Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Reason behind vector addition law What is the reason behind triangle law of vector addition, in other words, how is this really justified?
| A naïve answer...
If you go from A to B then from B to C, you can represent your displacement from A to B as an arrow and from B to C to another arrow. Clearly your displacement from A to C can be represented by an arrow going from A to C or by the two arrows already mentioned, placed with the tail of the second touching the head of the first. This is the rule for adding displacements and arguably it is self-evident. The rule can be extended to any number of displacements.
Velocity is displacement per unit time and so velocities must add as displacements.
Momentum and acceleration are defined in terms of velocity, so momenta and accelerations must add as displacements.
The argument can be extended, via Newton's second law, to forces and field strengths.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/488889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Power generation by extracting heat from the environment? It is chilly outside now and I am kept warm by a heat pump with COP 5.5: eating only 0.6 kW, it steals 2.6 kW from the outside air (making it even colder) and delivers the resulting 3.2 kW into my house.
Which makes me wonder: why isn't it practicable to generate electricity by drawing thermal energy from the environment? Do the laws of physics prevent the heat that heat pumps deliver from being used for electricity generation by whatever means (e.g. steam turbine)?
| This is all about entropy and the difference between heat and work. Electrical power is a form of work. To get work output using heat transfer, the best one can do is such that there is no net change in entropy of the system and its surroundings. So you run a reversible heat engine (power stations in practice do their best to get close to this ideal case). In such a heat engine, the entropy drawn out of the hot side $\Delta S_1 = Q_1/T_1$ is equal to the entropy delivered to the cold side $\Delta S_2 = Q_2/T_2$ so the heats exchanged are in the ratio
$$
\frac{Q_1}{Q_2} = \frac{T_1}{T_2}
$$
The work obtained, which you can consider the electrical energy generated, is
$$
W = Q_1 - Q_2
$$
So the efficiency, measured as work out per amount of heat supplied, is
$$
\frac{W}{Q_1} = 1 - \frac{Q_1}{Q_2} = 1 - \frac{T_1}{T_2} .
$$
This is all completely standard thermodynamics of heat engines. Your heat pump works by the same principle. In the heat pump, $Q_1$ is the heat delivered to the hotter place, and $W$ is the work supplied in order to make this possible. Overall, therefore, if you use a perfect heat pump to drive a perfect heat engine, then the net ratio of total electrical energy out to work energy supplied is $1$. You just break even and get no overall benefit. In practice, owing to imperfection such as friction and heat conduction, one won't achieve unit efficiency so instead there is a net loss of organised energy and a generation of heat and entropy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/489126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Coulomb's law experiment, why consider the distance between the body's centers? In $F_e=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$, why did Coulomb consider the distance between the center of the charges? Why not the distance between the immediate outer surface between the two bodies? Did he assume that this "electric ability" is perhaps condensed in the center?
Does this law present a deficiency?
| Coulomb’s Law as you wrote it applies to point charges.
When you aren’t dealing with point charges, you can treat a continuous charge distribution as a collection of infinitesimal volume elements that can be treated as point charges, where $dq=\rho dV$, with $\rho$ being the charge density. You can then integrate the infinitesimal forces between all the infinitesimal charges.
When you have spherically symmetric charge distributions, such as two spheres with charge distributed uniformly over their surfaces, this integration happens to produce a result that looks just like Coulomb’s Law for point charges! The integral gives the product of the total charge on each sphere, divided by the square of the distance between their centers, as if the charge of each sphere were concentrated at its center. This remarkable mathematical result, which applies to for any inverse square force, is called the shell theorem.
It’s the same in electroststics as in gravity, where the Earth and Sun attract each other as if all their mass were concentrated at their centers. But what is really happening is that every atom in the Sun is attracting every atom in the Earth.
If you put charge on two conducting spheres, it is free to move around, and it will not be uniformly distributed. For example, if both spheres are positively charged, the charge will tend to move toward the side of each sphere that faces away from the other sphere. The spheres won’t have a spherically symmetric charge distribution. In this case, the force will not be the product of the total charge on each sphere divided by the square of the distance between their centers, because the shell theorem will not apply.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/489216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How did Einstein know the speed of light was constant? I often hear the story of how Einstein came up to the conclusion that time would slow down the faster you move, because the speed of light has to remain the same.
My question is, how did Einstein know that measuring the speed of light wouldn't be affected by the speed at which you are moving. Was this common knowledge already before Einstein published his paper on special relativity? If not, what led him to that conclusion?
| In 1887 the Michelson Morley experiment provided evidence that the speed of light was independent of the direction of travel of the observer (in their case, they used the movement of the Earth around the sun).
Morley conducted additional experiments with Dayton Miller from 1902 to 1904 which confirmed the results.
Einstein was aware of this (at least later), but it is not clear (and he himself seems to have been unsure) what influence it had, if any, on his 1905 paper. On this matter, there is more detail given in the answers to the question linked in the comment above, which supports the idea that Einstein was more influenced by theoretical arguments.
The fact that the experiment yielded a negative result (they were looking for evidence of the "Ether") may have made it less noteworthy at the time. But its influence was indirect in that it did not invalidate the theories that relied on a constant speed of light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/489291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "92",
"answer_count": 6,
"answer_id": 2
} |
Why are massive particles clocks? Or are they not? I recently came across a public lecture "Dark matter decay?" by Sir Roger Penrose.
In his lecture he states that the two equations
$E=mc^2$ and $E=h\nu$ can be combined to form a formula for the frequency $$\nu=mc^2/h$$
He also states that "massive particles are clocks". I didn't quite understand this statement. I understand the derivation but why does this equation show that "massive particles are clocks". What is the concept behind this statement?
| Suppose you annihilate an electron and an antielectron, producing two gamma rays. Your equation then gives the frequency of the gamma rays. This is a sort of universal time unit. If we get in contact with extraterrestrials through SETI, one way of correlating our time units would be this.
On the other hand, suppose we lived in a universe without massive particles, only massless ones such as gluons, photons, and gravitons. Then there would be no numbers or standards to start with in defining such an absolute scale of time. The technical terminology is that under these circumstances, the laws of physics are conformally invariant. This is a universe in which time becomes kind of a timey-wimey thing that lacks any absolute meaning.
Penrose had a failed theory based on this called CCC.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/489585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Irradiance Measurement of Incandescent Lamps I have two spectroradiometers that use the same detector from two different manufacturers an Oceaan Optics FX and a Gamma Scientific GS1220. They each have their own cal lamps.
After calibrating the FX I can measure the absolute irradiance of that lamp and it looks great, just like the cal file. When I measure the GS1220 lamp with the FX the the curve is not "smooth" there is sort of a knee in the 700 to 750 nm range. The peak irradiance is pretty close to the cal for the lamp.
Could this be because the FX cal lamp has a color temp of 2800K and the GS1220's is 2600K?
| It looks to me that the root of the problem lies in the different reflectance properties of the two lamps holders and interfaces to the integrating sphere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/489721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Numerical calculation of skyrmion number I'm having trouble finding a numerical method to calculate the skyrmion number for a certain system. I want to calculate
$$ n = \frac{1}{4\pi}\int \textbf{M} \cdot \left( \frac{\partial \textbf{M}}{\partial x} \times \frac{\partial \textbf{M}}{\partial y} \right) dxdy$$
for a simulation I performed. The simulations results are in an archive with the following format
($x_1$,$y_1$,$M_{x1}$,$M_{y1}$,$M_{z1}$)
($x_2$,$y_2$,$M_{x2}$,$M_{y2}$,$M_{z2}$)
.
.
($x_N$,$y_N$,$M_{xN}$,$M_{yN}$,$M_{zN}$).
I had an idea of using cubic splines to interpolate the points and then integrate with any method. That way, with the splines i should have the value of the partial derivatives. The problem is I can't find lots of information about these splines for multivariable functions, just for one variable. I would like to know if there's another -and more efficient- method or should I keep going with the idea I have, in which case I would really appreciate some sources or tips to achieve it. Thanks.
| You can compute the skyrmion number discretising the integrals using finite differences, and then using the midpoint rule for the numerical integration. It's a rough approximation but it gives reasonable results, specially if your FD mesh has a decent resolution.
I wrote a tool to compute the sk number from OOMMF files a while ago, you can give it a try or check the documentation of the function
https://github.com/davidcortesortuno/oommfpy
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/489987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Help me distinguish between an Amperian loop and a surface
This image is from the Resnick-Halliday-Krane textbook.
These images have been used to see that a correction was needed in Ampere's circuital law.
What I am confused about is, how are the two end surfaces of a 3D surface used as Amperian loops and how do the different results from application of Ampere's law present any anomaly? To me, the 2D surface inside the capacitor should be independent of any result that comes from the 2D surface outside the capacitor.
| By Stoke's theorem you are free to use any surface bounded by your Amperian loop. That surface does not need to be contained in a single plane.
The problem that then arises is that Ampere's law as just
$$\int\mathbf B\cdot\text d\mathbf l=\mu_0 I_{enc}$$
doesn't work here because the right hand side for each case is different for the two surfaces ($I_{enc}=i$ for the top, and $I_{enc}=0$ for the bottom), but the left hand side is the same for each case (same field and same Amperian loop).
This is why you need the displacement current. To make the $I_{enc}$ the same in both cases so that Ampere's law (Stoke's theorem) is valid.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/490278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
2 dimensional massless scalar field propagator in position space I have been trying to calculate the massless scalar field propagator in position space by directly Fourier transforming the momentum space propagator. $$\int{d^2p\frac{1}{(p^0)^2-(p^1)^2}e^{-i(p^0t-p^1x)}}$$
Upon referring to multiple sources (linked below), I realize that the answer is actually proportional to $ln|x|$ but I don't see how this integral will result in that answer. All of these sources obtain that answer by finding the massive propagator and then taking the $m\rightarrow 0$ limit. I don't see what I am missing by directly doing doing the above integral.
To see how the above integral does not give $ln|x|$:
Evaluate the $dp^0$ integral using the Feynman prescription for avoiding the poles and this will give: $$\int\frac{i}{2\pi p^1}e^{-ip^1 (t-x)}dp^1$$This integral is actually a constant multipled by a step function.
I also head into a similar problem in the (1+3)-D case where a direct Fourier transform gives a different answer from the known propagator and from the answer got by taking the limit on the massive case. So, what am I missing by directly Fourier transforming the propagator from momentum space?
Sources:
*
*http://max2.physics.sunysb.edu/~rastelli/HW4Solutions.pdf
*H. Zhang, K. Feng, S. Qiu, A. Zhao and X. Li, "On analytic formulas of Feynman propagators in position space", Chinese Phys. C 34 (2010) 1576, arXiv:0811.1261.
*Phys.SE Q: Two-point function of massless scalar theory in 2d CFT
*Phys.SE Q: Massless limit of the Klein-Gordon propagator
| The idea for this kind of computation is the following. Firstly, add a mass term to the propagator. This will yield
$$
\int \frac{d^2p}{(2\pi)^2}\frac{1}{p^2-m^2}e^{ip\cdot x}.
$$
This integral can be evaluated provided we make the rotation $p_0\rightarrow ip_0$ that yields
$$
i\int \frac{d^2p}{(2\pi)^2}\frac{1}{p^2+m^2}e^{ip\cdot x}.
$$
Now, one us $d^2p=pdpd\theta$ and $p\cdot x = pr\cos\theta$ and one has to evaluate the integral
$$
\frac{1}{4\pi^2}\int_0^\infty dp\int_0^{2\pi}d\theta\frac{p}{p^2+m^2}e^{ipr\cos\theta}.
$$
Firtstly, we integrate on $\theta$. This can be done remembering that
$$
e^{ia\cos\theta}=\sum_{n=0}^\infty i^nJ_n(a)e^{in\theta}
$$
being $J_n$ the Bessel functions of the first kind of integer order. Integration in $\theta$ leaves just $J_0$ and so, our integral becomes
$$
-\frac{1}{4\pi^2i}\int_0^\infty dp\frac{p}{p^2+m^2}J_0(pr).
$$
This integral can be evaluated with techniques in complex integration, with a proper choice of the integration path, yielding
$$
G(r)=-\frac{1}{2\pi}K_0(mr)
$$
being $K_0$ the modified Bessel function of 0 order. This is the point where the references you cite bring you. The next step is to note that, for $m\rightarrow 0$, the massless limit,
$$
K_0(mr)\sim -\ln r
$$
and you are done. Note the presence of an infinite constant, $\ln m$, that is generally omitted taking the massless limit. The reason is that, in the massless limit, one can always add an arbitrary constant to the propagator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/490445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Electromagnetic waves according to Maxwell If a variable Electric field creates a variable magnetic field and VICE VERSA (according to Maxwell's equations), then why don't we enter a loop where E vector and B vector keep creating one another until they reach infinite magnitudes?
| The fields are vectors with (signed) direction. In a wave, the $\mathbf{B}$ field "creates" $\mathbf{E}$ field components, but they are, at some times at least, opposite to the currently present $\mathbf{E}$ field and therefore reduce the total field. And vice versa. This manifests via the relative minus sign between Faraday's law
$$ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$
and
Ampere's law
$$ \nabla \times \mathbf{B} = \frac{\partial \mathbf{E}}{\partial t}$$
(shown here in units where $c=1$ and in vacuum).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/490669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Newton's third law in magnetic fields Say I have a charged particle moving through a magnetic field perpendicular to it. It will experience a force, but according to Newton third law
Every force has an equal and opposite reaction.
So what is the opposite reaction/force of this magnetic force.
Which body experiences this force?
| @Jon Custer is right if a magnet producing the magnetic field is present.
But there is more to learn of this question: As Hertz famously discovered, there are so called electro-magnetic waves.
These waves are made up of alternating electric and magnetic fields, that are unrelated to any physical object in the classical newtonian sense. This is different to the magnetic field of the magnet.
Since Newton's Third Law is very much equivalent to conservation of momentum I will concentrate on this formulation of Newton's theory.
*
*The downfall of classical conservation of momentum: These fields can of course exhibit force on a charged particle with
non-zero mass, very much like the magnetic field of the magnet.
Therefore the fields are changing the momentum of the particle. This
is the downfall of the classical concept of conservation of
momentum, since there is no other particle that can assert for the overall change of momentum of the entire system. By classical I mean that momentum is just \begin{equation}
\mathbf{p}=m\mathbf{v} \end{equation} and therefore only associated
with mass. This is the Newtonian view on monentum.
*Why momentum is still conserved in a broader sense: Experiments have
shown that the fields themselves or the electromagnetic wave for
that purpose carry momentum themselves. So the change in momentum of
the carged particle is compensated by the change of momentum of the
electromagnetic wave. To fully understand this concept you shouldy
study Maxwell's theory.
Remark: I edited large parts of this answer, as it didn't meet my quality standard anymore and caused misunderstandings in the comment section.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/490909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do the thermal conductivity and specific heat influence the amount of heat transferred between two objects Through research on the topic I found that the one with the higher thermal effusivity would be able to transfer more heat. I was wondering what the effect of one material being in liquid state vs solid state would have on the transfer of heat if any at all?
| One essential difference between a solid and a liquid is that the liquid is mobile, so it can transfer heat via convection in addition to thermal conductivity. How much of a difference that makes will depend on a number of factors, such as the viscosity of the liquid or initial speed at which the liquid is moving.
Another important factor is the contact area. If we assume perfect contact with no gaps, there will be no difference between solids and liquids, but in practice the surface finish of the solids will play a very important role in the heat transfer, as will the wettability of the solid by the liquid, and what the gaps will be filled with.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/491052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $X\otimes X$ not the simultaneous position operator? I had thought that $X\otimes X$ would be the operator on $H_1\otimes H_2$ to simultaneously measure the x-positions of two particles.
But there seems to be something wrong with this -- for a given eigenvalue $z$, there is an entire subspace $\mathrm{Span}\left(|x\rangle\otimes|z/x\rangle\right)$ associated with it, so we don't get precise positions from measuring it, just the product of positions $x_1x_2$.
If so, what's the right operator representing "simultaneous measurement" of the x-positions? Is that even possible -- to have "vector eigenvalues"? Or do we just need a spacefilling curve or something?
| Let $\Omega$ denote any set of one or more mutually commuting self-adjoint operators, such as the observables corresponding to the $x$-coordinates of two particles. Let $\Omega'$ denote the commutant, which is the set of all oeprators that commute with everything in $\Omega$. Then let $\Omega''$ denote the double commutant, which is the set of all operators that commute with everything that commutes with everything in $\Omega$. (That's not a typo.) We can also describe $\Omega''$ as the (commutative) von Neumann algebra generated by $\Omega$.
(Technically, a von Neumann algebra contains only bounded operators, and operators like $X$ are unbounded, but that technicality doesn't affect the spirit of this answer.)
If the operators in $\Omega$ qualify as observables in the given model, then the algebra $\Omega''$ contains all of the projection operators that we need to characterize the possible outcomes of a simultaneous measurement of the observables in $\Omega$.
Here's the key: Every commutative von Neumann algebra is generated by a single self-adjoint operator [1][2]. That proves the existence of a single self-adjoint operator representing the simultaneous measurement of all of the observables in $\Omega$. Actually constructing such an operator is a different problem, and it probably wouldn't be useful. (As explained in knzhou's answer, $X\otimes X$ doesn't work.) Using separate operators, one for each coordinate of each particle, is more convenient.
References:
[1] EP10 on page 23 in Jones (2009), "Von Neumann Algebras," https://math.berkeley.edu/~vfr/VonNeumann2009.pdf
[2] Lemma 1 in Suzuki and Saitô (1963), "On the operators which generate continuous von Neumann algebras," https://projecteuclid.org/euclid.tmj/1178243811
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/491383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Do objects besides strings, ropes, and rods have tension? Why do we define tension only in strings and ropes and rods and such? Shouldn't every object experience tension force? Like when you pull a paper from opposite sides, it gets taut, and experiences what seems like a state of tension. If every object does experience tension, can you define tension?
| Pick up a brick. Try pulling it apart. What stops you from pulling it apart? It's the tension in the brick. Where does the tension in the brick come from? It comes from the inter-molecular forces inside the brick.
The same goes for strings. Here's a second example. If you look closely at a thick rope, it's made of many small strings wrapped around each other. Why is this? So that the friction in between the strings increases the available tension in the entire rope. In other words, the inter-molecular forces (in this case, friction) increase within the rope, making the rope "want" to stay together more; the more a rope wants to stay together, the more force it can handle when you try to pull it apart.
It's just like how normal force is the force from electron shells repelling each other (due to the Pauli Exclusion Principle), tension comes from a micro-scale force as well.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/491524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
The specific heat changes based on the actual temperature. How do I calculate the thermodynamic equilibrium when this value changes? I have 2 substances with huge differences in temperature, with the specific heat of one starting at 0.6 J/kgC and ending at about 2.8 J/kgC. It goes from -200°C to 0°C. With this huge difference in specific heat, how can I calculate the thermal equilibrium with another substance? All I know is a formula for the fixed specific heat, because that usually doesn't change as much. But is there a formula for it changing as much as it does?
| Instead of using $C(T_f-T_i)$ you use $$\int_{T_i}^{T_f}{C(T)dT}$$ To find the final temperature, this might then require you to solve a non-linear algebraic equation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/491809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trouble with Math in Physics I am a current high school student and I am very interested in physics, especially particle physics (that stuff is super cool!). Unfortunately, my school only teaches classical physics, so I have to continue my study at home. I've read several books and watched videos and online lectures on quantum mechanics and have gotten a basic overview of the big ideas, but when I try to dig even just a little bit deeper the math immediately gets too confusing to handle. Beyond basic summaries and oversimplifications, what other resources can I use to continue studying physics but avoid getting frustrated by math that I don't understand?
EDIT
Based on feedback so far, I think the question I should really be asking is "What are some resources that can help teach me the prerequisite math I need to know for quantum mechanics?"
| Ah I remember my high school days of yearning to learn complex physics (it was only 10 years ago lol). I’d start at one Wikipedia page and go from page to page trying to understand something but having to try to understand something else first. It was a lack of prerequisites as Allure said. If that happens, don’t let it frustrate you. I also recommend Griffiths but yes, you’ll need to understand linear algebra and calculus to get through it (integrals and partial derivatives are everywhere in quantum)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/492058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How does alternating current provide energy? In my head, direct current makes complete sense; the electrons carry energy around the circuit to something being powered losing its potential and then return to the battery or whatnot to have their potential raised again.
This is probably wrong, so I would like an explanation of how a direct current actually transfers energy and also how an alternating current does (I have zero intuition for this).
| Imagine a current going through a resistor and generating heat.
Does it matter which way the current goes? No, it doesn't; you get heat either way. So reversing the current many times a second, as AC, still generates heat.
Sometimes people get tangled up because they somehow think that electrons are "used up" in electric circuits. But they're not. The power source gives them energy and sends them through the circuit, where the energy is lost. That still works if the power source sends them through the other way.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/492136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Common potential in Capacitors If two isolated charged capacitors (of different capacitance) are connected in parallel to each other they acquire a common potential. But suppose if i connect positive plate of one capacitor to negative plate of another capacitor will they still acquire a common potential or will the charge acquired by two capacitors be same as the circuit looks like that the capacitors are connected in series in which -ve plate of capacitors is connected to + ve plate of another capacitors
| If you are asking about connecting positive to negative and negative to positive in a closed circuit with two charged capacitors, and nothing else, then you are creating a direct short which will drain both, heat the entire circuit, and may damage capacitors, and, or wires.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/492614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is carrying a heavy object more taxing on the body than pushing the same object on wheels? Where is the "extra help" coming from when rolling the object on wheels?
| Carrying a heavy weight feels more strenuous and uses more energy because your muscles are resisting the force of gravity on the load, as well as supplying the forward momentum for it. When you put the load on wheels, as the Viet Cong did with their bicycles on the Hoe Chi Minh trail, then provided you have a smooth, flat surface to travel on, you only have to provide the forward momentum. The energy you would otherwise use in resisting the force of gravity on the load is no longer provided by your muscles. That accounts for the load, but of course you still have the force of gravity on your body to deal with.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/492712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the difference between uniform velocity and constant velocity? I think that uniform velocity implies constant speed but not constant direction. while constant velocity implies constant speed without any changes in direction.
Both tell us that there's no acceleration (since magnitude of velocity is constant).
The same goes for acceleration: both imply constant magnitude, but only constant acceleration means that there's no change in its direction.
However, a lot of people on the Internet argue that whether it's the other way around or that there's no difference at all. Who's right and who's wrong?
| I don't believe the distinction between "uniform" and "constant" in this context is important: I would use them interchangeably. I certainly have not encountered any serious technical usage of these terms in this context that relied on an implicit knowledge of any such difference. In general, I would take both "uniform velocity" and "constant velocity" to mean a velocity vector that is not changing in magnitude or direction. The same goes for acceleration. If this is not the case in a certain situation and the difference between "constant velocity" and "constant speed" is important, you can expect the meaning to be clear from the context, or stated explicitly.
As for your third sentence, a constant magnitude of velocity does not mean there is no acceleration. Any body rotating in a circle at constant speed has a non-zero (centripetal) acceleration.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/492797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Momentum operator in QM - scalar or vector? The momentum operator for one spatial dimension is $-i \hbar\frac{\mathrm d}{\mathrm dx}$ (which isn't a vector operator) but for 3 spatial dimensions is $-i\hbar\nabla$ which is a vector operator. So is it a vector or a scalar operator?
| Momentum is a vector operator. Period.
When restricted to one-dimensional problems, momentum becomes a one-dimensional vector, which coincides with scalars in that space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/492919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
K theory and Holography I have a general or overview question related to charges on D- Branes lies in the K theory of Spacetime. We normally think charges of D branes lies in the Cohomology like $D_0$ branes couple to RR-1 form and so on.
The whole idea behind AdS/CFT duality (Maldacena' paper) is based on the fact that D-branes couple to RR form and it curves the spacetime. Can we think of this duality in terms of K-theory? By thinking in terms of K theory will give something new in this Holography business.
| This is a good question which -- as far as I am aware -- has received little to no attention (but see below).
On the one hand it is clear why the question is outside the scope of traditional discussions:
AdS/CFT is commonly practiced in the large-$N$ limit of a huge (in fact humongous) number $N$ of branes. In this limit the branes are classical, in fact they are well-described by the eponymous asymptotic AdS-throats in (super-)gravity. But the stable branes that K-theory sees beyond this limit are quantum states, as witnessed by the fact that they are torsion subgroup elements in K-theory -- meaning that some finite multiple of them vanishes, so that a large $N$-limit does not even make sense for them.
On the other hand, the AdS/CFT correspondence may be and originally was conjectured to hold (see here) also for small $N$ (large $1/N$), in which case, however, it no longer involves (super-)gravity but full-blown strongly-coupled string theory aka "M-theory". It was fashionable at some point to hence claim that AdS/CFT defines M-theory in the small $N$-limit, but not much technical discussion of what this entails has ensued.
Of course this case of small $N$ -- in which, yes, D-branes will have to be described by K-theory -- is the case ultimately of interest in AdS/QCD: here $N = 3 \ll \infty$.
One place that comments on K-theory in non-perturbative holography is a note we are preparing, see Remark 2.8 at Anyonic topological order in TED K-theory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/493059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the only absolute difference between types of light frequency? Probably a bad question but for some reason, it seems too simple in my head that anyone at home could theoretically create anything from radio waves to gamma waves by generating electrical signals at different frequencies.
Say I had a electronic frequency generator that was able to produce a signal at any frequency, and for illustrative purposes, say there was a diode hooked up this generator that could receive its signals.
If it created a signal at $10^{12}$ Hz, the diode would give off infrared radiation.
If I increased the signal to $10^{20}$ Hz, the diode would give off gamma radiation.
I’m using this example just to emphasize my question, is frequency the absolute and only differentiator in types of light on the electromagnetic spectrum?
| There are no different types of light in the elementary sense. Whys should we distinguish types of light based on frequency or spin, as we don't do this for any other particle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/493394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
In QFT, are forces made out of multiple fields? I’ve been reading about 1,5 books about quantum physics and I’ve also watched a few YouTube videos. In one book, I learnt that there are fields, such as the electromagnetic field, which carries forces between particles (vibrations in the field) via virtual particles. But in another book and on YouTube, they instead say that every elementary particle (quarks, photons, electrons etc.) have their own fields. But what I can’t understand is what the relation of the fields I read about that carries forces to the ones with the particles? Are these forces made up of interactions between multiple fields?
| Short answer: forces are particles (except maybe gravity). But not all particles carry a force!
Look for the Standard Model. A particle carrying a force is simply a "special" type of particle called a "force carrier". The photon is the most common example of a particle being also a force carrier (the electromagnetic interaction).
In QFT, there is an equivalence between particles and fields (or field excitations, to be more accurate).
Thus, the following statements are equivalent (and equivalently approximative in their phrasing):
*
*Fields carry forces between particles
*Particles carry forces between fields
*Fields carry forces between fields
*Particles carry forces between particles
Now, when considering a particle vs. considering a force, there can be some subtle differences — as for example the occurrence of virtual particles (as you pointed out) in some interactions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/493512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Applying Kirchhoff Rule to an LC circuit In a simple LC circuit (just one capacitor and one inductor), using the Kirchhoff rule has $$\frac{Q}{C} + L\frac{dI}{dt} = 0.$$ But isn't the voltage drop across an inductor $-L\frac{dI}{dt}$? What happened to the negative sign?
| You have to remember that, when a capacitor is discharging and the current on the inductor is increasing, then:
$$q=q_o-it$$
Therefore:
$$\frac{dq}{dt}=-i
\quad\Rightarrow\quad\frac{d^2q}{dt^2}=-\frac{di}{dt}$$
Upon doing the loop rule, you get:
$$-L\frac{di}{dt}+\frac{q}{C}=0
\quad\Rightarrow\quad
L\frac{d^2q}{dt^2}+\frac{q}{C}=0$$
That last equation is the equation we were looking for.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/493616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to find distance given candela values on a coordinate system? Consider this diagram:
Let's say that some lamp at 1.5D, 2R has a luminous intensity of 35,000 cd. I want to find the maximum distance at which this lamp can put 5 lux on the road.
I understand that candela, lux, and distance are related as such:
$$ E_v = 10.763 × \frac {Iv}{d^2} $$
Solving the above equation gives ~275 feet. 35,000 candela can put out 5 lux of illumination at 275 feet.
Now, let's say I want to calculate the 5 lux illumination distance of this theoretical lamp while it's mounted on a vehicle.
Would the 275 feet value be the hypotenuse of the right triangle I set up below?
| No. The base of this right triangle (you have labelled "road") will be the distance at which that point in the beam, 1.5D, intersects the road surface. With a more or less typical passenger car lamp mount height of 0.69m, your 35,000-candela hot spot at 1.5D will strike the road at 26.35 m, or just shy of 86.5 feet.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/493952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the difference between position, displacement, and distance traveled? Suppose the question is somewhat like this:
If $v=8-4t$ and the position at time $t= 0\ \rm s$ is $2\ \rm m$, find the distance traveled, displacement, and final position at $t=3\ \rm s$
Since $\text dx/\text dt=v=8-4t$, then $\text dx=(8-4t)\text dt$. After integrating we find $x(t)-2=8t-2t^2$, and substituting the value of $t=3\ \rm s$ we get $x(3)=8\ \rm m$.
Is the answer that I found displacement, position or distance?
It can't be distance. I am sure of this. But is it position or displacement?
| If we go from point A to point B we undergo a displacement. This is the distance from A to B, together with its direction, and is the archetypal vector quantity.
We can give the position of a point as a displacement from some agreed datum point or origin, O.
In your question – which in my opinion is not nicely worded – the motion is presumably along a straight line. Saying that the position at time 0 is 2 m implies a displacement of +2 m from some origin. Your x is the position of the particle at time t, that is its displacement from the origin at time t.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/494096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
How to understand observables in quantum field theory I am reading a paper about quantum field theory, something that I am new to. I have some experience with quantum mechanics. In the paper, it explains how a field is a function from a spacetime manifold(M) to a vector space(V):
Φ:M→V
The paper then explains the form of the observables. It explains how an observable is the value of Φ at x∈M, but our calculations will work out better if we instead look at average values of Φ in small regions of space, as infinitely much energy is required to know the value of a field at a point. To calculate the average values of the field in a small region of space, and thus our observable, the paper defines a Schwartz space, S, as a space of Schwartz functions of the form f:M→V*. It then explains that the observables are of the form:
$O_f (Φ) = \int_M 〈f,Φ〉$
So at this point I am confused. How does this integral lead to an average value of the field within a region of space? I would really appreciate any help showing me how the integral above is indeed the form of an observable for the average value of the field within a small region in space. Specifically, how does one carry out the calculation that gives them a value for $O_f$? Thanks in advance for any help.
| The following is a Schwartz function, for any $\epsilon > 0 $:
$$ f_\epsilon (x) = N \int e^{-(\|\vec{x}-\vec{y}\|^2 + (x^0 -y^0)^2)/\epsilon^2} \theta(\|\vec{y}\|-R) \theta(|y^0|-T/2) d^4 y $$
This is a smoothened version of the function which is $1$ in a ball of radius $R$ in space and for a time duration $T$, and zero outside of that. Integrating an observable against $f_\epsilon(x)$ thus, heuristically, gives the average of the observable in the ball of radius $R$, in a time interval of length $T$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/494213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does magnetic moment change under inversion symmetry? Since magnetic moment can be view as a small electric current circle. Pictorially, when apply inversion operation, the current direction is reversed, so I think the $\vec{m}\to -\vec{m}$ under inversion symmetry operation.
On the other hand, the formula for the magnetic moment is $\vec{m}=\int_V \vec{r}\times\vec{j}\mathrm{d}V$. Under the inversion symmetry operation $\vec{r}\to -\vec{r}$ and $\vec{j}\to -\vec{j}$, therefore $\vec{m}$ is unchanged.
The above two reasoning must have one being wrong, which one and what is the flaw of the reasoning?
| Under a full parity inversion, the magnetic moment is unchanged. In your first paragraph, the current is reversed, but the location of the current is also changed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/494488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Motion between two particles in a relative manner Suppose a particle A is travelling in east direction with velocity of x m/s and another particle B is travelling with velocity y m/s in the west direction. Why does the the particle B appears to move towards A with a velocity of x+y and not just y m/s?
| Suppose body $A$ is going to the right with a speed $x$ and another body $B$ is going to the left at a speed $y$.
The motion of the bodies can be represented as vector diagram 1.
To both motions add a velocity to the left of equal magnitude to that of the velocity of body $A$, ie stopping body $A$, as shown in vector diagram 2.
On adding the two vectors body $A$ is at rest and body $B$ is moving at speed $x+y$ to the left as shown in vector diagram 3 and this is the velocity of body $B$ relative to body $A$.
In symbols let $\hat l$ and $\hat r$ be unit vectors in the left and right direction such that $\hat l= - \hat r$.
Step 1 - The velocity of body $A$ is $x\hat r$ and than of body $B$ is $y\hat l$.
Step 2 - To both motions add a velocity to the left of equal magnitude to that of the velocity of body $A$ $(x\hat l)$
Step 3 - Velocity of $A$ is $x\hat r + x \hat l = x(-\hat l) + x \hat l =0$ and velocity of $B$ is $y\hat l + x \hat l = (x+y) \hat l$ and this is the velocity of body $B$ relative to body $A$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/494630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Resistors are used to reduce current to prevent light bulbs from "exploding" but it's also said that "current remains same at all points in a circuit" Resistors are used to reduce current in order to prevent light bulbs and other electrical components from “exploding”, but it is also said that “current remains the same at all points in a series circuit”.
Then what's the point of using resistor when the current will remain the same, and as it would remain the same, the light bulb would still receive that high current which could cause it to explode.
So, again, why are we using resistors? If current is still going to be the same, what's the point? The light bulb would still explode if it will not be receiving a lower current. Also, I'm talking about the series circuit, not a parallel one.
| You seem to be proposing that all series circuits have the same current. That isn't true.
The equivalent resistance of a set of resistor in series is $$R_{eq}=\sum_iR_i$$
i.e. you just find the sum of all the resistances. If they are all ohmic then the current that flows through the circuit with voltage source $V$ is given by
$$I=\frac{V}{R_{eq}}$$
Therefore, the more resistors you add the smaller the current through the circuit will be.
The power dissipated by a lightbulb in this circuit is given by $P=I^2R$, where $R$ is the lightbulb's resistance. Therefore, a lower current means less power dissipation, and less of a chance of exploding.
The current is constant throughout a series circuit (i.e. at all points in the circuit). That doesn't mean all series circuits have the same current no matter what. Certainly adding more resistance changes the current. That is why is is called resistance. It resists the flow of current.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/494828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Lorentz boost in light-cone coordinates Consider a particle with momentum $p^{\mu}=(p^+,p^-,p_{\perp})$, where the momentum is written in light cone coordinates defined as,
\begin{align}
n^{\mu}&=(1,0,0,1)& \bar{n}^{\mu}=(1,0,0,-1)
\end{align}
and $p^+\equiv n.p$ and $p^-\equiv \bar{n}.p$. I am not able to understand how the light cone coordinated transform under Lorentz transformation. If the particle is boosted along $-z$ direction by amount $\kappa$, then $p^-\to \kappa p^-$ and $p^+\to p^+/\kappa$. This is basically equation 2.9 of link.
Please explain or share useful link.
| In the usual coordinate system where $p_t^2-(p_x^2+p_y^2+p_z^2)$ is Lorentz invariant, a boost in the $t$-$z$ plane is
\begin{gather}
p_t\to p_t\cosh\theta+p_z\sinh\theta
\\
p_z\to p_z\cosh\theta+p_t\sinh\theta
\end{gather}
with
$$
\cosh^2\theta-\sinh^2\theta = 1.
$$
In light-cone coordinates with $p_\pm = p_t\pm p_z$, this implies
\begin{align}
p_+&\to (\cosh\theta+\sinh\theta)p_+ = \kappa p_+
\\
p_-&\to (\cosh\theta-\sinh\theta)p_- = (1/\kappa) p_-
\end{align}
with $\kappa=e^\theta$. This follows from the definitions of $\cosh\theta$ and $\sinh\theta$, namely
$$
\cosh\theta=\frac{e^\theta+e^{-\theta}}{2}
\hskip2cm
\sinh\theta=\frac{e^\theta-e^{-\theta}}{2}.
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/495509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can the flow be irrotational if the viscous forces act on fluid? I tried to answer the question only using the definitions and the Navier-Stokes equation:
$$\rho \frac{Dv}{Dt} = -\nabla P +\rho g -\mu[\nabla \times(\nabla \times v)] $$
In my opinion if the vorticity is zero, then the fluid is irrotational, regardless of presence of the viscous forces, thus $\mu$ can have a non-zero value which implies the existence of viskeuze forces, while the $\nabla \times v = 0$.
| The answer to the title question is yes, it is possible for the flow to be irrotational if there are nonzero viscous forces acting on fluid.
As an example, let us consider a simple yet physically meaningful example of such flow: radial, spherically symmetric flow in a viscous incompressible fluid (without gravity). A physical realization for such flow is a bubble of gas expanding into the space filled with viscous fluid.
Assuming that the fluid velocity is purely radial $\mathbf{v}= \hat{\mathbf r}\,v_r(r,t)$, the continuity equation gives us:
$$
\mathbf{v}= \frac{\hat{\mathbf r} f(t)}{r^2}.
$$
This flow is irrotational and Navier–Stokes equation (with viscous term identically zero) could be solved for the pressure.
So, while viscosity does not enter the equations, the viscous stress tensor is nonzero, and there would be energy dissipation within the fluid volume. Consequently, viscosity would enter the solutions either through energy balance equation or through boundary conditions. For example, at the fluid–gas interface of above mentioned spherical bubble we would have (ignoring gas viscosity and surface tension):
$$
- p_\text{gas} = -p_\text{fluid}+2 \mu \frac{\partial v_r}{\partial r}
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/495738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Equivalence of POVM and projective measurement Suppose I have a POVM whose elements are given by $\{M_i^\dagger M_i\}$ such that $\sum_i M_i^\dagger M_i = I_A$. Let it act on some state $\rho_A$. Everything here happens in the Hilbert space $A$.
By Neumark's theorem, it is known that one can write this POVM as a PVM by using
*
*An extension of the state to include an ancilla i.e. $\rho_A\otimes \vert 0_B\rangle\langle 0_B\vert$
*A unitary operator $U_{AB}$
*Projective measurement on the ancilla.
The unitary operator and the POVM elements are related in the following way
$$M_i = \langle 0_B\vert U_{AB}\vert i_B \rangle$$
How does one show that the unitarity of $U$ guarantees that $\sum_i M^\dagger_i M_i = I_A$? My attempt below is stuck at the first step and I'm not sure how to proceed.
$$\sum_i M^\dagger_i M_i = \sum_i \langle i_B\vert U^\dagger_{AB}\vert 0_B\rangle\langle 0_B\vert U_{AB}\vert i_B \rangle$$
| There is a mistake (or rather: inconsistency) in how you define the $M_i$ (which makes the condition you want to prove incorrect, so there cannot be a proof!).
To be consistent, given the Kraus representation $\rho\mapsto \sum M_i\rho M_i^\dagger$, you need to define
$$
M_i = \langle i_B| U |0_B\rangle
$$
(with $U$ the unitary in the Stinespring dilation and $|i_B\rangle$ the outcome of the projective measurement).
In that case, the trace-preserving condition corresponds to
$$
\sum M_i^\dagger M_i = I\ .
$$
This can then indeed be immediately proven from
\begin{align}
\sum_i M_i^\dagger M_i &= \sum_i \langle 0_B| U^\dagger |i_B\rangle \langle i_B| U |0_B\rangle
\\
&= \langle 0_B| U^\dagger U |0_B\rangle
\\
&= \langle 0_B| I_{AB} |0_B\rangle = I_A\ .
\end{align}
Note that with the convention you chose above, the condition $\sum M_i^\dagger M_i=I$ (with my convention, $\sum M_i M_i^\dagger = I$) corresponds to a unital channel, and is thus not satisfied in general!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/496048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can force be applied without accelerating? When I push against a wall, I am applying force on the wall and the wall applies an equal force against mine therefore the wall doesn't move and neither does my hand. But isn't acceleration required to apply force? My hand is not accelerating when I am applyin the force. Still let's assume that the muscle fibres are accelerating, but how is the wall accelerating to apply an opposite force. So are the atoms accelerating somehow?
|
But isn't acceleration required to apply force?
In fact, there's a branch of mechanics concerned with (static) applied forces that (vector) sum to zero so that there's no net force and, thus, no acceleration.
For example, consider the analysis of a truss in a steady state condition.
Image credit
Note that there are applied forces and yet the truss is static.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/496138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Meaning of time derivative of the Lorentz factor $\gamma$? This question about the Lorentz factor $\gamma$ in special relativity. I know what $\gamma$ means and how to drive. I'm wondering if I have time derivative of $\gamma$, what dose it mean conceptually?
| For a test particle, it's essentially the power being delivered to a test particle by a force, because $E=m\gamma$ (or $E=m\gamma c^2$ in units where $c\ne1$).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/496246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Normalization constant of a planar wave As we know for the plane waves ( $ae^{i k x}+b e^{-i k x}$), the normalization constant can be easily obtained from the integral $\int^{x_{2}}_{x_{1}}\psi^{*}\psi dx=1$ by the relation $|a|^{2}+|b|^{2}=1$. But what happens if the parameter $k$ is imaginary, i.e. $k=i \kappa$ where $\kappa$ is real. Do we have the same relation for the normalization?
| Plane waves can't be normalised, because they don't represent physically realisable states. It doesn't make sense to normalise a function like $ \psi = ae^{ikx} + be^{-ikx} $ over the boundary $(x_1, x_2)$ unless the particle is bounded, in which case the wavefunction will have a different solution. Another way to think about this: "There's no such thing as a free particle with a definite energy." See Griffiths intro to QM section 2.4
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/496385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Are the electrons' orbitals the same for all atoms? Are the electronic orbitals of an atom always quantified in the same way (i.e. the same energy required to reach the next level), or does each atom have its own values for each level?
If the quantification is universal, then the creation of photons (due to the deexcitation of the electrons) at the wavelength / color corresponding to the transition should be more abundant in the universe than all the other frequency. Except one detects no more photon of a given wavelength than of another.
So where is my reasoning error?
| To add to @john-rennie 's answer: things get even worse as the environment of the atom makes additional modifications to the energy levels (ranging in the meV). For example, a carbon atom bound to an oxygen will have a slight energy shift compared with a free carbon atom, or a carbon atom bound in a diamond structure. These shifts are very useful in identifying molecular species.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/496531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why are you doing no work when carrying a body through a horizontal distance? Work done on an object is equal to $$FD\cos(\mathrm{angle}).$$
So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.
However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.
But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?
Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction,
| You do work on the book. To carry the book in the horizontal direction you move with a force equal to Mass of the person plus mass of book times acceleration. F = Ma where M is the total mass. Hence work done is the force times the distance traveled. You have to keep applying a force to overcome atmospheric resistance and the friction of the ground.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/496644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What happens if a traveling sound wave encounters vacuum? Suppose a sound wave is emitted by an object in a medium like a gas so it travels in some direction. If the wave meets a rigid object, for example a wall, it reflects back as one should expect; and if it encounters another medium, like a denser one, it will be transmitted to the other side. In those cases the sound wave keeps traveling, and of course it can loose energy in its path and even be absorbed by some object, so it can be converted into another form of energy.
Now, I'm aware that sound waves can't travel in vacuum, so my question is: What happens to a sound wave that is traveling in some medium and encounters vacuum ?. Here's a drawing of the situation:
I'm thinking about the usual wave phenomenoma and reflection is not a logical option since there is not a defined object that can work like a wall, and transmission doesn't make any sense at all since sound waves can't travel through vacuum. So what happens with that wave ? Where does its energy go ?.
| It would get reflected due to the impedance transition from free air to vacuum. The exact nature of reflection will depend on the exact way how the air is separated from the vacuum. Whatever means this may be, it would have to prevent air molecules to enter the vacuum, i.e the normal velocity component at the boundary must be zero.
In other words: you need a some sort of a wall and the boundary would behave acoustically like a wall does.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/496791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is kinetic energy always conserved in an elastic collision/impact? While working out through some problems I encountered this problem :
A ball moving with a velocity $v$ hits a massive wall moving towards the ball with a velocity $u$. An elastic impact lasts for a time $\Delta t$
Now I have to answer whether the Kinetic energy of ball increases or remains same after collision.
In the theory books which I read, it is mentioned that Kinetic energy is conserved before and after in an elastic collision.
So that way for the above question Kinetic energy should be conserved.
But the answer given is that Kinetic energy increases.
So my question is how is it possible for Kinetic energy to increase after an elastic impact ? Is it because of the time interval $\Delta t$?
|
In the theory books which I read, it is mentioned that Kinetic energy is conserved before and after in an elastic collision.
Yes, but keep in mind this is the total kinetic energy. i.e. it's the sum of kinetic energy of both the ball and the wall.
So my question is how is it possible for Kinetic energy to increase after an elastic impact ? Is it because of the time interval Δt?
The total kinetic energy is constant, by the definition of elastic collision. However, your question is asking about just the ball. If the ball's kinetic energy increases, then the wall's kinetic energy must decrease.
Therefore, it looks like your confusion lies in what is being talked about when. The question is talking about just the ball. When we talk about kinetic energy being conserved in elastic collisions, we are talking about the entire system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/496923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to figure out the distribution of Normal Force on a body? I was solving a problem and came across some confusion regarding the point of application of Normal Force.
In Classical Mechanics 101, we had always treated the Normal Force as acting on a point (which can be called as the "centre" of the Normal Force), and it was calculated by applying Newton's Laws (Balancing forces and torques).
But how to I determine the exact distribution of the Normal Force on the surface of a body if the contact is not at points?
Because there can be a lot of distributions that may result in the same centre for the Normal force.
So, is this model for Normal force incomplete?
Edit
The problem I was solving was this.
Imagine a book kept on a piece of wood such that the book covers only half of the piece. I can figure out the centre of Normal Reaction but not how it's distributed over the surface.
| The distribution of a force over the surface of a body (as opposed to a "point") is called stress. If the force is normal (perpendicular) to the body, it is called normal stress and is given as
$$σ=\frac{F}{A}$$
When you deal with forces on deformable bodies (and all bodies are deformable) the force is never applied to a "point". A point has no area. From the equation you can see that if $A$ is zero, you would have an infinite applied stress.
The deformation δ for uniaxial loading due to the applied stress is given by:
$$δ=\frac{PL}{AE}$$
where $P$ is the load (normal force), $L$ is the length of the member, $A$ is the cross sectional area and $E$ is the modulus of elasticity of the material which is a measure of its stiffness and $\frac{P}{A}$ is the normal stress.
So if the load were applied to a point (zero Area) the body would need to be infinitely stiff (an idea rigid body) in order to have no deformation. There is no such thing as an ideal rigid body, at least at the macroscopic level.
Hope this helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/497332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Total force on upper block in two block system If a block $m$ is placed on another block $M$ and a force $F$ is applied on bolck $M$. Then how many forces are acting on block $m$.(Friction is non zero)
The image is taken from this site.
Is pseudo force acting on block $m$ or not?
| Pseudo-forces only appear if you work in a non-inertial frame of reference.
In this case, you could work in a frame of reference in which the lower block is stationary. Because we know that the lower block is actually being accelerated by the force $F$, this must be a non-inertial frame of reference, so pseudo-forces will appear.
In particular, if the upper block does not slip then it is also stationary in this non-inertial frame of reference. In order to "pretend" that the upper block is in equilibrium, we need to introduce a horizontal force that is equal and opposite to the force of friction that the lower block exerts on the upper block. This fictitious force is a pseudo-force - it is only introduced to make the forces on the upper block appear to balance.
As pointed out in other answers, it is actually much simpler to work in an inertial frame of reference, and then you don't need to introduce pseudo-forces at all.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/497577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Schwarzschild Radius of a Galaxy If an ultra compact/dense Galaxy has a Schwarzschild radius same as it is own radius, how can it be observed from the outside of the Galaxy?
| Such Galaxy would probably look more like one single mass-body, no empty spaces left in between, would you still call it a galaxy?
Gravitational effects as usual can be detected,so its presence would still be inferred indirectly. If you are speaking about detecting it through light, then light bending such as the lensing produced might provide you with indirect detection.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/497769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Symmetry reason why magnetic dipole transitions are suppressed In the theory of light-matter interaction, electric dipole transitions between two atomic states of same parity are forbidden. This is because the Hamiltonian conserves parity. Is there a symmetry reason why magnetic dipole transitions happen but their amplitudes are extremely suppressed?
| No. As a counterexample, magnetic dipole transitions in nuclei are often quite strong, competing against E1 and E2 transitions on fairly even terms. Since nuclei and the relevant forces (strong and electromagnetic) don't break any symmetry that is unbroken for atoms, any relative weakness of magnetic dipole transitions in atoms can't be due to symmetry.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/498039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why do all fields in a QFT transform like *irreducible* representations of some group? Emphasis is on the irreducible. I get what's special about them. But is there some principle that I'm missing, that says it can only be irreducible representations? Or is it just 'more beautiful' and usually the first thing people tried?
Whenever I'm reading about some GUT ($SU(5)$, $SO(10)$, you name it) people usually consider some irreducible rep as a candidate field. Also, the SM Lagrangian is constructed in this way. (Here, experimental evidence of course suggests it.)
| Irreducible representations are always determined by some numbers, labeling the representation, which correspond to the eigenvalues of some observables which are invariant under the (unitary) action of the Lie group.
If the group represents physical transformations connecting different reference frames (Lorentz, Poincare',...), these numbers are therefore viewed as observables which do not depend on the reference frame so that they define some intrinsic property of the elementary physical system one is considering.
If the group represents gauge transformations, these numbers correspond to quantities which are gauge invariant. In this sense they are physical quantities.
Finally, it turns out that in many cases (always if the Lie group is compact), generic unitary representations are constructed as direct sums of irreducible representations. This mathematical fact reflects the physical idea that physical objects are made of elementary physical objects (described by irreducible representations)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/498124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 3
} |
Complex rabi frequency I am learning quantum optics.
In the book written by Scully includes complex rabi frequency, when dealing with $\Delta$ level configuration, the Hamiltonian including laser-atom interacting concludes complex rabi frequency.
However, from the course I took as undergraduate, where rabi frequency are treated as real numbers (vectors), I cannot see why rabi frequency has imaginary part.
Could anyone help me with this? Why it has imaginary part? And how to deduce it?
| The Rabi frequency is defined as
$$ \Omega_{i,j} = \frac{\vec d_{i,j}\cdot \vec E_0}{\hbar} $$
where $\vec d_{i,j}$ is the transition dipole moment for the $i\to j$ transition and $\vec E_{0}={\hat {\epsilon }}E_{0}$ is the vector electric field amplitude. $\hat\epsilon$ is the polarization of the light, which can be complex (e.g. for circular polarized light). But I think more importantly, $E_0$ is in general complex: $E_0 = \text{e}^{\text{i}\varphi}|E_0|$.
More information can be found in chapter 3.2.2 of this document (PDF link).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/498300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If an earthquake can destroy buildings why it cant kill us according to physics? Most earthquakes with magnitude 5.5 and higher can damage or destroy buildings. However, according to my knowledge and experience, I have never seen someone dying from an earthquake itself. Rather, they die from an associated tsunami, damaged buildings, etc.
This seems counter-intuitive, since you need much more force to destroy a building or damage it than to break the human femur or cause similar damage to other species.
| As an addendum to other answers, pressure and volumetric energy density may be more useful in explaining the total effect on a small body vs large structure.
Since reasonably-sized buildings and people on the surface will be a small fraction of the overall earth mass being moved in a quake, the effective pressure $$P=F/L^2$$ in a localized region will be uniform. In other words, the effective force transmitted from the ground to an object will be proportional to the contact area of the object. A person's feet (or even a person lying prostrate on the ground) has a much smaller area than that of a building's rigid foundation.
Energy density has the same units as pressure, like $$[E/V] = [E/L^3] = [{FL}/{L^3}] = [P]$$ Similar calculations would show that the total energy transmitted to each object will be proportional to its mass, density and displacement. The conclusion is the same... that a small body (with less mass and volume) will receive less overall energy from the quaking ground than an entire rigid building of much greater mass. It does not matter that the human body is standing adjacent to or even leaning up against the larger building. The human body will still only receive its small portion of energy directly from the quaking ground.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/498528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Why would entropy of a system be fixed if it can exchange energy with its environment? Entropy maximization and energy minimization are equivalent statements of the same thing, as I understand it. If the internal energy is fixed, entropy is maximized because of statiatical reasons. If the entropy is fixed, and the system can exchange energy with its environment, then the system will give energy to its environment to maximize the environments entropy (and hence total entropy).
But I haven't been able to understand why we would assume here that the entropy of a system is fixed. What physical mechanism causes this?
| The energy maximisation or equivalently entropy maximisation is a property of an equilibrium state. So in the case of when there is energy exchange with the surrounding, clearly the system isn’t in equilibrium. Once all the exchange is complete, the energy/entropy is maximised.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/498742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Heisenberg's uncertainty principle: what is the correct interpretation? In this video: https://www.youtube.com/watch?v=xsnTrAEiyHg
Prof. Walter Lewin showed when a laser beam passes through a very narrow slit the projection of it becomes wider. He claims it is because of the uncertainty principle in action: as we know the position of the light/photons very precisely the momentum becomes uncertain. That implies the direction of the light is no longer determined. But isn't such sense of direction ($p=m \bar{v}$) taken from classical mechanics?
In quantum mechanics the momentum is defined as $p=h/\lambda$. That implies if the momentum becomes uncertain we should get a spread in the wavelength. So instead of the light spreading out on the screen, should we expect some color change?
| I'm not an expert here but having watched the video (very good) I would guess the change in momentum is due to direction change only. Likely the momentum imparted is from the slit itself. So total momentum is conserved but the light has spread. Likely the momentum spread is larger in slit but after the slit it is conserved to the original. We do know wavelength changes changes during refraction as c changes and the frequency remains constant. Being in the slit is similar to refraction in that photons are required to interact with the material.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/498891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Conservation of linear momentum with mass defect Suppose we have an insulating container, like a perfect black body, which absorbs all the radiation coming from a radioactive element placed in the center (or some equivalent process like matter annihilation). Assuming a spherically symmetric emission. It will transform all radiation into thermal energy until reaching thermal equilibrium.
If the $\Delta$m, due to mass defect, doesn't "pop out" from the body, like the example of the rocket, how is momentum preserved?
Does the body accelerate????
$\sum_{i} F_i = \frac{dp}{dt}=\frac{d(vm)}{dt}$
| From the spherical symmetry it is clear that the body cannot accelerate. (Which direction would it go, everything is symmetric). However, let’s look in a little more detail.
As you say, the center radiates and therefore loses mass. The mass lost is equal to the energy of the radiation. By the problem setup the container absorbs all of the radiated energy and therefore the container gains the same mass that was lost by the center. So again, there is no acceleration, but let’s look in a little more detail.
Specifically, let’s consider the situation in an inertial reference frame where the container and center are moving. In this frame, the mass of the center is decreasing so it is losing momentum and the mass of the container is increasing so it is gaining momentum. Since the mass decrease of the center is equal to the mass increase of the container there is no violation of the conservation of momentum, but how does the momentum transfer without causing acceleration?
Although the radiation is spherically symmetric in the rest frame, in this frame the forward traveling radiation is blue-shifted due to the Doppler effect, and the backward radiation is redshifted. The overall radiation therefore carries momentum from the center to the container. The amount of momentum thus carried is exactly equal to the momentum lost due to the mass decrease of the center and the mass gain of the container. Thus the center and the container do not accelerate because all of the momentum transferred by the radiation is already accounted for in the mass change.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/499386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is the statement that $U(x)$ is quadratic for simple harmonic motion equally strong as the statement that $F(x)$ is linear? Is the statement
"If the potential energy of a particle under oscillatory motion is directly
proportional to the second power of displacement from the mean
position, the particle performs a simple harmonic motion."
as strong as saying it in a more famous form of
Simple harmonic motion is a special type of
periodic motion or oscillation where the restoring force is directly
proportional to the displacement and acts in the direction opposite to
that of displacement.
I think it cannot be regarded as equally strong, since if we derive the expression for potential energy for any oscillatory motion, using taylor expansion, we get
only for small oscillations that $$U(x) = \frac{1}{2} U''(x_0) \ x^2$$ (where $U(x_0)$ is the potential energy at mean position) and we only get this result after certain approximations. Can someone please confirm this? Isn't the second statement stronger or more general? What I mean is, is the first statement always true?
| Both of your statements are true for SHM and can be seen as true for approximations as well. Each statement is essentially saying the same thing, because $F=-\text dU/\text dx$, and we can choose $U(0)=0$. Whether or not you use approximations to make these statements applicable to your system doesn't change that.
In other words, if you have approximated $U\propto x^2$, then you are also approximating $F\propto -x$. Both of your statements say the same thing. Therefore, they are equally "strong".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/499574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
From volume density to surface density Let us say we have a volume $V$ having a volumetric mass(charge) density $\rho(x,y,z)$.
Suppose we have a smooth surface $S$ within the volume with a surface mass(charge) density $\sigma(x,y,z)$, if possible, how can one obtain $\sigma (x,y,z)$ from $\rho (x,y,z)$?
If it is an integral, it must be of the form:
$$\sigma = \int \rho dx$$
where $x$ is some spatial coordinate. But a simple example made me doubt of the veracity of the expression above: Consider a solid sphere with mass $M$ and radius $R$, supposing that the formula above is correct and using the spherical symmetry of the problem, the surface density for a surface sphere within the solid sphere will be:
$$\sigma =\int \rho dr$$
since $\rho$ is constant, we can extract it from the integral:$$\sigma =\rho\int dr$$
Also, since we are considering a spherical sheet, the limits for $r$ are $r_{0}-\delta r$ and $r_{0} +\delta r$, where $r_{0}$ is the radius of our sphere, thus:
$$\sigma(r_{0})= \rho \int_{r_{0}-\delta r}^{r_{0}+\delta r} dr=\rho \times 2\delta r$$
taking the limit $\delta r \rightarrow 0$ we finally get:
$$\sigma(r_{0}) =0 ~C/m^2$$
this last equality must certainly be incorrect, for it states that there are no charges on $S$, which is not true.
| Modelling a true volumetric distribution of charge by a surfacic distribution usually only makes sense if one of the dimensions of the distribution is small (yet finite) compared to the others. It would be the case for a shell with a small thickness (difference between outer and inner radii) compared to the surface of the shell. In this case, you could integrate your distribution along the thickness, as you do, but without letting your thickness tend to zero. The implicit assumption while performing this is that you are able to compute the total charge as $ Q= \int \rho(r) S(r)dr \approx S \int \rho(r) dr$, that is $S$ remains approximately constant constant and $\int \rho(r) dr$ becomes your surfacic charge. For a sphere,
the only thing that could make sense is to compute the total charge $Q$ and divide it by the surface of the sphere to obtain the surfacic charge $\sigma = Q/ S$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/499717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\partial_i A_j - \partial_j A_i = \epsilon_{ijk}B_k$ Let us start from $\textbf{B}=\nabla \times \textbf{A}$ and write its components $B_k=\epsilon_{ijk}\partial_i A_j$.
I want to show that $\partial_i A_j - \partial_j A_i = \epsilon_{ijk}B_k$. I can sense that it works, but I want to see it directly. How should I start?
| Using the identity $\epsilon_{kij}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$,$$\epsilon_{ijk}B_k=\epsilon_{ijk}\epsilon_{klm}\partial_lA_m=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_lA_m=\partial_iA_j-\partial_jA_i=F_{ij}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/499817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Some digital watches display digits in black on a white background and in the dark the backlight is reversed. How is this possible? Some digital watches display digits in black on a white background and in the dark the backlight is reversed, blue on a black background. How is this possible?
Here is an example:
https://images-na.ssl-images-amazon.com/images/I/7196xuw29OL.UL1500.jpg
Usually the backlight is black on a blue background and it's very easy to explain:
https://en.wikipedia.org/wiki/Liquid-crystal_display
https://en.wikipedia.org/wiki/Polarizer
https://www.youtube.com/watch?v=MzRCDLre1b4
But even after searching everywhere I did not find an explanation to the case mentioned
The only hypothesis I have (and which I do not really believe)
is that the electroluminescent backlight is located between the LCD and the second polarizer
and that the light it emits is polarized perpendicularly to the second polarizer
and that it is transparent when it is off
Edit:
I may have mistaken the title
In reality the background does not really change, it is the digits that go from black to light (light blue)
Besides, under an external light the effect is clearly visible
Here's a video that shows the effect I'm talking about:
around 2:36
https://www.youtube.com/watch?v=CEpXyujMHaM
By analyzing frame by frame when the backlight switch
we see that the digits change without the background is really changed
It does not seem to be a reversal of the screen when the backlight is switched on
since the bottom does not change
Sorry for the mistake
| In a typical LCD panel with two linear polarizers, rotating either linear polarizer by 90 degrees will invert the colors. For instance, if the linear polarizers are originally cross-oriented (as in the Wikipedia article) so that applying a bias across the liquid crystal prevents the system from transmitting light, rotating one of the polarizers so that the polarization axes are parallel will cause the light to be transmitted almost entirely. The situation without bias is also similarly flipped. Here is a video demonstrating this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/499932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why are the left- and right-hand sides of a differential equation with two separated variables equal to a constant? While deriving the Time Independent Schrodinger Equation, my book mentioned this line.
So time and position of a particle are two independent variables. If they are equal to one another for all values of $t$ & $r$, then why should they be equal to a constant?
Can't we have other solutions to this other than treating both the sides as a constant?
| There are two logical options when you vary $t$: either the value of the left-hand side changes, or it doesn't. If it changes, then the right side must change as well, since they are equal. But the right-hand side can't change when you vary $t$, since it is not a function of $t$! Therefore, since varying $t$ produces no change in the left-hand-side, then the left-hand side must be constant. And since it is equal to the right-hand side, then they are both (the same) constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/500067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Differences between charge quantity and electric charge As a senior middle school from China mainland, I am teaching physics about electric field. I work with my workmates, and we got a problem now. We cannot get an agreement. There are three viewpoints.
The first is that: electric charge is physical attribute and a physical quantity. It means electric charge is a physical quantity. The unit of electrical charge is the coulomb (symbolized C).
The second is that: electric charge is physical attribute. Charge quantity is a physical quantity. The unit of charge quantity is the coulomb (symbolized C). Electric charge has no unit.
The third is that: electric charge is physical attribute and a physical quantity. Charge quantity is a physical quantity too. The unit of electrical charge is the coulomb (symbolized C). The unit of charge quantity is the coulomb (symbolized C) too.
| If you look up the NIST website for the SI units
https://physics.nist.gov/cuu/Units/units.html
you will see that they use both expressions when it comes to charge:"electric charge, quantity of electricity". They are both measured in coulombs so both can be used to designate a physical quantity. At least in the US.
Of course, "electric charge" is also a property, an attribute. This is quite common in the common language, not just in physics jargon. Just another example from Physics, "Current" is both the phenomenon and the physical quantity, even though the quantity in some languages is more precisely defined by the equialent of "intensity of the current". So what is the point to argue about it? An expression is whatever people using it want it to mean. As long as you explain what you mean by the expression, the communication is clear and this is the purpose of language.
In the end, I think it does not matter much from the point of view of undertsanding physics even though is interesting from the point of view of language. And it's no point to argue about language; there is no way to settle such arguments by doing an experiment, as we would do to settle arguments about physics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/500189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The answer to this question is not close to the age of the universe, then why do we say it is? I saw this question in our textbook
A great physicist of the century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics ($c$, $e$, mass of electron, mass of proton) and the gravitational constant $G$. He could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants, try to see if you too can construct this number (or any other interesting number you can think of). If it's coincidence with the age of universe were significant, what would this imply for the constancy of fundamental constants?
The answer comes out to be 6 billion years, which is not close to the age of the universe yet in many places it is said that it is approximately the age of the universe.
I've also talked to people but no one has yet given me an answer that made me understand this.
Please give me an explanation about this.
Also which equation is this? I haven't been able to find it online.
| From the table of fundamental constants, Dirac derived the following:
\begin{aligned}
&\left(\frac{1}{4 \pi \varepsilon_{0}}\right)^{2} \times \frac{e^{4}}{m_{e}^{2} m_{p} c^{3} G} \\
=& \frac{\left(9 \times 10^{9}\right)^{2} \times\left(1.6 \times 10^{-19}\right)^{4}}{\left(9.1 \times 10^{-31}\right)^{2} \times 1.67 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{3} \times 6.67 \times 10^{-11}} \\
=& 2.13 \times 10^{16} \mathrm{~s}
\end{aligned} which does not seem to be of the order of the age of the universe. Both the physical constants and the age of the universe have changed since the time of Dirac. Therefore Dirac hypothesis is obsolete. (https://en.m.wikipedia.org/wiki/Dirac_large_numbers_hypothesis)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/500316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why are some mistakes counted as "experimental error"? I have always thought (and been told) that experimental errors and uncertainties do not include mistakes which can be fixed by just conducting the experiment more carefully. Yet we still include parallax error and zero error when talking about experimental error. Can these two not simply be removed by conducting the experiment carefully? Or is it that these two might happen despite our care?
| Zero error can be usually fixed by re-calibrating measurement device or changing it into a newer not damaged device. It's one-time fix. However story is far more complex about parallax error - you can't be sure that you are looking at device readings at a right angle. Besides usually parallax error will be lost in other more fundamental error source which is half-value of measurement device smallest scale interval. For example. Consider that you are measuring square's perimeter with a ruler which has smallest interval of $1 mm$. Your ruler readings are (10 + 10 + 10 + 10) mm. Then because measurement errors have tendency to add-up, your total measurement error of perimeter is
$$ \frac{1}{2}mm+\frac{1}{2}mm+\frac{1}{2}mm+\frac{1}{2}mm = 2mm $$
Thus your perimeters measurement result would be $$40mm\pm2mm$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/500484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the equivalence of mass and energy a direct consequence of SR or not? Special relativity gives us the invariance of four-vectors. Consistency with Newtonian physics implies the conservation of four-momentum.
The spatial part of four-momentum is
$P^0=m\gamma(v)$
which can be expanded to second order
$P^0 =\frac{1}{c}\left(mc^2+\frac{1}{2}mv^2\right)$
This gives good reason to suspect the equivalence of mass and energy, but is a further hypothesis required to suggest that rest energy, $mc^2$, is interchangeable with other forms of energy, as is seen in nuclear fission, etc.?
| It is a direct consequence of Special Relativity. The equation
$$E^2-(\mathbf{p}c)^2=(mc^2)^2,$$
which expresses the invariance of the length of the energy-momentum four-vector, makes the relationship between mass and energy when $\mathbf{p}=0$ transparently obvious.
I wouldn’t call that relationship “equivalence”. Mass has energy, but energy doesn’t necessarily have mass. For example, a single photon has energy and momentum but no mass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/500585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to relate the Schwarzschild time $t$ to the status of the wormhole featured in the Kruskal diagram? The maximally extended Schwarzschild spacetime as described by a Kruskal diagram features a wormhole which is dynamical. As Kruskal time $T$ elapses, a spacelike hypersurface hits the past singularity, then the wormhole opens up and the hypersurface extends to the parallel universe, eventually the wormhole pinches off and the hypersurface hits the future singularity.
However, it is not clear to me how the Schwarzschild time $t$ relates to the status of the wormhole.
My question is:
If we consider for instance the black hole in the center of our galaxy and our “now” as observers, which is the status of the (hypothetical) wormhole? Is it still to open up, or is it open “now”, or has it already pinched off?
| In the Kruskal diagram, this "wormhole" consists of a single point in the diagram (i.e. a sphere in the spacetime). Hence it exists for only a single moment in Kruskal time. In the Krusal diagram, lines of constant Scharzschild time $t$ are straight lines passing through this point. So the "wormhole" exists for all values of the Schwarzschild time.
Note that in general relativity there is no well defined notion of "now" for a single observer; any spacelike surface between the past and future lightcones can be considered "now" for that observer. In particular, there is nothing special about Schwarzschild time (in this respect).
Note 2: The hypothetical "wormhole" exists only for eternal black holes (i.e. ones that have also have a past horizon). Dynamically formed black holes (like presumably the one in the center of our galaxy) will not have a "wormhole".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/501128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is a piezoelectric material a capacitor? I am reading about piezoelectric elements and I am not sure why a piezoelectric material is like a capacitor. So a mechanical stres on the material will displace the electric charge distribution creating a potential difference. But current cant flow through ? So why is it a capacitor ?
| Because that is exactly the effect a capacitor gives: It causes current to flow in other parts of the circuit around it due to attraction/repulsion of charges on either side of it - but no current can actually flow through it.
*
*In essence, in the piezoelectric element you have an atomic structure of the unit cell which skews the charge distribution towards one side when it is mechanically "squeezed". See this nice illustration from this source:
*
*A capacitor on the other hand is "filled up" with charges when connected to a battery or other voltage source. Charges of opposite sign will place themselves on either of the capacitor's plates. Illustration from this source:
The result in both of these is now that you have a positive net charge on one side and a negative net charge on the other, while there is no conducting material in-between through which charge can move. So, current can't flow through them, but they nevertheless are able to drive current as a voltage source because of their net voltage across their oppositely charged sides - much like a battery, albeit their voltages will vary very differently over time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/502258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Does light have mass? Why? I've been wondering whether light has mass. Yet given the wave-particle duality of light, the statement seems to be affirmative. With that, how to calculate it?
| In quantum field theory, a photon's rest mass is proven to be zero. But relativistically, the photon's energy leads to the relativistic mass $m=\frac{h\nu}{c^2}$.
Related link: http://www.desy.de/user/projects/Physics/Relativity/SR/light_mass.html
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/502412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Ward identity prohibits mass of photon On wikipedia one can read the following statement:
The photon and gluon do not get a mass through renormalization because gauge symmetry protects them from getting a mass. This is a consequence of the Ward identity.
Can someone briefly outline why the forbidden mass is a consequence from the Ward identity?
| Without gauge invariance, the masses of vector bosons would be affected by contributions from higher order Feynman diagrams. As a result, even if the bosons have zero mass in the fundamental theory, their masses can become nonzero due to these contributions.
In the presence of gauge invariance, those diagrams that would have contributed to these changes in the mass cannot contribute thanks to the Ward-Takahashi identities. They cause those diagrams to become zero.
For more detail, one would need to look at the self-energy diagrams that would produce contributions to the boson mass and then see what the effect of the Ward-Takahashi identities are on those diagrams.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/502632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why do things cool down? What I've heard from books and other materials is that heat is nothing but the sum of the movement of molecules. So, as you all know, one common myth breaker was "Unlike in movies, you don't get frozen right away when you get thrown into space".
But the thing that bugs me is that things in the universe eventually cool down. How is that possible when there are no other things around to which the molecules can transfer their heat?
| In the present epoch the universe is a long way from thermal equilibrium. It consists of a large number of isolated hot spots (a.k.a. stars) in a sea of background radiation which has an average temperature of just $2.7$ K.
But stars have finite lifetimes (although this can be a very long time for white dwarf start) so eventually all the stars in the universe will run out of fuel and will cool down until they reach the average temperature of the universe. On very, very long timescales even the protons in stars will may decay into lighter particles.
And the universe is expanding, and cooling as it expands because the wavelength of the photons in the background radiation increase as the universe expands. So eventually the universe will reach thermal equilibrium, when everything that remains in the universe has a temperature just a fraction of a degree above absolute zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/503023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 3
} |
Is the Earth a gyro? Due to rotation and low friction, can the Earth be considered a gyroscope? If so, any interesting implications to this? Thanks
| A spinning celestial body can indeed be considered to be a gyroscope.
Part of the motion pattern of the Earth is that it is subject to a gyroscopic precession with a period of about 26.000 years. The common name for this gyroscopic precession is 'precession of the equinoxes'. The precession of the equinoxes was noticed many hundreds of years before it was recognised that it is a case of gyroscopic precession.
The word 'gyroscope' doesn't necessarily have a sharp definition, but generaaly 'gyroscope' is taken to mean an axially symmetric spinning body, suspended in such a way that the effective point of suspension coincides with the center of mass.
(The expression 'spinning top' on the other hand is commonly used to refer on a spinning object that is standing on a tip, so that the point of suspension (the point where the weight of the object is supported) does not coincide with the center of mass.)
So in all, yeah, the Earth is a gyroscope.
The center of mass arises from the inertial mass of the object, and the center of gravitational attraction arises from the gravitational mass. As we know, gravitational mass and inertial mass are equal, therefore in a uniform gravitational field the center of gravitional attraction and the center of mass will coincide.
When a celestial body is non-spinning, and large enough so that its own gravity pulls it into a perfectly sphere, then the center of gravitational attraction and the center of mass will coincide perfectly.
In the case of the Earth: due to its spinning motion there is an equatorial bulge. Whenever an object is not perfectly spherical, the center of gravitatinal attraction and the center of mass to not quite coincide.
In the case of the Earth the gyroscopic precession is due to the gravitational influences of the Sun and the Moon.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/503159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Why does gradually increasing refractive index coating reduce reflection? Why does gradually increasing refractive index coating reduce reflection?
EDIT:
As @Michael Seifert nicely describes in his answer, reflection only occurs if there is an abrupt change in refractive index. Since for a gradually increasing refractive index, there is no abrupt change, it makes sense that there is very little reflection. However, what I don't understand is how thick this layer needs to be ? ...because I guess that if the gradient of the refractive index is too steep then it will almost look like a abrubt change, right ?
Any literatur reference is also appreciated.
| The question itself isn't entirely clear, but I believe this can be understood in terms of the Fresnel equations. Whenever a light wave hits an interface, some of the wave is reflected and some is transmitted. If the media are very similar to each other, then very little of the wave is reflected; in the limit that the indices of refraction are identical, there is no reflected wave at all. A continuously varying index of refraction can be understood as a succession of interfaces between thin layers, each with a slightly larger index of refraction; if the change in the index of refraction between each "layer" is sufficiently small, then there will be very little reflection at each layer, and (hopefully) very little reflection overall.
An analogy would be water waves coming in towards a beach. The phase velocity of surface waves in shallow water is proportional to the square root of the depth of the water ($c = \sqrt{gh}$.) This means that as the waves travel towards shore, their phase velocity is decreasing, just as the phase velocity of the light decreases as the index of refraction increases. But because the increase in the phase velocity is gradual, the waves never "reflect" off of the shallower water and head back out to sea.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/503260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
In de Sitter space, does the cosmic horizon change its shape for fast-moving observer? If an observer moves at a speed close to the speed of light, will the horizon deviate from spherical shape?
If no, will it be the same horizon as for stationary observer (at the same position)?
| De Sitter space is a maximally symmetric space, which means that all its points and all 4-velocities at each point are equivalent and could be related via an isometry. This means that any inertial observer will experience de Sitter space in the same way, in particular cosmological horizon for each observer would have the same geometry.
Cosmological horizons are observer dependent and so different observers would have different horizons even when their worldlines intersect.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/503639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Spring mass damper system: Distance from equilibrium after applying velocity to mass I have a spring fixed to a wall on one end and a mass object on the other end in its natural resting position. The question is how far does the spring stretch when a velocity $v_0$ is applied to it, assuming there is no friction.
My idea was that the spring will be stretched until the velocity $$v_0=0$$ and the kinetic energy $$k=0$$ resulting in the max $E_{pot}$.
However I can't figure out a relationship between $E_{kin}$ and $E_{pot}$ other than $E_{kin} + E_{pot} = E_{total}$ and therefore I don't know how to continue from here on.
| With the spring attached to a wall (and not the ceiling) one might assume that the mass is sliding on a friction-less horizontal surface. Then the starting kinetic energy = the final potential energy: (½) m vo^2 = (½) k x^2 where k is the spring constant and x is the stretch. If the spring is hanging vertically, the rest position occurs where its force matches gravity. If it is stretched further down, this matching force is added to, linearly. The result is that gravity can be ignored (with upward motion treated as compression). The energy equation remains the same.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/503818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Seeming violation-wave travelling faster than speed of light Consider the basic relation $$E=\sqrt{(pc)^2+(mc^2)^2}.$$
Every particle possesses a wave nature and it depends on the situation in which one among the two is perceptible...
Consider a particle with rest mass $m$. If we consider the speed of De-Broglie Waves, as usual for a wave $$v_{wave}=\nu \lambda.$$ And since we are taking relativistic effects into account, let's write $$\lambda =\frac{h}{\gamma mv}$$ where $\gamma$ denotes the Lorentz factor $\gamma =1/\sqrt{1-(v/c)^2}$, and $v$ the speed of the particle. Now clearly the energy of the wave could be written as $E=h \nu$. And for the particle, Energy is equal to $\gamma mc^2$. So clearly
$$h \nu =\gamma mc^2.$$
Now plugging into $v_{wave}=\nu \lambda$, we get
$$v_{wave}=\frac{\gamma mc^2}{h}\frac{h}{\gamma mv},$$
or
$$v_{wave}=\frac{c^2}{v}.$$ Doesn't this seem to go against what we know, that the velocity of the wave is less than or equal to $c$?
So can anyone point out what's the mistake here? Does this have anything to do with phase or group velocity?
| What you have calculated is the phase velocity, $v_p$, of the de Broglie wave associated with the particle. The phase velocity can be greater than $c$, and indeed it is always greater than $c$.
The velocity of the particle is the group velocity, $v_g$, and as you have demonstrated the two are linked by:
$$ v_p v_g = c^2 $$
The group velocity must always be less than $c$ and that implies the phase veocity must always be greater than $c$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/503967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Motivation for introducing quantum field theory in particle physics Why is it so that because particles can be destroyed and recreated we introduce QFT? I read at the begining of some textbook that this is so. My main problem is not the rest of the book but the first motivation for introducing QFT for modeling.
My thinking...till now every quantum operator id est observable was attached to a particle in question but when you have variable number of particles you can not do this so you imagine that there is a more fundamental thing which we observe and one observable is also the number of particles.
Also, somewhere I read that because we are now in relativistic regime we have to define observables that are spacelike separated to commute. And because of that also we have to define observables as functions of spacetime points.I dont see it.
| To give a short answer, hoping it will be of some use to some other poor soul roaming the physics land. Lets look at it this way: In reinterpreting electrodynamics as a quantum theory we arrive at real relativistic quantum theory of fields. This is done from the obvious reason to explain the fact that EM radiation comes in quanta. This was known from black body radiation and from photoelectric effect. So there is no mystery in trying to quantize EM field theory. This gave good predictions and was a success. On the other hand, trying to make single fermionic particle wave equation relativistic turns out to be hard. All sorts of problems come out. So, it is not good. Also, we cant explain creation and destruction of the particles. SO, we would need some multiparticle theory. Knowing about EM fields we naturaly think of field theory. So we try to model these processes with quantum fields. Main connection here is, I think, quantization of EM field which assures us that idea of field quantization is good.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/504303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Climate Change Review paper Could anybody recommend me some review paper on climate change or tell me some major names in that field? I work in a completely different area but I would like to learn about the current state of our knowledge regarding this subject. In particular I am interested in currently accepted forecasts, the human impact on the current trends etc..
| Of course the most authoritative review paper is the IPCC report.
You may find the latest report, the fifth revision, AR5, at the webpage: https://www.ipcc.ch/reports/
Depending on your background / time, you may like to first read the Synthesis Report and even Summary for Policy Makers. In the main report, consisting of many volumes, you may find technical explanation and references.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/504438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Standard Model Lagrangian and Euler-Lagrange Equations First off, note that I only know physics through quantum mechanics, so forgive me if this is a foolish question. I've seen the Standard Model's Lagrangian density written out in full. My question is, could one send this through the Euler-Lagrange equations to get a system of equations describing the evolution of the fields, or does quantum field theory have different ways to extract information from it?
| There are several formalisms in which you can work out quantum field theory, and one of them does indeed involve using the Euler-Lagrange equation in some capacity. This is the Schwinger–Dyson equation.
As all quantum formalisms should be roughly equivalent, you can probably prove this quite generally, but it is simplest to show using the path integral formulation (quite simply because it's the one where the action appears directly). Recall that in the path integral formalism of QFT, we have, roughly speaking,
\begin{equation}
\langle 0 | F[\phi(x)] | 0 \rangle = \int \mathscr{D}\phi(x) F[x] e^{i S[\phi(x)]}
\end{equation}
Using some manner of equivalent of the fundamental theorem of calculus, we have that the path integral of this total derivative is zero :
\begin{equation}
\int \mathscr{D}\phi(x) \frac{\delta}{\delta \phi(x)} \left[ F[x] e^{i S[\phi(x)]} \right] = 0
\end{equation}
And therefore, we get that
\begin{equation}
\int \mathscr{D}\phi(x) e^{i S[\phi(x)]} \left[ \frac{\delta F[x]}{\delta \phi(x)} + i F[x] \frac{\delta S}{\delta \phi(x)} \right] = 0
\end{equation}
This leads to the Schwinger-Dyson equation :
\begin{equation}
\left\langle \frac{\delta F[x]}{\delta \phi(x)} \right\rangle = - i \left\langle F[x] \frac{\delta S}{\delta \phi(x)} \right\rangle
\end{equation}
In particular, if we consider the raw quantity for $F = 1$, we simply get
\begin{equation}
\left\langle \frac{\delta S}{\delta \phi(x)} \right\rangle = 0
\end{equation}
This is the equivalent of the Euler-Lagrange equation for quantum field theory, and this is why we can use the Klein-Gordon or whatever else equation to work out the field operators of our theory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/504526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Electron cloud and Quantum Physics Is it possible to detect the Electron Cloud? Also, it it possible for the Electron Cloud itself to contain any mass?
| The electron cloud is just a simplified and somewhat inaccurate way to describe electron orbitals around the atom. The electron cloud is not a physical thing, it's just something to convey the idea that the electrons do not exist at any point in space at some point in time. Rather, there is a probability of detecting an electron at a certain position in space. It is this idea of probability that the electron cloud is trying to capture. It is also used to contrast earlier models of the atom where the electrons existed in orbitals at fixed distances from the nucleus.
Therefore, there is no way to detect an electron cloud, and it certainly doesn't have any mass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/504774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Arnold's holonomic constraints being limits of potential energy The following quote comes from Arnold's "Mathematical methods in mechanics" book:
"We consider potential energy $U_N = Nq_2^2 + U_0(q_1, q_2) $, depending
on parameter $N$ (which we will tend to infinity). We consider the initial conditions on $\gamma$ [supposed to be path in $q_1, q_2$ coordinates]:
$ q_1(0) = q_1^0, \dot{q}_1(0) = \dot{q}_1^0, q_2(0) = 0, \dot{q}_2 (0)= 0 $.
Denote by $q_1 = \phi(t,N)$ the evolution of $q_1 $ under a motion with these initial conditions in the field $U_N$."
Then he mentions a theorem but without a proof. I would be happy if someone could provide argument for believing the theorem.
"The following limit exists as $N \to \infty$: $\lim \limits_{N \to \infty}\phi(t,N) = \psi(t)$. The limit $q_1 = \psi(t)$ satisfies Lagrange equation [...] where [new Lagrangian] $L_*(q_1, \dot{q}_1) = T|_{q_2=0, \dot{q}_2=0} - U_0|_{q_2=0}$".
In the previous quote, $T$ is kinetic energy term.
| *
*The 2nd particle $q_2$ is effectively attached to a spring with coupling constant $2N$. From mechanical energy conservation, $|q_2|\leq\sqrt{E/N}$. In the stiff spring limit $N\to\infty$, the 2nd particle $q_2$ becomes confined to stay in the origin, thereby enforcing the holonomic constraint $q_2= 0$.
*Meanwhile the 1st particle $q_1$ will go about doing its business
$$ m_1\ddot{q}_1~=~-\frac{\partial U_0(q_1,q_2)}{\partial q_1},\tag{1}$$
and interact with the confined 2nd particle at $q_2=0$. (Formally, we need continuity of the solution $q_1$ to the ODE (1) wrt. the parameter $q_2$. This is guaranteed by imposing certain regularity conditions on $U_0(q_1,q_2).$)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/504985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
About the pressure of a confined gas According to fluid mechanics, we have Pascal's principle $P_2 = P_1 + \rho gh$. So, the pressure of a confined gas is different depending on the depth.
However, in thermodynamics, we have another formula $PV = nRT$. They use a confined gas to use this formula. Here, what is the $P$? Does $P$ refer to the pressure exerted onto the piston by a gas or the average pressure?
Or, since the density $\rho$ of the confined gas is so low that we can neglect the pressure difference?
Also, on the microscopic level, they say the reason why the (ideal) gas exerts on the walls is that every molecule collides with the walls. In this case, how can we explain the difference of the pressure depending on the depth of the wall?
| The atmospheric pressure drops about 11.3 pascals per meter the first 1000 meters above sea level. The pressure at sea level is 101,325 pascals. The fractional difference between the bottom and top of a cubic meter of air is therefore only 0.00011152
Hope this helps
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/505165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
On diagonal terms in the Coulomb matrix I have seen many machine learning algorithms for prediction of quantum chemistry properties that use Coulomb matrix as their input. Coulomb matrix is defined as
$$\boldsymbol{M}_{i j}^{\mathrm{Coulomb}}=\left\{\begin{array}{cc}{0.5 Z_{i}^{2.4}} & {\text { for } i=j} \\ {\frac{Z_{i} Z_{j}}{R_{i j}}} & {\text { for } i \neq j}\end{array}\right.$$
where the diagonal terms:
... can be seen as the interaction of an atom with itself and are
essentially a polynomial fit of the atomic energies to the nuclear
charge.
I was trying to read more about these diagonal terms and how exactly this equation is fitted. It seems that the model was first utilized by Rupp et al Phys. Rev. Lett.,108(5):058301, 2012, but there is not much more than this one sentence there. Essentially, I'm looking for something that says: ". . . and from there we get $0.5Z^{2.4}$" and if these energies can be extended to bonds as well, as the off-diagonal term in Coulomb matrix represent repulsion forces rather than potentials. Any references would be appreciated as well.
| I know nothing about machine learning algorithms for quantum chemistry, however it is easy to guess what that $0.5 Z^{2.4}$ term means: it is an approximation to the total energy of atoms.
Consider, for example, the following ground-state energies of neutral atoms computed at the Dirac-Hartree-Fock level:
L. Visscher and K.G. Dyall, Dirac-Fock atomic electronic structure calculations using different nuclear charge distributions, Atom. Data Nucl. Data Tabl., 67, (1997), 207.
Data are available as Table II at
http://dirac.chem.sdu.dk/doc/FiniteNuclei/FiniteNuclei.shtml
A fit of those energies (using the values computed with a point-charge nucleus) to a functional form of the type $E=0.5 Z^\alpha$ for $Z=1$ to $Z=109$ results (my calculations) in $\alpha =2.42$, and the typical fitting error is about 3%. In all probability the constant is fixed to $0.5$ to reproduce exactly the energy of the hydrogen atom.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/505317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Notion of Present Can't I sync all watches in spacetime and call this time slice the present? In Carlo Rovelli's book he tried to explain that the notion of the present is local only, which I could not follow.
| In special relativity, you can pick an inertial reference frame, and then in that frame you can do essentially what you describe: place clocks all over space (not spacetime) and synchronize them all. The synchronization can be done by various equivalent methods, such as transporting clocks slowly or Einstein synchronization.
In general relativity, this doesn't work anymore for a general spacetime. It works only in a static spacetime, which is one that is not changing over time and is not rotating. In a non-static spacetime, Einstein synchronization is not transitive, so synchronizing clock A with clock B and B with C does not mean that A is synchronized with C.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/505506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Is there a way to inspect thin metal weld (2 mm - 3mm) by ultrasonic testing? I am working with an ultrasonic device to inspect welds. So far, I have learned that the minimum thickness of the metal sheet for this inspection is 6mm - 8mm. But the product of mine has 3mm thick welds:
I have basic knowledge working with angle beam transducer to inspect thick metal welds. How can I inpsect thin metal welds by using conventional ultrasonic device?
| The following is just a speculation, but maybe you can place a thicker plate under your thin plate and use reflections from the bottom of the thicker plate. Reflections from the bottom of the thin plate will be a problem, so maybe some oil should be placed between the thin and thick plates to make sure more ultrasonic power gets into the thicker plate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/505906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Differentiability of wave function at boundary in infinite square well I was told in class that a wave function should have the following properties:
*
*Finite and single-valued
*Continuous
*Differentiable
*Square integrable
But if we consider the wave function in an infinite square well, the wave function isn't differentiable at the boundaries since $\Psi (x)$ is:
$$\Psi (x) =\begin{cases}
\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L}), & 0<x<L \\
0, & \text{elsewhere}
\end{cases}$$
This violates one of the properties of wave functions. So how is this an acceptable wave function?
| Like with a plane wave, the wave function you are giving is only a "limiting case".
In reality you should start with finitely high walls and impose all four requirements that you list.
Then, as you take the limit of higher and higher walls, you will obtain your wave function. However, think of it only as a limiting case. In reality, walls will always have finite height.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/506007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Two-coupled oscillator: Doubt in finding normal modes and natural frequency I want to find the natural frequency of a two coupled oscillator system like this-
My book does it this way but I don't really get it.
The equations of motion for the pendula are-
$$I\frac{d^2\theta_1}{dt^2}=−M_\text{eff}\ gL\sin \theta_1− \kappa l^2(\sin \theta_1−\sin \theta_2)$$
$$I\frac{d^2\theta_2}{dt^2}=−M_\text{eff}\ gL\sin \theta_2+ \kappa l^2(\sin \theta_1−\sin\theta_2)$$
To find the natural frequencies of the system, we take the sum and subtraction of the equations and we obtain (Using small angle approximation):
$$I\left(\frac{d^2\theta_1}{dt^2}+\frac{d^2\theta_2}{dt^2}\right)=−M_\text{eff}\ gL(\theta_1+\theta_2)$$
and
$$I\left(\frac{d^2\theta_1}{dt^2}-\frac{d^2\theta_2}{dt^2}\right)=−M_\text{eff}\ gL(\theta_1−\theta_2)−2\kappa l^2(\theta_1−\theta_2)$$
The two equations above are uncoupled and represent the two normal modes of the coupled system. The $\theta_1+\theta_2$ mode or ‘+’ mode represent the in-phase motion of the pendula where both the pendula are moving with same phase (same direction). The $\theta_1−\theta_2$ mode or ‘−’ mode represent the out-of-phase motion of the pendula where the pendula are moving with opposite phase (opposite direction).
I have marked the portions I don't understand in bold above.
Doubts:
*
*What is meant by uncoupled?
*Why does the two equations represent the normal modes?
*Why does $\theta_1+\theta_2$ represent in-phase and $\theta_1-\theta_2$ represent out of phase motion?
| *
*Suppose $\alpha = \theta_{1} +\theta_{2}$ and $\beta = \theta_{1}-\theta_{2}$ then the first equation only depends on single variable $\alpha$ and the second on $\beta$, thus the equations are uncoupled.
*Stating that the equations are uncoupled, or that the matrix of the system is diagonal, or that the equations define the eigenvectors and eigenvalues, or that the equations define the normal modes and natural frequencies, it all means the same, as far as I know.
*$\theta_{1}=(\alpha + \beta)/2$, $\theta_{2}=(\alpha-\beta)/2$, if the system oscilates in its first normal mode only then $\beta=0$, meaning $\theta_{1}=\theta_{2}$, the two masses have an in-phase motion. With second mode only $\alpha=0$, $\theta_{1}=-\theta_{2}$, the two masses have an out-of-phase motion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/506201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Work done by friction in a complicated path
A block of mass $M$ is taken from point $A$ to point $B$ in a complex path by a force $F$ which is always tangential to the path. We also have coefficient of friction as $K$. What will be the work done by force $F$ when it reaches point $B$ from point $A$? Given that the vertical displacement from $A$ to $B$ is $h$ and the horizontal displacement from $A$ to $B$ is $l$.
In this question, I tried solving the problem using conservation of energy, we know that the total energy will remain constant. So with that, we will have,
$$\Delta U_{gravity}+W_{friction}+W_{F} = O$$ But how do you calculate the work done by friction in this case?
Moreover, in the answer, the work done by friction is only dependent on l!!
EDIT:
1.The body is moved very slowly.
| Friction is not a force from a scalar potential. As such the work done is path dependent so there is not enough information to answer the question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/506329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Deriving the speed of light from Maxwell's equations? Relationship between speed of light and EM force?
Can it be said that Maxwell used measurements of the "strength of electric force and strength of magnetic force", to derive the value for the speed of light?
Explicitly, is Maxwells work fundamentally based upon measurement of "forces?
I'm investigating "force" for the fundamental role it plays in the worlds operation. If Maxwell derived or "founded" his work based upon measures of force, then I think that is something important to keep in mind. From where and how and in what order we derive our understandings of the world is important context, if a fundemental understanding of the worlds operation is the ultimate goal. I think thats a fair assumption
| It is true that Maxwell was able to derive an extra equation from his four equations of electromagnetism which allowed him to calculate the speed of electromagnetic waves, which turned out to be equal to the known speed of light.
That equation contained two physical constants (which had previously been measured in the lab) which expressed how easily 1) a magnetic field and 2) an electric field can propagate through empty space.
Regarding your question "Explicitly, is Maxwells work fundamentally based upon measurement of "forces"?", I am not sure I understand the question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/506675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What does convergence to equilibrium for the Fokker-Planck equation mean? I am a math major who recently started to study thermodynamics seriously. I have some confusing points while studying it, so I'd appreciate it if you'd correct me and give me some answers.
(1) As far as I know, a small particle immersed in a fluid is in Brownian motion, which takes place in thermal equilibrium (if the temperature is maintained constant).
(2) The special Fokker-Plank equation describes this Brownian motion.
(For example, one dimensional Fokker-Plank equation is $\frac{\partial W}{\partial t} = \gamma \frac{\partial (vW)}{\partial v} + \gamma \frac{kT}{m}\frac{\partial^2 W}{\partial v^2}$, where $W(v,t)$ is the distribution for the particle.)
However, the solution of the (general) Fokker-Planck equation $W(v,t)$ is not in equilibrium? If so, what sense of equilibrium is this?
(3) I've seen the phrase "convergence to equilibrium" in the literature about the Fokker-Planck equation. Here, even though $W(v,t)$ is not in equilibrium, it converges to equilibrium?
I am confused about the terminology 'equilibrium' here. Please help me!
| The stationary solution of your Fokker-Planck equation is an equilibrium distribution.
Here, assuming that $v$ is the variable for the velocity, that mean a Maxwell-Boltzmann distribution.
So if your initial condition for the distribution $W(v,t)$ is not an equilibrium distribution, your system will not be at equilibrium. However the evolution of your distribution via the Fokker-Planck equation will brings it towards an equilibrium distribution, hence the convergence towards the equilibrium.
Fokker-Planck equation that do not contain non-conservative forces are also called equilibrium Fokker-Planck equation, as theirs stationary solution are equilibrium distribution, and the evolution is a relaxation towards an equilibrium distribution.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/507503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The potential at a point According to my book, 'The potential at a point is said to be 1 volt when 1 joule of work is done in bringing 1 coulomb charge from infinity to that point.' But I wonder how it is possible. As the charge is being brought from infinity, the work done = force * infinity, thus, the work done would be infinity indeed. Please help me out.
| The force between two point charges is an example of an inverse square law where the force is proportional to the reciprocal of the separation squared, $F\propto \frac {1}{r^2}$, as illustrated below.
As the separation increases so does the force decrease.
The area under a force against separation graph is the work done and as long as the area under the graph from $r=R$ to $r\, (=) \, \infty$ is finite, which it is, there is no problem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/507661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How can I prove Clebsch-Gordan coefficients are real? I know it is convention to taken Clebsch-Gordan $\color{blue}{<j_1,j_2;m_1,m_2|j_1,j_2;j,m>}$coefficients are real.If I want to make proof it's reality from any physical defination in backward then how can I do that?
| Mmmm... I don't think there is a proof as such, since it's simply a choice of phase for the individual highest weight states $|j,j\rangle$ states in the decomposition . It would be like asking for a proof that the coefficients $\sqrt{j(j+1)-m(m-1)}$ in the action of the ladder operator $J_-$ are real. This latter reality is merely a choice of phase for the states $|j,m\rangle$ relative to that of $|j,j\rangle$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/507770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Understanding simple LC circuits I'm trying understand the physics of simple inductor-capacitor circuits such that there is just an inductor L and a cacpacitor C and a switch.
Imagine first that the capacitor is fully charged and the switch is then closed.
I do not understand why the current increases from an initial low value as the charge difference between the plates DECREASES because this is in direct contradiction to how a capacitor discharges in isolation.
I know the solution lies in the inductor being present but I can't seem to follow the physics of cause and effect to understand it properly.
Any illumination would be appreciated.
| The instant the switch is closed the current in the circuit $I$ is zero because the changing current in the circuit $\frac{dI}{dt}$ induces an emf in the inductor $L\frac {dI}{dt}$ which opposes the exactly "opposes" the voltage across the capacitor $V = \frac QC$.
As time progress the voltage across the capacitor decreases and so must the the emf induced by the inductor $L\frac {dI}{dt}$ which in turn means that the rate of change of current in the circuit must decrease.
$L\frac {dI}{dt} = CV$ so as the voltage across the capacitor decreases so does the rate of change of current in the circuit whilst at the same time the current is increasing.
When you discharge a capacitor through a resistor the resistor does not induce an emf in opposition the the voltage across the capacitor, indeed as you have stated the current almost instantaneously reaches a maximum value and the decreases.
You will note that I used the words "almost instantaneously" because in the real world the capacitor and resistor circuit would also have inductance as the circuit is a loop.
This very, very small inductance would prevent the current having a finite value at the start but in practice you would not notice that because the change from zero current to maximum current would take place over a time which was much, much shorter than the time constant of capacitor-resistor circuit, $CR$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/507982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Do the oil and mercury in this manometer affect the pressure in A and B? One of my friends came to me with this homework question:
We've worked out that the height difference between A and B is 50 mm. That makes the difference in pressure between them
$$
\Delta P = \rho g h
=1000 \times 9.81 \times 0.05
=490.5 \mathrm{Pa}
$$
Our question is, do the oil and mercury make any difference to this pressure? My first guess is no, since the pressure comes only from the column of water above this height, but this seems too simple for 10 marks (although maybe the hard part is working out the difference in height). Are we missing anything?
| Yes the oil and mercury do make a difference, and have to be taken into account.
When you go up in a fluid there is a drop in pressure. When you go down there is an increase in pressure. The change in pressure depends on the density of the fluid as well as the distance.
Working from A round the manometer to B, there is -
*
*an increase in pressure of 250 mm of water
*a drop of 75 mm of mercury
*an increase of 100 mm of oil
*a drop of 125 mm of mercury and
*a drop of 200 mm of water.
Convert these values to kPa and combine, then you will have your answer.
This is similar to working out the potential difference between two points in an electrical circuit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/508240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Do photons generate gravitational waves? I guess I’m trying to understand the difference between a rock orbiting earth, that would radiate gravitational waves. And say a photon orbiting a black hole that is just following a straight line path. Why does one radiate and the other doesn’t? Best Regards, Andy
| Questions of quantum gravity aside, viewed as moving blob of energy-momentum, a photon moving around a black hole should produce gravitational waves just like a massive particle would. In practice, this is a negligible amount. The fraction of the photon's (wavepacket's) energy that gets converted to gravitational waves is proportional to the ratio of the photon's energy to the mass of the black hole ($c=1$ obviously). For any realistic photon/black hole pair this is a mind boggling small number.
What exactly happens when view the photon as a quantum mechanical particle, requires understanding of how quantum mechanics interacts with gravity, i.e. a theory of quantum gravity. I kind of suspect that this should have a somewhat universal answer within perturbative quantum gravity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/508391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why violet light bends the most? I have read about refraction through a triangular prism,rainbow formation and other natural phenomena.They all told about the strong bending of violet light .
I know that it has a shorter wavelength and high frequency.From the sources I searched I get to a conclusion that it is because of the above mentioned property.I also know that it has a smaller critical angle.what accounts for such properties?
| when light travels through a medium, it interacts with atoms or molecules in the medium. It brings temporary deformation of electron clouds near atoms or molecules and the energy retained is re-emitted without energy loss unless being absorbed by atoms or molecules in the medium. This interaction takes a little time and that's why light has different speed in different medium.
Then why a violet light bends more stronger than other? this is because refractive index varies with wavelength(or frequency). As we know, light has same speed 'c' in the vacuum(refractive index=1), but has different speed in different medium depending on the wavelength. I recommend you to look up 'dispersion curve'.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/508535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does water boil harder when you push a ladle to the bottom of the pot? I noticed this today while cooking. When I push a ladle to the bottom of my pot on my stove top, the boiling sound gets louder, and the bubbles rise to the top more aggressively. Can someone explain?
| There’s either one or two factors at play, depending on what, “… push my ladle to the bottom of my pot…” means.
If you were pushing the ladle hard on a wonky old electric coil stove, then it could be what someone else eluded to: the force pushing down on the pot can effect how much of the pot is in direct contact with the heating element, more direct contact equates to more/quicker heat transfer, and lastly, better heat transfer equates to a stronger boil.
If you were not pushing the ladle down hard on the bottom of the pot (or you don’t have a wonky old electric coil stove), you could still notice the boil getting stronger if you depress something large enough deeper into the water. Obviously, it takes more time to get a pot full of water to boil, than a pot with just an inch of water. By pushing something into the pot, you are displacing some of the water farther away from the heat source. Less water getting more direct heat will change it’s state of matter faster (boiling is just H2O going from a liquid to a gaseous water vapor) which makes the whole pot look like it’s boiling harder.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/508661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can someone explain what is the force the ball will exert? If a ball is falling under free fall then the force exerted by the ball on the ground would be $mg$. But that's not the case in real life ball would hit with more force. But when i draw free body diagram there is only one force that is acting on it $mg$
Can someone explain what is the force the ball will exert ?
| The ground must do work on the falling ball to bring it to a stop. From the work energy theorem the net work done on the ball equals its change in kinetic energy or
$$F_{Ave}d=\frac{mv^2}{2}=mgh$$
$$F_{Ave}=\frac{mgh}{d}$$
Where $F_{Ave}$ = average impact force, $d$ = stopping distance after impacting the ground due to deformation of the ground and/or ball, and $h$ = height from which the ball dropped. Assumes $d$ << $h$ to neglect the change in potential energy after impact
Hope this helps
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/508808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Is dark matter inside galaxies different from dark matter in intergalactic space? I just read a text about astronomy and when talking about dark matter the author says:
[...], the dark matter responsible for the orbits of the stars in the Milky Way is probably different from the dark matter responsible for the orbit of the Milky Way within the local super-cluster of galaxies.
Is this true? How would it be different? And why?
For context, this is the whole paragraph:
Since the 1930's astronomers have measured the orbits of galaxies in clusters of galaxies, clusters of galaxies in clusters of clusters, and so forth. They have found similar anomalies in the angular velocity of galaxies at these larger scales. Again the anomalous high angular velocities of the galaxies and clusters of galaxies may be explained by postulating a mysterious dark matter that fills the universe. It is doubtful that one can explain the anomalous angular velocities at different scales by the same type of dark matter. Thus, the dark matter responsible for the orbits of the stars in the Milky Way is probably different from the dark matter responsible for the orbit of the Milky Way within the local super-cluster of galaxies.
And the article can be found here:
https://mathblog.com/wp-content/uploads/2009/12/Keplers-New-Astronomy.pdf
| Nobody knows what dark matter is, or even whether it exists (see for instance this link). So asking how many kinds of dark matter there are is like asking how many angels can dance on the head of a pin.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/508909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Justification for excluding gravitational energy from the stress-energy tensor I did general relativity years ago at Uni and was just revising with the aid of Dirac''s brilliant book; the beauty of this book is that it is so thin and concise. On reading this book I find that I have a few questions regarding energy.
One thing I had not appreciated before was that the energy in the energy tensor only includes all energy excluding gravitational energy. Is this true? What is the evidence for this position? How could we know that this energy term actually excludes gravitational energy?
The only argument that I can see is that the energy in the energy tensor is not fully conserved, so you could infer that there is a missing energy term and that that energy is gravitational energy. But if you take the missing quantity and call it the gravitational energy, this quantity turns out not to be a tensor.
Hence, its form will in general always look different in at least some different coordinate systems regardless of whichever quantities you use to write it out in. This latter point might only have mathematical consequences, but does it have physical consequences?
| The simplest way to see this is that by the equivalence principle, we can always let the gravitational field at any point have any value we like, including zero. Therefore there is no possibility of defining an energy density of the gravitational field at one point.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/509036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
Cold fusion with ionized hydrogen and neutron radiation Would bombarding ionized hydrogen gas with neutron radiation cause cold fusion as the neutrons have no electrostatic repulsion to overcome therefore making it much easier to get within range of the strong force?
| You don't have to ionize the hydrogen, because the neutrons (to excellent approximation) do not see the electrons. Low-energy neutrons in hydrogen are captured to form deuterium. Because deuterium doesn't have any bound excited states, each capture releases a single 2.2 MeV photon.
In building a practical apparatus, it's important to know that the probability of n-p scattering is much higher in any individual interaction than the probability of capture. This is especially true if the hydrogen is warm or if the neutron energy is above 15 milli-eV, which is the energy of the first rotational excitation of the hydrogen molecule.
Most people who say "cold fusion" are interested in energy production, and are disappointed that there is no stable isotope from which neutrons can be liberated for less than 2.2 MeV. The complete cycle requires substantial energy input.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/509248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do the expressions for an object rolling down an incline not depend on the coefficient of static friction? In my physics course, we are doing an experiment rolling disks and spheres down an incline (assuming there is no slipping). In doing the derivations (assuming a moment of inertia of $\frac25mR^2$ for sphere and $\frac12mR^2$ for disk) I have derived that the final velocity should be $\sqrt{\frac{4gh}{3}}$ for the disk and its acceleration should be $\frac{2g\sin{\theta}}{3}$. I have derived similar equations for the sphere. My question relates to the coefficient of static friction.
Logically, I think that when the coefficient of static friction increases, it should increase the force that is applying torque to the rolling object and thus increase the final speed at which the object is rolling. This should result in a lower final velocity since more of the initial energy ($mgh$) is "allocated" to the rotational kinetic energy as opposed to the translational kinetic energy.
Yet, according to the derivations, the final velocity does not depend on the coefficient of static friction. Why is this? I believe my equations are correct so I know that the final velocity is entirely determined by the object's moment of inertia, and the angle of the incline, but I intuitively think that a higher coefficient of friction should change how much energy is "allocated" between translational and rotational kinetic energy.
| For rolling without slipping the force of friction does no work. The force of friction provides the constraint that keeps the point of contact of the rolling object at rest with respect to the surface the object is rolling on. The magnitude of the frictional force while rolling without slipping can be determined by a force and torque (about the center of mass) balance. For example, see Halliday and Resnick , Physics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/509358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Momentum conservation in spontaneous pair creation and annihilation I know that in free space, photon cannot decay into an electron and a positron since momentum is frame dependent for massive particles while invariant for a photon.
Given this, how is spontaneous pair creation and annihilation possible? Can someone shed some light on it? Or is it that it is actually impossible, and that I had a wrong comprehension?
| Pair creation
You are right: In free space a photon cannot
decay into an electron-positron pair
(because it would violate energy/momentum conservation).
However, near an atomic nucleus a photon can decay
into an electron-positron pair.
In this process the atomic nucleus receives some recoil.
See also Wikipedia: Pair production.
The process can be visualized by a Feynman diagram like below.
image from Wikipedia: Pair production
Pair annihilation
You are also right, that an electron-positron pair in free space cannot decay
into a photon (again because it would violate energy/momentum conservation).
However, an electron-positron pair can decay into
two photons. The two gamma ray photons depart
in roughly opposite directions.
See also Wikipedia: Electron–positron annihilation.
The process can be visualized by the Feynman diagram below.
image from Feynman diagram for annihilation
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/509453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Interpretation of the photon scattering rate? The photon scattering rate $\Gamma$ describes the rate at which photons scatter off an atom$^1$. In a two-level system, the ansatz for the photon scattering rate often is given by
\begin{equation}
\Gamma = \rho_{22}\gamma
\end{equation}
where $\rho_{22}$ is the probability to find the atom in the excited state and $\gamma$ is the rate of spontaneous decay. However, I don't see the connection between the ansatz above and what the photon scattering rate is physically meant to be.
$^1$In my imagination, the photon scattering rate is the absorption rate for photons at a certain frequency $\omega$. Hence $\Gamma(\omega)$ shows the saturation broadened Lorentzian absorption line of the atom, centered around a resonance frequency.
| Why do you imagine the photon scattering rate to the be absorption rate? The scattering rate is the rate at which an atom absorbs AND re-emits incident photons. $\rho_{22}$ captures how excited an atom becomes for a particular incident field and $\gamma$ captures how quickly the atom decays and re-emits the excitation. Thus the scattering rate is $\Gamma = \rho_{22} \gamma$.
Note that $\rho_{22}$ depends on the intensity and frequency of the incident light. This is what gives the amplitude and frequency dependence of the scattering rate. If the light you're driving the atom with doesn't excite the atom at all we don't expect the atom to scatter any photons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/509576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How do you pick the wavenumber at which the group velocitiy is evaluated? The equation for group velocity is
$
v_g(k) = \frac{d\omega}{dk}.$ This is obviously a function of $k$ but typically the word is used as if there is a single group velocity and not a whole function. How do you assign a group velocity to a "group" ? Is there any clear cut mathematical definition which gives another condition or is there a special $\omega$ at which the function is typically evaluated ? Or is the word a misnomer and there isn't a single group velocity which can be assigned to a wavepacket ?
| It depends a bit on context. For transmission of signals in optical cables, the bandwidth is very small compared to the frequencies of the laser modes. So the meaning of group velocity is unambiguous. (But over a long distance, there will be noticeable dispersion, which limits the data rate.)
In a case like water waves, the situation is different. But there one can say that for deep water the group velocity is half the phase velocity. Both depend on wavelength:
$$c_\phi = \sqrt{\frac{g\lambda}{2\pi}} = \frac{g}{2\pi}T.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/509796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there a black hole information paradox in string theory? In string theory there is, for several black hole-like objects, a precise way to account for the microstates which make up the macroscopic Bekenstein-Hawking entropy. For instance for the black brane made of a bound state of $Q_1$ D1- and $Q_5$ D5-branes wrapped on $T^4 \times S^1$ the (1,5) and (5,1) open strings provide the leading contribution to the entropy at large charges, which is exactly $A/4$.
How, if at all, is the information paradox solved in this example or examples like this? What is the mechanism by which a pure in-state evolves into a thermal density matrix out-state-- and what happens in between?
| No. There is no information paradox in string theory.
The example you cite (the D1-D5 system) is a stable BPS configuration, so no evaporation is taking place and all the black hole microstates (correctly computed by string theory) are evolving unitarily because string theory does not modify the rules of quantum mechanics ( see the beautiful Motl's post Why string theory is quantum mechanics on steroids).
The fact that we perceive a black hole with a unique macroscopic state in the supergravity approximation as follows from higher-dimensional no-hair theorems (sometimes violated in string theory) is just an artifact produced by the fact that higher string modes are integrated out (by definition of zero-slope limit) but after all stringy modes are taken into account, no loss of coherence is possible.
But even if you allow evaporating black holes in string theory, information is preseverved because all the objects in string theory evolve unitarily. Even in time dependent situations, string theory predict the correct entropy computation (beautiful example). Even if you take into account non-perturbative corrections , or you analize Calabi-Yau black holes outside the OSV regime (as was done in here and here) you will never find loss of coherence.
Arguably the most clear examples of the absence of information-like paradoxes in string theory are black holes realized within M(atrix) theory (example) because the BFSS theory is manifestly an "ordinarily" quantum mechanical model.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/509933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"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.