~\int\! dt~ L(q(t), \dot{q}(t),\ddot{q}(t),\dddot{q}(t),\ldots,t) \tag{1} $$
with a vertical infinitesimal quasi-symmetry
$$ \delta q^i~=~\varepsilon Y^i(q, \dot{q},\ddot{q},\dddot{q},\ldots,t) , \tag{2} $$
the bare Noether charge is
$$Q~=~ \sum_{k\geq 1} \left(\frac{d}{dt} \right)^{k-1}\left(Y^i \sum_{m\geq k} \begin{pmatrix} m \cr k \end{pmatrix} \left(-\frac{d}{dt} \right)^{m-k}\frac{\partial L}{\partial q^{i(m)}} \right).\tag{3}$$
To unpack formula (3) for a second-order Lagrangian, see e.g. my Phys.SE answer here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/387320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Gravitational potential at the center of a uniform sphere Feynman in second Messenger Lecture said that potential at the center of ball with small radius $a$ is equal to average potential on surface of ball minus $G$ times mass inside the ball divided by $2a$. You can see it on the picture. I don't understand that. I had an idea that it is from Taylor series of potential at the centre, but I can't figure it out. Can you help me?
| Late answer but I'll bite.
Feynman's talking about a ball, which means that he is talking about a solid sphere, with uniform density, which I shall call $\rho$. You can apply Gauss's law for gravity to then calculate the potential.
Gauss's law states that:
$$ \int FdA = -4\pi GM$$
where F is the g-field, A is a surface area and M is the mass enclosed by our Gaussian surface.
Let's say that our ball has radius $a$. We can imagine a Gaussian sphere, of radius $r < a$ which encloses some mass $\rho V$ where V is the volume of our Gaussian sphere (we can use multiplication instead of integration here because it's a uniform sphere). Call A the Gaussian area and by symmetry, $\int FdA = FA$,
so that we can write, for our Gaussian sphere:
$$F = \frac {-4πG\rho V}{A} $$
For our Gaussian sphere, $V = \frac{4}{3}\pi r^3$ and $A = 4\pi r^2$ so that:
$$\frac VA = \frac r3$$
leading to:
$$F = \frac{-4πG\rho r}{3} $$
This tells us the field at any point in the sphere. Replacing r with a tells us the field at the surface of the sphere.
Anyway, we want the potential at the surface of the sphere, which is the negative work done per unit mass when moving from the center of the sphere to its surface i.e from $r=0$ to $r=a$ (assuming that our reference point is set at the center of the sphere, so that the center has zero potential energy). This is the same thing as the integral of the field along this distance, since the field is the force per unit mass. Thus the potential at the surface of the sphere is:
$$\int_0^a Fdr =\int_0^a \frac{4πG\rho r dr}{3}$$
which is:
$$ \frac{4πG\rho a^2}{6} $$
The volume of the entire sphere is $ V = \frac 43 \pi a^3$, so that $\frac {4\pi a^2}{6} = \frac {V}{2a}$. Since $M$, the mass enclosed by the sphere is $\rho V$, where this time V is the volume of the entire sphere:
$$ \frac{4πG\rho a^2}{6} = \frac{G\rho V}{2a} = \frac {GM}{2a}$$
So the work done in moving from the center to the surface of a ball is given by $\frac {GM}{2a}$, provided the ball is uniform. To get Feynman's equation just change the reference point for the potential energy, so that the potential energy at the center is no longer $0$.
Edit:
The above result holds only if the field due to matter outside the sphere is the same at the center and on the surface. If not the work done will depend on where on the surface you move to, since every point on the surface is not equidistant from external matter. If the ball is small enough, a criterion which Feynman mentions, then there's not much difference between different points on the surface or at the center, so you can approximate the field as constant on the ball.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/387439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trouble in understanding AM modulation Amplitude modulation is in fact, superimposing the low frequency transmission signal into a high frequency carrier signal, right?
So, if the transmission signal can be represented as $c(t)=A_c \sin (\omega_ct )$ and the carrier wave can be represented as $c(t)=A_m \sin (\omega_mt)$, then upon superimposing and simplifying the equation using trigonometric identities we will have
$$
c_m(t)=A_c \sin (\omega_ct )+\frac{\mu A_c}{2}\cos(\omega_c-\omega_m)t-\frac{\mu A_c}{2}\cos(\omega_c+\omega_m)t,
$$
but this equation has the carrier wave and two sinusoidal waves with frequencies $\omega_c-\omega_m$ and $\omega_c+\omega_m$. So where is the signal which is transmitted? What we've got is a corrupted signal with uneven frequencies i.e. $\omega_c-\omega_m$ and $\omega_c+\omega_m$
Correct me if I'm wrong at any point.
| You are right that by superimposing a constant modulation on the carrier wave you cannot transmit a signal because you get a constant single frequency carrier wave plus two constant sideband frequencies. For the transmission of a signal, you need a modulation changing in time (frequency and/or amplitude) like in speech transmission in AM radio or in Morse signals transmission in amateur shortwave radio.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/387564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How is semicolon derivative notation defined for multiple derivatives? I have a covector $\eta_\mu$. Then I have some notation which says $$\eta_{\alpha;\beta\gamma}$$ What does this mean? I understand that given a vector $A^\alpha$, that $$A^\alpha_{;\beta}=\nabla_\beta A^\alpha=(\nabla A)^\alpha_\beta$$This makes sense to me. However I do not understand what happens when two indices follow the semi-colon. If I had to guess, I would go with $$\eta_{\alpha;\beta\gamma}=\nabla_\beta\nabla_\gamma\eta_\alpha$$But I am not sure of this.
| Actually, $$\eta_{\alpha;\beta \gamma} = \eta_{\alpha;\beta;\gamma} = \nabla_{\gamma} \nabla_{\beta} \eta_{\alpha}$$ Your result is only the same in the flat spacetime, where the Riemann tensor (which is related to the commutator of covariant derivatives) vanishes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/387689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the double slit experiment performed measuring single photons? As a hobby physicist I don't understand how the double slit experiment is performed in terms of single photons. Is the experiment really conducted by sending and measuring single photons? Is there not always interference of some kind (except for some device like the LHC)?
If yes: what would happen if the experiment is conducted over a prolonged period of time, for instance: 1 photon every n hours. Would they theoretically interfere with "themselves"? (don't know if i'm still making sense here).
If no: why is the interference considered to be originating from the sent photons and not from an external influence of some sort?
Thanks!
| Yes, there are single photon through a double slit experiments , as also single electron ones.
Here is one single photon at a time:
. Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.
You can see the single photons hitting on the left frame, the hits seem random. By the time the accumulation has reached 1000 frames(about 50.000 photons counting the ones in the single frame) the interference pattern starts appearing, and it is the classical interference of light in the far right (~25.000.000 photons).
The external influence is the boundary condition: two slits d1 width each, d2 distance apart. Photon hits the boundary conditions. One sees the probability distribution for a single photon to scatter off these specific double slits. Each photon follows that probability distribution, except individually it looks random. In the accumulation the wave nature of the photon wavefunction is demonstrated.
The timing would have no influence on the pattern if the experimental geometry remains fixed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/388026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do convective clouds form? What are necessary conditions to form convective clouds with their flat cloud base, like this one:
I know that convective instalibity at the ground is at start of the process, but I have realized that I don't understand what exactly determines the height of the cloud base. Usually, it is told that the rising bubble of air expands, cools and finally its temperature drops below dew point and water droplets form. But the bubble must be warmer than the surrounding air to ascend, so is warmer also at cloud base and even so water condenses inside the bubble and not outside it.
| There is no bubble and Convective instability isn't used to explain the cloud base.
It starts with a mixture (wet air) of dry air and water vapor. Hot mixture moves up due to buoyancy, is cooled, and moves down, establishing a vertical convective circular flow.
Temperature above the earth surface decreases over height. When the condition of temperature and water mixture fraction is such that dew occurs, liquid phase of water (droplet) starts to form which is what we call cloud. The droplet grows and may rise until it loses it kinetic energy and falls down due to gravity.
The cloud base is determined by the dew condition (temperature and water mixture fraction). It should be flat in a small region because of the temperature. Its height is determined by local temperature, i.e. in hot weather, the cloud or cloud base is higher in the sky, whereas in cold and wet days, the base is lower.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/388154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Definition of stress-energy tensor
The image from the wiki article on the stress energy tensor gives $T_{00}$ as $1/c^2$ times the energy density. I believe this is incorrect and that the $1/c^2$ factor should be dropped. Am I missing something?
| The stress-energy tensor can be written with $T_{00}$ as an energy density or a mass density. The latter is of course just $E/c^2$ in accordance with Einstein's famous equation $E=mc^2$. Both forms are used and neither is more correct that the other.
In any case general relativists usually choose units where $c=1$ and the distinction disappears.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/388442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Are all waves either transverse or longitudinal? So I recently searched up "em wave transverse proof", and I understood it pretty well enough I think.
After that, I just started to wonder if all waves are either transverse/longitudinal. If there are waves that are neither one of them, how do we put that in mathematical notation?
| Short answer: no.
For example, gravity waves (i.e. ripples) on the surface of a liquid have both transverse and longitudinal motion, so they are not purely either.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/388543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why is the internal energy of a real gas a function of pressure and temperature only? While studying thermodynamics, I read that the internal energy of an ideal gas is a function of temperature only. On searching the internet, i found an article which stated that the internal energy of a real gas is a function of temperature and pressure only. I could not find a proper reason for this.
So my question is: why is the internal energy of an ideal gas a function of temperature only and that of a real gas a function of temperature and pressure only?
Is this property of ideal gases and real gases derivable through any equation?
| First off, let's start with a fundamental assumption: the internal energy of an gas consists of only its kinetic energy. This is the basic assumption of kinetic gas theory, and is a reasonable assumption for real gases. The temperature of any system is simply the average kinetic energy. Therefore, the kinetic energy of any one particle in a system is proportional to the temperature.
K ∝ T
This is the energy of a single particle. The energy of the entire system is simply the average kinetic energy times the number of particles, so:
K ∝ nT
So clearly, kinetic energy is a function of the number of moles of gas and their temperature. But by the ideal gas law, PV = nRT. This means this statement can be rewritten as:
K ∝ PV
As you can see, it is true that both a higher pressure and a higher temperature indicate a higher internal energy. However, they are functions of each other, not two components of the same equation.
In my opinion, stating that kinetic energy is a function of temperature and the number of particles better expresses the essential concepts of the derivation. It might rather be said the pressure and volume are alternate ways of expressing those quantities.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/389127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 2
} |
What is the shortest distance between electron and positron before they are annihilated? I just want to know how does an electron felt the presence of a positron before they are converted into energy? Also how does the electron tell if it is positron or proton if this makes any difference?
| I don't think there is a simple answer to this.
The problem is that in quantum electrodynamics the particles we call electrons and positrons are the states described by the quantum field in the limit of negligible interactions i.e. when the particle is too far from any other particles for any significant interaction to occur. In this limit the particles are described by the free field states i.e. Fock states.
The problem is that when interactions become significant the states of the quantum field are perturbed away from the free field states. We can calculate this perturbation, using (unsurprisingly) perturbation theory, to calculate scattering probabilities, but we don't actually know what the states are. So in the annihilation process it isn't really the case that there is one electron and one positron present because the state of the quantum field cannot simply be described as an electron state and a positron state. If we insist on trying to describe it using the free field states then we have to conclude there are other particles present as well i.e. the virtual particles.
The point of all this is that during the annihilation process the electron-positron separation isn't well defined so it doesn't make sense to ask how close the two get before they are annihilated.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/389213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
$Q$ value of beta plus decay (positron emission) I am unable to understand Q value for positron emission. The general reaction is as follows:
$$p \to n + e^+ + \nu$$
$$ ^A_ZX \to ^A_{Z-1}Y+ e^+ + \nu \tag{1a}$$
This reaction $(1a)$ was giving in my text. First question is that where did one electron go? We began with $Z$ electrons but on the right side it seems only $Z-1$ electrons are present. Maybe this is the cause of confusion.
So I rather took this reaction $(2a)$ for positron emission:
$$^A_ZX \to ^A_{Z-1}Y+ e^+ + e^-+ \nu \tag{2a}$$
Now writing the Q value, we find the mass defect
$$[m_n(^A_ZX)-(m_n(^A_{Z-1}Y)+m_{e^+}+m_{e^-} +m_\nu)]$$
here $m_n(^A_{Z}X)$ are mass of nuclei. Now we can rewrite in terms of atomic mass numbers $m_a(^A_{Z}X)$ as
$$\Delta m=[(m_a(^A_ZX)-Zm_e)-((m_a(^A_{Z-1}Y)-(Z-1)m_e)+m_{e^+} + m_{e^-}+m_\nu)] $$
$$\Delta m= (m_a(^A_ZX)-m_a(^A_{Z-1}Y) - 3m_e -m_\nu) \tag{2b} $$
but as you can see, if we use reaction $(1a)$ then we will get
$$\Delta m= (m_a(^A_ZX)-m_a(^A_{Z-1}Y) - 2m_e -m_\nu) \tag{1b} $$
Am I wrong somewhere? I have highlighted key points(mistakes) in italic.
| You should write the reaction in terms of atom and ions as
$$^{\rm A}_{\rm Z }\rm X \Rightarrow ^{\,\,\,\,\,\rm A}_{\rm Z-1 } Y ^-+ e^+$$
noting that the daughter is a negative ion which has a net negative charge because it still has $Z$ elections orbiting the nucleus.
Think of it as a change happening in the nucleus resulting in the emission of a positron which is nothing to do with the orbiting $Z$ electrons although of course those $Z$ electrons will have to move into new orbitals but there will still be $Z$ electrons.
The masses given in tables are the masses of the neutral atoms.
So if the equation is to be written in terms of neutral atoms it can be written as follows
$$^{\rm A}_{\rm Z }\rm X \Rightarrow \left (^{\,\,\,\,\,\rm A}_{\rm Z-1 } Y +e^- \right )+ e^+$$
with both sides of the equation having a net zero charge.
In essence in order for this reaction to occur two electron masses have to be created and you have to include those two electron masses when you calculate the change in mass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/389442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is potential energy a type of energy at all? Is potential energy, whether it be that of a charge in an electric field or a mass in a gravitational field or anything like that, actually an energy that the particle itself contains, like kinetic energy? Or is it just a measure of its ability to do work?
Is it the case that instead of integrating conservative forces over distances to find work done, we use the fact that the work done by a conservative force doesn't depend on path and hence we just use the notion of 'potential energy' and it's variation with distance, and just take the difference between the potential energy at two points to easily find the work done? And hence, is potential energy nothing but a tool to calculate work done by conservative forces?
| Maybe we can say that a particle "contains" chemical energy, since we have to break its chemical bonds to access to chemical energy, and maybe we can say that a particle "contains" nuclear energy, since we have to break nucleons to access to it.
Kinetic energy, however, depends on movement, which means that it depends on the frame of reference of the observer. So for sure we cannot say that a particle "contains" kinetic energy. Think for example about a ball sitting still in a moving train: how much kinetic energy does it have from the point of view of someone standing on the platform? And for someone sitting inside the train?
In a similar way, potential energy depends on the position of the particle relative to something. If we move the particle, its potential energy is generally going to change. So we cannot say that a particle "contains" potential energy either.
However, this doesn't mean that kinetic or potential energy are not "real", since we can always use them to perform work on or to heat something, which it's what really matters in the end.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/389496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
What is the reason for the edge effect in capacitors? The electric field lines bend at the edges of the capacitors like this:
What is the reason for this? Any quick explanation as to why they bend?
| From answer:
This means that the electric field near the edges of the plates is actually larger than the electric field between the plates
Possible correction: the electric field just above the curvature ($E_1$) is the same as the electric field between the parallel plates ($E_2$). $E_1$ from the same plate reduces like a sphere of charge ($E=(kQ)/R$), while $E_2$ above the same plate remains ideally constant. The combined $E_1$ from both plates is constant, but is weaker than $E_2$. ($d_1$) is the $E_1$ bowed field line. ($d_2$) is the $E_2$ unbowed distance between the plates. For the combined plates: ($E_1 < E_2$) and ($d_1 > d_2$). The voltage is same: $V=(E_1)(d_1)=(E_2)(d_2)$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/389766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Why is the singlet state for two spin 1/2 particles anti-symmetric? For two spin 1/2 particles I understand that the triplet states ($S = 1$) are:
$\newcommand\ket[1]{\left|{#1}\right>}
\newcommand\up\uparrow
\newcommand\dn\downarrow
\newcommand\lf\leftarrow
\newcommand\rt\rightarrow
$
\begin{align}
\ket{1,1} &= \ket{\up\up}
\\
\ket{1,0} &= \frac{\ket{\up\dn} + \ket{\dn\up}}{\sqrt2}
\\
\ket{1,-1} &= \ket{\dn\dn}
\end{align}
And that the singlet state ($S = 0$) is:
$$
\ket{0,0} = \frac{\ket{\up\dn} - \ket{\dn\up}}{\sqrt2}
$$
What I'm not too sure about is why the singlet state cannot be $\ket{0,0}=(\ket{↑↓} + \ket{↓↑})/\sqrt2$ while one of the triplet states can then be $(\ket{↑↓} - \ket{↓↑})/\sqrt2$. I know they must be orthogonal, but why are they defined the way they are?
| There are at least 2 approaches. One is just show that it is symmetric by applying the lower operator for total spin to the maximal $S_z$ state which should satisfy ($\hbar=1$):
$$ S_-|1,1\rangle=\sqrt 2 |1, 0\rangle$$
so
$$|1, 0\rangle =\frac{1}{\sqrt 2}S_-|1,1\rangle=\frac{1}{\sqrt 2}(S_{1-}+S_{2-})\uparrow_1\uparrow_2$$
$$\frac{1}{\sqrt 2}[(S_{1-}\uparrow_1)\uparrow_2+\uparrow_1(S_{2-}\uparrow_2)] $$
$$ =\frac 1 {\sqrt 2}(\uparrow_1\downarrow_2 + \downarrow_1\uparrow_2)$$
That gives a nice demonstration of how to work with the ladder operators, but there is a much deeper reason it must be symmetric.
To find the rotationally invariant subspaces of a tensor product of $N$ states with dimension $d$ you do the following (this just a sketch of the procedure):
Find $N$. It is $N=2$, now we partition $2$ in every way possible:
$$ 2 = 2 $$
and
$$ 2 = 1 + 1 $$
For each of these partitions we draw the Young diagrams and connect those with irreducible representations of the permutation group on $N=2$ letters. This is called the Robinson-Schensted Correspondence.
Take the $2=2$ diagram an make a normal Young Tableau and then compute the Young symmetrizer. In this case, you get the purely symmetric operator: $S=(1 + e_{2,1})/\sqrt 2$
For $2=1+1$, you do the same an get the antisymmetric operator: $A=(1 - e_{2,1})/\sqrt 2$.
Schur-Weyl Duality tells us that applying these to the indices (here, particle labels), will tell us the rotationally invariant subspaces of this tensor product space; moreover, the remarkable Hook-Length Formula tells us the dimensions of the subspace, and the result for $d=2$ is the symmetric one has ${\bf 3}$ dimensions, and antisymmetric is ${\bf 1}$.
This is written:
$$ {\bf 2} \otimes {\bf 2} = {\bf 3}_S + {\bf 1}_A $$
So it simply has to be that all the states in the triplet have the same exchange symmetry.
Note that can add another spin $\frac 1 2$, and the whole procedure will show you that:
$$ {\bf 2} \otimes {\bf 2} \otimes {\bf 2}= {\bf 4}_S + {\bf 2}_M + {\bf 2}_M$$
which means the four $S=\frac 3 2$ states are symmetric and there are two doublet $S=\frac 1 2$ states with mixed symmetry, corresponding to the partitions:
$$ 3 = 3$$
and
$$ 3 = 2 + 1 $$
Note that the hook length formula for:
$$ 3 = 1 + 1 + 1 $$
yields a subspace of dimension zero: there is no antisymmetric combination of 3 spins.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/389946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
How is thermal energy and dipole related? My textbook states :
The extent of polarization of dielectrics depends on the relative strength of two mutually opposite factors - the dipole potential energy tending to align the dipole with the external field and thermal energy tending to disrupt the alignment.
I know that thermal energy arises due to random motion of particles in a substances. It's high for gases and low for solids as the molecules only have vibrational motion in the latter.
I am confused as to how they'll affect dipole.
| Thermal energy CAUSES the random motion of particles. The more thermal energy there is, the more the dipoles will jiggle around randomly. This makes their orientation more random, so less line up with the external electric field. This effectively lowers the polarization.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/390018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Where can I find values for apparent brightness of stars? I'm in high school and doing a project. I want to calculate the distance to stars using their luminosity and apparent brightness, from the equation $b=\frac{L}{4 \pi d^2}$. I have found values for luminosity and apparent magnitude from the Hipparcos dataset. However, I cannot find how to convert these values of apparent magnitude to apparent brightness anywhere. I need the values to be in $\frac{W}{m^2}$ for the equation to work. Is there a way to do this? Is my idea completely wrong? Any guidance you can give me will be greatly appreciated!
| First, you need to use distance modules derive the absolute magnitude M (like in v band) for the star using its apparent magnitude m (in the same band like v):
$$\underbrace{m-M}_\text{Distance Modulus} = 5\log_{10}\left(\frac{r}{10}\right)$$
where r is the distance (d) and then you need to use the driven absolute magnitude (M) to get the L (luminosity) from the below relation taking the sun as the reference
$$L / L_\odot = 10^{0.4(M_\odot-M)}$$
here M for the sun is +4.75.
It gives you a luminosity ratio compared to the sun.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/390148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the definition of the charge conjugation? I seem to have troubles finding definitions of the charge conjugation operator that are independant of the theory considered.
Weinberg defined it as the operator mapping particle types to antiparticles :
$$\operatorname C \Psi^{\pm}_{p_1 \sigma_1 n_1;p_2 \sigma_2 n_2; ...} = \xi_{n_1} \xi_{n_2} ... \Psi^{\pm}_{p_1 \sigma_1 n_1^c;p_2 \sigma_2 n_2^c; ...}$$
He does not really seem to specify what he means by "antiparticles" around there, but I'm guessing this is the one-particle state that is conjugate to this one. This assumes that it is possible to decompose everything into one-particle states.
Wightman seems to go with $C \gamma^\mu C^{-1} = \bar \gamma^\mu$, which isn't terribly satisfying and also only works for spinor fields.
I've seen thrown around that the $C$ conjugation corresponds roughly to the notion of complex conjugation on the wavefunction but never really expanded upon.
Is there a generic definition of charge conjugation that does not depend on how the theory is constructed? The CPT theorem in AQFT indeed seems to not have any of those extraneous constructions, but the action of the different symmetries is a bit hidden as
$$(\Psi_0, \phi(x_1) ... \phi(x_n) \Psi_0) = (\Psi_0, \phi(-x_n) ... \phi(-x_1) \Psi_0)$$
Is the action of $C$ symmetry $\Psi' = C \Psi$ just a state such that for any operator $A$,
$$(\Psi, A \Psi) = (\Psi', A^\dagger \Psi')$$
or something to that effect? From some parts seems like it may just be $C \phi C^{-1} = \phi^*$.
| There is no natural definition of charge conjugation that works for all QFTs. Rather, you should understand the CPT theorem instead as a combination of reflection-positivity and Wick rotation. See this paper, Appendix A.2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/390342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What experimental bounds do we have on big $G$? I know that there has been a large amount of controversy surrounding the exact value of the gravitational constant $G$, but I know that there is not a substantial difference in the measured value. So I was wondering what experimental bounds we have on it so far?
| This page shows the range of some measurements from 1982 to 2014. Basically, it looks like arguments could be made for any value between -460 to +250 parts per million below or above the commonly accepted value of 6.67408 x 10^-11 m^3 kg^-1 s^-2.
http://iupap.org/working-groups/wg13-newtonian-constant-of-gravitation/
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/390541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
When wood absorbs water, expands, and breaks stones, how is conservation of energy working? There is a traditional stone cutting method consisting of making a series of small holes in the stone, then inserting wood into the holes, and then adding water to the wood. When the wood expands, it will break the stone.
How is energy being conserved, when the stone is being broken by the expansion of the wood?
| I think the deleted answer by @naiad (wet wood has higher entropy than [water] and [dry wood] separately) may make sense. It is well-known that wood dimensions change depending on the humidity of the surrounding air. If water is concentrated in some area, its contribution to entropy is lower than if it is dispersed. Therefore, energy can be supplied from the environment, which is maintained at some relatively constant temperature.
Another possible explanation: capillary action (absorption of water by wood) is caused by intermolecular forces, so the potential energy of intermolecular forces decreases in the process.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/391141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does the kinetic energy of a photo-electron vary? Provided it is above the threshold frequency of the metal, when electromagnetic radiation is shone onto a metals surface photo-electrons are emitted. This occurs because 1 photon is absorbed by 1 electron giving it enough energy to be ejected.
We know that the energy of the incident photons are all equal from the equation E = hf. If this is so why does the kinetic energy of the emitted photons vary? Why is there a maximum kinetic energy, is it not the same amount every time?
|
"We know that the energy of the incident photons are all equal"
True only if the light shone is monochromatic. Look at the plot used to determine the photoelectric effect:
it uses the maximum electron energy for a given frequency, so as to not depend on the kinematic dispersion discussed in the chosen answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/391464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What can be the simplest way to find the thickness of a soap bubble? Is it possible to measure the thickness of a soap bubble without using any sophisticated instruments such that anyone can do it?
| One can measure temperature and pressure of the air used to blow a bubble, the diameter of the bubble, and its vertical speed in air (using a camera). Then one can use the Stokes law, calculate the mass of the bubble and the mass of the air in the bubble, their difference is the mass of soap water, then one can calculate the thickness of the bubble. For example, if the densities of soap water and air are respectively 1000 kg/m^3 and 1.29 kg/m^3, the diameter and the thickness of the bubble are respectively 1 cm and 1 micron, the mass of the air in the bubble is about 0.7 mg and 0.3 mg, so they are of the same order of magnitude.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/391563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Melting of ice on applying pressure Can someone explain why ice melts on increasing pressure based on Le Chatelier's principle? By the principle, ice will try to oppose increase in pressure. It can do so by increasing volume right? But on melting volume decreases
| You almost answered your question, except that you deviated a bit. What you got wrong was the condition for equilibrium.
Equilibrium is that state at which any the system will try to counteract a an external change to the system. That is the primary definition of Le Chatelier's Principle also. So when we apply an external pressure to the ice-water system, the system will try to increase the pressure on the external agent, to keep the system, let's say, pressure-balanced. Like, when you do work on a system (like pushing a piston in a gaseous system that is in equilibrium), the system does work on you to keep it in equilibrium (the piston is pushed back). Same thing here. Now to increase the pressure on the external agent, the ice-water system can decrease its volume, and since water occupies a lower volume than ice, more ice is converted into water.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/391717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do I calculate Holevo information when information is knowable in principle, but unknown? I am trying to better understand the meaning of Holevo information $\chi$. Suppose Alice starts with data encoded on qubits in the $\{0,1\}$ basis. She takes one of these qubits, originally in pure state $\lvert0\rangle$; randomly chooses from a uniform distribution of SU(2) transformations with which to encrypt the qubit; sends the qubit to Bob.
I have the formula $\chi=S(\sum_j p_j \rho_j)-\sum_j p_j S(\rho_j)$, where $\rho_j$ are all the encrypted states Alice chooses from, with corresponding probabilities $p_j$. The first term is clearly equal to $1$ (the entropy of a completely mixed qubit state); the second term is zero, since Alice always sends pure states (albeit unknown ones); so $\chi=1$.
Have I done this correctly? It makes sense to me, but it's not the way I see others handling the calculation.
edit: note that Bob does not know Alice's key. See comment for clarification.
| One should consider the states associated with each of the plaintext messages Alice could send, in this case 0 and 1. For both of these cases, the state that reaches Bob is effectively mixed: even though it is actually pure, it is indistinguishable from a mixed state (e.g., one qubit of an entangled pair).
I found Preskill's notes helpful: http://www.theory.caltech.edu/~preskill/ph219/chap10_6A.pdf
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/391938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why is Helium 4 so stable? I've been looking at stuff to do with binding energies and was wondering why Helium 4 is so stable. The fact everything up to carbon is less stable seems a bit odd. Is there a reason for this or another, that's how the universe works?
| If one regards the nucleus as a potential well for nucleons, there is one lowest level. It can contain:
*
*a spin-up proton
*a spin-down proton
*a spin-up neutron
*a spin-down neutron
Then that lowest level is full. Any additional nucleons can only find a place in levels with higher kinetic energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/392175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What is a "decade" as a unit of measure (ex. a decade of the EM spectrum)? Reading through papers and online sources about radio galaxies, I kept stumbling across a term--a "decade" of the electromagnetic spectrum. Radio galaxy emission encompasses "11 decades of the EM spectrum". Or this quote from NASA:
Astronomers have made observations of electromagnetic radiation from cosmic sources that cover a range of more than 21 decades in wavelength (or, equivalently in frequency or energy)!
Source.
What exactly does this term correspond to?
Note: I used the electromagnetism tag because of the context, but I am not sure if the unit can be used outside of the field. Feel free to edit away!
| A decade is a factor of $10$, so it's a way of assigning a unit to the common logarithm ($\mathrm{log}_{10}$). It's also frequently assigned the unit symbol $\mathrm{dex}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/392508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 3,
"answer_id": 2
} |
Why can't the Lagrangian for a free point particle depend on distance? I have looked thorough the derivation of Lagrange equations in Landau and Lifshitz, Vol 1, $\S 3$, p.5. They argues that the lagrangian of a free particle cannot explicitly depend on position vector $\vec{r}$ because of the isotropy and homogeneity of space. But why can't it depend on $r^2$? (They do not explain why this cannot be possible.) Or say a power of $r^2$?
| Because of the homogeneity of space.
If it depended on $r^2$, then the origin with respect to which you're taking that distance $r$ would have a privileged status in the theory. For a free particle this cannot happen, so you cannot have that dependence.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/392635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Magnetic field due to a current carrying wire at the wire Why do we take zero magnetic field at a point on the axis of a current carrying wire.
When i use $B =\frac{\mu_0 I}{2\pi r}$ formula for magnetic field calculation at a distance $r$ from the wire then for $r\to 0$, $B$ becomes infinity. What I am missing here?
I am assuming that wire has zero thickness which we normally use for problem solving.
| Complementing jim's answer in order to explicitly address the questions that possibly motivated the original post:
*
*does the field intensity diverge at the wire?
*where does it point to?
The answer is that in practice we have a current density ($I(A)/A$); and, as long as this density is finite, considering a position $r\to 0$ necessarily leads to the magnetic field at this position being that generated by a vanishing amount of current and, thus, having vanishing intensity: $\mathbf{B}\to 0$.
That is, instead of diverging, the magnetic field is zero and, in particular, has no defined direction.
What about $B =\frac{\mu_0 I}{2\pi r}$, then? Well, if you try to apply it for $r\to 0$, you approach the limit of the wire being a mathematical line, with zero radius and area, and that implies, if the current is finite, an infinite current density (finite $I$ going through a vanishing cross-section area). So you're assuming a diverging physical situation, it's then not surprising you get other divergences.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/394016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Torricelli's Law and number of holes Trying to determine if the number of holes at the bottom of a bucket will change time it takes for water to empty the bucket.
Looking at the equation, it would seem that as long as the area of the hole(s) are the same, the time for the water to flow out should be the same?
Bucket 1 with one hole of a certain area vs Bucket 2 with two holes, but each hole is half the area of the hole in bucket 1 vs Bucket 3 with four holes, each hole is a quarter of the area of the hole in bucket 1.
Is it the size of the hole the water flows out, regardless of how many there are? So, half the size (but twice as many) would still produce a time 1.41 times longer? Quarter the size (but with 4 holes) would still produce a time 2 times longer?
| In a world without friction, where water is a perfect fluid, as long as the total area of the holes in the bucket is the same, the rate of water flowing out would be the same. However, since friction exists, and water is far from being a perfect liquid, the bucket with the smaller holes would have a lower rate of water flowing out since the water would be experiencing energy losses.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/394134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Electricity generator We know in an electricity generator , electrons move from negative voltage to positive voltage of the stator winding and we can use that electric power on its way , so is it possible for the stator to loss all of its electrons((because we consumed it )) so it can not produce more voltage and electric current in the generator?i mean does it finally turn to positive ions ?
| An electricity generator does not create electrons nor do appliances consume them. The electricity generator creates an electromotive force (emf) by moving the generator rotor's conductors in a magnetic field provided by the stator. This emf drives electrons. The resulting electron motion is called current. Without sufficient emf, the electrons are not able to power anything, but they do not 'disappear'.
When something is being powered, the electrons are always moving around a circuit, or between earth (or ground) and some circuit. If electrons are moving from earth to the circuit at one point, somewhere else, electrons are moving from the electrical circuit to earth, unless you are discharging static electricity or a battery, which is a transitory situation. The electrons are not consumed in the average electrical circuit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/394359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What motivates the term "State function"? What motivates the term "State function"? Or,in other terms, what is the relation between, potential theory, gradient fields and exact and inexact differentials?
| A rigorous answer would require a good chunk of a calculus course. Assuming that you are quite familiar with exact differentials, the idea behind a State function is that, given certain conditions (coordinates over a manifold), the scalar value of the State function is always the same. That is, if I have a quantity $S$ that depends on $T$, then $S$ is a state function if $S$ only depends on $T$, and not on the way you got to $T$. If you allow, in this oversimplified picture, the temperature $T$ to fully characterise your physical states, then $S$ depends only on the actual state of your system, hence $S$ is a State function.
Mathematically, the property of not depending on the path that leads to the state $T$ translates into the notion of exact differential. If $\gamma$ is any closed path that starts at $T$ (and ends at $T$, so perhaps it is now best to imagine that the state $T$ is an element of a multidimensional manifold rather than just $\mathbb R$), you have
$$\oint_\gamma\text d S = S(T)-S(T) = 0$$
provided that $S$ is a state function. Conversely, if you have a 1-form $\alpha$ for which
$$\oint_\gamma\alpha = 0$$
for any closed curve $\gamma$, then there exists a scalar-valued function $f$ for which $\alpha =\text df$. You can then call $f$ the potential of the field $\alpha$, which has the properties of a State function.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/394783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Velocity of light in Galilean transformation What is the velocity of light in Galilean transformation? Is it infinity?
| Let's say a light beam travels at a speed of $c$ in one reference frame $R_0$. Now, consider another reference frame $R_1$ where it moves in parallel to the light beam with velocity $v$, then the speed of the light beam will be $c-v$ in the frame $R_1$.
So under Galilean transformation, we see that the speed of light is not invariant.
You can naively set $c=\infty$, so that $\infty=\infty-v$. But this is not exactly correct, because the concept of an infinite speed is not well-defined.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/394890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Can you vibrate an object and split its molecules? Let’s say I have paper (or, I don't know, any other material). Can I with some sort of device vibrate the molecules of those atoms so fast that they separate, and as a result I don’t have paper anymore but just air?
If so what is this called?
Please try to simplify your answer.
| What you're describing is the end result of increasing thermalization, or the addition of energy to the mechanical degrees of freedom of a molecule. These degrees of freedom include rotational or translational ones as well. Energy stored in these degrees of freedom is called heat.
So you're not describing a process that is at all exotic. As Chris, one commenter, put it:
A match would suffice :)
The breaking of bonds as the end result of energy addition is very common in chemical reactions. The endings "-lysis" "-lyzation" from Greek "Luein" to unbind (c.f. German lösen (unravel, solve, dissolve), English "loosen" and many other modern Indo-european derivatives) in names for chemical and biological processes often (but not always) betoken exactly this kind of thing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/395088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Concept regarding Venturi Tube-Bernoulli application I was recently studying applications of Bernoulli Equation and came across the Venturi tube. This is diagram I have used to analyse the venturimeter. I understand how we obtain the first equation using bernoulli theorem which is
$$P_1 - P_2 =(1/2)ρ(v_2^2 - v_1^2) \tag{1}$$
and also the continuity equation
$$A_1 v_1 = A_2 v_2. \tag{2}$$
However, I am unable to process how to obtain the following third equation
$$ P_1 - P_2 = ρgh $$
where $h$ is difference in the heights of the liquid level in the two tubes and $ρ$ is density of fluid.
Lets say the atmospheric pressure at the top of each tube is $P$. Now since the fluid in the two VERTICAL tubes are at rest and not moving, their velocities are 0. Hence, if I proceeded by applying Bernoulli equation.
$$P_1 +ρ(v_1^2)/2 = ρg(h_1) + P \tag{3}$$ and
$$P_2 + ρ(v_2^2)/2 = ρg(h_2) + P \tag{4}$$
Here $P_1$ and $P_2$ are the pressures at the points in the tube and constriction respectively and the points are at the SAME HORIZONTAL LEVEL.
Subtracting (3) and (4) and even using (1) does not yield me
$$P_1 - P_2 = ρgh $$
rather gives me $0 = ρg(h)$ which makes no sense whatsoever.
What am I missing here and how do I obtain the right result ?
| You cannot apply Bernoulli's theorem through the two lateral pressure taps (where the vertical tubes are linked to the main tube). In general, it is assumed that the pressure is continuous and the speed is clearly discontinuous at this point. (To justify the continuity of the pressure, one would have to look in detail at the nature of the flow around the hole).
For a unidirectional flow, we can show that the pressure varies as in statics in a direction perpendicular to the flow (We prove this by projecting the Euler equation perpendicular to the flow).
So you can write, as for a static fluid $P_1=P+\rho g h_1$ and $P_2=P+\rho g h_2$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/395372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Paradox in special relativity involving capacitor In the laboratory reference frame (LRF), a horizontally moving (with constant speed) flat capacitor would have a different size of plates therefore resulting in different capacity $C'$, namely
$$
C' = \frac{1}{\gamma}C,
$$
where C is a capacity in its own reference frame. The energy of a capacitor is
$$
W' = \frac{q^2}{2C'} = \gamma W.
$$
Here I assumed the value of a charges remain constant in different inertial reference frames (otherwise we could distinguish one reference frame from another). So if the capacitor is closed on a resistor in his reference frame then for me in LRF would be seen like there was more heat produced on a resistor since $Q=W'$ no matter what current was. So, wouldn't it be the way I distinguish one inertial reference frame from another?
| If we somehow contract the plates of a charged capacitor, the charges get closer to each other and the repulsive Coulomb forces between charges become stronger. The voltage of the capacitor increases. The capacitance of the capacitor decreases.
If we somehow contract the plates of a charged capacitor, and also the fields of the charges, the repulsive Coulomb forces between charges stay the same. The voltage of the capacitor stays the same. The capacitance of the capacitor stays the same. This is what happens when we accelerate or boost a capacitor.
Let's say we have a spherical fleet of spaceships, every ship is positive charged. Adding a positive ship to the fleet takes lot of energy. Now let's say the fleet accelerates to high speed keeping it's dimensions unchanged, like Bell's spaceships. Now it's easy to add a ship to the end of the formation, because the electric fields are contracted, and it's easy to add a ship to the side of the formation because magnetic attraction is helping. The sphere seems to have a larger capacitance when it moves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/395464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How to show that $\int\nabla^2\psi_n (x)\overline{\psi_m (x)}dx=0$ Let us consider the three-dimensional time-dependent Schrödinger equation that has the general solution $\psi(x,t)=\sum_n c_n\psi_n(x)e^{-iE_nt/\hbar},$ where the functions $\psi_n$ are orthogonal.
How to show that
$$\int[\nabla^2\psi_n (x)]\overline{\psi_m (x)}dx=0.$$
| I don’t think the result is true in general. It is certainly not true for the harmonic oscillator ground state where $\psi_0^{\prime\prime}(x)$ is proportional to $\psi_0(x)$ and $\psi_2(x)$.
More generally, recall that the kinetic energy $T=-\frac{1}{2m}\nabla^2$
and that, by assumption of your problem,
$$
H\psi_n(x)=E_n\psi_n(x)=T\psi_n(x)+V\psi_n(x)
$$
so that
\begin{align}
T\psi_n(x)&= E_n\psi_n(x)-V\psi_n(x)\\
T_{mn}&=\int dx\,\psi^*_m T\psi_n(x)= E_n\delta_{mn} - V_{mn}
\end{align}
so unless the potential is diagonal, the kinetic energy will not be diagonal either, and thus $\psi^{\prime\prime}_n(x)$ will not be a multiple of itself, implying that
$$
\int dx\,\psi^*_m(x) \left(\frac{d^2}{dx^2}\psi_n(x)\right)\ne 0\, .
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/395644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Griffith's vector potential identity I am working through some problems in Griffith's text "Introduction to Electrodynamics", and am having some trouble with an identity he uses. In Example 2.7 (and many times throughout the book) he says something along the lines of (with a sphere of radius $R$ constant and $z$ varying either inside or outside the sphere)
$$
\sqrt{(R-z)^2} = z-R
$$
for $z>R$, and for points $z<R$,
$$
\sqrt{(R-z)^2} = R-z
$$
I'm guessing this is something simple but I don't see it. I thought about making an argument that the overall term is negative in one case and lets me pull a minus sign outside $\sqrt{(R-z)^2} = \sqrt{(-(z-R))^2} = \sqrt{(z-R)^2} = z-R$ but I don't think this is right, I could do it for either.
| There is no mysterious "vector potential identity" involved. As the square root is supposed to be a positive distance, you have to take the appropriate sign of the square root when R-z or z-R is positive.
PS: In example 2.7, Griffiths points this out explicitely himself!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/395893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple explanation for what a torsor is I am studying Chris Elliott's notes on Line and Surface Operators in Gauge Theories (available here).
In the notes, there's a mention of the fact that (for $G = U(1)$),
$$W_{\gamma, n}(A) = e^{in\oint_{\gamma}A}.$$
The gauge field $A$ is not actually a 1-form, but upon choosing a principal $U(1)$-bundle the connections on that bundle become a torsor for $\Omega^1(X)$.
Is there an intuitive way to understand the idea of a torsor in this context?
EDIT: I found a nice post by John Baez from 2009: http://math.ucr.edu/home/baez/torsors.html, which explains a few things.
| Oversimplified & in a nutshell:
*
*Recall the slogan
A $G$-torsor is like the group $G$ that has forgotten its neutral element.
*Example: An affine space $A$ is a torsor for a vector space $V$.
*The space of $U(1)$ gauge fields is an affine space, while $\Omega^1(X)$ is a vector space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/396082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What does self-closing bra-ket mean in Robetson-Schrodinger Uncertainty Relation? I was reading:
https://en.wikipedia.org/wiki/Heisenberg%27s_uncertainty_principle#Robertson–Schrödinger_uncertainty_relations
Where an inequality is presented:
$$ \sigma_A \sigma_B = | \frac{1}{2} \langle \lbrace \hat{A}, \hat{B} \rbrace \rangle - \langle{\hat{A}}\rangle \langle \hat{B} \rangle | ^2 + | \frac{1}{2i} \langle [ \hat{A}, \hat{B} ] \rangle ^2 $$
I found the notation hard to understand:
$$ \langle \lbrace \hat{A}, \hat{B} \rbrace \rangle$$
Is bra-ket expression, but it contains a single expression inside$\lbrace \hat{A}, \hat{B} \rbrace $ (so it can't be a dot product). How do I interpret this? Similar problem arises here:
$$ \langle{\hat{A}}\rangle \langle \hat{B} \rangle$$
where the $A, B$ are single-entities contained in a closed bra-ket expression.
| The angled brackets here are not bra-kets. For any operator $\hat{O}$, the expression $\langle \hat{O}\rangle$ denotes the expectation value of the operator with respect to a state - here implicitly assumed to be arbitrary, but the same state for all such expressions occurring.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/396286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Can all waves interfere with each other? What conditions must two waves have such that they interfere? Do they need to have the same frequency or amplitude? Should they pass through a given space at the same time? Should they have the same sources?
| The conditions are as follows:
1) The first condition is obvious, but should be stated for completeness. The waves must overlap in space and time- a wave crossing Pond A will not interfere with a wave crossing Pond B, nor will a wave on Pond A today interfere with a wave crossing Pond A tomorrow.
2) The waves must not be polarised at right angles to each other. The interference effect occurs in connection with components of displacement in the same direction.
3) Two or more waves will interfere with each other regardless of their respective frequencies and wavelengths- interference simply means that their amplitudes at any point where they overlap is the sum of their individual amplitudes.
4) Random interference of the sort implied by point 3) will produce a random pattern, which is what you observe on the surface of the sea, for example. Random interference in light is not usually noticeable because the effects cancel each other out on average, and happen so quickly (the time is related to the frequencies of the waves involved) and over such tiny areas that your eye has no chance of catching them.
5) If by 'interfere' you meant 'produce a meaningful interference pattern', then there has to be some stable relationship between the frequencies of the waves involved. The frequencies do not need to be exactly the same- indeed the idea of 'beats' in sound arises from waves with slightly different frequencies. However, the simplest case to model is where the waves have the same frequency and a fixed phase relationship.
6) Finally, the waves must be fundamentally capable of interacting with each other. A water wave will not meaningfully interact with a light wave in a manner that would normally be thought of as interference.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/396418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How does holding a glass prevent it from falling? When I hold a glass of water,
$\hspace{1.5cm}$,
I am applying a force horizontally, but its weight acts downwards. Should it not fall?
How do you describe the equilibrium?
|
"I am applying a force horizontally ..."
Absolutely, totally wrong.
In the photos, you are applying an upwards force to the mass. Just as you would expect.
The only purpose of the finger pinch force is to increase static friction. Note that if you simply dry the glass and glue the glass to your fingers (or use, say, gloves with velcro and velcro on the glass) you of course can completely stop the "pinching" force. For example, very simply if the glass is dry, the horizontal "pinching" force need be much less - and if the glass is sticky rubber (like a reusable Starbucks mug) the horizontal "pinching" force is almost nothing.
The method of holding the glass (friction, pinching, velcro, whatever) is totally irrelevant.
You are right now in the photo applying an upwards force to the mass. - note that, extremely simply, if you slowly decrease the upwards force you are applying, the glass will, of course, move down towards the floor. Similarly if you, say, stand on a chair as in the photo, and jump off the chair, the glass (and you, and your hand) will of course move straight down. No matter how much pinching force you apply, the pinching force will obviously make utterly no difference to it moving down.
One small detail. As any video game programmer will instantly observe, actually purely regarding the pinching force: you absolutely are not applying a horizontal force. The various forces from your fingers/thumb cancel out as seen from overhead. The force in the X axis is, supplied by "you", is of course precisely zero. (If it was not zero, you would be ... throwing the glass! it would be moving in the direction of that horizontal force! You absolutely are not applying a horizontal force.)
To help with the embarrassing downvoters, here's a very basic explantion:
and here's all horizontal forces on the glass:
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/396588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Tangent vector of photon If I had some line element such as minkowski line element:
$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $
And assuming this photon or beam of photons travels in the x direction, how would one find the components of the tangent vector?
Actual answer should be $k^\mu = B(1,1,0,0)$, but how is this obtained?
| For an inertial observer with 4-velocity $t^a$ such that $g_{ab}t^a t^b=t_bt^b=-1$,
let $x^a$ be the unit spacelike-vector orthogonal to $t^a$. So, $g_{ab}x^a t^b=x_bt^b=0$ and $g_{ab}x^a x^b=x_b x^b=1$.
(In this frame, $t^a=(1,0,0,0)$ and $x^a=(0,1,0,0)$.)
So, $k^a=A t^a+B x^a$.
Since the photon tangent vector must satisfy $g_{ab}k^a k^b=k_b k^b=0$,
we have:
$\begin{align}
0&=k_b k^b\\
&=(A t_b+B x_b)(A t^b+B x^b)\\
&=A^2t_bt^b+B^2x_bx^b\\
&=-A^2+B^2
\end{align}$
So, $|A|=|B|$.
Thus, for a future-directed, forward-pointing tangent vector, we have $k^a=Bt^a+Bx^a$, where $B>0$. In coordinates, this is $k^a=(B,B,0,0)$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/397079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Problem understanding Fermi - Dirac distribution function I am studying Statistical Mechanics by R K Pathria. There the author tries to calculate the partition function of in canonical ensemble as:
$$Q_N(V,T) = \sum_{[n_\epsilon]} (e^{-\beta\sum_{\epsilon}n_\epsilon \epsilon})$$
(In above the expression the summation runs over the occupation sets complying to total particle number restriction and also restriction on $n_\epsilon$)
This is easily evaluated for Boltzmann-Maxwell distribution, but due to restricted values of $n_\epsilon$ that can be taken in case FD distribution ($0$ and $1$), he argues that it is difficult to evaluate. So he jumps to grand canonical partition function and there he doesn't care about the values that can be taken by $n_\epsilon$.
Why is that in case of Grand Canonical Partition Function, there are no such restrictions, after all we are still dealing with Fermi-Dirac statistics?
| You are confusing the two restrictions. The restriction the author is referring to which makes the derivation difficult using the canonical ensemble is the restriction on total particle number $N_0$ in the canonical ensemble, not the restriction on the individual occupancies $n_{\epsilon}$ to be $0$ or $1$.
$$
\sum_{\epsilon} n_{\epsilon} = N_0
$$
In the grand canonical ensemble it is still true that each $n_{\epsilon}$ can only take on the values $0$ or $1$, but there is no constraint on the sum of all of the $n_{\epsilon}$. The author certainly uses the fact that each $n_{\epsilon}$ can only be $0$ or $1$ in the derivation of the Fermi-Dirac distribution using the grand canonical ensemble.
The reason we are licensed to use the grand canonical ensemble instead of the canonical ensemble is that in the thermodynamic limit the statistics of a system with a fixed number of particles, $N_0$, are the same as the statistics of a system which can exchange particles with a reservoir so long as the average number of particles, $\langle N \rangle$ is equal to the same number of particles as the isolated system: $\langle N \rangle = N_0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/397254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Strong force gets weaker at small distances yet approximated by -1/r potential My particle physics textbook (by Martin and Shaw) has confused me, it states in ch.7 that the strong force gets weaker at small distances, and that it can be approximated by $V(r) = -\frac{4 \alpha_s}{3 r}$ for $r<0.1fm$, so doesn't this potential suggest the opposite? $Force = -\nabla V = \frac{4 \alpha_s}{3 r^2}$ so the force goes to infinity as r goes to zero. I know that the strong coupling constant $\alpha_s$ is not really constant, but according to the book it's approximately constant when $r<0.1fm$, so what's going on?
| I understand the strong force to be the result of a competition between the Coulomb electrostatic forces between charged particles, and the magnetic forces resulting from the high speed motion of these charges. Two charges moving parallel to each other develop a magnetic attraction force and the ratio between the two is given by(see this ref http://www.physnet.org/modules/pdf_modules/m124.pdf); fm/fe=(v/c)^2. Thus when v is very near to c, the two forces are in balance and you have asymptotic freedom like situation. But far away the electrostatic force is the dominant one. The force can change from attraction to repulsion by changes of the sing of the charges or the direction of motion.
Due to such balancing, the forces are negligible inside and outside and are effective only around the size/radius of a particle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/397471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does a wire bend the electric field? The electrons in the current in a wire are constantly colliding with the metal cations which means there must be a constant cause of acceleration: an electric field. How, why is it that the electric field is identical in shape to that of the wire?
| The electric field has to be provided by a battery for example. There is no electric field within a conductor by itself. If a voltage is applied to a bent conductor, surface charges accumulate in curved parts which leads to a net electric field in axial direction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/397606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If light could bend around a person ,would that make the person invisible? If we place an object such as a coin in front of the light source ,e.g a candle we cant see the candle through the coin but if the size of the coin would be comparable to the wavelength of visible light the we can see the candle through the coin because light would bend around it ... So my question is , can a device like this be designed
a device that generates electromagnetic waves whose wavelength initially is in the visible region then whose wavelength continuously increases then bends around the person then its wavelength sort of decreases till it reaches the eye of the receiver as visible light .So the person would be invisible to the reciever. Is such a device possible? Is my idea stupid? And if at all we are able to make such a device would it happen just like i said or is my idea just wrong?
| Well, light can only travel in straight linear paths, so that machine would be implausible and also it can't be plausible because light would have to travel around the person (it wouldn't work as your drawing suggested because the light would have to change its frequency to become a higher wave) and because it can only travel in a straight line (as I said) not plausible!
However this is a great mind experiment.
If we could somehow create circular (going in a circe) light then it would be possible.
BUT if any other light is present then I don't know if it could work.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/397693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Pauli's first paper about the spin Wikipedia states, that the spin degree of freedom was first formulated by Pauli in 1924:
In 1924 Wolfgang Pauli introduced what he called a "two-valued quantum degree of freedom" associated with the electron in the outermost shell.
However, there is no reference to this quote. Does someone know the publication where Pauli introduced this quantum number for the first time? The only paper I know this one, where he formulates his exclusion principle. But this is from 1925, so one year later.
| Pauli himself talked about this in his Nobel Lecture: https://www.nobelprize.org/uploads/2018/06/pauli-lecture.pdf:
In the autumn of 1924 I published some arguments against this point of view, which I definitely rejected as incorrect and proposed instead of it the assumption of a new quantum theoretic property of the electron, which I called a « two-valuedness not describable classically »3.
However, the reference 3 is to a paper published in January 1925.
Reference
W. Pauli, Z. Physik, 31 (1925) 765
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/397824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to make a laser by sending sunlight through an optical apparatus? 1) Is it possible to create a laser from focused sun light by separating and using only one wavelength of light as a laser and using the proper mechanism to polarize it and make it coherent?
2) If so, would it be possible to use some type of a wavelength filter enabling it to focus different wavelengths for different applications?
| Even with perfect filtering (answer of S. McGrew), one can still distinguish filtered sunlight from true laser light by using the 2nd-order autocorrelation function $g^2(τ)$.
For a laser: $g^2(0) = 1$.
For thermal light (e.g. sunlight): $g^2(0) = 2$
In words: While the temporal arrival time of laser photons is randomly distributed (poissonian distribution), thermal photons arrive in small bunches (Bose-Einstein distribution). This can be measured with a Hanbury-Brown-Twiss experiment.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/398002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Are materials which are bad at conducting heat always bad at conducting electricity also? When defining a material's conductivity, we usually consider its conductivity of heat and conductivity of electricity separately. However, I realize that materials like metal conduct both heat and electricity well. In contrast, materials like wood and glass conduct both heat and electricity poorly. Therefore can we conclude that if a material is bad at conducting one kind of "flow of energy", then it will also be bad at conducting another kind of "flow of energy"? Thanks a lot.
| The ratio of thermal conductivity to electric conductivity is a constant at a given temperature. This relation is mathematically defined by the Wiedmann-Franz law.
There are exceptions to this. The human body is a bad conductor of heat but is an excellent conductor of electrical.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/398266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Is it possible to harvest the energy from the movements of a satellite in orbit? I was thinking about how energy is harvested on Earth from movements of certain forces like wind and ocean currents. Could similar principles be applied in space?
Satellites are virtually in perpetual motion when orbiting the Earth. Is there kinetic energy that can be extracted from this orbital motion and harvested for use on Earth?
| Energy is 'harvested' from satellites all the time. Air drag converts the satellite's energy into heat in the atmosphere. This is, of course, not a perpetual motion machine. The orbit of the satellite will get closer to Earth over time, until the satellite crashes back down, as the Tiangong-1 did last week.
It should also be clear that you can't get more energy out of the satellite than you put in the rocket fuel in the first place. Even if you could throw a lasso around a satellite to take its energy, such a process would be incredibly inefficient. Similarly the energy harvested from wind is not because the wind is in "perpetual motion", instead it comes from the sun.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/398397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Is Yellow a monochromatic light? I have got a serious doubt.
I have read,
"yellow light from a low pressure sodium vapour is monochromatic"
How can it be monochromatic when yellow light is a combination of red and green primary colours?
| You are right that yellow is a secondary colour in light - where RGB is primary. But you are slightly confusing the difference between how RGB works to create colours in a screen compared to sodium vapour.
You can tell whether an apparently yellow source is made up of a combination of red and green, or is just yellow by using a prism: different wavelengths of light have different refractive indexes for the same material, which is why a prism splits up colours. Sodium happens to produce (almost) only light in a single frequency/wavelength, but a different yellow source that looks to be the same colour could well be a combination of red and green wavelengths.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/398501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Where does rest mass come from? Firstly I'll say that I know the current standard terminology is to just refer to "mass", but I wanted to be clear about what I was talking about.
I've heard that in standard model of physics (which is compatible with special relativity), it is necessary to treat mass as an emergent property of a particle. I've done some quantum mechanics and some special relativity, but no QFT yet. So where does rest mass come from? What is it fundamentally? We've all seen that
$E=mc^2$ (or $E^2 = m^2c^4 + p^2c^2$) , but most intuitions I have for energy come from the Newtonian domain. Is energy (or momentum-energy) the new first class citizen in relativity from which all else is derived, or is something more subtle going on here? What are the grounding base concepts?
I have a decent understanding of 4-vectors and the geometry of spacetime, it's just once we start putting things into it that the confusion starts.
| In field theory, each elementary particle is associated with a field of its own. One writes down a Lagrangian (density) $\mathscr{L}$ for the field. When you quantize the theory you obtain particles. The positive coefficient ($\mu^2>0$) of the term in $\mathscr{L}$ which is quadratic in the field becomes proportional to $m^2$ in the relation $E^2=p^2c^2+m^2c^4$ of the particle.
However, sometimes it may so happen that the Lagrangian $\mathscr{L}$ does have a term quadratic in the field but $\mu^2<0$. In the early universe, the Lagrangian Higgs field was such that $\mu^2<0$, and hence cannot be directly associated with the mass of the particle. As the Universe cooled down, $\mu^2$ (being a function of temperature) turned positive and the Lagrangian broke some symmetry called spontaneous breakdown of symmetry with $\mu^2>0$. Since in the Standard Model (SM), the Higgs field is coupled to other particles, its vacuum expectation value gives masses to them (except photon and neutrino).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/398665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Sending a Signal inside the Photon sphere You and a friend are inside the Photon sphere of black holes(Non rotating and uncharged).
At what angles and speed(Momentum) you need to throw a stone(or anything else )so that it reach him According to the General relativity(schwarzschild metric).
(if he is on the other side with the same distance separation from the photon sphere).
| The problem with your question is that a physical observer can only be on time-like orbits, whereas the photon sphere is defined by null-like orbits. So you couldn't really exist on the photon sphere.
However, if you (and your friend) somehow accelerate to nearly the speed of light, it might be good enough to stay very close to the photon sphere for a finite amount of time. Then every photon you emit will essentially be going in the same direction that you are, with an enormous blueshift as seen by a stationary observer. If your friend is moving in any direction different than you, she will be vaporized by the beam.
But let's say you are both orbiting at the equator at the same speed, 180deg apart in phase, and moving at 0.999999c. From the point of view of a distant observer, it would take a million orbits for the photon to catch your friend. But from your point of view, it would happen in a single light-crossing time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/398972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Where does the energy of the expansion of frozen water come from? As you cool a system, you are removing energy, yet as water transitions to a solid, it expands, exerting sufficient force to rip through metal, for example in residential copper water pipes that freeze. Where does that energy come from?
| The change from liquid to solid releases some energy as (stronger) bonds are made between the water molecules - latent heat of fusion.
The expansion pushes back the surroundings and work is done at the expense of some of the released energy when the bonds are being made.
The internal energy (potential energy) of the water decreases as heat is abstracted from the water and the water (ice) does work on the surroundings in expanding.
As the bonds are formed the molecules are closer together than their equilibrium separation ie there is “elastic” potential energy stored in bonds and when the containing vessel is ruptured those compressed bonds expand just like a compressed spring releasing energy when it expands.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/399166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A doubt related to Significant Digits Could someone please explain this statement to me
"Reporting the result of measurement that includes more digits than significant digits is superfluous and also misleading since it would give a wrong idea about the precision of measurement."
Also, shouldn't the word accuracy be used instead of precision because isn't precision the closeness of various measurements for the same quantity and only one measurement is being talked of here.
| Suppose I have a ruler marked in millimeters, and I can estimate lengths measured with my ruler to about half a millimetre by estimating by eye how far between marks my length is. So when I measure a length my precision is $\pm 0.5$ mm.
If I measure a length then report it as $3.14159$ mm this would be be misleading because it would imply that I have measured it to an accuracy of one part in the last digit i.e. $\pm 0.00001$ mm. The correct way for me to report it would be $3.0 \pm 0.5$ mm. That's what the statement means.
Note that I used the word precision in my first paragraph. That's because the precision refers to how accurately I can use my measuring equipment i.e. my ruler. But suppose it turns out my one metre rule was actually only $90$ cm long due to a manufacturing error. That means no matter how precisely I do my measurement it's still going to have an inaccuracy of $10$%. The accuracy refers to the comparison of my reported result with the real length while the precision refers only to how precisely I can use my equipment.
In this case my experiment has a systematic error due to my badly manufactured ruler. Systematic errors are the bane of an experimenter's life because they are not easily detected. The precision of my measurement can be estimated just by doing the measurement lots of times and seeing how much variation there is in my results, but a systematic error affects all the measurements in the same way so it isn't easily detected.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/399313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is it easier to apply torque via short bursts There are two popular tools I use to apply torque to a fastener (bolt, screw, etc.): an impact driver and a drill.
The drill is a motor hooked up to some gears and eventually a bit that fits over the fastener. If I want to apply 40 lb/ft of torque, I feel as though I have to brace myself for that amount of torque, like by using both hands and my body.
The impact driver is a similar motor to the drill, but there is a spring-loaded mechanism that applies the same amount of energy, but in short bursts rather than continuously. I can easily apply 40 lb/ft of torque with my wrist barely moving; using two hands or bracing myself doesn't really make a difference.
Why is this the case? Why is there no equivalent force on my wrist when using the impact driver?
This may be similar to using a hammer to drive a fastener into the ground: if I generate force by swinging very fast with a hammer, why isn't there an equivalent force that lifts me off the ground?
| the reason you can sustain torque bursts easier with the rotary impact tool than steady torque loading from the nonimpact tool with the same average power rating is the rotary inertia of the impact tool reflects most of the impact shock into the tool bit before it can get transmitted out of the tool and into your hands (i.e., it is inertially clamped). Demolition tools like pneumatic and electric jackhammers operate on the same principle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/399457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How is the centripetal force of a car when turning distributed over the wheels?
The centripetal force can easily be calculated as: $F = (M*v^2)/R = (M*v^2)*sin(\delta)/L$. But how is this force distributed over the (front and rear) wheels? My initial thought was to just divide it by 4 for each wheel, but when you turn your front wheels 90 degrees, there will be no force over the rear wheels. So when simply dividing by 4 is wrong, then how is the distribution in reality?
Is it also safe to assume the forces on the front wheels are equal to each other, and also the same for the rear wheels?
| The two-wheel model shown in the question is good start for determining the loading on the tires. But this would only result with the total load on the front tires, and the total load on the back tires, without any further details on how these loads are distributed left-to-right. To get there, you also need to consider the height of the center of mass relative to the wheel axles.
In any case, I will proceed with the two-wheel model below.
The laws of motion provide with a balance of forces and well as a balance of torques for you to consider. Maybe doing a free body diagram first, might help. I hope you are familiar with high school physics, trigonometry, and the concept of vectors.
Here centerline of the front wheels is at A, the centerline of the rear wheels at B and the center of mass at C. The side forces at the rear are $\vec{F}_B$ directed towards the center of rotation O, the side forces at the front are $\vec{F}_A$ also directed towards O, driving force on the front is $\vec{P}_A$ (assume front wheel drive car), and the centrifugal force $\vec{F}_C$ is located at C pointing away from O.
Notice also that the radius of turn $R$ is located where the rear wheels are, describing an arc with radius equal to $$R = L \cot \delta$$ where $L$ is the distance $\overline{AB}$.
The rotational speed of the vehicle $$\Omega = \frac{v_B}{R} = \frac{v_B^2 \tan \delta}{L}$$ where $v_B$ is the speed on the rear wheels. Also the radius of the center of mass is $$R_C = \sqrt{R^2 + c^2} = \sqrt{\left( c^2 + L^2 \cot^2 \delta\right)} $$
The arc radius of the front wheels is needed also $R_A = \sqrt{L^2+R^2}$
So the centrifugal force magnitude is $$F_C = m \Omega^2 R_C = m \frac{v_B^2 \tan \delta}{L} \sqrt{\left( 1 + \frac{c^2 \tan^2 \delta}{L^2}\right)}$$
But the car might also accelerate or decelerate changing its rear-wheel speed with acceleration $a_B$ causing rotational acceleration $$\dot \Omega = \frac{a_B}{R}$$ and inertial force on the center of mass $$P_C = m \dot \Omega R_c = m a_B \sqrt{ \left( 1+ \frac{c^2 \tan^2 \delta}{L^2} \right)}$$
Additionally, the change in rotation requires a net torque $M_C$ to be applied to the center of mass that equals to $$M_C = I_C \dot \Omega = I_C \frac{a_B \tan \delta}{L}$$ where $I_C$ is the mass moment of inertia of the car about the center of mass. This can be estimated with $I_C \approx m c (L-c)$ by approximating the car as a lumped mass of $\tfrac{L-c}{L} m$ on the rear wheels and $\tfrac{c}{L}m$ on the front wheels.
And now for the balance of forces and torques (about the center of mass)
$$\begin{aligned}
F_B + \tfrac{R}{R_A} F_A + \tfrac{c}{R_C} P_A - (m \dot \Omega R_C) \tfrac{c}{R_C} - (m \Omega^2 R_C) \tfrac{R}{R_A} & = 0 \\
P_A \tfrac{R}{R_C} - F_A \tfrac{L}{R_A} - (m \dot \Omega R_C) \tfrac{R}{R_C} + (m \Omega^2 R_C) \tfrac{L}{R_A} & = 0 \\
c \left( F_B \right) - (L-c) \left( F_A \tfrac{R}{R_A} + P_A \tfrac{L}{R_A} \right) - I_C \dot \Omega & = 0
\end{aligned}$$
Where front-wheel forces act along the $\delta = \tan^{-1} \left( \tfrac{L}{R} \right)$ angle and center-of-mass forces act along the $\delta_c = \tan^{-1} \left( \tfrac{c}{R} \right)$ angle. The above is to be solved for $F_A$, $F_B$ and $P_A$ given a specific rotational acceleration $\dot \Omega$.
The results are very complex, but they can be simplified further with some assumptions. For example, the constant speed assumption, $\dot \Omega = 0 $ has the following solution
$$ \begin{aligned}
F_A & = (m \Omega^2) \frac{L^3 R_C-L^2 R_C c +L R_A c^2 + R^2 R_A c}{L(L^2-L c+R_A R_C)} \\
F_B & = (m \Omega^2) \frac{R (L-c)(L\,c+R^2)}{L(L^2-L c+R_A R_C)} \\
P_A & = (m \Omega^2) \frac{R\,R_C (c-L)}{L^2-L c+R_A R_C} \\
\end{aligned}$$
To see if this is even remotely correct, I created a small angle approximation for the general case (small steering angle $\delta$) to find that
$$ \begin{aligned}
F_A & \approx m \frac{c v_B^2-(L^2-c^2) a_B}{L^2} \sin \delta \\
F_B & \approx m \frac{(L-c) v_B^2 + (L^2-c^2) a_B}{L^2} \sin \delta \\
P_A & \approx m a_B \cos \delta \\
\end{aligned}$$
which kind of makes intuitive sense, in terms of how speed $v_B$ and acceleration $a_B$ might affect the tire forces.
and for the non-accelerating case
$$ \begin{aligned}
F_A & \approx m \frac{c\, v_B^2}{L^2} \sin \delta \\
F_B & \approx m \frac{(L-c) v_B^2 }{L^2} \sin \delta \\
P_A & \approx 0 \\
\end{aligned}$$
which is a result that matches the general rule of the best handling cars are the ones with $c = L/2$. Use this above to find $F_A \approx F_B$, which means the tires are equally loaded resulting in the best handling limit (This is the topic of different discussion).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/399761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Question regarding vector direction and Unit Circle
Find the electric field at the origin for a line of charge density λ on the $y > 0$ portion of the unit circle if λ is constant.
The solution provided was:
$$\int_{0}^{\pi}\frac{\lambda d \phi}{4\pi e_0}(-\hat\rho).$$
If it is $y>0$, why is the direction ($-\hat\rho$) negative?
| Note that the electric field points from positive to negative, so if $\lambda > 0$, the electric field points away from the line, towards the origin (negative radial direction). Then the integral is just summing up these contributions. Also, the value of $\hat{\rho}$ depends implicitly on $\phi$, as the radial direction changes as you move on the semi-circle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/399895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Arranging solenoids into a sphere to create a magnetic monopole? [I believe that this question is very different than the question about arranging magnets into a sphere. Solenoids are ruled by Ampere's Law and magnets are not. Ampere's Law is what could make it work.]
I am wondering if Ampere's Law can be a loophole around Gauss's law of magnetism (Gauss's Law not allowing for magnetic monopoles). With solenoids arranged in a sphere, particularly if you had two of these solenoid-sphere's (one positive and one negative), couldn't the field lines be induced to extend out rather than loop back around on themselves?
Arrange solenoids into a sphere so that the outside of the sphere is made only of the positive ends of the solenoids, and the inside of the sphere only has the negative end of the solenoids (or vice-versa). Doesn't Ampere's Law suggest that there would be no place for the field lines to reconnect? The contour integrals have a maximum capacity that is used up by the field lines moving towards (or away from) the center. Could this not create a magnetic monopole?
Whether or not this would work, surely someone has done the experiment, right? I cannot find evidence of this experiment. I really want to know how this experiment turned out, does anyone know?
| If your solenoids were thin enough to behave approximately as Dirac strings, with an effective monopole at each end, your sphere would have an equal number of negative and positive magnetic charges. A Gaussian surface that enclosed the entire sphere would still have zero magnetic flux. Symmetry considerations might suggest you'd have a monopole-like magnetic field in the region between the negative and positive shells, as in the case for electric charges, but the presence of the many solenoids extending from the sphere of negative ends to the sphere of positive ends would make a physical implementation quite complicated --- nothing at all like the symmetric case for electric fields, where a "small" charge like a nanocoulomb is made of $10^{10}$ fundamental charges.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/400061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What exactly is deterministic in Schrödinger's equation? I have read the following on Wikipedia but I can't understand it:
In quantum mechanics, the Schrödinger equation, which describes the continuous time evolution of a system's wave function, is deterministic. However, the relationship between a system's wave function and the observable properties of the system appears to be non-deterministic.
–"Deterministic system", Wikipedia [links omitted]
How can a system be deterministic and not deterministic at the same time?
Can anyone explain simply?
| TL;DR Schrödinger's equation determines the wave function. The wave function determines probabilities. But a particle's position, for example, is not determined by a probability.
If you know the wave function $\Psi(x,t)$ at some time $t_0$, Schrödinger's equation will give you the wave function for all $t>t_0$. This means that the equation determines the evolution of the wave function$\dagger$.
If we want to know, say, the position of a particle, QM won't give us a function $x(t)$ defined for all $t$. What QM can give us is the probability of the particle being in some interval $(a,b)$ at time $t$. But a probability alone doesn't determines the position of a particle. The particle may as well be outside $(a,b)$. Thus the theory isn't deterministic.
$\dagger$ Note however that measurements will collapse the wave function in a non-deterministic way.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/400162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 9,
"answer_id": 7
} |
Why polarization filter do not dim the light completely? In a circle there's infinite amount of degrees (eg. 0 deg, 0.00000000000...1 deg etc.) In a ground school we are thought that there's 360 degrees in a circle.
A landscape behind my window is incoherent light source, so it randomly emits photons with all polarization directions.
When I put a polarizer between landscape and my eye... i can still see the everything. But how is that possible if the polarizer transmits only $1/\infty$ of all photons (since there's infinite amount of directions of polarization)?
Even if we assume that there's just 360 degrees in circle... The landscape behind my window is not 360 times darker when I observe it through filter (eg. polarization glasses).
Why won't polarizer dim the light severely?
| A polarizer does not work by transmitting one very specific polarization angle and blocking all other - even only very slightly different - polarizations completely.
Instead a polarizer obeys Malus' law: If the angle between the axis of polarization of the incident light and the polarizer's polarization angle is $\theta$, then the fraction of transmitted intensity to incident intensity is $\cos^2(\theta)$. Or, equivalently speaking in terms of photons, the probability that a photon with definite polarization at such an angle is transmitted and not absorbed is $\cos^2(\theta)$.
The reason for this is that any incident polarization can always be seen as a linear superposition of the polarizer's polarization and the one orthogonal to it. The orthogonal polarization is completely blocked and the parallel polarization is completely transmitted - this is the definition of what a perfect polarizer does. Now, an incident electric field vector $\vec E$ at angle $\theta$ is
$$ \vec E_0 = \cos(\theta) E_{||} + \sin(\theta)E_{\perp}$$
in terms of the field in the parallel direction $E_{||}$ and the field in the orthogonal direction $E_\perp$. After the polarizer, all that is left is $\vec E_1 = \cos(\theta) E_{||}$. Therefore, the ratio of the intensities is
$$ \frac{I_1}{I_0} \propto \frac{(\vec E_1)^2}{(\vec E_0)^2} = \cos^2(\theta),$$
which shows Malus' law.
If the polarization angle of the the incident radiation is uniformly randomly distributed among the possible angles, then the average intensity will be
$$ \frac{1}{2\pi}\int^{2\pi}_0\cos^2(\theta)\mathrm{d}\theta = \frac{1}{2},$$
i.e. "totally unpolarized" photons have, on average, an equal chance to be absorbed by or transmitted through the polarizer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/400387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why do two-photon interactions only occur at extremely high energies? I've been reading for pleasure about two-photon interaction experiments, and one thing that confuses me is why, for example, two photons in the visible spectrum cannot interact. Is this indeed the case, and if so, why?
| Photons are electrically neutral, so their interaction is by vacuum polarisation. The involvement of virtual charged particles makes the interaction difficult at energies $\ll m_e c^2$, $m_e$ the electron mass. (Difficult, but not impossible; pair production, of course, would be.) Eq. (1) here, first derived in 1935, shows the cross-section is proportional to the sixth power of the photons' CM energy. (It's also proportional to $\alpha^4$, with $\alpha\approx 1/137$ the fine-structure constant, so obviously $\alpha^4$ is very small.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/400526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the drift velocity of an electron in a wire with constant current depend on the area of its cross section? Suppose current, $I=nevA$ with the electron density $n$, the cross section of the conductor $A$ and the drift velocity $v = \frac{V}{RneA}$ where $R$ is the resistance of the wire. This equation becomes $v/dlne$ where $d$ is the resistivity of the wire if we break the $R$ into the formula $R=\frac{dl} A$. So drift velocity should not depend on $A$ or area of cross section. Which way am I wrong?
| $V$ in your equation is the voltage drop over the restive element. It's not the voltage in the circuit.
It depends on the resistance of the element and cannot be presumed to be constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/400697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why two opposite circular polarization filters let light pass through? I've been playing with 3D glasses. I punched out one of the filters, stacked them with some space in between and looked through both the filters without flipping anything over. I expected the interaction to be like between linear polarization filters at 90° angle - no light should pass. If they were linear polarization filters at 90° all light that passed throught the first filter would be polarized according to the first filter and would not be able to pass through the second. So I thought all light that was able to pass through the first circular polarization filter would have, lets say clockwise polarization and it will not be able to pass through the second filter, because the second would only let counter-clockwise polarized light through. But actually a lot of light passes through both the filters - it only changes its tint between blue and yellow depending on the angle I twist them at.
I know the equation describing the chance of a linearly polarized photon to pass through a linear filter, but I don't know any such equations about circular filters.
| These glasses consist of two layers: an ordinary linear polaroid-type polarizer and a quarter-wave retarder plate. The $\lambda/4$ retarder is the front surface. It converts circularly polarized light to linear.
Whay happens then if you put a second analyzer for circular polarization behind the first one? The linearly polarized light from the first analyzer will then pass through a $\lambda/4$ retarder and be converted to circular polarized light. Half of this will be absorbed by the second linear polarizer, but half will pass through.
So the analyzers need to face eachother. One way to accomplish this without punching them out is to look at oneself in a mirror with these glasses.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/400769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can there be general relativity without special relativity? Can General Relativity be correct if Special Relativity is incorrect?
| As far as I know, and I'm not an expert or a physicist, Special Relativity developed the principle of Relativity which stipulates mainly two rules:
*
*All events follow the same laws of physics in every frame of reference.
*The independence of the speed of light, which means that light is always moving at a constant value (in the vacuum, at 299792458 m/s), even if one is in an inertial frame and another accelerating.
General Relativity also follows this principle, because the speed of light is always the same in vacuum for all observers, it doesn't change. This might be incorrect, but I think that also GR without SR it would not work. We can take the ISS as an example:
We know that the gravity becomes weaker the further that you go from the source of the gravitational field; in our case, the ISS is further from the earth, our source of gravity, than us. Because of the time dilation due to gravity, the further that you go from the source, the faster that time passes, so we might guess that in the ISS the time for the astronauts goes faster than for us. But we need to remember that the ISS is moving, and is moving so fast, fast enough for the relativity effects to be non-negligible, and we know that the fastest and object is moving, the time dilation increase due to the Lorentz factor, and the time gets slower. So having this two assumptions, the SR effects are bigger in the ISS than the GR effects, so the time there gets slower than for us, and the astronauts in the ISS are getting older slowly.
I hope that this could help you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/400910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
} |
Comparing work in thermodynamics with work done in mechanics Let us the consider a gas as our system enclosed in a cylinder with piston.
1st case(Expansion of gas):
Here force on the piston is exerted by the gas in upward direction and during expansion piston moves up. So, the work done here is positive(force and displacement in same direction). Also the relation W=PΔV (with their usual meanings) also satisfies the "positive" sense of work, since the volume increases during expansion.
2nd case(Compression of gas):
Here, the surrounding exerts the force on the piston and compresses the gas. Since, the direction of force by surrounding on the system and displacement of piston(both downwards) are in same direction, should not the work done by the surrounding on the system should be positive?
But, W=PΔV gives -ve work, since volume decreases during compression. Why does the mechanical concept of work and W=PΔV does not give same result?
(In Physics)We are usually told that work done by the system is positive and work done by the surrounding on the system is negative.
But
2nd picture shows exactly what I am confused with. During compression, it is the gas that does negative work not the surrounding does the work in the gas?
1st picture is screenshot of book University Physics.
2nd picture is from here.
| In mechanics as well as in thermodynamics, the work done by the system on the surroundings is equal to the force vector the system exerts on the surroundings dotted with the displacement vector. If the force vector exerted by the system on its surroundings is in the same direction as the displacement (as, for example, when the mechanical system is decelerating), the work done by the system on its surroundings has a positive sign (and the work done by the surroundings on the system has a negative sign). If the force vector exerted by the system on its surroundings is in the opposite direction of the displacement (as, for example, when the mechanical system is accelerating), the work done by the system on its surroundings has a negative sign (and the work done by the surroundings on the system has a positive sign).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/401134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Where does the buoyant force come from? If I place a cube in water, the force at the top of the cube, $F_1$ will be $Ah\rho_wg$.
Where,
$A =$ cross-sectional area
$h =$ height at the top
$\rho_w =$ density of water
$g$ = acceleration due to gravity
All this made sense, because this downward Force ($F_1$) is actually the weight of the water above the cube.....but where did the upward force come from? is it the opposite force of the weight of the cube? or something else?
| Lets assume you position a 10 cm cube one meter under the water.
The water pressure on the four sides cancel each other. So the cube is experiencing the difference between opposing pressures on the top and bottom surfaces.
The top pressure is $h g\times\rho_{water}.\times 100cm $
The bottom pressure is $(h+10cm)g\times\rho_{water}\times 100cm $
That excess pressure on the bottom surface is the buoyancy.
$$ Buoyancy = (h+10-h) g\times\rho\times 100cm = (10cm\times100cm)\times g\rho_{water} = cube.volume\times g\rho_{water} $$
If its less than the weight of cube the cube will sink to the bottom but will weigh les. If it is more than the weight of cube it will push the cube up above the surface of water until the volume of submerged part is offering enough buoyancy to keep the floating equilibrium.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/401229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
are redshift and Doppler effect identity with speed of light? we know that redshift and Doppler effect say wavelength and energy are relative to the speed of cause body, so how can I understand the special relativity which said that the speed of light is same regardless of the speed of body emitting light?
| *
*redshift happens to a photon when it is coming towards us from a far away galaxy that is receding from us, in expanding space
*as the photon travels in expanding space, its wavelength gets bigger, so it gets redshifted
*the speed of the photons is c when measured locally, and it does not depend on the emitting body, you are right
*when the photons are coming from a receding galaxy, in expanding space, their wavelength gets bigger, their frequency smaller, so they lose energy, you are right
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/401381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.