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Why does twisting a cork make it easier to remove from a bottle? When we want to remove a cork from a bottle first we turn the cork. Turning in one direction makes it easier to remove in the axial direction.
Does anyone know something more about this?
| As Steeven said, kinetic friction is a smaller force than static friction. Once the cork is moving in any direction, it is easier to move in the direction you want.
As Anna V said, bonds may be broken that make it easier to move a second time after it has moved once.
So why is rotation easier than longitudinal movement? Think of a screwdriver. A large diameter handle allows you to exert a large torque on the screw. That is, small forces a large distance from the axis apply large forces a small distance away. So the large handle on the corkscrew helps.
| {
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What's the difference between a perfect fluid and an ideal gas? This is how I understand it at the moment:
*
*A perfect fluid is a collection of non-interacting particles, which are as a whole characterised by energy and pressure.
*An ideal gas is also a collection of non-interacting particles, but here the ideal gas law holds. There, we have pressure, volume and temperature (let's assume a fixed number of particles for both cases), s.t. by applying the ideal gas law, again two parameters remain.
Furthermore, the stress-energy tensor of a perfect fluid can be seen e.g. here on Wikipedia, but I haven't found the stress-energy tensor of an ideal gas.
| A perfect, or ideal fluid is also incompressible, but an ideal gas is not. Pressure and volume changes on an ideal gas can cause changes in its density.
A perfect fluid is described by an irrotational velocity vector field $\bf v$, so that $$\nabla \times \bf{v} =0$$ and this is not necessarily true for
ideal gases.
The molecules of an ideal gas interact with each other via elastic collisions, and the motion of these molecules is random. The behavior of ideal fluids cannot be described by the ideal gas equation.
While perfect fluids do not have interactions between their molecules, these molecules do not move randomly, but are held together to a fixed volume.
Hence, the behavior of ideal gases and ideal fluids will be different.
Also, in the context of the stress-energy tensor mentioned in your question, perhaps an ideal gas is taken as a simple example of a perfect fluid. An ideal gas can be modelled by a stress-energy tensor which is one described by the energy density and pressure components only. Example of such modelling.
| {
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Brachistochrone to a vertical line Just for fun, I am working through some problems in Mathematics of Classical and Quantum Physics by Byron and Fuller. Problem 2.13 reads:
Prove that a particle moving under gravity in a plane from a fixed point $P$ to a vertical line $L$ will reach the line in minimum time by following the cycloid from $P$ to $L$ that intersects $L$ at right angles.
Solving the Brachistochrone is more or less straightforward calculus of variations, taking $P$ to be the origin wlog. The action to be minimized is given by (ignoring the optimization irrelevant leading factor of $\frac{1}{\sqrt{2g}}$):
$$
f(x, y, y') = \sqrt{\frac{1 + {\left(\frac{dy}{dx}\right)}^2}{y}}
$$
And the derivation of the resulting cycloid, which can be found literally everywhere, gives the parametric function:
\begin{align*}
x &= a(\theta - \sin \theta) \\
y &= a(1 - \cos \theta)
\end{align*}
Where $a$ is a free parameter allowed to vary. (Note: gravity is taken to be in the y-positive direction to eliminate a pair of redundant minus signs.) If $L$ is taken to be the line $x = x_b$, then in theory it ought to be possible to minimize the cost function with respect to $a$.
If the cycloid intersects the line at a right angle, then it follows that
$\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \ne 0$ at the point of intersection.
\begin{align*}
\frac{dy}{d\theta} = 0 &= a \sin(\theta) \\
\theta &= n\pi : n \in \mathbb{Z} \\
\frac{dx}{d\theta} \ne 0 &= a(1 - \cos \theta) \\
\theta &\ne 2n\pi : n \in \mathbb{Z}
\end{align*}
Because only the first solution could possibly be the curve of fastest descent,
the value of $\theta$ must be $\pi$, which gives $a = x_b / \pi$. So, if we take the derivative of the cost function with respect to $a$ and set it equal to zero, it ought to be the case that we find the equal yields this value. Unfortunately, this is not what I find. Instead, I think I might be being too careless with what is a function of what.
The cost function is given by:
$$
I = \int_0^{x_b} \sqrt{\frac{x'^2 + y'^2}{y}} dx
$$
Where $x$ and $y$ are functions of $\theta$, and their primed derivatives are with respect to $\theta$.
The most problematic part is going to be finding the upper limit ($\theta_b$),
which we only know implicitly:
$$
x_b = a(\theta_b - \sin \theta_b)
$$
Because $\theta_b$ is also a function of $a$, it is necessarily to make this
explicit. So let $\theta_b(a)$ be the unique solution to the aforementioned
implicit equation. With some algebraic manipulation $I$ becomes:
\begin{align*}
I &= \int_0^{x_b} \sqrt{\frac{{x'(\theta)}^2 + {y'(\theta)}^2}{y(\theta)}} dx \\
&= \int_0^{\theta_b(a)} a(1 - \cos \theta) \sqrt{\frac{a^2{(1 - \cos \theta)}^2 + a^2{(\sin \theta)}^2}{a(1-\cos \theta)}} d\theta \\
&= \int_0^{\theta_b(a)} a(1 - \cos \theta) \sqrt{2a} d\theta \\
&= {\left[\sqrt{2} a^{3/2} (\theta - \sin \theta)\right]}_{0}^{\theta_b(a)} \\
&= \sqrt{2} a^{3/2} (\theta_b(a) - \sin \theta_b(a)) \\
\end{align*}
Next, take the derivative and set to 0.
\begin{align*}
\frac{dI}{da} = 0 &= \frac1{\sqrt{2a}} (3a(\theta_b(a) - \sin \theta_b(a)) + 2a^2 \theta_b'(a) (1 - \cos \theta_b(a))) \\
&= \frac1{\sqrt{2a}}(3x_b + 2a^2 \theta_b'(a) (1 - \cos \theta_b(a)))
\end{align*}
From here, we have to take care to find $\theta_b'(a)$.
\begin{align*}
x_b &= a(\theta_b(a) - \sin \theta_b(a)) \\
\frac{d}{da} x_b &= \frac{d}{da} (a(\theta_b(a) - \sin \theta_b(a))) \\
0 &= \theta_b(a) - \sin \theta_b(a) + a \theta_b'(a) (1 - \cos \theta_b(a)) \\
\theta_b'(a) &= - \frac{\theta_b(a) - \sin \theta_b(a)}{a (1 - \cos \theta_b(a))} \\
\end{align*}
Now to continue solving:
\begin{align*}
0 &= \frac1{\sqrt{2a}}(3x_b + 2a^2 \theta_b'(a) (1 - \cos \theta_b(a))) \\
&= \frac1{\sqrt{2a}}(3x_b - 2x_b) \\
&= \frac{x_b}{\sqrt{2a}}
\end{align*}
Which seems to imply that the critical point lies at $a = +\infty$, which is clearly wrong (given this would imply an infinite distance to travel), and also not equal to the previously found value $a = x_b/\pi$. Which begs the question, where is my error? And is there a better approach for this problem?
| Hints:
*
*To have a well-defined variational problem, we have to impose adequate boundary conditions (BC). Recall in particular that BCs are necessary$^1$ for the proof of the Euler-Lagrange (EL) equation.
*The initial BC is an essential/Dirichlet BC $y(x\!=\!x_i)=y_i$. However, the final BC is clearly not an essential/Dirichlet BC. It must then for consistency reasons be a natural BC $p(x\!=\!x_f)=0$. This turns out to be a Neumann BC $y^{\prime}(x\!=\!x_f)=0$, i.e. the final tangent is horizontal/forms a right angle with the vertical line.
*Since the EL equation of the modified brachistochrone problem is unchanged, the solution is still a cycloid.
--
$^1$ Conversely, if we haven't fixed the BCs, we cannot assume the EL equation.
| {
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Is it possible to take long-distance X-ray images at a high resolution in Earth's atmosphere? It's my understanding that different wavelengths of EM radiation are affected by scattering and attenuation to varying degrees. Would a camera using only X-rays be able to take a picture of something (for instance) a few miles away with similar detail and clarity to visible light?
This might cross over a little with Science Fiction SE due to a fictional situation prompting the question, and me not needing a mathematically super-precise answer.
| X-ray telescopes (XRTs) need to be above the atmosphere as it is opaque to X-rays. So long range X-ray photography in the atmosphere, unless possibly at very high altitudes, would not be very feasible.
| {
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Why do little chips break off so easily from strong neodymium magnets? I have some strong toy neodymium magnets. Typically after a while little chips start breaking off, unlike from most other small metal objects, like in this image.
It could of course be that neodymium is more brittle than metals used for other objects, or that they often hit each other much harder than in a fall due to their magnetism, or that they are just low-quality, but I was wondering if it could have to do with internal tensions that are not present in non-magnetic objects, maybe due to adjacent domains of different magnetization?
Does anyone know what could cause this?
| These rare earth magnets are very brittle as you said.
From Wikipedia:
There are two types: neodymium magnets and samarium–cobalt magnets. Rare-earth magnets are extremely brittle and also vulnerable to corrosion, so they are usually plated or coated to protect them from breaking, chipping, or crumbling into powder.
| {
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If I have $N$ particles which move in one dimensions, and that collide elastically, is there a way to solve for their trajectories? There are $N$ particles on a line, and I know each of their masses and initial velocities and positions, and that the total energy and momentum is conserved when they collide.
Is there a way to solve for their positions as a function of time exactly? Or is there a way to do it statistically? I'm very interested to find out.
| It depends on how realistic and how precise you want your model to be.
The first and most obvious simplification is to work in a reference frame in which the centre of mass of the particles is at rest.
After making that simplification, the very simplest model is to model each particle as a point mass, and assume each collision takes zero time.
If all of the particle masses are equal, this model allows a further elegant simplification. When two point particles with equal masses collide elastically they simply exchange velocities, so you can actually model the particles as passing straight through each other and dispense with the collisions.
However, if the particle masses are not equal this simplification does not work. You will have to calculate the time of the first collision and the velocities of the particles after that collision, and then go on to find the time of the second collision etc. This can become complex if you have a large number of particles, but it can be handled with a fairly straight forward computer simulation.
If you want a more realistic model then you could give each particle a size., and take this into account when modelling each collision. Collisions will still take zero time if you assume the particles are rigid.
An even more realistic model would treat each particle as a small spring which contracts in each collision, and then expands again. Collisions now take non-zero time, and you have to take account of the deformation of each particle during each collision.
| {
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Bremsstrahlung Radiation A thought experiment.
Consider an electron falling into a black hole. From an external observer to the electron and the black hole, the electron accelerates, and should give off Bremsstrahlung radiation
From the electron's frame of reference, it is travelling along a geodesic in free fall, and so is not accelerating at all so doesn't generate Bremsstrahlung radiation.
Which is the correct situation and why?
| Philip's answer is great, I'd just add a couple of things to it.
Generally, it's important to point out that radiation is not an observer-independent phenomenon. For example, ignoring gravity, an accelerating particle will radiate according to an inertial observer, but not according to an observer who is co-accelerating. Maxwell's equations are not invariant under a transformation from an inertial to an accelerating frame.
To expand on the reason to this phenomenon a little bit, the accelerating observer experiences a Rindler horizon - anything behind that horizon cannot reach them as long as they continue to accelerate. A co-accelerating charge emits radiation into the spacetime region behind the observer's horizon, as explained in this paper (https://arxiv.org/abs/physics/0506049).
Analogously, a particle in free fall will not radiate according to a freely falling observer but will do so according to one who is stationary in the gravitational field. Meanwhile a stationary charge in that field will radiate according to the freely falling observer but not according to a similarly stationary observer.
| {
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Has a true Superconductor ever been produced? Having a conversation with a friend, we where discussing about the conductivity and resistance changes of superconductors.
He insisted that available superconductors produced by universities or corporate entities that are being produced today worldwide can in practice maintain a Zero resistance state in specific laboratory conditions.
I doubted this to be the case, claiming that maintaining a conductor in a R=0 state, in practice, is nothing more than an ideal experiments are trying to achieve, the same way maintaining mass at 0ºK would. I was also claiming that experiments that would indicate that the material is in fact in a R=0 state would do this due to lack of the necessary hardware sensibility to detect the minuscule deviation from true Zero resistance and that respectable sources would make account of this documenting the experimental error.
I am unaware of experiments that show reliable methodology to indicate the viability of a true "superconducting phase". I understand that superconductivity, by definition, is a current flowing without resistance, and that this is only possible if the voltage across the junction is 0, otherwise there would necessarily be both current and voltage and therefore an effective resistance =/. My suspicion is that the superconducting state is more akin to an () function, thus my doubt of extensive documentation and testing on true Zero resistance materials being produced, which leads me to my initial question:
Has a true Superconductor ever been produced AND experimentally proven to be a true superconductor?
If so, where could I find the documentation for these experimental procedures and the methodology claiming the veracity of the experimental potential of achieving this physical state in laboratory conditions? If not, to what degree of experimental error has Superconductivity been recorded?
In other words, are materials called "superconductors" popularly misnamed and/or misrepresented?
Or, is this specific matter still unknown due to the lack of the necessary sensibility of equipment, making the knowledge on the subject no more than hypotheses and speculation?
Thank you for your time.
| Yes, in 1911 by Gilles Holst and Heike Kamerlingh Onnes in Leiden. Kamerlingh Onnes received the Nobel prize in 1913.
| {
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What causes things to roll into potential wells? If an object is sitting on a potential slop, why must there be a force to push it into the well?
| Exactly, it is the force. If there is a conservative force, a potential can be defined as a function $U(\vec{r})$ that satisfies
$$\vec{F}(\vec{r}) = -\nabla U(\vec{r})$$
Therefore the force pushes towards the minimum of the potential.
The common image of a potential "well" into which the objects "fall" is a great analogy because if you consider the gravitational force field here on Earth, $\vec{F} = -mg\hat{z}$ and so $U = mgz$. In this case a "physical well" (like a hole in the ground) is also a "potential well", because $U$ is proportional to $z$.
But one should be careful not to bring this analogy too far. In general a potential is just a way to express the fact that there is a force, and has nothing to do with height and falling.
| {
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Need help with knowing when I should use $KE = \frac12mv^2$ or $V_f^2 = V_0^2 + 2ad$, in order to find final velocity I'm looking at this pdf for question 15.b) and I figure out $V_f$ by using $V_f^2 = V_0^2 + 2ad$, but the answer says $W = \Delta PE = \frac12mV_f^2$, and our answers ending up totally different, can someone please explain to me whats going on?
| The $v_f^2 = v_0^2+2ad$ equation only works under constant acceleration.
Based on the graph, you can see that the force $F$ keeps changing, so the kinematic equation above does not apply, since the acceleration $a$ will not be constant.
The work-kinetic energy theorem (or conservation of energy) $W_\text{net}=\Delta \rm KE$ is true for any acceleration. And since you found out the net work (work by spring force), you can find the change in kinetic energy, and subsequently, the speed.
As a rule of thumb in intro physics, you always want to check if the problem is solvable using conservation of energy/work-KE theorem since these are very simple and easy methods compared to the alternatives.
| {
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Amount of force required it to tip over a cone Say I have a cone of height $h$, radius $r$, and mass $m$.
How can I determine the amount of force required to tip it over (to have it fall completely to the other side), say exerted (horizontally) at the top of the cone? And in addition, how does the position at which I exert the force affect the amount of force I will need to tip it over?
Any advice on how I would aproach this problem is appreciated.
| Hint: Tipping occurs if there is a net moment about a point on the rim of the cone. The net moment is the sum of the moments due to the horizontally applied force and the weight of the cone acting through its center of gravity.
Hope this helps.
| {
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Pressure exerted by an ideal gas according to kinetic theory of gases In my textbook and Wikipedia, I have observed that force exerted on a wall of the container by one molecule is taken into account. Such that $F=\frac{mu} {\Delta t}$ where ${\Delta t}=\frac{2l}{u}$. But this change in time is the time required for a molecule to move from one wall to the opposite. In a gas container, each gas molecule doesn't get to move this freely. Then why do we assume ${\Delta t}=\frac{2l}{u}$? Is it that the molecules remain in random motion and tends to maintain constant density all over the place for which the statistical value of $\Delta t$ turns out to be the same?
Another small question, were polyatomic molecules also considered as one sphere each in the kinetic theory of gas? Or was it each atom resembled a sphere but not a molecule?
| Addition to Al Browns answer.
The change in momentum would be given by $$F\Delta t=2mv_\text{rms}$$ or $$F= \frac{2mv_\text{rms}}{\Delta t}$$ where $v_{\text{rms}}$ is the root-mean-square velocity, usually given by $$v_{rms} = \sqrt{\overline{v}^2} = \sqrt{\dfrac{3k_BT}{m}}$$ where $\bar v$ is the average speed.
The time is given by $$\Delta t=\frac{2l}{v_\text{rms}}$$ where $l$ is the distance between each side of the container.
were polyatomic molecules also considered as one sphere each in the kinetic theory of gas? Or was it each atom resembled a sphere but not a molecule?
Yes. Each molecule is modelled as a hard sphere, and not a point particle, and all collisions are assumed to be elastic.
| {
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How to calculate gas equation (not necessarily ideal) when the internal energy depends solely on the temperature? I know that for an ideal gas it can be shown through the equation of state and the Maxwell relationships, that the energy of the gas depends only on temperature.
But my question is how can I calculate a state equation when I know that the energy of specific real gas depends only on the temperature?
| It is unnecessary to use Maxwell relations to show that if a gas obeys the equation of state of the perfect gas, the energy depends only on the temperature (and the number of moles).
Indeed, from $pV=nRT$, one gets
$$
\frac{p}{T}=\left.\frac{\partial{S}}{\partial{V}} \right|_{U,n}= \frac{nR}{V}
$$
which can be integrated to give
$$
S=nR\log V + f(U,n)
$$
Therefore, the inverse temperature
$$
\frac{1}{T}=\left.\frac{\partial{S}}{\partial{U}} \right|_{V,n}=
\left.\frac{\partial{f}}{\partial{U}} \right|_{V,n}(U,n).
$$
From this last equation, by using the inverse function theorem, we get $U(n,T)$ using the inverse function theorem. Notice that unless additional information is provided, it is impossible to obtain the unknown function $f(U,n)$.
If we know that the internal energy is a function of $T$ (and n) but does not depend on $V$, this implies that
$$
S(U,V,n)=\phi(V,n)+f(U,n)
$$
(the dependence of the internal energy on $T$ and $n$ can be derived analogously to the previous case). However, we cannot go beyond the result that.
$$
\frac{p}{T}=\left.\frac{\partial{S}}{\partial{V}} \right|_{U,n}= \left.\frac{\partial{\phi}}{\partial{V}} \right|_{n}.
$$
There is no way to get the unknown function $\phi(V,n)$ without additional information.
| {
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Do the quark types differ from each other in ways other than charge and mass? I've read things online here and there that seemed to hint that there's more to quark type than mass and charge. Is this true?
For clarity's sake, I'm not asking about properties individual quarks have other than mass and charge, such as spin and color charge. I'm asking about the properties of quark types.
| If the six flavors ($d,u,s,c,b,t$) all had the same mass and charge, they'd all still be distinguishable in the sense that the model still includes observables that (if measured) would detect one species and not the others.
Consider quantum chromodynamics (QCD) with $N$ quark flavors of equal mass, without the weak and electromagnetic and Higgs interactions. This model has a symmetry that mixes all of the flavors together, but it still has observables that detect one of the $N$ flavors without detecting the others. The observables are related to each other by symmetry, but they're still distinct observables. (I haven't thought about how one would actually measure such an observable, but the theory has those observables mathematically.)
This is analogous to the fact that we can build detectors that detect only horizontally-polarized photons or only vertically-polarized photons, even though they're related to each other by rotation symmetry. We don't call different photon polarizations different species, because they're related to each other by a spacetime symmetry. In contrast, the different quarks in $N$-flavor equal-mass QCD are related to each other by an internal symmetry. Whether we call them different species or not is irrelevant: whatever we call them, they are distinguishable in the sense that I described above.
| {
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Question about instant axis of rotation The given question is:
A man is rotating a stone of mass 10 kg tied at the end of a light rope in a circle of radius 1m. To
do this, he continuously moves his hand in a circle of radius 0.6 m. Assume, both circular motions
to be occurring in the same horizontal plane. What is the maximum speed with which he can throw
the stone, if he can exert a pull not exceeding 1250 N on the string?
Now I am lacking some conceptual clarity in this area, So could someone explain the logic behind this question so that I can also get the conceptual clarity I need in this area? The answer given is 10m/s.
| If the stone is moving in a circle - as we are told - then the net horizontal force on the stone must be directed towards the centre of the circle. And the only horizontal force on the stone is the tension in the rope. So the centre of the stone's circle must lie somewhere along the line of the rope i.e. along the line between the stone and the man's hand (including its extension).
So if we call the centre of the circle O, the position of the man's hand A and the position of the stone B then O, A and B always lie on a straight line. We know that the distance OA is $0.6$ m and OB is $1$ m. The order is either OAB - in which case the rope has a length of $0.4$m - or AOB - in which case the rope has a length of $1.6$ m.
In either case the tension in the rope is $T=\frac {mv^2}{r}$ where $m=10$ kg, $r=1$ m and $v$ is the speed of the stone.
(Note that this means the given answer of $v=10$ m/s is incorrect)
| {
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How do geodesics explain two identical balls thrown up at the different speeds? As stated in the title, two identical balls, both thrown directly upward, but at different speeds. The slower ball will reverse direction at a lower height than the faster ball. But the curvature of spacetime that they are passing through would be nearly identical.
How do geodesics explain these two paths?
Another way of looking at the same issue: Two identical balls dropped, but from different heights. Both balls travel straight down but hit the ground at different velocities. Each ball should be passing through the same spacetime curvature at the point of impact. Since neither ball experienced a force or acceleration and their motion is purely a product of the curvature of spacetime, why are they traveling a different speeds at impact?
| The curvature of the spacetime through which the balls are moving is identical, but the balls set off through it in different directions. Whereas in space alone they appear to be heading in the same direction, in four dimensions they are taking different paths. To see this, simply plot two dimensions, z and t, using everyday units of m and s- you will see that the balls are launched in quite different directions through spacetime. If you use the unit ct for the time axis, the difference in the angles of launch of the two balls will then seem tiny, but they are moving at such a huge speed using those units that the overall effect on the trajectory is exactly the same.
| {
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Is there a relationship between quantum physics and chaos theory on a classical scale? Im a complete physics lay person and I read somewhere that chaotic systems are subject to tiny differences in initial conditions and that the brain is a chaotic system.
Does that mean our thoughts are subject to quantum randomness?
| I would say that the brain as a physical system is affected by quantum randomness.
But our thoughts are a different thing, thoughts are immaterial, non-physical processes. Mental processes are not governed by causality or any other law of physics.
But I do believe that our minds do use thermal or electromagnetic noise in the brain in the creation of new ideas.
| {
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When does the interference pattern of DSE disappear as size of "projectile" is increased? In trying to learn about Quantum mechanics (QM) from popular science books and Stack Exchange (I of course expect my knowledge to be anything but complete) I regularly come up with seemingly childish questions where I wish I could interrupt the author or visit them during office hours.
I'd prefer low level answers (beginning undergrad in maths with no formal physics whatsoever).
Regarding the double slit experiment (DSE):
Q1: I've heard about Feynman's view that the electron somehow takes every possible route from start to finish. For example, in Brian Greene's "The Elegant Universe" (TEU): "Feynman argued that in traveling from the source to a given point on the phosphorescent screen each individual electron actually traverses every possible trajectory simultaneously."
How can this argument be made without violating FTL?
Q2: When does the interference pattern result of DSE disappear as you increase the size of the "chunks" fired? E.g. if you fire protons or even larger ensambles. And, does it happen for all discovered particles? Is there a sharp point at when this happens or does rather the interference pattern disappear gradually?
| What Feynman is pointing out is actually a general wave property. For example, light passing through a slit can also be seen as a a superposition of all possible directions, note however, with an intensity equal to the slit's Fourier transform. At a distant, in terms of wavelength, point P all of these waves arrive, but only the wave with k-value pointing from the slit to P survives destructive interference. This is one of the fundaments of Fraunhofer approximation, the other being the assumption of small angular extension of the slit at P.
| {
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What are quantum fields made up of? If quantum fields are mathematical entities made up to explain nature, what they explain is definitely something physical and is made up of something. So why can’t there be an answer to what these mathematical quantum fields are made up of?
I mean, if physicists are making them up, then they definitely would have certain criteria as to what these mathematical fields are made up of, won't they?
| "The book of nature is written in mathematics", said Galileo. It is hard to sing a song about a book that cannot be read; here is a song.
Quantum fields are made up of quantum oscillators, an infinity-of-infinities of them. These oscillators are little gadgets, everywhere, that spew out and consume quanta, the building blocks of our world (the notes/tones of a song?).
Their understanding has been around for virtually a century (Jordan), but still hard to intuit: that's why it takes years of physics training to handle them blind. These quanta are photons, matter particles, gravitons, and such.
They are more physical and may produce more accurate answers than engineering and any other physical theory. As a matter of fact, most of the engineer's/layman's intuition about the physical world is an elaborate empirical but flawed summary of quantum fields, and a real challenge is how that arises out of the more-real-than-real quantum fields. What you/we see in your day is but a warped image/bluff of collectivities of them, but you/we are spoiled since birth to assume it is more real.
As Steve Weinberg, the late (just) superhero of understanding such quantum fields said, "The universe is an enormous direct product of representations of symmetry groups”. These representations are quantum fields.
In principle, they undergird everything, but working out how our cattywampus macroscopic world arises out of them, "emergence", is sometimes a challenge, and rarely not. But there are hardly credible, complete, efficient, logical alternatives to them.
A better song is this one.
| {
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Fields as sections of fibre bundles versus functions on spacetime Why should fields be thought of as sections on a bundle? In particular, what is the problem with thinking of them as functions (with additional conditions on target domain and smoothness) on spacetime?
I could come up with one example where there is a clear difference - sections on the Möbius strip with base space the unit circle and fibres the interval $[-1,1]$. Here a smooth section is forced to take the value $0$ on at least one point of the circle, while general smooth functions on the unit circle do not have to satisfy this.
This example seemed to me however not very physically motivated - are there physical examples where this difference between sections and smooth functions on the base space becomes more pronounced? Or is there some other motivation for defining fields as sections over bundles? For what it is worth, I have encountered this concept only during my studies of general relativity, maybe this distinction only becomes apparent in QFT?
| The distinction between functions and sections is only contentful when dealing with a non trivial bundle. One might then wonder when non trivial bundles actually arise in physics. The most famous example is that of a magnetic monopole. However, how did that non trivial bundle "get there"? Well, the answer is, with our current paradigms, spontaneous symmetry breaking. You could have some larger non abelian symmetry group which is broken to U(1). Given some fairly general conditions on the gauge group, you could possibly have monopole configurations. These are, in a sense, not "true" non trivial bundles, but low energy "effective" non trivial bundles, where the ground state has come to rest in some topologically protected non trivial configuration.
| {
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Can we feel heat in outer space? Is there air outside of earth atmosphere? If not, could we feel heat coming from sun?
| I would say yes, one should be able to feel heat in outer space. Heat transfer can be due to contact, convection, or radiation. Although there is practically no air, in outer space there is light, or more precisely, electromagnetic radiation (visible, infrared, etc.) from the sun. So in principle at least, one should be able to feel heat from the sun in outer space.
| {
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Single slit diffraction from Feynman's rotating amplitudes ("Little Arrows") In Feynman's NZ lectures (and consequent book) “QED – The Strange Theory of Light and Matter”, he gives a model for optics.
He describes a probability amplitude for a photon to be detected after being emitted from a source. The amplitude is a complex number, whose angle rotates at a constant rate (depending on the photon frequency), and whose modulus is proportional to $1/l$, where $l$ is the path length. The total amplitude is the sum of amplitudes from different paths. The probability is the total amplitude's square-length. This is a simplified model for the Path Integral.
I have build a Mathematica simulation for this method. I tried to simulate a single-slit experiment: a source (at the origin), a slit (at x-position $d$, y-positions $-yrange \to yrange$), and a detector at varying positions $(1,h)$. For each detector, I run over different paths (like the blue and yellow paths below). Each path is two straight-lines: origin to some middle point $(d,y)$, and from the slit to the detector. I sum over all paths with $y$ as a parameter. The photon wavenumber is $k$. The probability is not normalized in this method.
For $k=20$:
As you can see, I do not get a $Sinc^2$. What am I missing?
| It seems that the slit was too close to the source and to the screen. I have also made the wavenumber higher.
When both distances are $50$, and $k=1,000$:
For two slits (source to slits $=500$, slits to screen $=500$, $k=500$, slit-width $=0.2$, slit distance $=1$):
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Magnetic moment and angular momentum of electron I recently got to know about something really interesting.
These are as follows:
1: The magnetic moment of an electron is, $\cfrac{ev}{2πr}$, where $e$ is the charge of the electron, $v$ is its velocity, and $r$ is the radius of the orbit it revolves.
2: The direction of the magnetic moment of the electron is anti-parallel to the direction of angular momentum.
3: The ratio $\cfrac ML$, where M is the magnetic moment, and L is the angular momentum, is constant $\cfrac e{2m}$.
Are these facts somewhat or in some way related to the spin quantum number?
| Yes and no. While nothing you listed is relativistic, L and M will take on only discrete values based on quantum effects in the bohr model. The orbits are integer numbers of wavelengths of the electron’s debroglie matter-wave. In that sense, yes quantum.
But the formulae considered are all classical Maxwell stuff. The field for the dipole comes from the classical (non-relativistic) Bio-savart Law for a point charge:
$$\vec{B} = \frac {\mu_0 q \vec{v} \times \vec{r}}{4 \pi |r|^2}, \text{ } q=e<0$$
(The following article is easy and quick to pan through, see Moving Charges Create Magnetic Fields: https://www.school-for-champions.com/science/magnetic_field_moving_charges.htm )
This is how the magnetic field is created. $\vec{v} \times \vec{r}$ by the right-hand rule points a direction we know define as up, making $\vec{B}$ down due to $q<0$.
The field maximum is inside the electron loop on the plane of orbit. A closed-current loop is commonly referred to as a magnetic dipole with $IA$ its magnetic dipole moment $\vec{M}$, direction down as of $\vec{B}$, where $A$ is enclosed area, and current $I=\tfrac{dq}{dt}=fe$ , frequency times charge (the electron’s charge times how many times the electron passes per unit time). $\implies M = feA$:
$$M = feA = (\tfrac{v}{2 \pi r})e(\pi r^2) = \tfrac{1}{2} evr$$
$$\vec{L}= I \vec{\omega} = (m|r|^2) (\vec{v} \times \tfrac{\vec{r}}{|r|^2})=m\vec{v} \times \vec{r}$$
Where $\vec{L}$ points up by $v \times r$ as noted in the question.
$$\implies \frac{M}{L} = \frac{e}{2m}<0$$
| {
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Is there a way to precisely quantify entanglement in general? Entanglement is among the most remarkable features of quantum mechanics. It is pointed out by many as the responsible for breaking Bell inequalities and numerous other surprising aspects of quantum theory. My issue is that I do not understand how to quantify entanglement in general quantum systems.
I know that the Entanglement Entropy is a very good quantifier for pure states. However, most physical states are not pure. For mixed states in bipartite two-level systems, I know that the Negativity (or Concurrence) are good entanglement quantifiers and are equivalent, making it seem that there is a "unique" way of quantifying entanglement in these systems.
My question is: What quantifiers can be used for mixed states in more general setups with multipartite systems and general Hilbert spaces? Is there a "unique", or best way of quantifying entanglement in these setups?
| Determining whether or not a given mixed state was proven to be an NP-hard problem by Gurvits. Given that, it's quite challenging to have an easy formula for quantifying entanglement, even for bipartite systems! It is worth recalling that nonzero negativity is only a sufficient condition for bipartite entanglement in most dimensions: there are entangled states with zero negativity.
In general, there are infinitely many witnesses of entanglement, each of which gives a sufficient condition for the existence of entanglement; to prove entanglement/separability, one has to check many different witnesses. So, in that sense, there is neither a unique nor a best way of quantifying entanglement in general. There may be specific purposes for which specific types of entanglement are useful and so that can always inform your choice of an entanglement quantifier.
| {
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A Special Relativity question My problem is simple. imagine you have a proton at rest at the origin and an electron traveling along the x-axis with the speed of $0.9c$. (this is our lab frame) now I want to calculate the force on the electron when the electron is at the $x=+d$. I faced a contradiction:
one way of calculating the force is simply $F=Eq$ in which:
$$E=\frac{e}{4\pi\varepsilon d^2 }$$
the other way is going to the electron's frame: in the new frame, we have an electrical field, which is smaller than the previous field by a factor of $1/\gamma^2$ (because of length contraction); and for moving to lab view we don't need anything because the longitudinal component of force won't change.
so now we have two different values for force, why? what's my mistake?
| In order to abide by the conventions used in Resnick's book, I had to assume that the primed frame is attached to the proton so that we have:
$$x'=d'\space , \space E'_x=\frac{e}{4\pi\epsilon_0 d'^2}$$
According to Resnick, the electric field of the proton (in the place of the electron) is calculated as follows from the viewpoint of the electron:
$$E_x=\frac{e\gamma (x-ut)}{4\pi\epsilon_0\left [\gamma^2(x-ut)^2+y^2+z^2 \right]^{\frac{3}{2}}}\space.$$
Since $y=z=0$, we have:
$$E_x=\frac{e}{4\pi\epsilon_0\gamma^2(x-ut)^2}\space.\tag{*}$$
On the other hand, the Lorentz transformation asserts that:
$$x=\gamma(x'+ut')\space,\space t=\gamma(t'+ux'/c^2)\space.$$
Using $x'=d'$ and $t'=0$, we get:
$$x=\gamma d'\space,\space t=\gamma ud'/c^2\space.$$
Substituting the above terms in (*) gives:
$$E_x=\frac{e}{4\pi\epsilon_0\gamma^2(\gamma d'-\gamma d'u^2/c^2)^2}=\frac{e}{4\pi\epsilon_0 d'^2}=E'_x\space.$$
Therefore, the forces are measured the same.
| {
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Deriving the equivalent capacitance in a series circuit formula When we derive the formula for the effective capacitance in series, we say:
$$Q/C_{eqv} = Q/C_1 + Q/C_2 + Q/C_3$$ (if there were 3 capacitors in this case). We would then cancel $Q$ to obtain the formula.
I understand why each capacitor has the same charge, but why does the effective capacitor have the same charge as each individual capacitor? I'd expect the effective capacitor to store a total charge of 3Q (in the given example), not Q?
When the capacitor discharges, would the overall amount of charge released not be 3Q (i.e. the overall charge of the capacitors)?
I saw a similar question on here, and it was answered by explaining that the 'inner capacitors' are isolated from the rest of the circuit, and the +Q and -Q charges cancel? But even so, the isolated charges can trigger electron flow from the 'outer capacitors' during discharge.
If anyone can clear up these doubts, I would be grateful.
|
I understand why each capacitor has the same charge, but why does the
effective capacitor have the same charge as each individual capacitor?
I'd expect the effective capacitor to store a total charge of 3Q (in
the given example), not Q?
It is because each capacitor has a charge $-Q$ on one plate and a charge $+Q$ on the other plate. When you connect them in series, the charges on the interior capacitors cancel. See Fig 1 of the following link:
https://courses.lumenlearning.com/physics/chapter/19-6-capacitors-in-series-and-parallel/
Hope this helps.
| {
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How to add two Matrix Product States of different bond dimensions? If I have the MPS representation of two quantum states, how do I add them? Note that the bond -dimensions need not be the same for the two MPSs.
| Say you have two mps to be added together called $\psi$ and $\phi$. At each site, each mps has a 3-tensor with 2 bond indices and 1 physical index. Lets call the 3-tensors of $\psi$ to be M and of $\phi$ to be N. For each physical index value we can view these M and N as matrices because they have 2 indices, namely the bond indices. So if the physical index dimension is two, we can say we have 2 matrices per site for each mps.
When we add the two mps together, we end up with a new mps, lets call it $\chi$. Now $\chi$ has two matrices per site lets call them H. For addition of mps, H is the direct sum of M with N, this means M resides in the top left block of H and N resides in the bottom right block of H, such that H is block diagonal. The resulting bond dimension of $\chi$ is the addition of the bond dimensions of $\psi$ and $\phi$. This is because the bond dimension would be the dimension of the matrices on each site, since these matrices have two indices which are the bond indices, and given we form H matrices by stacking in block diagonal form the M and N matrices, we end up with the addition of the initial bond dimensions.
Here is a reference (section 4.3): https://arxiv.org/abs/1008.3477
Note: Above I am saying for example matrices M and what I mean is matrices $M^{\sigma_i}$, where $\sigma_i$ is the physical index that would label the two matrices on each site when the physical index dimension is 2, eg $\sigma_i = 1, \sigma_i = 2$ and i labels which site of the mps we are on.
| {
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Where does pseudo force act at? It is known that, to apply Newton's laws in a non-inertial frame, we use the concept of pseudo force. We also know that force is a bound vector. Hence, is there a general way to determine where the pseudo force vector would be located at?
| Pseudo-forces are determined by the acceleration of the reference frame and generally act at all points in space. For a linearly accelerating frame, the resulting pseudo-force is uniform. For a rotating reference frame the force depends on things such as the distance from the rotation axis, the speed of the rotation. The Coriolis force acting on a point particle also depends on the velocity of the particle relative to the rotating frame, but this effect still exists at all points in the rotating reference frame.
If you want to "condense" these forces to single points and moments for an extended object then you just have to find the weighted average of these quantities like you would for any other distributed load. Note that you need to do this separately for forces and torques; it's not true in general that the weighted average of the torques is equal to the torque of the weighted average of the forces.
| {
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It more efficient to generate heat burning electrolyzed hydrogen, or through an electric resistance? It's convenient and simple to use electricity to generate heat, doubtlessly, say in en electric kettle or boiler or heater, but is it more efficient to generate heat burning electrolyzed hydrogen, or through the resistance (kettle / boiler / heater)? Assume the same amount of electricity was used.
| Sorry for my poor english. My native language is french.
I think there is something missing in your question. The current in the resistance is not enough to find the energy. You also need voltage. 1 A at 1 V is different from 1 A at 1000 V. So I assume the same voltage : the minimum voltage to produce a certain amount of hydrogen and oxygen.
It suppose that we produce the hydrogen and oxygen reversibly (very slowly, at the minimum necessary voltage). In this case, the minimum electrical work is equal to the variation of Gibbs free energy : $W=∆G$
Since I assume the same voltage and the same current, this energy would be that transformed into heat if it was supplied it to a resistance.
Now, the question is whether the combustion of hydrogen produces more or less heat than the energy required to produce it. The thermal energy released is this time the variation of enthalpy $W'=∆H$.
The relation between the two is $∆G=∆H-T∆S$ with evidently $∆S>0$ for liquid water electrolysis. So we find $W<W'$ : If I am not mistaken, you gain more heat when you produce hydrogen and then burn it.
The difference does not seem negligible since we have a Gibbs standard free energy of 237 KJ / mol and a standard enthalpy of 285 KJ / mol.
Obviously all this is theoretical and does not take into account the efficiency lower than 1.
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Is it possible to get $\langle AB \rangle$ from $\langle A \rangle$ and $\langle B \rangle$, and vice versa? Assuming that $A$ and $B$ are operators (not necessarily observables) which do not commute and that the quantum system in an arbitrary state $| \psi \rangle$, then ist it possible to get $\langle AB \rangle$ from $\langle A \rangle$ and $\langle B \rangle$, and vice versa ?
Take the example of a quantum harmonic oscillator where the operators involved are ladder operators. I'm using Ehrenfest theorem
$$
\frac{d}{dt} \langle A \rangle = \frac{i}{\hbar}\langle [H,A] \rangle + \langle \frac{d}{dt} A \rangle
$$
which yields ordinary differential equations for the expectation values of the operator $A$, no states are mentioned. So $A$ in the last equation above is to be $a^\dagger, a, a^\dagger a$ and $a^2$. So for example if one has that the initial values for $\langle Q \rangle$ and $\langle P \rangle$, it is easy to find the those for $\langle a \rangle$ and $\langle a^\dagger \rangle$. But is it possible from those initial conditions to get the corresponding initial expectation value $\langle T \rangle$, where $T = \frac{1}{2m}P^2$ is the kinetic energy, to get the time evolution of $\langle T \rangle$?
| No, it is not possible for an arbitrary state: Let $\{|n\rangle\}_{n\in \mathbb N_0}$ denote the set of (normalized) eigenvectors of $N\equiv a^\dagger a$. From the ladder operator algebra we find that
$$\langle n|a|n\rangle =\langle n|a^\dagger|n\rangle = 0 \quad , $$
for all $n$. On the other hand, it holds that
$$\langle n|N|n\rangle = n \quad . $$
Consequently, it is not possible to find the expectation value of $N$ from the knowledge of the expectation values of $a$ and $a^\dagger$ with respect to these states, since the latter are always zero, while the former can be any natural number.
However, note that here in this special case (of being in an eigenstate) the converse is true: Given the expectation value of $N$ with respect to its eigenstates, we trivially know the corresponding expectation values of $a$ and $a^\dagger$.
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Can electromagnetic standing waves be used to move particulates? As detailed in this paper:
https://www.researchgate.net/publication/3941396_Ultrasonic_separation_of_suspended_particles
The researchers use the formation of acoustic standing waves to deposit particulates into the nodes of the waves. Can this me achieved with electromagnetic standing waves and if yes, what qualities in terms of mass, volume or any material property in relation with the incident light should there be?
| Yes they can, see here for a review of optical dipole traps, page 19 for standing wave traps. The trapping potential for a focused retroretlected beam is
$$U\propto -\frac{\Gamma I}{\Delta}\cos^2(kz)\left(1-2\left(\frac{r}{\omega_0}\right)^2-\left(\frac{z}{z_R}\right)^2\right),$$
where $\Gamma$ is the damping rate, $I$ is the intensity of the beam at the focus, $\Delta=\omega_0-\omega$ the detuning of the beam at frequency $\omega$ from the transition $\omega_0$, $k$ is the wave number of light, $z$ the displacement along the direction of the beam from the focal point, $r$ the radial direction, $\omega_0$ the beam waist, and $z_R$ the Rayleigh length.
Optical forces can be used to trap atoms and small particles. For example optical tweezers are used to manipulate small particles. The relation between the material properties and potential depend on what you are trapping, the equation given above is derived for a two level atom. For larger particles the properties of the medium should be considered such as the refractive index.
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How does the small angle approximation lead to 0 here? I'm finding the equations of motion of a mass attached to four springs in a box. See picture:
In the prompt, we're instructed to use "the small-oscillations
approximation, and neglect terms of order $\frac{x^2}{a^2}$ , $\frac{y^2}{a^2}$ , and $\frac{xy}{a^2}$". This all makes perfect sense to me.
Using both force diagrams and the Lagrangian approach, I find the equations of motion. I have the solution, but I do not see how it is possible to reach that solution.
For example, let's find the x-component of the force from the spring at the "top" of the box. The length of the spring for an arbitrary x, y is $\sqrt{x^2 + (a-y)^2}$, and so our total force vector is $F_1$ = $K_2\left(a-\sqrt{x^2 + (a-y)^2}\right)$. And taking the x-component we have:
$$
F_{1x} = K_2\left(a-\sqrt{x^2 + (a-y)^2}\right) \frac{x}{\sqrt{x^2 + (a-y)^2}}
$$
And I am told from the solutions that $F_{1x} \approx 0$. I cannot see how this is possible. I've tried using the approximation $(1+x^2)^{-1/2} \approx (1-\frac{1}{2}x^2)$, but it seems no matter what I do I fail to reach 0.
Does anyone see how small angle approximation can lead to getting $F_{1x} = 0$ here?
| the force $F_{1x}$ is:
$$F_{1x}={\frac {K_{{1}} \left( \sqrt {{x}^{2}+{y}^{2}-2\,ya+{a}^{2}}-a
\right) x}{\sqrt {{x}^{2}+{y}^{2}-2\,ya+{a}^{2}}}}
$$
take the Taylor series for the denominator
$$\sqrt {{x}^{2}+{y}^{2}-2\,ya+{a}^{2}}\overset{\text{Taylor}}{\mapsto}=a$$
and for the nominator
$$K_{{1}} \left( \sqrt {{x}^{2}+{y}^{2}-2\,ya+{a}^{2}}-a \right) x\overset{\text{Taylor}}{\mapsto}=-K_{{1}}yx=0$$
thus $F_{1x}=\frac{0}{a}=0$
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Cauchy sequences through examples in Quantum Mechanics (at the level of the rigor of physicists) I have just read the definition of a Cauchy sequence:
A sequence ($\psi_n$) is a Cauchy sequence in a vector space $V$ when $||\psi_n-\psi_m||\to 0$ when $n,m\to\infty$. The limit of every Cauchy sequence $(\psi_n)$ converges to a definite element $\psi\in V$ i.e. $$\lim\limits_{n\to\infty}\psi_n=\psi.$$
But I cannot feel it completely unless I see an example of such a sequence. What is an example of a Cauchy sequence of vectors $(\psi_1,\psi_2,...)$ that we encounter in quantum mechanics?
| Here's a concrete example. For a particle in an infinite potential well of width $a$, the normalized energy eigenvectors are of the form
$$\psi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right)$$
Most wavefunctions - such as the $\Psi(x) = \frac{1}{\sqrt{a}}$, corresponding to a uniform spatial probability density throughout the well - cannot be written as a finite linear combination of energy eigenvectors. It can, however, be expressed as the sum of the convergent series
$$ \sum_{n=1}^\infty\frac{2\sqrt{2}}{(2n-1)\pi} \psi_{2n-1} \rightarrow \Psi(x)$$
as illustrated with the following plot of the first $n$ partial sums:
The sequence of partial sums $\Psi^{(n)}:= \sum_{k=1}^n\frac{2\sqrt{2}}{(2n-1)\pi}\psi_{2n-1}$ is indeed Cauchy (which can be verified as a nice exercise), as requested.
I have just read the definition of a Cauchy sequence [...]
Note that the proper definition of a Cauchy sequence is that, for any $\epsilon>0$, there exists some $N\in \mathbb N$ such that for all $n,m>N$, $\Vert \psi_n-\psi_m\Vert <\epsilon$. In words, given any arbitrarily small tolerance $\epsilon$, if we go far enough along in the sequence we can find a point $N$ past which all of the terms from the $N^{th}$ onward are within $\epsilon$ of one another.
The definition you provide is problematic, in my opinion. In what way do $n,m\rightarrow \infty$? Do you mean that we hold $n$ fixed, send $m\rightarrow \infty$, and then send $n\rightarrow \infty$ afterward? If that's the case, then we need the sequence to have a well-defined limit which is a priori not necessary for a generic Cauchy sequence. Or we send them both to infinity at the same time? If that's the case, it matters how exactly we do this.
| {
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Rigorous proof that a net force of zero guarantees zero linear acceleration in rigid bodies I've never found a rigorous proof of this fact.
The center of mass' acceleration is not necessarily the linear acceleration, specially if the body is attached to a pin or another geometric constrain, then the center of mass spins like the rest of the body. So how can we find the linear acceleration of a body?
EDIT: ok, the pin or constrain seems to add an external force and thus is a bad example to ilustrate the zero transltational acceleration derived from zero net force.
Yet, the result is still seemly true.
| The statement in question is a direct result of the following two statements
*
*Linear momentum of a rigid body is defined as the total mass times the velocity of the center of mass.
$$ \boldsymbol{p} = m \, \boldsymbol{v} \tag{1}$$
*The net force acting on a body equals the rate of change of linear momentum (Newton's 2nd Law).
$$ \boldsymbol{F} = \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} \tag{2} $$
As far as a rigorous proof of the above,
*
*Can be obtained by summing up the individual linear momentum of each particle in a body $\boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i$. But this depends on a) the definition of momentum which is taken at face value, and b) the kinematics of a rigid body and how any affect due to rotation cancels themselves out and only the motion of the center of mass matters.
*While I have not read Principia myself I am hoping there is sufficient evidence and reasoning behind this law. Maybe someone else can chime in here and point me and the op in the correct direction in proving (2).
Note that the corollary to your statement is that
A rigid body under the influence of a pure torque will rotate about its center of mass.
| {
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Is a 6-quark particle viable? It is my understanding (which may be flawed) that protons and neutrons are stable because the 3 (R, G, and B) quarks form a "white" color singlet. Wouldn't 6 quarks or even 9 quarks create a white singlet? What about RGBGR?
| In principle, a hadron with any number of quarks can be formed provided that the overall color is neutral.
However, hadrons with more than three quarks (observed in particle accelerators) are unstable$^1$ and decay rapidly.
Tetraquarks and pentaquarks have been observed in high energy collisions, but rapidly decay. A particle of the form $\mid RGBGR\rangle$ does not appear to be viable since the overall color is not neutral.
However, something like $\mid RGBG\bar G\rangle$ where $\bar G$ means anti-green, could be a possible quark "molecule" or pentaquark, since we have $$\mid\underbrace{RGB}_{neutral} \ \underbrace{G\bar G}_{neutral}\rangle$$ or four quarks and one antiquark bound together.
As per the link, a tetraquark, or "meson molecule" (two mesons) the combination $$\mid\underbrace{q\bar q}_{neutral} +\underbrace{Q\bar Q}_{neutral}\rangle$$ is also possible, provided we have color-anti-color, or net color neutral (the bar here means anti-color and not antiparticle).
There has in fact been experimental evidence at CERN for a six-quark state, “dibaryon” or Hexaquark.
It would appear that provided we have enough energy, the synthesis of a fleeting hadron with any number of quarks is possible, once again, only if the net color of the combination is neutral.
As for a "nine-quark state", it could possibly be a "tribaryon" or something of the form $$\mid\underbrace{RGB}_{neutral} \ \underbrace{RGB}_{neutral}\ \underbrace{RGB}_{neutral}\rangle$$ but you would need a lot of energy to synthesize this particle, and I do not think anyone has observed a particle with greater quark number than a dibaryon (six).
$^1$ This is because there are more stable lower mass hadron states that they can decay into.
| {
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Can we define operators like $\dfrac{1}{a^\dagger a}$? I was recently reading this paper on Enhancement of Few Photon Optomechanical Effects and could not quite understand eq.(2). The author has written an operator like this:
$$\hat \xi=\dfrac{g_oa^\dagger a}{w_m-g_{cK}a^\dagger a}$$
I don't understand how I am supposed to interpret that number operator in the denominator. My guess is that the operator $\hat \xi$ is defined such that its product with $(w_m-g_{cK}a^\dagger a)$ gives $g_oa^\dagger a$. But I am not sure.
Any help is appreciated.
| To supplement Andrew's nice answer, note that vectors in the occupation number basis are eigenvectors of $a^\dagger a$, i.e.
$$a^\dagger a |n\rangle = n|n\rangle$$
As a result, the action of your operator $\hat \xi$ on such a vector becomes
$$\hat \xi|n\rangle = \frac{g_0 n}{w_m - g_c n}|n\rangle$$
which extends by linearity to any state $|\psi\rangle = \sum_n c_n|n\rangle$. This basis can also be used to rewrite the operator as
$$\hat\xi = \sum_n \left(\frac{g_0 n}{w_m - g_cn}\right) |n\rangle\langle n|$$
This so-called spectral decomposition is the means by which most "functions of operators" are actually defined.
As a technical note, the trouble with power series expansions of operators is that they generally only converge properly when the operator in question is bounded. As an example, one is tempted to write
$$\frac{1}{w_m - g_c a^\dagger a}\simeq \frac{1}{w_m} \sum_k \left(\frac{g_c}{w_m}a^\dagger a\right)^k = \frac{1}{w_m}\left(1+\frac{g_c}{w_m}a^\dagger a + \ldots \right)$$
The problem with this expansion should be clear if we try to apply it to the vector $|n\rangle$, at which point we have
$$\frac{1}{w_m - g_c a^\dagger a}|n\rangle = \left[\frac{1}{w_m} \sum_k \left(\frac{g_c}{w_m}n \right)^k\right]|n\rangle$$
The prefactor clearly goes to infinity whenever $n> \left| w_m/g_c\right|$.
If instead we use the spectral definition, we simply obtain
$$\frac{1}{w_m - g_c a^\dagger a}|n\rangle = \frac{1}{w_m - g_c n}|n\rangle$$
which (a) is perfectly well-defined for all $n$, and (b) can easily be shown to agree with the power series expansion, provided the latter converges.
| {
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Basics of centripetal force Suppose an object is moving in a circular path. We know that the net force that is working on that object is the centripetal force towards the center. But the object should have gone closer towards the center in that case due to the radially inward force working on it, but it doesn't. Why does the object remain on the circular path instead of going closer towards the center?
For people who would be introducing centrifugal force in this case, i have a doubt on this too. Centrifugal is a pseudo force that only works when we are in the frame of the rotating object meaning we experience a pseudo force that pushes us radially outward. When we are in this frame, does centripetal and centrifugal both work on us?
But let us stay in ground frame as of now. Then what is the cause of the object not being pushed radially inward due to the effect of centripetal force? I am asking this question to clear out my doubts for strengthening my basic concept of physics. Hope the physics lovers will find this question relevant.
| That's cause centripetal force is exactly the amount of twist needed for the momentum vector of the particle at an instant so that it sticks tangent to the path at the next.
This works for unit speed, and unit mass, the force is exactly equal to the curvature of the path.
| {
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How does a system with two black bodies connected via a circulator avoid violating the second law of thermodynamics Consider two different temperature black bodies connected by an ideal circulator (or similar passive and directional device) such that thermal radiation emitted from the cooler black body is able to reach the hotter black body, but thermal radiation from the hotter body is redirected in some other direction and not able to reach the cooler black body (image provided to illustrate system).
If the black bodies were replaced with (ideal) electronic transmitters having identical emission characteristics to the black bodies, there would be no issue with an ideal circulator directionally passing the radiated signals as described. However, the second law of thermodynamics indicates that the system can not function as described when the thermal radiation sources are black bodies, since it implies that the passive circulator is capable of decreasing entropy. Since a circulator is not capable of distinguishing between electromagnetic radiation that originates in a black body or a transmitter, it's not clear how the circulator's behaviour could change to ensure that the second law of thermodynamics isn't violated.
There must be some aspect of the system description, the behaviour of circulators or the second law of thermodynamics that avoids the apparent contradiction. What is it?
This question is an extension of questions 661991 and 136083, but substantively different.
| Let us assume the perfect passive circulator.
We do not only have 2 heat baths (blue and red). We have a third, at the bottom of the circulator. It surely has some temperature.
Over time, all three heat baths will reach the same temperature.
Suppose T < B < R. Then B is cooling as it sends radiation to R, and R is cooling as it sends radiation to T, and T is sending less radiation to B which allows B and R to cool because they ultimately contact T.
Suppose B < T < R. Then although B cools towards R, it receives more from T, so B, the coolest blackbody, is warming up in net, as you expect.
Suppose B < R < T. Then B is warming up in net, as you expect, due to T.
The net result of the three bodies coming to equilibrium is an increase in entropy.
| {
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How much energy was used to charge the inductor? Electrical systems question I am a math major, and I am taking a mandatory class that looks more like physics and the professor gave us an assignment with a question that I don't have any idea on how to solve it. The question is originally in portuguese, so, I will try my best to translate.
In t = 0, a battery is connected to an inductor with inductance L. At the instant T, the current in the inductor is constant. How much energy was used to charge the inductor?
The question is general, without any explicit number. If anyone can help me, I would be grateful. I feel like it's simple, but I don't even know how to start.
The professor is using the following book: Modern Control Engineering by Katsuhiko Ogata (this problem is not here).
| For an inductor (with inductance $L$) the defining relation
between voltage ($V$) and current ($I$) is:
$$V(t)=L\frac{dI(t)}{dt} \tag{1}$$
To find the total energy $E$ fed into the inductor
you need to integrate the power ($VI$) over time $t$ (from $0$ to $T$).
Then use equation (1), and finally do the integration by substitution.
$$E=\int_0^T V(t)\ I(t)\ dt
= \int_0^T L\frac{dI(t)}{dt}\ I(t)\ dt
= \int_0^{I(T)} L\ I\ dI
= \frac{1}{2}L\ I(T)^2$$
So the energy depends only on the final current,
but not on the explicit time course of the current $I(t)$.
| {
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What causes the diffuse reflection of an object onto a nearby object to be colored the same color as that of the object? Assuming a light source, emitting white light,a red colored diffuse sphere lit up by the light, and a white plane below the sphere acting as the 'floor', we find the diffuse reflection of the sphere on the floor to be reddish in color, just like the sphere itself
Why does it have the exact same color as the sphere? The emitted light was white in color, so when the photons of the light source, collide and reflect from the sphere, should they not still be white, causing the diffuse reflection to instead be white? What causes the photons to take on a reddish hue?
In the case, of a black sphere as opposed to a red sphere, can we expect the same thing to happen? Will the diffuse reflection be grey-black in color as well? If, on the spot where the reflection is cast,a light grey soft shadow, lighter than the diffuse reflection were to exist, would the reflection of the sphere, 'overpower' that of the shadow, making the region darker as a result?
| When you shine white light at a colored object, the light that scatters off it is missing some of its wavelengths i.e., it is tinted: this is why it appears colored to our eyes.
So now you have colored light radiating away from that object, and when it strikes a nearby object and is scattered off, the scattered light is of course still missing the same wavelengths that got absorbed by the first object.
This tints the second object to a color similar to that of the first object.
| {
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Is it possible to create a magnetic environment where it pushes and pulls at the same time, making the target “levitate”? To better understand why I ask this, the backstory is I’m getting more and more annoyed by rolling office chairs rotating, hitting my ankles, hard to switch directions when wheels having ~90 degrees differences.
Imagine a ball, and put a bowl on it (upside down). The ball is metal, and the bowl is a magnet that pushes the ball, so it doesn’t touch it (because of the shape), and pulls it at the same time, so if you lift the bowl up the ball goes with it. This used as wheelchair castors could prevent my, and possibly others’ ankles to be hit.
Is it possible to create this environment that the magnets are in perfect balance in every direction, both pushing and pulling?
|
Is it possible to create this environment that the magnets are in
perfect balance in every direction, both pushing and pulling?
If by magnets you mean fixed magnetic moments in a static configuration, then the answer is no per Eanrshaw's theorem.
If the magnets are electromagnets or at least can change their magnetic fields dynamically, then maybe something can be done. With gravity you can build systems that are pull out of the ground (a metallic surface) due to induction while gravity pushes them down.
Another possibility is to use superconductors. With the right material and magnet, you can have a magnet that is locked at a certain distance from a superconductor (it can go up or down) due to flux pinning. Of course, this would only work at very low temperatures.
| {
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Does relativity mean that the crew of a relativistic rocket would experience less acceleration than in our frame of reference? I have been told regarding a 1 g rocket that "the amount you accelerate would be less due to relativity".
Does that mean that from the crew's time dilated perspective, they would experience less acceleration than we observe in our frame of reference? Could a ship be accelerated at say, 10 g from our frame of reference on Earth, while the crew of the ship only experiences 1 g of acceleration in theirs?
If this were possible, how far can we take this, and how quickly?
| There is no frame of reference in SR in which an acceleration can stay constant at a non zero value for an infinite amount of time. This would inevitably lead to a velocity that is greater than the speed of light.
It is well-known that when two frames $S$ and $S'$ move relative to each other with velocity $v$ and a velocity is $w$ in $S$ then in $S'$ it is
$$
w'=\frac{w-v}{1-vw/c^2}\,.
$$
To calculate the relations for the accelerations is straightforward:
$$
\dot{w}'=\dot{w}\frac{1-v^2/c^2}{(1-vw/c^2)^2}\,.
$$
Note that $\dot w'=dw'/dt$ and therefore, $dw'/dt'=\gamma\,dw'/dt=\frac{dw'/dt}{\sqrt{1-v^2/c^2}}\,.$ It follows that
$$
\frac{dw'}{dt'}=\frac{dw}{dt}\frac{\sqrt{1-v^2/c^2}}{(1-vw/c^2)^2}\,.
$$
It is easy to see that for very small $w$ the acceleration in $S'$ is less than the one in $S$ which is probably what your citation meant.
| {
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Is it possible to calculate angular acceleration from a measured linear acceleration? I am trying to use an accelerometer to measure the angular acceleration of a robotic arm.
From rigid body kinematics, the following relation is known
\begin{align*} {^{i} {\boldsymbol{a}}_m} & = {^{i} {\boldsymbol{a}}_l} + ^{i} \dot{{\boldsymbol{\omega }}}_{i} \times {^{i} {{\boldsymbol{X}}}_{S_m}} + {^{i} {{\boldsymbol{\omega }}}_{i}} \times \left({^{i} {{\boldsymbol{\omega }}}_{i}} \times {^{i} {{\boldsymbol{X}}}_{S_m}} \right) \; \end{align*}
where $\dot{\omega}$ is the angular acceleration and $ {^{i} {\boldsymbol{a}}_m}$ is the measured acceration at a point $X$ along the arm.
The problem is that this equation is not solvable for $\dot{\omega}$ because, it's in a cross product with the position vector of the sensor. ($a \times (b +ka) = a \times b + k(a \times a) = a \times b$)
This paper claims that they used an extended Kalman filter and this relation to estimate the angular acceleration, but I have no idea how that is possible when this equation does not have a unique solution for $\dot{\omega}$.
Can anyone point out if I'm missing something which can help me solve this problem?
| I'm not clear on what your equation indicates or exactly what your accelerometer is measuring, but if you need an angular acceleration for an arm, that could be only horizontal or vertical or a vector sum of the two. In each case α = a/r.
| {
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Doubt in derivation of pressure exerted by ideal gas In books the derivation of pressure exerted by a gas on a closed container is derived for a cubical geometry, but since it contains number of molecules per volume (denote it by $n$) the books say it's independent of geometry. My question is how they say it's independent of geometry because we get this $n$ for cube only, how can we be so sure that this holds for any super complicated and distorted volume?
| You said it yourself , it depends on the number of molecules per volume not the total number of molecules. And by saying so it means you are only considering about the density of gas contained in the closed objest whatever it may be. And also density of gas will not change by changing the shape of container.also the main formula for net pressure has total mass and volume as final factors which can change depending on the shape of the container.
| {
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Simplest exactly solved model displaying a phase transition? The classical example of an exactly solved model which displays a phase transition is the 2D Ising model. However, all the proofs I've seen of this have been very long and complicated.
So, I wanted to know whether there were any other exactly solved models with phase transition, which were easier to solve, or that the 2D Ising model is the simplest such model that we know of.
| The quantum Ising model in a transverse field
$$
H=\sum_n \left(\hat \sigma_{z,n} \hat \sigma_{z, n+1} +\lambda \hat \sigma_{x,n}\right)
$$
is easily solvable and has a phase transition. It's a one-dimensional quantum model but, through the usual quantum-classical map that takes the hamiltonian to the transfer matrix, it is equivalent to the classical Ising model in 2d.
| {
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When a car accelerates relative to earth, why can't we say earth accelerates relative to car? When a car moves away from a standstill, why do we say that the car has accelerated? Isn't it equally correct to say that the earth has accelerated in the reference frame of the car? What breaks the symmetry here? Do the forces applied to the car have special significance in determining which frame is inertial and which one is not?
Please explain in simple terms.
|
What breaks the symmetry here?
The accelerations are not symmetric because (proper) acceleration itself is not relative (frame variant). A simple accelerometer can measure the asymmetry. The car’s accelerometer measures a large acceleration. The earth’s does not. The measured asymmetry in the acceleration is due to the asymmetry in the mass under equal (but opposite) forces.
Note, the accelerometer measures proper acceleration. It is possible to discuss coordinate acceleration instead. Coordinate acceleration is relative, but it is also not particularly physical. No physical experiment can depend on coordinate acceleration.
| {
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Mandl & Shaw QFT chapter 1 question Page 3 of Mandl & Shaw claims that, given a vector $\pmb{A}(\pmb{x},t)=\pmb{A}_{0}e^{i(\pmb{k}\pmb{\cdot} \pmb{x} - \omega t)}$, $\pmb{\nabla} \pmb{\cdot} \pmb{A} = 0$ (eq. 1.6) implies $\pmb{k} \pmb{\cdot}
\pmb{A} = 0$ (eq. 1.7).
I'm having trouble figuring out why (1.6) implies (1.7).
| $$\mathbf{\nabla}\cdot\mathbf{A}=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}=i\mathbf{k}\cdot\mathbf{A}=0$$
| {
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What is the relationship between the Galilean group and the Poincaré group? What is the relationship between the Galilean group and the Poincaré group?
Are they siblings within the Lie group? Or does the Poincaré group contain the Galilean group as a subgroup?
I'm not so much interested in the Galilean group being the limit of the Poincaré group for c -> inf.
| It is assumed you have appreciated Inönü, E.; Wigner, E. P. (1953), "On the Contraction of Groups and Their Representations" Proc. Natl. Acad. Sci. 39 (6): 510–24, and the super-helpful Gilmore text in Group contraction.
Very crudely, the Poincaré Lie algebra,
$$
[J_m,P_n] = i \epsilon_{mnk} P_k ~, \qquad [J_i, P_0] = 0 ~, \\
[K_i,P_k] = i \delta_{ik} P_0 ~, \\
[P_0,P_i]=0 \qquad [P_i,P_j]=0 \qquad [K_i, P_0] = -i P_i ~, \\
[J_m,J_n] = i \epsilon_{mnk} J_k ~, \qquad
[J_m,K_n] = i \epsilon_{mnk} K_k ~, \\
[K_m,K_n] = -i \epsilon_{mnk} J_k ~,
$$
given relabelings $E=-cP_0$ and $K_i=cC_i$ contracts upon $c\to \infty$ to the Galilean algebra,
$$
[J_m,P_n] = i \epsilon_{mnk} P_k ~, \qquad
[J_i,E]=0 \\
[C_i,P_j]= 0,~\\
[E,P_i]=0, \qquad
[P_i,P_j]=0, \qquad
[C_i,E]=i P_i \\
[J_m,J_n] = i \epsilon_{mnk} J_k ~, \qquad
[J_m,C_n] = i \epsilon_{mnk} C_k ~, \\
[C_i,C_j]=0 .
$$
There are a few subtleties and wrinkles, extensions, to be sure, which I gather you are not focussing on, but, crudely, the third and the last commutation relations trivialized/collapsed. (There is more, but I am oversimplifying...).
This collapse/amputation is the Lie algebraic manifestation of a group contraction.
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Why won't all voltage be used up on first resistor in series? tldr: I am having trouble conceptually understanding voltage between resistors in series, even though I know how to calculate it using Ohm's law. How do the electrons "know" there are more resistors after going through the first one? Why doesn't it use up all the 10V so it's 10V -> resistor 1 -> 0V?
Case A: If a circuit has 1 resistor, the voltage on each side will be completely used up. For example, with a 10V battery and $5\Omega$ resistor, the voltage will be 10V -> resistor -> 0V, with a current of 2A.
Case B: However, if we attach another $5\Omega$ resistor, then the voltage will be 10V -> resistor 1 -> 5V -> resistor 2 -> 0, with a current of 1A.
From the perspective of the electrons, they don't know that there is another resistor down the line.
So why does the voltage drop across the first resistor 10V in Case A but only 5V in Case B?
I'm probably not understanding something very simple. This is my first circuits class and it's a lot to wrap my mind around! Thanks in advance.
| Think of it this way.
One battery terminal sits at $10V$, the other at $0V$. The first resistor has one terminal connected to the $10V$ pole of the battery.
Now assume, as you did, that the voltage is "all used up" after the first resistor (i.e. if $2A$ flows through it). In that case the voltage would be $0V$ at that point. That means that the second resistor has both terminals at $0V$. Hence there could be no current flowing through that resistor. However, since the resistors are in series, that automatically means the current through the first resistor also has to be $0A$.
This is in contradiction with our assumption that $2A$ flows through the first resistor. The only way to fix the contradiction is to accept that the same current flows through both, i.e. a total of $1A$
| {
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Maximum value of angular acceleration In a video of Prof. Walter Lewin, in which he is talking about the maximum value of angular acceleration, he says that it does not imply that the angular acceleration is zero if the angular velocity is zero and that when the object conducting some angular motion is stationary the angular acceleration is maximum for the object—or something like that; at least that's what I understood from it. Can anyone please give and explanation as to what he is trying the say.
| It's similar to SHM. When an object performs SHM the displacement from the equilibrium is
$$x=A\sin (\omega t) $$
by differentiating, there are equations for velocity and acceleration
$$v=A\omega\cos (\omega t) $$
$$a= - A\omega^2\sin (\omega t) $$
from graphs of those functions it can be seen that at positions where $v$ is zero the acceleration is maximum. That's because the zero velocity happens furthest from the equilibrium position, where the restoring force is largest.
The Professor on the video was talking about a similar thing for angular quantities.
| {
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Mechanism of Relativistic Momentum The formula for relativistic momentum is $\vec{p}=\gamma m\vec{v}$.
To derive this formula, one analyzes a collision while assuming the principle of relativity and the conservation of momentum principle are correct:
https://www.feynmanlectures.caltech.edu/I_16.html
I'm fine with all that. What I want to know is the mechanism which causes this. At speed $0.0001c$, force applied is approximately proportional to acceleration. At speed $0.89c$, the same force applied changes the speed of the object very little. Acceleration due to a force is a precise function of the mass and the pre-existing velocity. What is causing this, is there something unseen which 'acts' upon the moving object to render this phenomenon? Does it have something to do with the Higgs field? Is it because of time dilation? Thanks.
| I think what you are looking for is an understanding of 4-force, that is the vector which describes force in relativity. Without getting into the details (which are easily accessible by a google search), the key aspect is that the relativistic four force for a given classical force depends on the velocity of the object that it acts upon. The four force is in some sense a consequence of time dilation, or maybe more exactly that objects experience their own proper time, which is different from that of other objects in different frames.
As for a specific mechanism for objects when they are accelerated at relativistic speeds, there cannot be one, for consider Einstein’s postulate that (paraphrased) all inertial observers agree on what happens in the universe. Now consider a observer who sees the object stationary, and another who sees it at a relativistic speed, and consider the object to be accelerating. If the relativistic observer sees some special mechanism that affects its acceleration, then the stationary observer must see this also, but then that would make the idea of the relativistic mechanism redundant. Perhaps I have misunderstood your question in this part though?
| {
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How come black body have more emissivity and more absorbtivity(a) at same time? We have two definitions to look at
Absorbtivity(a): the ratio of absorbed energy and incident energy on a body
$a_{BlackBody} = 1$
so if i have a tourch light that gives red light, in a dark room I point this tourch light at this black body, then according to above definition, I should see black color
Good absorbers are good emitters from kirchoffs law, and also stefans law
$$\frac{d\theta}{dt} = \sigma AeT^4$$
Now according to that statement, if we reconduct same experiment, when I point my red torch light towards black body it should now emmit red color
Aren't both definitions contradictory to each other or did I misinterpret something
| You are mixing up two concepts, namely the emission of light as a consequence of temperature and the reflection of incident light.
When you look at objects in daylight, their apparent colour is caused by the fact that they absorb some of the incident daylight and reflect the rest. It is the combination of the frequencies of the reflected portion that determines their colour. A matt black object will absorb most of the incident light and reflect little of it, which is why it seems very dark.
Light emitted by a body as a result of its temperature is caused by a different mechanism, namely internal thermal agitation.
If you have a piece of ebony and, say, a piece of birchwood and you look at them both in daylight, the ebony will appear very dark and the birch very light, the difference being that one reflects less of the incident daylight than the other. If you put both pieces in a fire so that they burn, they will both appear orange/red, because that light is being generated internally, rather than being reflected daylight.
| {
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Conservation of angular momentum in an inelastic collision
I have a question about the second method used to solve the problem above.
The moment of inertia with respect to the stick's midpoint after the collision is $ml^2/12 + ml^2/4$ or $ml^2/3$ so the angular momentum with respect to the stick's center after the collision is $ml^2/3*w$. Therefore, equation 8.56 becomes $mv_0l/2 = ml^2/3*w$ but this doesn't give the same w value. Can someone explain why?
| The CM of the system (the mass and the stick together) moves in a straight line and every point of the system rotates about the CM.
I think the mistake arises from considering the mass m and the stick together as a whole system and writing the conservation of angular momentum about a point (middle of the stick) which is not the CM and this point, itself, is rotating around the CM. You can do that, but have to take the angular momentum of the CM about the middle point into calculation, as it is done in the second method.
I considered the stick as a system (without the mass m) and wrote angular momentum conservation around its CM (the middle point of the stick) and got same answer as 8.56
| {
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Eigenvalues of Product of 2 hermitian operators Let $A$ and $B$ be two Hermitian operators. Let $C$ be another operator such that $C = AB$. What can we say about Eigenvalues of $C$? Will they be real/imaginary/complex? What I did was to search for examples. The following were examples (in matrix representation) I looked for:
$ A =
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$ and $ B = \begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}$ to get a hermitian matrix and so real eigenvalues.
Next I tried:
$ A = \begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}$ and $ B = \begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}$ to get Anti-Hermitian matrix and so imaginary eigenvalues.
Is there a more concrete way of solving this? Can we have a general complex number as eigenvalues for the product of the Hermitian Matrices?
| In general, we can say that $C=AB$ will have real, imaginary and complex eigenvalues (complex of the form $z=a+ib$ where and $\{a,b\in \mathbb{R}\mid a,b \ne 0\}$ as shown in the comments by Mark and Qmechanic's answer). For example, if
$$A=\begin{bmatrix}
0 &1 \\
1& 0
\end{bmatrix}\ \ \text{and}\ \ B=\begin{bmatrix}
1 & 0\\
0& -1
\end{bmatrix}$$ where $$AB=\begin{bmatrix}
0 &-1 \\
1& 0
\end{bmatrix}$$ will not have real, but imaginary eigenvalues.
However, one thing we can say is that if $A$ and $B$ commute then $C=AB$ will always have real eigenvalues, since the eigenvalues of all Hermitian operators are real.
So if $$C=AB$$ then $$C^\dagger =(AB)^\dagger =B^\dagger A^\dagger =BA$$ since $A$ and $B$ are Hermitian, and clearly $$C^\dagger =C$$ if $$[A,B]=AB-BA=0\rightarrow AB=BA$$ This means that $C^\dagger =C$ only if $A$ and $B$ commute in which case $C$ will have real eigenvalues.
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A lens with a high permeability In real life materials with high $\mu$ values are not realistic, so lenses are made with high $\epsilon$ materials. But what would be the impact of achieving the same refractive index by increasing $\mu_r$ too.
For instance, if we made a lens with $\epsilon_r=\mu_r$ the wave impedance of free space would be the same as the lens, right? So would the lens just work with much much much less reflections? How big of an impact would this make?
Or what if we used a lens with $\epsilon_r=1,\mu_r>1$ , how different would different ways of achieving the same refractive index be? Which would be the best pick?
| The refractive index is $\nu = \sqrt{\epsilon_r \mu_r}$ because the speed of light is $c/\nu$ in that medium. At this level of idealization it does not matter what the relative permittivity or relative permeability is, only their product matters when it comes to ideal propagation in a homogeneous medium.
If the two media have different permittivities, $\epsilon_r, \epsilon'_r$, and permeabilites, $\mu_r, \mu'_r$ , then the Fresnel reflection formulas will also change from the commonly known ones as in https://en.wikipedia.org/wiki/Fresnel_equations because you now have to match both the E and H fields at the interface. For example, in the case of the incident E field parallel with the interface the reflectivity $\mathcal R$ becomes, see Jackson 7.39:
$$\mathcal R = \frac{2\nu cos {\mathcal i}}{\nu cos i + \frac{\mu_r}{\mu'_r}\sqrt{\nu'^2-\nu^2 sin^2 i}}$$
If now you assume that the angle of incidence is $i=0$ and also the special case $\mu_r/\epsilon_r=\mu_r/\epsilon_r$ you do get $\mathcal R=0$, as expected.
| {
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If the probability of a point (photon) hitting another point (electron) is zero why do they collide? If the probability of a point (photon) hitting another point (electron) is zero why do they collide? To have a probability greater than zero almost one of them should be not a point. Correct me, please if I am wrong.
| In popular presentations of particle physics you often find the statement that photons and electrons are "point particles", while protons and other composite entities are not. However to call a photon or an electron a "point particle" is quite misleading unless you immediately add that we are dealing with quantum physics and a quantum physical "particle" is always spread out over some range of position and momentum.
When you see Feynman diagrams it looks a lot as if one little point-like thing comes along and absorbs or emits another little point-like thing at a vertex in the diagram, but this is wrong. The lines in the diagram usually represent states of well-defined momentum and energy, which means the position of each entity (e.g. electron or photon) is totally spread out, so these states are as far removed from "point-like" as they could possibly be!
In practice what happens is that you have states that are intermediate: neither completely spread out, not completely focused at a point. Such states are called wave-packets.
The basic example of a photon interacting with an electron is the process called the Compton effect. If the incoming photon and electron were each focused in such a way that their wave-packets never overlapped then indeed they would not interact! The Compton effect is observed when the wave-packets do overlap.
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What is light cone? Explain to mathematicians who understand the Lorentz group but not light cone Mathematically the Lorentz group is precisely the $O(1,3)$ is the 4-vector rotation preserving the inner product of 4-vector under this metric
$$
\eta_{\mu \nu}=(+1,-1,-1,-1).
$$
There are four distinct sectors of this $O(1,3)$.
Say the 4-vector is $A$ and $B$, then for the Lorentz transformation $R_{n \nu} \in O(1,3)$ on the 4-vector, we have the following invariant inner product of 4-vectors under the Lorentz transformation:
$$
A^\mu \eta_{\mu \nu} B^{\nu} ={A^\mu}' \eta_{\mu \nu} {B^\nu}'=A^\mu R^{T}_{\mu m} \eta_{m n} R_{n \nu} B^{\nu}
$$
Lorentz group $O(1,3)$ explains the symmetry group of the spacetime at any fixed point.
Built in on this data of Lorentz group $O(1,3)$, how do we explain what is light cone to mathematicians?
Namely what is light cone? mathematically? How to explain to mathematicians who understand $O(1,3)$, but not light cone?
What exactly is this cone of light cone defined mathematically?
| Given a four-vector $A^\mu$, define the “interval” associated with $A$ as
$$
\Delta s_A = \eta_{\mu\nu}A^\mu A^\nu = \left(A^0\right)^2 -\vec A{}^2
$$
We say that $A$ is
*
*“spacelike” if $\Delta s_A < 0$. An example is $(0, \vec A)$.
*“timelike” if $\Delta s_A > 0$. An example is $(A^0, \vec 0)$.
*“lightlike” if $\Delta s_A = 0$.
The “light cone” is the surface formed by all light-like four-vectors. The interval $\Delta s_A$ is preserved by Lorentz transformations.
| {
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Provided a unit vector and Force, how can I calculate it's components? Say I have a $F=kQ_{1}Q_{2}/r^{2}$ and a direction vector $(x, y, z).$ How can I find the component forces $F_{x}$, $F_{y}$, and $F_{z}$?
| You need to know the direction of the force as well as its magnitude. The force’s component along the $x$ axis is then
$F_x = |\vec F| \cos \theta$
where $\theta$ is the angle between $\vec F$ and the $x$ axis etc.
If the force is radial i.e. $\vec F = |\vec F| \vec {\hat r}$ then its components at $(x,y,z)$ are
$\displaystyle F_x = |\vec F| \frac x r$
etc.
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Does transmutation of nuclear waste need more energy than it originally produced? Nuclear transmutation has been proposed as a method to reduce nuclear waste from nuclear power plants:
*
*https://en.wikipedia.org/wiki/Nuclear_transmutation
I recall reading somewhere that this is economically not very attractive, because it basically needs more energy to transmute the nuclear waste than the burning of the nuclear material produced in the first place. However I couldn't find any good source for this claim, so maybe this is wrong. It might also depend on the method of transmutation (with fast breeders or accelerator driven systems). Does anyone know how much energy transmutation needs in comparison to the energy produced by the nuclear material? How much would transmutation on industrial scale lower the total efficiency of energy production in nuclear power plants?
| This article describes some of the reasons we don't use nuclear waste as fuel.
The biggest problem is that we have plenty of uranium (and other energy sources) and burying the waste is not very expensive so there is not a large economical incentive to burn the rest of the fuel. This in turn leads to less interest in developing a reactor that can do what you suggest. So even though more energy may come out than you put in, it may not be worth the monetary investment to research and develop these reactors.
| {
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Inverse Square vs Exponential I feel a little foolish asking this, but I keep reading sources which say that for an inverse square law relationship, e.g. light intensity vs distance from source, the intensity decays exponentially.
Are inverse square and exponential the same? I would think not, as I cannot find an algebraic way of writing $I \sim 1/r^2$ in the form of $I \sim \exp(ar)$. I do know exponential is the same as "geometric" increase (or decrease), as you keep multiplying by the same number every time.
This may be a symptom of the modern informal usage where people say something changing "exponentially" just means "a lot, very quickly" (don't get me started...). But I know the folks here can set me straight.
| Inverse square is not the same as exponential dropoff. Any source which says this is using "exponentially" in a colloquial way. Hopefully they don't then try to do mathematics immediately afterwards!
There are some exponential dropoffs in physics, such as the intensity of evanescent fields, but the drop off of normal light is decidedly an inverse square law and not an exponential law.
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Do electromagnetic waves contain electrons? I understand that EM waves are oscillating electric and magnetic fields. But doesn't this mean that the wave itself contains charged particles that generate the fields?
| Weirdly, they don't contain electrons, but are actually made up of photons - even those waves that are not in a frequency we know as visible light.
Here's a source from Nasa.gov -
https://imagine.gsfc.nasa.gov/science/toolbox/emspectrum1.html#:~:text=Electromagnetic%20radiation%20can%20be%20described,energy%20found%20in%20the%20photons.
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What does it mean for a field theory to be invariant? In this paper A. N. Schellekens, Conformal field theory p.8 they mention the following
If a field theory has a conserved, traceless energy momentum tensor, it is invariant
both under general coordinate transformations and Weyl transformations.
What is meant with 'a field theory is invariant'? Is the action invariant? The fields? The equations of motion? All of the above?
| The action, yes. The equations of motion, therefore also yes. But the fields themselves, no. That's like expecting a rotation of axes to preserve Cartesian coordinates in Newtonian physics.
| {
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Understanding Space-time intervals and its types I am taking Introduction to Modern Physics class. There, we were studying spacetime intervals as a subsection of Lorentz's transformation. My professor said that $\Delta x^2-c^2\Delta t^2$ is invariant, and then he said us that it is a lightlike event if $\Delta x^2-c^2\Delta t^2=0$, timeline if $\Delta x^2-c^2\Delta t^2<0$ and spacelike if $\Delta x^2-c^2\Delta t^2>0$. I understand that a spacelike event is when different reference frames do not agree on the order of the events. That is because we used Lorentz's transformation to show that, if say A and B are space-like events, A and B happen simultaneously in one frame, A happens before B in one reference frame, and B can happen before A in another reference frame. These are all mathematical notions for these events. I do not understand when these events can occur in reality. Can someone give me some examples so that I can understand them?
| "Events" are akin to [localized] points.
It's not the "events" that are spacelike, timelike, or lightlike.
Rather, it is the "relationship between pairs of events" that are spacelike, timelike,or lightlike.
The pair of events "A and B" is timelike related if there is a timelike path in spacetime (say the worldline of a particle with nonzero mass) from A to B.
It's possible for events "B and C" to spacelike related to each other,
while each being timelike related to a third event A. So, it's about the relation of
a pair of events.
In special relativity, the character of the "interval between a pair of events A and B" is given by the sign of the square-norm of the spacetime-displacement-vector from A to B.
To get a feeling for these relations, it's best to draw a spacetime diagram.
Then draw in the lightcones of each event...
and see which events lie inside the lightcones of others.
Such events are timelike-related to each other.
Note that there is a transitivity property for future-timelike-related events.
If B is in the timelike-future of A, and C is in the timelike-future of B,
then C is in the timelike-future of A.
You can extend this notion to "causal-futures" (future-timelike-or-future-lightlike-related).
But this doesn't work for lightlike-relations or spacelike-relations.
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How to simplify this complex circuit? I am new to circuit solving and I tried to simplify this circuit, but I am unable to do so. I can't figure out which resistances are in series and which resistances are in parallel.
Do I have to use star-delta conversion here or the circuit can be solved without conversion?
I just need a simplified circuit diagram and the rest I can solve. Thanks for any help!
| I would define 5 current loops (making sure that each resistor is in at least one loop). Then write 5 voltage loop equations (summing voltage drops). Solve for the currents.
| {
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How to demonstrate that light carries angular momentum by making an object rotate? Electromagnetic fields, carry angular momentum. However, I want to demonstrate by an experiment and convince a bunch of high school students, that electromagnetic fields do carry angular momentum. To that end, can we design an experiment in which the angular momentum of the electromagnetic field be transferred to a mechanical object causing it to rotate?
| One example can be found in Introduction to Electrodynamics by Griffiths, example 7.8.
Horizontally suspend a uniformly-charged insulating ring from its center (perhaps by attaching spokes from the ring to its center, and then attaching a string to the center). Place a solenoidal electromagnet directly below the center of the ring with orientation perpendicular to the plane of the ring (i.e. vertically).
Switching the electromagnet on or off will impart angular momentum to the ring and cause it to rotate, even though no mechanical torque is acting on the ring.
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How can a person be hit by a high-energy proton beam? There is this somewhat famous story of a Russian particle physics Ph.D. student from the 70s, who stuck his head into a particle collider and got hit by a beam of high-energy protons. For more details see here, here or here.
I am confused about how exactly this is even possible. As far as I understand one needs to create an ultra-high vacuum in a particle collider to let different particle beams collide. If there is "air" in the collider, which there certainly would be some in the case of maintenance work, wouldn't the high-energy protons just interact with the surrounding air and ionize it? This would also happen in a relatively short distance (this is intuition, not sure if it is true and how one could check...), so I'm not really sure how a person can be "hit" by a proton beam...
| Here is a slide that I found by Goggling "Proton beam in air"
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Linearization of 1D maps about a fixed unstable point Recently, I was going through the paper Controlling Chemical Chaos in a three variable autocatalator system, by Peng et al. Here are the references
Although I have been introduced to 1D maps and the logistic map as well, but a certain point has been bugging me for some time.
In the above picture, I don't seem to understand how are they trying to linearize the map in the way mentioned, and I even don't know the dependence of $s$ on $k$ or $n$. If $\beta _{n+1}$ is some function of $\beta _{n}$, then it can be expanded about the fixed unstable point for small perturbations. But how do they arrive at the above mentioned stage, I'm totally unaware of. Can someone help me out in this?
| It's a simple Taylor expansion, but their notation maybe is not great.
When they write $f(\beta_n -\beta_s)$ it might look like $f$ is a function and $\beta_n -\beta_s$ its argument, when actually $f$ is essentially a constant number that is multiplying $\Delta\beta=\beta_n -\beta_s$, which is the distance from the $k$-periodic point at $\beta_s$.
This follows directly from the Taylor expansion truncated to the linear term:
$$g(x) \approx g(a) + g'(a)\cdot(x-a),$$
with
$$x\to \beta_n, \quad g(x)\to \beta_{n+k}, \quad g(a)=a\to \beta_s, \quad g'(a)\to f,$$
resulting in their expression
$$\beta_{n+k} = \beta_s + f\cdot(\beta_n -\beta_s).$$
As the page you've included explains:
$f$ is the slope of the map at $\beta_s$, the position of the unstable point
which means that $\beta_s$ and $f$ implicitly depend on $k$ — after all, the curve on the return plot (Fig. 2 of OP) in general depends on it being the 1st, the 2nd or the $k$-th return.
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Would a pressurized container move by itself if opposite edges have different size surface area? If inside a closed container there is gas with higher pressure than outside the container, and one edge of the container has a larger surface area than the opposite side, would the container move by itself? Wouldn't there be a net force in one direction since one surface area is bigger than the opposite side while the pressure is the same?
| No, you are forgetting that the pressure on the two other slanted faces also contribute a component of force opposite to the largest face. The net forces still balance.
This is because the vector area (the integral of all infinitesimal area vectors) of any closed surface is zero.
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How is Newton's third law working here? I understand that if I were, for example, to throw a bowling ball in outer space then the ball would move away from me by the force I generated. But the ball would also exert a force on me in the opposite direction and move me away in the opposite direction of the bowling ball.
I don't understand how Newton's third law works in the next scenario though: I saw on YouTube an MIT professor let the gas out of a fire extinguisher on the back of his bicycle which then propelled him forward on the bicycle. What's exactly causing that motion? How is Newton's third law working here?
The gas molecules are pushing on neighboring gas molecules which is ultimately causing gas to escape from the nozzle. The gas molecules are pushing on one another like my bowling ball in space example, right? But where is the gas pushing on the fire extinguisher? Is there more pressure on one side of the fire extinguisher?
| The gas molecules next to the metal of the fire extinguisher push on it. When these molecuoes bounce off the metal surface, the rest of the gas molecules in the container push them back towards the metal, keeping the pressure on.
| {
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Why does light have multiple frequencies? The wavelength of visible light ranges from 750 - 400 nm, and so do the corresponding frequencies. However, a photon only has one frequency, given by $E =h\nu$, at a given time, and it can’t be changed unless the photon gets energy from somewhere, which isn’t possible in the vacuum; i.e. once it leaves the source it can’t possibly get enough packets of energy to become excited or even lose the energy to some other particle, as space is empty.
So, why does light come with so many wavelengths and frequencies if a single photon can only have one frequency at a time and are emitted from the same source?
A single photon goes with its individual oscillating electric and magnetic fields, right? So multiple photons mean multiple fields. Won’t these different fields affect the adjacent fields in any way and change their properties?
| Light comes with so many wavelengths because it is made of so many photons. A typical lightbulb puts out something on the order of $1\ \mathrm{W}$ of power in the visible spectrum, while individual photons in the visible spectrum each have an energy on the order of $10^{-19}\ \mathrm{J}$. Therefore, a typical lightbulb will produce something on the order of $10^{19}$ photons per second. While each photon has one specific frequency, there are so many of them that the ensemble of photons appears to have a continuous range of frequencies.
| {
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Temperature rise in IC engine without heat flow In an IC engine, the air fuel mixture is ignited in a cylinder resulting in temperature rise of the mixture. However, there is no heat flowing into the engine cylinder from some heat reservoir. Yet, in the P-V graph below, it says that heat $Q_H$ is flowing into the cylinder.
Question:
From what heat reservoir is $Q_H$ coming?
| The part of the cycle where the red arrow is pointing is the 'ignition'.
That's when the new fuel/air mixture taken into the engine is ignited and provides the energy input. So the heat comes from the ignition of the new fuel.
| {
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What does $F$ in the equation of surface tension $(T=\frac{F}l)$ mean? As far as I knew or assumed: If an imaginary line is thought to be present on a surface, then molecules of the surface will attract or exert equal forces throughout the length of the Line. So in a nutshell let say the centre of mass or the centre of the line faces a force equal to $F$ on one side, so unit length of the line will face $\frac{F} l$ amount of force and by definition this is supposed to be the surface tension (force per unit length). The line is supposed to be static so it faces equal force on the opposite side but tension is same throught surface, hence it wouldn't change anything I presume.
If the concept I mentioned is correct,then what does $F$ stand for the equation of surface tension of a ring? I mean in case of line I could say that equal small forces sum upto a larger force acting on one side of tha centre of the line, perpendicular to the line. And those smaller forces are actually the tension. But can we assume the same thing for a ring or a disc? I am asking this question because of not being able to visualize the same thing for a curvature. Could it be that this time tension equals to the ratio of net force acting on a small straight linear fragment of circumference to the lenght of that fragment? But I have seen in some textbooks for discs tension is considered to be equal to the ratio of force and the circumference which for me didn't make any sense at all.
| For a disc or ring
this time tension equals to the ratio of net force acting on a small straight linear fragment of circumference to the length of that fragment
seems like a reasonable definition.
It's true that the total force has no resultant direction, but it's often the case that energy is used to deal with such problems - i.e the energy needed to expand the ring from radius $r$ to $r+\delta r$ would be (from force x distance)
$$E= T\times 2 \pi r\times \delta r$$
Some textbooks would use the above concept to define the magnitude of the net force for a disc, hence define surface tension as the ratio of the 'net force' and the circumference.
| {
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Direction of supersonic shockwave on a bullet with lateral velocity When a bullet travels at supersonic speeds, it generates a shock-wave. For a bullet traveling in a straight path, the shock-wave is oriented in the same direction as the bullet. But if a bullet is being deflected by the wind, the nose of the bullet will orient itself into the wind (the opposite of the direction of flight, which is counter-intuitive).
My question: is the shockwave oriented to the nose of the bullet, or is it oriented to the direction of flight?
| A shock wave will obey the Rankine–Hugoniot relations, which means the normally incident flow direction is the relevant direction.
My question: is the shockwave oriented to the nose of the bullet, or is it oriented to the direction of flight?
What you are looking for is called aberration. So first find the angle between the velocity of the fluid incident on the bullet in the absence of wind and the velocity of the wind. That is, imagine you are sitting on the bullet traveling through a perfectly calm, uniform fluid background. The incident flow is perfectly anti-parallel to the velocity of the bullet in the lab frame. If you now add wind, then the incident flow is canted at an angle defined by:
$$
\alpha = \tan^{-1}{\left( \frac{ v_{wind} }{ v_{b} } \right)} \tag{0}
$$
where $v_{wind}$ is y-component of the velocity of the wind and $v_{b}$ is the x-component of the velocity of the bullet relative to some lab rest frame. You can then construct a transformation matrix, $\mathcal{R}$, to transform from the lab rest frame to the aberrated frame, where:
$$
\mathcal{R} =
\begin{pmatrix}
\cos{\alpha} & -\sin{\alpha} & 0 \\
\sin{\alpha} & \cos{\alpha} & 0 \\
0 & 0 & 1
\end{pmatrix} \tag{1}
$$
You can then apply $\mathcal{R}$ to any of the lab frame vectors to get the aberrated coordinate basis vectors.
But if a bullet is being deflected by the wind, the nose of the bullet will orient itself into the wind (the opposite of the direction of flight, which is counter-intuitive).
Most bullets, early in their flight, will have their spin axis mostly aligned with their direction of flight unless the wind is really strong or the bullet is a low velocity type (e.g., the .22 caliber rounds like those used in biathlon are influenced by even ~5 mph winds). The more massive and fast the round, the smaller the influence of the wind.
| {
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Absolute zero: how close can we go? Kinetic theory says that the minimal temperature is zero kelvin, at which every motion is stopped.
However, we do know that quantum theory says that we have unavoidable quantum fluctuations, so: what is the minimal quantum temperature that can be reach? Is there any known closed formula (for either bosons or fermions)?
Remark: we do know that quantum black holes have a temperature is inversely proportional to black hole mass, and if we assume a $10^{11}-10^{12}$ solar mass black hole, that would give a minimal mass about 10 yoctokelvin. But this mass is macroscopic, is there any microscopic known limit on quantum solids?
| Approaching absolute zero means subtracting smaller and smaller amounts of energy from the system. Using the energy-time uncertainty relation, $\Delta E\Delta t\geq \hbar/2$ we conclude that the minimum attainable temperature is greater than:
$$
T_{min}> \frac{\Delta E}{k_B}> \frac{\hbar}{2k_B t_{universe}},
$$
where $t_{universe}$ is the lifetime of the universe, although for amore appropriate extimate one should probably use the time since the mankind exists, or the duration of the current civilization, or, optimistically, the expected time till the end of humanity.
| {
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Derivation of anomalous commutators of currents in Fradkin's book I am trying to understand the derivation of the anomalous commutators of the left- (and right)-moving currents in Fradkin's book (see e.g. here). I am not sure I understand how (6.71) leads to (6.72).
My understanding is that we can work out identities with the Delta-Distribution for test functions. Taking $\epsilon,\epsilon'=0$ we have
$$ \int \left(\frac{\delta(x-x')}{x-x'} - \frac{\delta(x-x')}{x'-x}\right) f(x) dx = \int \frac{\delta(x) f(x+x')}{x} - \frac{\delta(x)f(x'-x)}{x} dx$$
which almost looks like $-f(x)\partial_x \delta(x-x') \simeq \partial_x f(x)|_{x=x'} = \lim_{h\to0} \frac{f(x'+h)-f(x'-h)}{2 h}$, but I seem to be missing a crucial factor of $2$ here. I am not sure if I am missing something more formal here. (I am actually trying to get a consistent convention of bosonization identities for myself, so it seems rather crucial.
| Note that these equations are (5.244) and (5.245) in the book version, compared to the online notes. Luckily I still have my own notes scribbled in this section of my book copy. Here it is:
$$
\begin{split}
[j_+(x), j_+(x')] &= \lim_{\epsilon, \epsilon' \rightarrow 0} \left( \frac{i \delta(x' - x + \epsilon' + \epsilon)}{2\pi(x - x' + \epsilon + \epsilon')} - \frac{i \delta(x - x' + \epsilon' + \epsilon)}{2\pi(x' - x + \epsilon + \epsilon')} \right)\\
&= \lim_{\epsilon, \epsilon' \rightarrow 0} \left( \frac{i \delta(x' - x + \epsilon' + \epsilon)}{2\pi(\epsilon' +\epsilon + \epsilon + \epsilon')} - \frac{i \delta(x - x' + \epsilon' + \epsilon)}{2\pi(\epsilon + \epsilon' + \epsilon + \epsilon')} \right)\\
&= \lim_{\epsilon \rightarrow 0} \left(\frac{i \delta(x' - x + \epsilon)}{4\pi \epsilon} - \frac{i \delta(x-x'+\epsilon)}{4\pi \epsilon} \right) \\
&= -\frac{i}{2\pi} \partial_x \delta(x-x').
\end{split}
$$
Essentially, all I did was use the delta functions to replace $x-x'$ in the denominator with epsilons, which gives the desired factor of two.
| {
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Graph for Coulomb Force vs $1/r$ My teacher told me that the graph for the coulomb force $F$ vs $1/r$ where $r$ is the distance between the 2 charges should be parabolic but I can't seem to understand why. I am aware that equations of the form $y^2=4ax$ are parabolic but why should $F$ vs $1/r$ graph be parabolic?
| Assume $F$ to be $y$ and $1/r$ to be $x$
Then, according to coulomb's law
$$y = cx^2$$
where $c = \frac{q_1 q_2}{4 \pi \varepsilon_0}$
Now you can rearrange the equation as follows.
$$x^2 = c' y$$
where $c' = \frac{1}{c}$
Now you can take $c' = 4a$, and this will give you your familiar expression for a parabola with the directrix parallel to the $x$-axis
$$x^2 = 4ay$$
Note that there was no need to do all this. A quadratic expression is always a parabola.
| {
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What is wrong with the high-school definition of a vector? Why is the high-school definition of a vector as "a quantity with a magnitude and a direction" incomplete? For example, Griffiths Introduction to Electrodynamics book says:
The definition of a vector as "a quantity with a magnitude and direction" is not altogether satisfactory.
(section 1.1.5 "How vectors transform")
However, I am not very satisfied with his chain of arguments.
| The definition is not satisfactory because vector spaces do not come equipped with a notion of direction or magnitude. The matrix
$$
\begin{pmatrix}
1 & 1 \\ 0&1
\end{pmatrix}
$$
defines an automorphism of $\mathbb{R}^2$ as a vector space, but preserves neither magnitude nor direction (understood as the angle between vectors).
To have a notion of magnitude and direction, we need a metric (inner product), i.e. a non-degnerate symmetric bilinear form.
The confusion comes from always thinking of vector spaces as having a natural inner product. But often there is no natural or even meaningful inner product. For example, the set of pairs $(\text{\$ spent on apples}, \text{\$ spent on oranges})$ is a perfectly good and even useful vector space. What should the magnitude of such a vector be? There's no meaningful answer.
| {
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Direct experimental observation of magnetic orbital quantum number $m_l$ Is there an experimental way to observe magnetic quantum number $m_l$ values directly, the way electron spin was detected by Stern Gerlach experiment or proton's spin by nuclear magnetic resonance experiments? The Zeeman effect comes to mind, but in the Zeeman effect one cannot ignore the electron spin. In short, how can one experimentally or spectroscopically see that if $l=2$,, $m_l$ will be -2, -1, 0, +1, and +2?
| A hint to the literature of fine-structure spectroscopists: an $ℓ=2$ state is sometimes known as a “quintet” because of its splitting into five sublevels. See also singlet, doublet, triplet, etc.
You suggest in a comment that we consider the famous sodium doublet, which is visible without any magnetic field. That doublet is a spin-orbit effect: the $3p$ first excited state in sodium can have $j=1/2$ or $j=3/2$, depending on the orientation of the electron spin relative to its orbit.
(source)
In a magnetic field, the energy degeneracy among the various $m_j$ is broken, and the two lines in the sodium doublet split in ways that reveal their multiplicity. This diagram suggests that the $\frac12 \to \frac12$ transition subdivides into two doublets.
The $\frac32\to\frac12$ transition divides into six pieces, rather than eight, because the transition from $m_j=-3/2$ to $m_j=+1/2$ would require the photon to carry away at least two units of angular momentum.
(It’s not immediately obvious to me why the six transitions in the $\frac32\to\frac12$ transition should be equally spaced, as sketched, but I’m prepared to believe they are.)
Magnetic splitting was actually discovered by Zeeman in this sodium transition, but this transition is an example of the “anomalous Zeeman effect.” One description.
| {
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Isn't it possible for a matter to have mass of an electron but opposite charge? I read that "Positrons" have the same mass as of an electron but it is oppositely charged but we don't call it a normal matter we have given a special name to it "anti matter". But why ? Why can't a matter particle have those properties ? Why is it classified as an "anti particle" and not a normal particle with lesser existence ?
| Because it's not just mass that's identical, all properties of an electron are shared by the positron except opposite sign for charge (and also handedness).
| {
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Basic question about orbital speed I was reading a Sci-Fi book recently and had a weird thought:
I know that objects closer to a gravitational well need to move faster to stay in orbit and objects further away move slower. But if you want to increase your orbit/escape the gravitational well you have to speed up while if you want to lower your orbit you have to decrease your speed.
In my mind this seems like a paradox. I'm sure I'm just thinking about it the wrong way but I can't figure out how to solve this. Can someone explain it to me, please?
| This is the well-known satellite orbit paradox. The key point is HOW you thrust in a cicular orbit. When you do a so-called impulsive thrust, i.e a thrust for a short time (small relative to the orbital period), you do it at that given position and the thrust energy goes only into the kinetic energy, i.e. you locally increase speed, which brings you from a circular orbit into an elliptic orbit. On this orbit the speed lowers the further you get out owing to the gravitational potential.
On the other hand, if you do a continuous small thrust along-track (this is where the paradox comes in), then you actually do NOT increase orbital speed. The thrust energy then goes into the orbit, which is determined by its orbital radius: The bigger the radius the bigger the orbital energy, because you work against the gravitational potential. In summary, by the type of thrusting you determine how the thrust energy is transfered into orbital energy.
| {
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Mathematical justification of the Born-Huang expansion in the derivation of the Born-Oppenheimer Approximation In the book K. Huang and M. Born, Dynamical Theory of Crystal Lattices (1954, Appendix VIII) and also in the Wikipedia article https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation#Derivation, the derivation of the Born-Oppenheimer approximation is given. In these derivations, the following expansion of the exact electron-nuclear wave function is used
\begin{equation}
\Psi\left(r,R\right) = \sum_{m} \phi_{m}\left(R\right) \chi_{m}\left(r,R\right).
\label{eq:BornHuangExpansion}
\end{equation}
Here, $r$ and $R$ refer to all electronic and nuclear variables, respectively. Further, $\Psi$ is the wave function satisfying the exact time-independent Schrödinger equation of the exact Hamiltonian of electrons and nuclei (comprising the kinetic energies and Coulombic interactions) and $\chi_{m}$ the wave function satisfying the Schrödinger equation for the exact Hamiltonian minus the nuclear kinetic energy. Please check some details from the Wikipedia article, if necessary.
I have not found any discussions from the literature in which this expansion is discussed in detail from a mathematical point of view. Does anyone see how to prove that this expansion is exact or see that it can't be exact?
| Let
$$
H = H_e + T_n
$$
Where $H$ is the full Hamiltonian and $T_n$ is the nuclear kinetic energy. Notice that the only dependence on $R$ in $H_e$ is through the position operator, i.e. there is no dependence on $\frac{\partial}{\partial R}$. This means we can treat $H_e(R)$ as a (Hermitian) operator on the space of functions of $r$ for a fixed value of $R$. Since $H_e$ is Hermitian it has, for any given value of $R$, a complete basis of eigenfunctions, $\chi_m(r,R)$ which can be used to write any other function of $r$. In particular they can be used to write the eigenfunctions of the full Hamiltonian, $H$, exactly as,
$$
\Psi(r,R)= \sum_m \phi_m(R)\chi_m(r,R)
$$
Note that since both $\Psi$ and $\chi_m$ are parameterized by $R$, so is the basis coefficient $\phi_m$, but the $r$ dependence is entirely absorbed into the basis functions $\chi_m$
| {
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Is the intensity of light dependent on number of photons per unit area? I was learning about the photoelectric effect of light and there it says more the intensity of light, the more number of electrons will be ejected from the metal surface given that the frequency of light is more than its threshold frequency.
Now what does intensity basically means?? Is it dependent on the number of photons per unit area over which the light falls..
Let us consider two light rays, $X$ and $Y$ which have same frequency and same wavelength and it is said that $X$ has more intensity than $Y$. The light rays fall over an area, $dA$. Does that mean that light ray $X$ will have more photons falling on that area $dA$ as compared to light ray $Y$ ??
| Number of photons per unit area is proportional to intensity. That’s because intensity = power/area = (energy/time)/area. A photon has energy $h\nu$, where $\nu$ is the frequency. So intensity is $N h\nu$/(time x area), where $N$ is the number of photons.
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If our solar system and galaxy are moving why do we not see differences in speed of light depending on direction? May be a silly and simple question, but I've been wondering if:
The speed of light is constant, and
*
*When we're moving in the same direction (where both the emitter and the receiver move with the light direction) we would be making it take more time for the light to reach the other end.
*Conversely when moving in the opposite direction we'd be shortening the time it takes.
Why do we not see a non-uniform speed of light caused by solar/galactic movement?
Or do we?
|
The speed of light is constant
is one of Einstein's postulates of relativity and says something profound that can take time to realise.
It means that the light is measured to have the same speed independent of the motion of the source.
So for example if a light was shone from a stationary spaceship we would receive it at $3\times 10^8$ m/s. If the same spaceship is moving towards or away from us at half the speed of light, and shines the same light towards us - we would still receive it at $3\times 10^8$ m/s.
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Why is there a single slit in the Young double-slit experiment? I am studying waves these days and my teacher just introduced Young double-slit experiment, which has always been brilliant (see picture below).
However, I have a question: why is there a single slit in the experiment?
My teacher says that the single slit is used to make sure the light at both of the double-slit are coherent (i.e. they have same frequency).
But what I think is that, since the light source is monochromatic (i.e. the light has only one frequency), theoretically, any two light rays from this source must be coherent. So it's very unnecessary to use the single slit. I believe there must be some other reason for this single slit.
Can anyone help me out?
| The single slit is used to create light from a single source.
Even what we regard as a single source of light e.g. a bulb, can have different parts to it, e.g. different parts of the filament, emitting light at different frequencies and amplitude and varying with time in different ways.
The single source is then split into two with the two slits. That creates two sources of the same frequency and constant phase difference - i.e. coherent.
If the light from the original source landed directly onto the two slits, the two slits are not guaranteed to be coherent sources.
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Confused about the Pauli exclusion principle I've been struggling to understand this: Let's say I have a gas of one million electrons. Does every single one of those electrons have a different energy (up to the degeneracy from the different momentum components)?
| Avogadro constant, which is a good estimate of the order of magnitude for a macroscopic number of particles, is much bigger than a million ($N_A\approx 10^{23}$). Yet, there is nothing difficult in giving every pair of electrons (with different spins) their own momentum state, for momentum is continuous - i.e., the number of available states is infinite.
If we confine the electrons (e.g., in a box) then the spectrum becomes discrete and, for sufficiently small sizes of the confining potential the spacing between levels start become noticeable and limit the number of electrons that can be put isnide of a container - this is routinely observed in nanoscale devices (quantum wires and quantum dots), although at such small scales one cannot consider the electrons as non-interacting - this leads to Coulomb blockage in quantum dots and Luttinger liquid behavior in quantum wires.
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Why does stimulated emission not contribute to linewidth? The rough quantum mechanical explanation for linewidth is that the lifetime $\tau$ of an excited level is associated with an uncertainty $\Delta E$ in its energy satisfying $$\Delta E\tau=\hbar$$ and so there is a linewidth $$\Delta \omega=\frac{\Delta E}{\hbar}=\frac{1}{\tau}.$$
However, under stimulated emission, the rate of depopulation of the upper level increases and so its lifetime effectively decreases. Why does the linewidth not increase correspondingly?
Classically, it is obvious that stimulated emission does not contribute because it is coherent whereas spontaneous emission is incoherent. However, quantum mechanically, I can't see why the reduced lifetime would not be included in the uncertainty principle.
| The short answer: the relevant lifetime is not the one of an individual excitation, but the first-order coherence time, the timescale at which the phase diffuses.
The longer answer is very interesting. On the quantum level, both the photons coming in and out of the cavity are described by a Lindblad process. For the gain, the jump operators are $\sqrt{R}a^\dagger$ (leading to rate $Raa^\dagger=R(n+1) $) and for the losses they are $\sqrt{\gamma}a$ (leading to rate $\gamma a^\dagger a=\gamma n $). The +1 for the gain is what is typically attributed to the spontaneous emissions.
Note that all the absorptions and emissions would invoke some kind of fluctuations, but they are much smaller.
The semiclassical picture, such as introduced in Theory of the linewidth of semiconductor lasers (Henry,'82) attributes all the fluctuations to the spontaneous emissions, but this makes them unphysically large. In this picture, each spontaneous emission doesn't increase the particle number deterministically with +1, but can change it either way up to the order of $\sqrt{n}$!
I puplished a paper related to this quantum-classical correspondence in a slightly different system, but should be related to this, especially sections III and IV.
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Matrix element and Dirac notation If
$$
T=
\left[
\begin{array}{cccc}
e^{\beta J} & e^{-\beta J} \\
e^{-\beta J} & e^{\beta J} \\
\end{array} \right]
$$
and
$$Z = \sum_{S_i=\pm 1} ... \sum_{S_N=\pm 1} \exp{\beta J(\vec{S_1}\vec{S_2}+\vec{S_2}\vec{S_3}+...+\vec{S_{N-1}}\vec{S_N}+\vec{S_N}\vec{S_1})}
$$
Then why can we say that
$$Z = \sum_{S_i=\pm 1} ... \sum_{S_N=\pm 1} \langle S_1|T|S_2\rangle\langle S_2|T|S_3\rangle...\langle S_N|T|S_1\rangle ?
$$
|
Then why can we say that:
Because, each $S_i$ can only take on two values: +1 or -1
For example, if $S_1 = +1$ and $S_2 = +1$ then the $e^{\beta J S1S2}$ is ${e^{\beta J}}$, which is exactly what the ++ matrix element of $T$ says.
As another example, If $S_1 = +1$ and $S_2 = -1$ then the $e^{\beta J S1S2}$ is ${e^{-\beta J}}$, which is exactly what the +- matrix element of $T$ says.
And so on.
Update:
In the Dirac bra/ket notation, $\left < S | T | S' \right >$ is a number that depends on S and S'. (Note: We could use a different symbol to denote $T$ as a stand-alone matrix vs $T$ when it is in the Dirac bra/key notation, but in this case we dont.)
For example, when S=+ and S'=+:
$$
\left < + | T | + \right > = e^{\beta J}
$$
For example, when S=+ and S'=-:
$$
\left < + | T | - \right > = e^{-\beta J}
$$
For example, when S=- and S'=+:
$$
\left < - | T | + \right > = e^{-\beta J}
$$
For example, when S=- and S'=-:
$$
\left < - | T | - \right > = e^{-\beta J}
$$
And we write these four possible values together in a matrix, whose indices span + and - in both directions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Which of these sentences best describes the equivalence between mass and energy?
Which of these sentences best describes the equivalence between mass and energy?
*
*mass is a form of energy.
*mass and energy are two manifestations of the same property: mass-energy.
*mass is energy confined to an object.
*energy acts as mass when momentum is zero.
*another different......
I discard "mass and energy are the same thing" since mass is the magnitude of the four-vector momentum-energy, while energy is the timelike component of said four-vector. Also, mass is invariant and energy is not.
| I would say 5. "$m^2 c^2 = E^2/c^2 - p^2$". This is closest to your 4., but includes what happens when $p\ne 0$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/673083",
"timestamp": "2023-03-29T00:00:00",
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Proof that you can't disentangle two parties if you only operate on one Let $A$ and $B$ be two entangled systems. Can someone prove or sketch a proof of why you cant unentangle $A$ and $B$ by only acting on $A$ or $B$ alone? i.e. by only applying $\mathbb{I}_A\otimes U_B$, with $U_B$ unitary.
| If there was some unitary operator factorized as $\mathbb I_A \otimes U_B$ that would send the entangled state $|\psi\rangle$ to a factorized state $|\phi_A\rangle\otimes |\phi_B\rangle$, ie :
$$|\phi_A\rangle\otimes |\phi_B\rangle = (\mathbb I_A\otimes U_B)|\psi\rangle$$
Then, we would have :
$$|\psi\rangle = (\mathbb I_A\otimes U_B^\dagger)(|\phi_A\rangle\otimes |\phi_B\rangle)= |\phi_A\rangle \otimes (U_B^\dagger|\phi_B\rangle)$$
contradicting the fact that $|\psi\rangle $ is entangled.
In short, factorized unitary operators preserve disentangled states, so they must preserve entangled states as well.
Edit : Entanglement is basis independent. A state is unentangled if it can be written as a product state in some basis, and entangled if it cannot be factorized in any basis.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Why bother buying efficient lights if you are already heating your house? Assume I live in a location where at any time of day and any time of year, I need to heat my house. Assume further that I have a room with no windows. In this case, does it make sense for me to buy efficient light bulbs, considering that any inefficiency in converting electricity to visible light simply leads to more heat being added to the room, which in turn, results in less heat being output by the heater to maintain constant room temperature.
Although these are somewhat idealized conditions, I don't think they are too far off from being realistic. For example, say you live near the arctic circle, it might be smart not to have many windows due to heat loss, and it seems reasonable that in such a climate, heating will be required at all times of the day and year. Assuming I haven't missed something, it seems to me, somewhat unintuitively, that buying efficient light bulbs is not a logical thing to do. Is this the case?
| Hot air rises
You'll have noticed the radiators around your house are close to the floor - under it if you're lucky enough. Your lighting, on the other hand, is close to the ceiling. The hot air they produce will do little to warm you where you actually are, usually seated and close to the floor.
| {
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"url": "https://physics.stackexchange.com/questions/673836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do stars produce energy if fusion reactions are not viable for us? From what I've learned, fusion reactions are not currently economically viable as of right now because the energy required to start the reaction is more than the energy actually released. However, in stars they have immense pressures and temperatures which are able to allow these reactions to take place. However, if these reactions are considered endothermic for us, how are they exothermic in stars? i.e. how are stars able to release energy?
Moreover, why are such fusion reactions for us endothermic in the first place? Given we are fusing elements smaller than iron, wouldn't the binding energy per nucleons products be higher and hence shouldn't energy be released?
|
However, if these reactions are considered endothermic for us, how are they exothermic in stars?
The reactions are still exothermic for us. In fact, they are very exothermic. The fact that they are not net energy producers is due to inefficiencies in our existing technologies for producing these reactions, not because the reactions themselves are endothermic. In other words, our fusion devices waste a lot of energy, heat that leaks out to the environment and so forth. So even a highly exothermic reaction does not compensate for all of the waste and inefficiencies.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/674089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Concept of Gravitational potential energy Change in Potential energy corresponding to a conservative force is defined as $$\Delta U = U_f - U_i=-W_f$$ and gravitational potential energy is $$\Delta U = U_f-U_i = -W_g $$ Suppose a mass $m_1$ is kept at a fixed point $A$ and a second mass $m_2$ is displaced from point $B$ to point $C$ such that $AB = r_1$ and $AC = r_2$.
$\therefore$ , $$\Delta U = -W_g = \int{\frac{Gm_1m_2}{r^2}}dr$$ $$U(r_2)-U(r_1) = Gm_1m_2\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$$ Now I am free to choose any reference point thus if I take potential energy at $U(r_1) = 0$ and $r_2 = \infty$ Then I will get potential energy at infinity as $$U(\infty) = \frac{Gm_1m_2}{r_1}$$ which I think is wrong as a reference point at $r_1$ the potential energy at infinity should be infinite.
So where I am wrong, is my concept of gravitational potential energy wrong itself.
| No the equation that you have derived isn't wrong. What actually is at fault here is the logic that potential energy at infinity must be infinite. This is completely incorrect. For example consider the following equation:
$$a-b = c$$
In this equation the only given information is the difference in magnitude of the two said quantities (a and b). Therefore if the value of b turns out to be 0 then $a=c$.
Since you have not provided any background as to why you think the potential must be infinite hence I cannot supply any intuitive answer for it.
Regarding the comment:
I thought that the potential energy at infinity should be infinite because as we increase the distance from the fixed object the gravitational potential energy increases
Consider that you have fixed the value of $U(r)$ at $r=r_0$ to be $c$. As you correctly state that Gravitational potential energy increases as we go away from that point. Therefore $U(r) > U(r_0)$ if $r>r_0$. But what you failed to notice is that the rate at which the difference $U(r)-U(r_0)$ grows gets slower and slower as we go away from $r_0$ because $F_g$ (Force of gravitation) gets smaller and smaller. This when evaluated as limit gives a finite value and hence the value is not infinite at infinity.
| {
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"url": "https://physics.stackexchange.com/questions/674392",
"timestamp": "2023-03-29T00:00:00",
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Local order parameter for confinement in gauge theories I would like some help clarifying what Zinn Justin is saying in his book "Quantum Field Theory and Critical Phenomena" p.805 on detecting confinement of gauge theories.
In particular, I understand Elitzur's theorem just states that there is no such thing as gauge symmetry breaking (since gauge is just a labeling redundancy of the theory). However I don't understand why I cannot find a local order parameter to detect confinement in gauge theories. I understand local observables cannot be charged (since charged operators don't commute with gauge transformations)
Could someone help understand why I cannot construct some local order parameter to detect gauge field confinement? (with equations would be nice).
I think one difficulty is I don't have a clear definition of what qualifies as a local order parameter. To prove the statement one would need to define what is a local order parameter.
| The reason is that non-abelian gauge theories are asymptotically free. This means that at short distance, one cannot distinguish between the confined and deconfined phase because the gauge interaction turns off. Therefore it is hard to imagine what kind of local observable could measure the confinement phase transition.
A local order parameter just means any physical observable one can use to detect a phase transition. It is a somewhat general definition, but phase transitions come in a large variety so it is hard to make a more precise definition.
| {
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"url": "https://physics.stackexchange.com/questions/674502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Locally-Flat Minkowski limit of Schwarzschild Metric I've seen questions and answers dealing with similar topics, but none that seem to provide what I'm looking for.
The Schwarzschild metric (and indeed any valid metric) should reduce to the Minkowski metric over a sufficiently small, linearized region.
I am trying to do this mathematically by Taylor expanding the Schwarzschild metric terms, but struggling a bit with the math, specifically, what value of $r$ I should center it at, presumably not $0$ or $\inf$, maybe $1$? And what terms to neglect.
Can someone help me with this derivation, or at least tell me if I am on the right track?
Note: this is NOT the same as the Newtonian limit, where $r$ goes to infinity, because the locally Minkowski property should hold even at very high curvatures, including inside the horizon.
| You can always choose locally-inertial coordinates to see that any metric (Schwarzschild, or otherwise) at a given point assumes the Minkowski form, its first derivative vanishes (therefore Christoffel connection vanishes too), but the second derivative does not vanish. The fact that the second derivative cannot be made to vanish is just another way of saying that even if you go to a local frame with $g_{\mu \nu} (p) = \eta_{\mu \nu}$ and $\partial_\alpha g_{\mu \nu} (p) = 0$ at a point $p$, there is still curvature characterized by the 20 independent components of the Riemann tensor (which has second derivative of the metric).
In this specific sense, you cannot really take a limit of Schwarzschild to Minkowski, since the former has non-zero Riemann curvature. You can read about locally inertial coordinates applicable in general (and not just for specific metrics like Schwarzschild) in Carroll's book in chapter 2.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why can a very large body of water not store summer heat? On this page, it states "The key disadvantage of using a very large body of water to achieve heat exchange with a relatively constant temperature is that you are not able to store summer heat in that body of water – to have the benefit of retrieving those higher temperatures in winter."
Why is it so? Is it because a very large body of water would have more heat exchange with the air and hence would lose the heat gathered in summer?
But "heat exchange with a relatively constant temperature" also points in the direction of having a large body of water, so I am a bit confused.
| I think you may be misinterpreting the statement. It is not claiming that heat of summer cannot enter a lake (and be stored in it), but rather that 'you' in your actions with the heat pump cannot 'store' useful energy into the lake.
If you had a large (say a few tens of cubic metres) insulated tank of water in your basement to use as your heat exchanger then during summer, when you don't need to heat your home, you could very efficiently pump heat from outside into the tank of water. This would raise its temperature and you could then extract the 'stored' heat (effectively for free) during winter to heat your home.
But if you use a 'large body' (or flowing body) of water for heat exchange, no amount of heat input from 'your heat exchanger' will have any significant effect on the bodys effective tempetaure. So you wont be able to extract that heat in winter.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/674852",
"timestamp": "2023-03-29T00:00:00",
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How do we measure time? I'm having a little trouble trying to put to words my problem and I apologize in advance for any causation of trouble in trying to interpret it.
We define periodic events as those events that occur over equal intervals of time. But, don't we use periodic events themselves to measure time (like a pendulum or the SI unit definition of transition frequency of Cesium)? Then how is it we know we have equal intervals of time?
Another way to put my problem would be:
We metaphorically describe time in terms of the physical idea of motion, i.e., 'time moves from a to b', but how do we deal with how fast it moves because to know how fast it moves, we must know its rate and to know its rate is like taking the ratio of time with time?
This is all very confusing. I apologize again for any problem in trying to understand.
| When we think about time, we naturally go to the processes of how we measure it. But time is not the ticking of a clock, or the oscillations fo an atom. Time is the construct we use to differentiate events. (No time based terms in the definition.) Just as space is the construct we use to differentiate objects. We need time to perceive order (1st, 2nd, 3rd, before, after, etc). So need the 3 dimensions of space and the time dimension are the constructs we need to describe a universe in motion. If the universe was static then time would be unnecessary (maybe not even exist) since there would be nothing to order that couldn't be accomplished with the 3 space dimensions.
So all the issues with methods of measuring movement with perfect accuracy as has been discussed are accurate.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is an expression of a quadrupole as an expansion of dipoles possible? Would it be possible to express a quadrupole as an expansion of dipoles? Because a possible definition of a quadrupole seems to be: an electric field equivalent to that produced by two electric dipoles.
| Actually, for any kind of magnetic field or electric field when a multipole expansion is done in different poles, as dipole, quadrupole, sextupole, octupole and so on, it is like a Taylor expansion:
$$ B_z(x) = \overbrace{B_{z0}}^{dipole} +\overbrace{\frac{dB_z}{dx} x}^{quadrupole} + \frac{1}{2!} \overbrace{\frac{d^2 B_z}{dx^2} x^2}^{sextupole} + \ldots$$
one observes that a dipole field is constant whereas a quadrupolar field depends linearly on the offset which the particle has with respect to the origin. From this point of view a quadrupole cannot be made of 2 dipoles.
Actually, you could put 2 dipoles one having an offset $R$ the other one having an offset $-R$ laterally to the origin. Assume that the distance $2R$ is just large enough to have some lateral space inbetween. Then you would rely on that the stray field between the 2 dipoles makes up a quadrupole.
The following comment has to be made in order to avoid confusion.
Actually a dipole with a stray field would not be a dipole in the strict sense, because a dipole in the strict sense has just constant field and would not allow for a stray field.
Of course a realistic dipole would never be a dipole in the strict sense and can (and will) have some stray field.
Probably you would get some quadrupolar component of these combined stray fields, but this combination field would actually be mixture of many different terms in the multipole development that one would never call that a quadrupole.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/675224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Second Kepler's law explanation What is the explanation for the second Kepler's law? Why is the law valid?
Is it that the total energy of a planet equals to the kinetic energy plus the potential energy?
| Kepler's second law (a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time) is a consequence of conservation of angular momentum. It applies not just to gravity, but also to motion under any central force i.e. a force that is always direct towards a fixed point.
Of course, Kepler himself derived his three laws empirically, by observing the motions of the known planets. It was Newton who proved that the Kepler's first and third laws are a consequence of the inverse-square nature of gravity, and that Kepler's second law applies more generally to any central force.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/675822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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What does rotational balance mean? For example, in many badminton rackets, it says that the balance point is rotational. How does this actually work?
| I am not sure but here is a guess. It may mean the racket is dynamically as well as statically balanced about the balance point. The racket is balanced statically in that the racket is stationary with a fulcrum at the balance point. The racket is dynamically balanced in that rotation (spinning the racket) about a principal axis through the balance point is steady without requiring a restraining torque.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/676123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doppler Effect and Relativity paradox Let, Alice is moving towards Bob at very high speed. Therefore, events in Bob's frame will appear to happen slowly in Alice's frame due to time dilation. Since velocity is relative, the same is also true for Bob.
Now, since Alice is moving towards Bob, Bob's light will be blueshifted towards him. Special relativity will cause it to blueshift even greater. That means, time between two light pulses, therefore two events on Bob's actually decreases as seen by Alice. Not to mention, the same thing will also happen for Bob.
So Alice sees the events at faster than normal speed even though they are happening at slower than normal speed. How is this patadox resolved?
| The Doppler effect and time dilation are two separate phenomena- the first will either reinforce or counteract the apparent effect of the other, depending on the direction of the relative motion concerned.
The Doppler effect is an apparent change in clock rate that results from changes to the distance light has to travel to reach an observer. The clock rate appears to increase when the distance is reducing (ie when the observer is moving towards the source), and to decrease when the distance is increasing.
Time dilation is a quite separate effect which is caused by the geometry of spacetime, and reflects the fact that different paths between two events can have different durations. If the path you follow between two events lasts for 10 minutes, while the path I follow between the same events lasts for 11 minutes, the elapsed time on you clock will be less than the elapsed time on mine- that is not because your clock has 'run slow', but because it has taken a shorter path through time.
If a clock is moving towards you, the Doppler effect will cause it to appear to speed up, thus countering the effect of time dilation. If the clock is moving away from you, the Doppler effect will cause the clock to appear to slow, thus adding to the effect of time dilation.
| {
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"url": "https://physics.stackexchange.com/questions/676214",
"timestamp": "2023-03-29T00:00:00",
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