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Estimation of greatest speed in a polyhedron in order to control velocities in a three dimensional volume, I look for a proof or a proof idea for the following assumption:
Given a non-empty solid polyhedron in 3D, all points inside this set have a speed than is lower or equal than the maximum of the speed of the vertices of this polyhedron.
This seems to be self-evident, but my inspirations how to show that are ugly and complicated and real messy and therefore unrepresentable. And I need a proof.
The same problem arises with ball and sphere: are the speeds inside the ball volume lower or equal than the speeds of the sphere?
| It seems OP is essentially asking about the speed of a convex combination
$$ {\bf r} ~=~ \sum_{i=1}^N \alpha_i {\bf r}_i,\qquad \sum_{i=1}^N \alpha_i~=~1, \qquad \alpha_i~\geq~0, $$
in a rigid body. The velocity is then
$$ {\bf v} ~=~ \sum_{i=1}^N \alpha_i {\bf v}_i. $$
Finally the speed is given by the 2-norm
$$\begin{align} |{\bf v}|~\stackrel{\text{triangle ineq.}}{\leq}& \sum_{i=1}^N \alpha_i |{\bf v}_i|\cr~\stackrel{\text{Hölder ineq.}}{\leq}&
\max(|{\bf v}_1|,\ldots, |{\bf v}_N|)\sum_{i=1}^N \alpha_i\cr
~=~~~~& \max(|{\bf v}_1|,\ldots, |{\bf v}_N|).\end{align} $$
| {
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Why do fans spin backwards slightly after they (should) stop? Today, I've decided to observe my PC fan as I shut the computer down. The fan slowly lost angular momentum over time. What I've found really interesting is the fact that the momentum vector change did not stop at the zero vector, but instead flipped its orientation and "went to the negatives", albeit very small in the absolute value compared to the powered spin; this caused the fan's angle to deviate by a few degrees (opposite to the powered spin rotation) compared to the observed angle when momentum was equal to the zero vector.
If I let $\overrightarrow{L}$ be the momentum vector, $\overrightarrow{L}_0$ be the momentum vector at $t_0$ (= poweroff time), and $\overrightarrow{L}(t) = y(t) * \overrightarrow{L}_0$ (with $y_0 = y(t_0) = 1$), then these are the plots of $y$ through the course of time.
Expected fan poweroff behavior:
Observed fan poweroff behavior:
Can anyone explain why may this happen?
| I also think the effect is cogging. This torque is position dependent and its periodicity per revolution depends on the number of magnetic poles and the number of teeth on the stator. Cogging torque is an undesirable component for the operation of such a motor. It is especially prominent at lower speeds, with the symptom of jerkiness.
Cogging torque results in torque as well as speed ripple; however, at high speed the motor moment of inertia filters out the effect of cogging torque.
| {
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How do you make more precise instruments while only using less precise instruments? I'm not sure where this question should go, but I think this site is as good as any.
When humankind started out, all we had was sticks and stones. Today we have electron microscopes, gigapixel cameras and atomic clocks. These instruments are many orders of magnitude more precise than what we started out with and they required other precision instruments in their making. But how did we get here? The way I understand it, errors only accumulate. The more you measure things and add or multiply those measurements, the greater your errors will become. And if you have a novel precision tool and it's the first one of its kind - then there's nothing to calibrate it against.
So how it is possible that the precision of humanity's tools keeps increasing?
| Your intuition that you can only get precision from precision is correct, but your conclusion that you need precise instruments to get precision is incorrect. The trick is always to use precision found in nature to get precision. You should actually look at how people make things to get an idea of where they are getting their desired precision from. For example, one can obtain a nearly perfect parabolic mirror by using the physics of a rotating liquid under uniform gravity. Another is to get a very precise measurement of time using the physics of atomic state transition, which is a clear example of using the central limit theorem to obtain a statistical guarantee on the precision of the output of the atomic clock.
| {
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Is there still an anomalous precession in Mercury's orbit that GR doesn't predict? I was reading through the Wiki pages on tests of relativity and I was curious about this section. Am I interpreting it correctly? Is there still about $1.21''/\text{ cy}$ of precession (Total Predicted: $575.31$ - Total Observed: $574.10$) that are not predicted by GR? The reference is from 1947. Does anyone have a more recent measurement and/or prediction?
Edit: My apologies. I transcribed the difference between the observed and predicted incorrectly. I've updated the question with the right values. Sorry for the inconvenience.
| The wikipedia section on the precession of Mercury's orbit is in serious need of an update. The presented observational data is centuries old. Its main reference is the Clemence paper from 1947. Since then the astronomical data have been updated to give a GRT prediction of 42.98"/cy. Clemence gives a slightly higher value. The other two relevant numbers are the Newton predictions on the precession without the GRT effect and the observed precession. The difference of these is the anomaly to be explained by e.g. GRT. The Clemence paper relies on Newtonian calculations from 1911 and observation data from 1863. A modern prediction of the total precession including the effect of GRT is given in this open access paper. The reported number (575.31"/cy) is also given in the wikipedia article (just added extra ref). I am still searching for modern observational data. The 43"/cy number is far too important to rely on very old observations. There must be more modern information. I hope the Bepi Colombo mission will also address this. I will update this answer when I find anything.
| {
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What does it mean that a free particle has no definite energy in quantum mechanics? In the quantum mechanics case of the infinite square well, the general solution to the Schrodinger equation is a linear combination of solutions with definite energy states. When you measure the particle, it will take one of these energy values.
I am now looking at the case of the free particle and I see that (because the stationary states are not normalizable) a free particle cannot exist in a stationary state. In his textbook, Griffiths states "there is no such thing as a free particle with a definite energy".
I'm struggling to see what this actually means; if you measure the energy of a free particle you will still get a value right? And won't this energy necessarily correspond to one of the infinite solutions for the free particle?
| If a particle is somewhere in a very large volume then it is described by wave function normalisation factor that tends to zero. Because it is limited to a finite volume it will have a band width. This bandwidth can be made arbitrarily small. Perhaps Griffiths intends to state this fact.
Plane waves are useful as basis functions and to describe particle bundles that are wide compared to a relevant length scale. For example, for atomic transitions one can use a plane wave for the EM field because at the transition energy the EM wavelength is much larger than the scale of the electronic orbital. This leads to Fermi's golden rule.
| {
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Contraction of Christoffel symbol and metric tensor How can I prove this contraction of Christoffel symbol with metric tensor?
$$
g^{k\ell} \Gamma^i_{\ \ k\ell} = \frac{-1}{\sqrt{|g|}}\frac{\partial\left(\sqrt{|g|}g^{ik}\right)}{\partial x^k}
$$
I know the relation for the Christoffel symbol contracted with itself and this one is similar, but I cannot find the clue.
I start from the definition of gamma:
$$
g^{k\ell} \Gamma^i_{\ \ k\ell} = \frac{1}{2}g^{kl}g^{ij}(\partial_k g_{jl} + \partial_l g_{jk} - \partial_j g_{kl}) =
\frac{1}{2}g^{ij}(2g^{kl}\partial_k g_{jl} - g^{kl}\partial_j g_{kl})
$$
Now I can see that I can use the relation for derivative of det(g) in the second term in bracket, but don't know what to do with the first term.
| The most important point about this computation is to use the formula for the derivative of the metric determinant
$$\frac{\partial_i g}{g} = g^{jk} \partial_i g_{jk} $$
The derivation of this identity can be found in the answer to this question. You can then derive the relationship between $g^{ij}{}_{,k}$ and $g_{ij,k}$ by taking a derivative of $\delta^i{}_{j} = g^{ik}g_{kj}$. Finally, you take the formula for the Christoffel symbols in terms of metric derivatives and after some algebra you get the result!
| {
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Electromagnets' relative strength of attraction & repulsion I have two small 12V electromagnets:
When powered oppositely they attract powerfully so that I cannot pull them apart.
But when they're connected the same way they repel only weakly and I can easily push them into contact.
Why the difference?
Thank-you,
| The difference is the magnetization of the iron core. When the directions are opposite, there is essentially no magnetic flux in the iron cores, almost no magnetization, so the iron does not contribute much to the force.
When the directions are such that the flux through the magnetic circuit is large, the iron is almost saturated, and it is hard to pull them apart. The holding power can be calculated by the energy required to open a small air gap.
| {
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What does "commuting with the Hamiltonian" mean? In quantum mechanics an observable or an attribute to a particle (like spin) is conserved if and only if it commutes with the Hamiltonian. What does this mean? What observables do not commute with the Hamiltonian?
| Apart from two other answers, consider the Heisenberg picture and the equation of motion where its written as
$$\frac{dA}{dt} = - \frac{i}{\hbar} [A , H]$$
So if an operator commutes with the Hamiltonian, from the above equation its obvious that
$$\frac{dA}{dt} = 0$$
So the quantity attributed to $A$ is conserved.
| {
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Conditions on $\phi$ and $\boldsymbol{A}$ for when $\boldsymbol{B}$ is uniform I'm reading "Classical Mechanics" (5ed) by Berkshire and Kibble, in the example for uniform magnetic field on pg.243 (Chapter 10 Lagrangian Mechanics) I came across this
A charged particle moves in a uniform static magnetic field B.
Find the solutions of the equations of motion in which ρ (axial radius, cylindrical coordinates) is constant.
For a uniform magnetic field, we may take
$$\phi=0 \text{ and } \boldsymbol{A}=\frac{1}{2}\boldsymbol{B}\times\boldsymbol{r}$$
The authors did not explain where these come from and I cannot understand why such conditions are imposed.
I'm particularly confused about the first condition (scalar potential=0),neither of the four Maxwell's equations require $\phi=0$ for when $\partial_tB^i=0$.
Is this purely a choice or am I missing something?
| The authors are just helping you out. It should take only a few moments to verify that those choices for $\phi$ and $\mathbf A$ yield the correct electric and magnetic fields.
Electromagnetism exhibits gauge invariance, so there are an infinity of other choices of $\phi$ and $\mathbf A$ which would also yield the correct fields, but the one they give you is simple and convenient.
If you want to be less convenient, pick any scalar function $\chi(x,t)$ and add $\frac{\partial \chi}{\partial t}$ to $\phi$ and subtract $\nabla \chi$ from $\mathbf A$. These potentials correspond to the same electric and magnetic fields as before so your final answer would be the same, but in the absence of additional motivation there's no reason to do this to yourself.
| {
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Question on Newton's first law In our physics course, Newton's first law was given as a definition of inertial reference frame. Now in order to use it we need to take some object which has zero interaction with other objects. But now in order to check it, we need to know which forces are acting on an object. But how can we be sure that the fact that an object, with zero net force on it, is accelerating because we are in a non−inertial reference frame and not because of the fact that we haven't accounted for some "real" (as opposed to those called "fictitious") forces. And in general, how can we determine that the force that acts on an object is "real" and not "fictitious" ? E.g. how can we determine that electromagnetic force is "real" and not "fictitious" ?
| Very good question. You would do many experiments with many different bodies and you would find that, for all of them, you would have to account for a mysterious force given in terms of a constant vector $\boldsymbol{f}_i=-m_i\boldsymbol{A}$ that must characterize your system kinematically, because it affects all moving bodies equally in proportion to their inertia. Your version of Newton's law would be,
$$
m_{i}\boldsymbol{a}_{i}=\boldsymbol{F}_{i}-m_{i}\boldsymbol{A}
$$
You would conclude --or it would be reasonable to conclude-- that $\boldsymbol{A}$ is something kinematical characterizing your frame.
| {
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Physics experiment with sound I am doing an experiment for a school project which consists of blowing into 6 different bottles to create different notes/harmonics. Each bottle is filled with different volumes of water to create a different sound. I have to calculate the theoretical frequency of each, but I do not know how.
| This is called a Helmholtz resonator, a kind of mass-spring oscillator where the mass is the mass of the air in the neck of the bottle, and the spring comes from compression of air in the main volume of the bottle. A formula for the resonant frequency is given here: https://en.m.wikipedia.org/wiki/Helmholtz_resonance
| {
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Why do fireproof safes "capture and hold in moisture"? From The Best Fireproof Document Safe | Reviews by Wirecutter:
Humidity can be an issue with these safes, as they tend to capture and hold in moisture, which can damage their contents.
Someone commented:
Is this really as good as it gets? Is the humidity concern a consequence of its fire rating or are these poorly designed? I would have appreciated if this was addressed in more detail because the amazon reviews are scathing and has me deeply concerned about this purchase. I don’t want to have to maintain my safe, I want to put stuff in and not touch it for months or years. Surely that’s possible?
I replied:
I think the humidity concern is a direct consequence of water resistance and the safes being made out of relatively thermally conductive materials (e.g. metal). The safes are roughly air/water tight so it's more likely that the moisture in the air inside the safe (slowly) condenses into liquid water over time. Or maybe moisture seeps inside the safes (via air) but, for some reason, can't similarly 'escape'?
Am I right? Or are these safes somehow acting as (slow) 'water pumps' and causing the humidity inside to increase over time? Or is there another explanation?
| Anything that is air-tight will cause trapped moisture to condense as the ambient temperature changes the temperature inside. I don't really see what else could be particularly special about fireproof safes.
| {
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Doubt in potential energy of sphere on inclined plane This is not a question but a general doubt. So if a sphere is kept on an inclined plane what would be its potential energy.Will it be mgh or mg(h+r).
I need to use it in derivation (final velocity at bottom when rolling) kindly help me. thanks
| If $h$ is the height of the point of contact P of the sphere with the ramp, and the angle of the ramp is $\theta$, then
$$
V = mgh+ m\,r\cos \theta
$$
as the centre of the mass O of the ball is not directly over $x$ because PO is perpendiculr to the ramp surface.
As @Jonas has said, the $r\cos \theta$ does not change with $h$ so if your are interested only in changes of $V$ it is irrelevelent.
| {
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Why do bullets shoot through water but not through sand? There are a few questions only on this site about this but none of them answer my question.
Can cannonballs go through water?
Why does a bullet bounce off water?
I find it hard to understand why bullets shoot through water at longer distances but stop in sand almost right away:
*
*Both water and sand are made up of smaller droplets/grains and both are relatively heavy elements (sand is only 1.5 times heavier per volume). Water molecules are bound by Van der Waals force into droplets, sand molecules are bound by covalent bonding into crystals
*At slow speed, I can put my hand through water and sand both. The droplets and grains can roll over and accommodate an object easily.
*At high speed, an airplane crashing onto water will fall into pieces because water acts in this case like a solid, because the molecules and droplets don't have enough time to rearrange to accommodate the object. Same with sand.
*Now in the case of a bullet, this argument seems not to work. In air, bullets reach speeds over 1800 mph. Bullets penetrate water, and can keep high speeds up to 10feet. On the other hand, bullets can't penetrate sand at all, they stop completely almost with no real penetration.
Bullets can keep high speeds up to 10 feet in water.
https://mythresults.com/episode34
Bullets in sand are completely stopped after 6 inches.
https://www.theboxotruth.com/the-box-o-truth-7-the-sands-o-truth/
Question:
*
*Why do bullets shoot through water but not through sand?
| They can go through sand--just not very much of it.
There's a case from IIRC Desert Storm. A US tank realized an Iraqi tank was hiding behind a dune from the heat of it's exhaust. It successfully engaged the tank through the sand dune.
| {
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Gauss' law from Hamiltonian density of electromagnetic field I am going through David Tong's QFT course, for which lecture notes and exercises are available online at http://www.damtp.cam.ac.uk/user/tong/qft.html.
In Question 1.8 we have the Lagrangian (density)
$$L = -\frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \frac{1}{2} m^2 C_\mu C^\mu,$$
$$F_{\mu \nu} = \partial_\mu C_\nu - \partial_\nu C_\mu,$$
which is like the standard electromagnetic field in the case $m=0$.
I eventually derive the conjugate momenta $\Pi_\mu$ to $C_\mu$ and convert the Lagrangian to a Hamiltonian
$$H = -\frac{1}{2} \Pi_i \Pi^i + \frac{1}{4}F^{ij}F_{ij} - \frac{1}{2} m^2 C^{\mu}C_{\mu} - \Pi_i \partial^i C_0,$$
answering the question.
However in a pdf of tutor's solutions I came across online (which I maybe shouldn't link), the tutor comments and interprets further:
they rearrange the last term,
$$H = -\frac{1}{2} \Pi_i \Pi^i + \frac{1}{4}F^{ij}F_{ij} - \frac{1}{2} m^2 C^{\mu}C_{\mu} - C^0(\partial_i \Pi^i) - \partial_i (\Pi^i C^0),$$
and comment
[the term] involves an irrelevant three-divergence term. Since the remainder of the Hamiltonian contains no derivatives in $C^0$, $C^0$ may be regarded as a multiplier that, in the $m=0$ theory, imposes the constraint $\nabla \cdot \Pi = m^2 C^0 = 0$, which is precisely Gauss' law.
Since we are back to examining the $m=0$ case, this is a statement about the standard electromagnetic field.
I don't understand either statement here.
How is $\partial_i (\Pi^i C^0)$ "irrelevant"? Can we just ignore this divergence, which as far as I can see has a nonzero value?
$- C^0(\partial_i \Pi^i)$ could be a (Lagrange) multiplier, how is it rearranged to include the $m^2$ term and (together) constrain to $\nabla \cdot \Pi=0$?
| What you found is actually the Hamiltonian Density $\mathcal{H}$. The Hamiltonian is the spatial integral of Hamiltonian Density
\begin{equation}
H=\int_\mathcal{M} d^Nx\,\mathcal{H}(x).
\end{equation}
If you have an spatial divergence in your Hamiltonian Density, its contribution to the Hamiltonian can be converted into a hypersurface integral using divergence theorem
\begin{equation}
\int_\mathcal{M} d^Nx\,\partial_i\Pi^i=\int_{\partial\mathcal{M}}d\sigma\,n_i \Pi^i,
\end{equation}
where $\partial\mathcal{M}$ is the boundary hypersurface, and $n_i$ is the normal vector to it. As that boundary is in infinite, where $\Pi^i$ must vanish, the contribution of the $\partial_i\Pi^i$ term to the Hamiltonian Density is nule.
The Lagrange multiplier would be just $C_0$ in the $m^2=0$ case, enforcing $\partial_i \Pi^i=0$. When $m^2\neq0$ that constraint is not true.
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Why is the radial velocity considered zero? I had recently come across a question which is stated as below:
A disc placed on a large horizontal floor is connected from a vertical cylinder of radius $r$ fixed on the floor with the help of a light inextensible cord of length $l$ as shown in the figure. Coefficient of friction between the disc and the floor is $\mu$. The disc is given a velocity $v$ parallel to the floor and perpendicular to the cord. How long will the disc slide on the floor before it hits the cylinder?
I thought hard for a few days but I couldn't solve it as the mathematics was terrible. Finally, while trying many other things, I tried considering the radial component of velocity to be zero and it worked! I got the answer.
But I am not able to understand the logic behind considering the radial velocity zero. Would someone please help me to understand it?
Edit: The figure is given as below:
| Let's say the disk has got a non-zero radial velocity. This then has $2$ possibilities. First, the radial velocity is outward along the string and second, the radial velocity is inward along the string.
The First case cannot happen because of the restriction given in the question, the string is inextensible.
For the Second case, if the disk has a velocity inward along the length of the string, the string will slacken after the disk moves a distance $dl$ which will then lead to the tension force, the force exerted by the string and the only force that can provide a angular motion to the disk to instantaneously become zero. So, in this case, the disk will keep moving in the same direction with decreasing speed until the string tightens again to start providing a angular motion.
| {
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Why does ice fragment vertically? The pool in my back yard froze over, and now that it's warmer it is starting to break up. But I notice that the ice is marred by fractures, and that they all seem to be completely vertical. Handling the ice indeed causes it to fragment into "sticks", for lack of better term.
This link points to an album of two short video clips to demonstrate.
Why does it do that?
| Those are some pretty unusual chunks of ice - as you can see, the boundaries of the 'columns' are visible before he breaks them up. So what you've got is a crystalline solid which clearly is the merging of many "source" crystal starting points, or "seeds" as mentioned in the comments.
I can say with certainty from my many years of shinny on frozen ponds that ice does not typically form that way, or at least not at that small scale. I'm going out on a limb (but not on thin ice :-) ) and suggesting that there was a lot of pollen or other small-size debris on your pool which acted as the seed locations for this ice.
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If we have a net negative charge in a spherical conductor, why negative charges goes to the surface? So imagine that we have a set of net negative charges, in physics books they said that this set of charges goes to the surface because they repel each other in such a way that this reach to an electrostatics equilibrium when this set of negative charges sits on the surface. This must create a negative potential energy in the surface of the sphere, so imagine that we have a negative charge in the middle or anywhere inside the sphere. By definition, negative charges tends to move to higher potential energy position, so why they are still going to the surface?
| "negative charges tends to move to higher potential energy position, so why they are still going to the surface?"
The surface is at no lower a potential than the rest of the sphere. We can show from the inverse square law that charges uniformly spread over the surface of a sphere produce zero resultant field at any point inside the sphere. So the potential is the same at all points inside the sphere. Any charges placed inside the sphere later will mutually repel and finish up on the surface.
This argument doesn't hold for the field inside a body bounded by a non-spherical surface. But if the body is made of a conducting material there is zero field after a very short time from placing of charges for a different reason. The charges when first introduced will indeed set up electric fields and free charges will move accordingly. But an equilibrium set-up will soon be reached, when charges have stopped moving (and reside on the surface). If there were any electric field inside the conducting body the free charges would be moving. So no (macroscopic) electric field in a conducting body in equilibrium. Nothing to stop new charges introduced from repelling each other and finishing up on the surface.
| {
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How does Inertial force arise? Consider the following scenario$-$
A ball sits on the floor of a bus, which was originally at rest w.r.t the ground. Suddenly it accelerates forward, and we observe the ball moving backwards. Well, originally the ball does not move, it is only the bus that moves forward. Why doesn't the ball simply accelerate with the bus? A common answer might be $-$
The body tries to maintain it's inertia of rest.
In this case, my question would be, how is this force generated that tries to oppose the motion of the ball with the bus?
P.S.- Please consider frictional force too, since I feel like I know the frictionless case.
| @joseph h
inertia is a natural property of the ball and all material objects
This may be seen from a different perspective. There are confusing uses of mass in which it can be interpreted as an emergent behavior (like temperature) or arguably fundamental, such as its association with the Higgs particle. It is not clear, at least to me, that we understand what mass is and its manifestation as inertia, although we do use the concept successfully to describe the dynamics of events.
| {
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Kepler's laws for circular orbits Kepler's first law states that planets revolve around the sun in an ellipse with the sun at one focus of this ellipse. (a special case would be a circular orbit with the sun at the center).
The second law states that the areal velocity is a constant. Thus we can write ($dA = c dt$). If we integrate over one complete cycle we find that the area of the orbit, which is proportional to the square of the radius, is proportional to the time period.
The third law states that the square of the time period is proportional to the cube of the semi-major axis of the elliptical orbit.
My question is, should we swap out "semi-major axis" and replace it with "radius", or is there something missing? if we can, that leads to a contradiction with the second law.
however, how can the result obtained from the second law be wrong for circular orbits? What am I missing?
| For circular orbits,
$$\frac{dA}{dt}=\frac{L}{2\mu}\Rightarrow \pi R^2=\frac{L}{2\mu}T$$
Further $$E=\frac{\mu C^2}{2L^2}\ \ \ \text{For circular orbits}$$
From now on I'm just going to keep track of proportionality.
$$R^2\propto LT\propto\frac{1}{\sqrt{E}}T$$
But further we known that $R\propto 1/E$
$$R^{3/2}\propto T\Rightarrow \boxed{R^3\propto T^2}$$
| {
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Relationship between entropy and the number of symmetries
*
*We count symmetries of a system by counting the number of transformations/operations under which a feature of the system remains unchanged.
*Entropy is a measure of the number of microstates that correspond to the observed macroscopic state (which remains unchanged when the change of microstates remain in a certain set, similar to the way symmetries work).
Is this a mere analogy or would it be possible to define entropy in terms of the concept of symmetry?
| @ali I'll take a feeble stab at this.
First, this is my guess at what you mean.
Here is a "system" microstate 00110. Here is another "system" microstate 11000. Distinct microstates and the operation was roughly interchange places 3,4 and 1,2.
A system macrostate property is the sum over "places" for a given microstate. In this case both microstates have the macrostate property of 2. Conceptually, total entropy for this system is the number of ways I can get macrostate 2 out of rearrangements of the two 1's and 3 0's.
If we accept the definition provided for symmetry, then the symmetric "feature" is 2 and the symmetry "transformations" are the rearrangements of 1's and 0's. Within the context of this simple example, your question is justifiable but I sure don't know how far it goes. It only takes one counter-example to disprove any equivalence type of statement.
| {
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Two boxes, identical other than mass and density, are pushed with a coefficient of kinetic friction $\mu$. Do they slow down at the same rate? Do they slow down at the same rate? And stop simultaneously?
I believe they do because $v_f = v_0 -\mu g t$ for both. But I'm not completely sure why that is.
I've also seen another equation for the slow down due to friction.
$x=\frac{v_o}{\mu}(1-e^{-\mu t})$
I've been told both are correct.
Neither equation involves mass.
| These two $\mu$s are of different origin. The second $\mu$ depends on the mass.
The first equation
$$ \tag{1}
v_f = v_0 - \mu g t.
$$
The is kinetic friction
$$
f_k = \mu N = \mu m g;\\
a = f_k / m = \mu g.
$$
The accelaration is independent of mass.
But the second equation:
$$ \tag{2}
x=\frac{v_o}{\mu}(1-e^{-\mu t})
$$
This was from the damping force $f_d = -\gamma v$ with damping coefficient $\gamma$
The corresponding equation of motion
$$
m \frac{dv}{dt} = - \gamma v;\\
\frac{dv}{dt} = -\frac{\gamma}{m} v \equiv - \mu v.
$$
The velocity
$$
v(t) = v_0 e^{-\mu t}.
$$
This integrates again $x(t) = \int v(t') dt' $ to get the position of Eq.(2). The $\mu = \gamma / m$ in equation 2 is mass dependent, and of diiferent nature from that in Eq.(1).
| {
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Interesting inertia problem Consider the following.
A car is accelerating with acceleration $a$. A string is attached to the roof of the car and to the bottom of the string, an object of mass $m$ is attached. Given $\theta$, the angle between the vertical and the string (which is not $90^\circ$ due to inertia of the object).
How to derive an expression for the acceleration $a$ of the car given $\theta$ and $m$?
And when does $\theta$ remain constant?
I found a similar question, but the answers to that post were too low quality (as is also evident by the fact that the user didn't accept any of those as solutions); so don't flag this post as a duplicate of that.
| In summary, when $\theta$ is constant we have
$T \sin \theta = ma
\\ T \cos \theta =mg
\\ \displaystyle \Rightarrow \tan \theta = \frac a g \text{ ; } T = m \sqrt{a^2+g^2}$
| {
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What does 3D buckling reconstruction mean? In condensed matter/material science papers, I often encountered phrases like structural reconstruction, 3D reconstruction, 3D buckling reconstruction, etc. What do these phrases mean (especially the last one)?
For example, this paper (and a non-paywalled link).
| In the "structural reconstruction" context it means that the crystal structure has changed -- often at a surface. In particular a "buckling reconstrction" means that the initially flat layer has changed shape -- buckling into the direction normal to the surface --- because of the Moire-modifield electronic structure.
| {
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How to define energy density and pressure for a scalar condensate? Consider a scalar field theory in flat spacetime whose Lagrangian reads
$$\mathcal{L}=\frac{1}{2}\eta^{\mu\nu}(\partial_\mu\varphi)\partial_\nu\varphi-V(\varphi).$$
Then the energy-momentum tensor reads (we use the signature (+,-,-,-))
$$T^{\mu\nu}=(\partial^\mu\varphi)\partial^\nu\varphi-\eta^{\mu\nu}\mathcal{L}.\tag{1}$$
Now we want to define the energy density and pressure for a scalar condensate from the above energy-momentum tensor. In Weinberg's book "Cosmology'', he defines these quantities as follows.
Comparing Eq.(1) with the energy-momentum tensor for a perfect fluid,
$$T^{\mu\nu}=(\rho+p)u^\mu u^\nu-\eta^{\mu\nu}p,$$
one may identify
$$p=\frac{1}{2}\eta^{\mu\nu}(\partial_\mu\varphi)\partial_\nu\varphi-V(\varphi),\tag{2}$$
$$\rho=\frac{1}{2}\eta^{\mu\nu}(\partial_\mu\varphi)\partial_\nu\varphi+V(\varphi),\tag{3}$$
and
$$u^\mu=\left(\eta^{\rho\sigma}\partial_\rho\varphi\partial_\sigma\varphi\right)^{-1/2}\partial^\mu\varphi.$$
However, there are some problems with the above definitions.
(i) The first problem is that, according to Eq. (3), one has
$$\rho=\frac{1}{2}\dot{\varphi}^2-\frac{1}{2}(\nabla\varphi)^2+V(\varphi).$$
Why the gradient energy has a negative contribution to the energy density?
(ii) The definitions are only valid for
$$\eta^{\rho\sigma}\partial_\rho\varphi\partial_\sigma\varphi\geq 0.$$
What if $\eta^{\rho\sigma}\partial_\rho\varphi\partial_\sigma\varphi<0$? Can't we define the energy density and pressure for a scalar condensate which is, say, space-dependent but time-independent?
With the above two problems, I would think that the definitions (2) and (3) are problematic. Then how to define in general the energy density and pressure for a scalar condensate?
| There's a mistake in your formulae (2) and (3). You should start from (1) and the following (for an isotropic field, since you want pressure of a perfect isotropic fluid):
\begin{align}\tag{1}
\rho &= T_{00}, &p &= T_{11} = T_{22} = T_{33}.
\end{align}
Then (notice the + sign in front of the squared gradient) $$\tag{2}
\rho = (\partial_0 \phi)^2 - \mathcal{L} = \frac{1}{2} \, \dot{\phi}^2 + \frac{1}{2} (\nabla \phi)^2 + \mathcal{V}(\phi).
$$
If the field is isotropic everywhere, then $\nabla \phi = 0$. The isotropic pressure is (with $\nabla \phi = 0$):
$$\tag{3}
p = T_{11} = \mathcal{L} = \frac{1}{2} \, \dot{\phi}^2 - \mathcal{V}(\phi).
$$
| {
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Reflection and transmission wave on three joined strings Suppose we have a system of three joined strings, of different linear mass densities and
subjected to a constant tension force, such that the velocity of propagation is $v_1$ in the
string 1 (at $x <0$) and string 3 (at $x> L$), and $v_2$ on string 2 (from $x = 0$ to $x = L$).
A plane wave of wave number $k_1$, frequency $ω$, and amplitude $A$ propagates to the right
on string 1.
We have to calculate the reflected wave's final amplitude $A_r$ (on string 1), and the transmission wave amplitude $A_t$ on string 3, supposing that at first there is only one incident wave moving towards the positive X axis on string 1. Then, once this part is done, we have to show that the reflection and transmission coefficients ($R = |A_r/A|^2$ and $T = |A_t/A|^2$, respectively) are:
$$R = \frac{(k_2^2-k_1^2)^2\sin^2(k_2L)}{4k_1^2k_2^2+(k_2^2-k_1^2)^2\sin^2(k_2L)}$$
$$T = \frac{4k_1^2k_2^2}{4k_1^2k_2^2+(k_2^2-k_1^2)^2\sin^2(k_2L)}$$
Attempt:
(a) In the $x<0$ region, we have an incident ondulatory movement (OM) and a reflected one at $x=0$, that spread with velocity $v_1$:
$$y_1(x,t) = Ae^{i(k_1x-\omega t)} + Be^{-i(k_1x+\omega t)}$$
(b) In the $0<x<L$ region, we have a transmitted OM and a reflected one at $x=L$, that spread with velocity $v_2$:
$$y_2(x,t) = Ce^{i(k_2x-\omega t)} + De^{-i(k_2x+\omega t)}$$
(c) In the $L<x$ region, we have a transmitted OM that spreads with velocity $v_1$:
$$y_3(x,t) = Ee^{i(k_1x-\omega t)}$$
Where $B = A_r$ and $E = A_t$
The boundary conditions at $x=0$ and $x=L$ are:
1 - The string is continuous in $x=0$:
$$y_1(0,t)=y_2(0,t) \Longrightarrow A+B=C+D$$
2 -
$$\left[ \frac{\partial{y_1}}{\partial{x}} \right]_{x=0} = \left[ \frac{\partial{y_2}}{\partial{x}} \right]_{x=0} \Longrightarrow k_1(A-B) = k_2(C-D)$$
3 - The string is continuous in $x=L$:
$$y_2(L,t)=y_3(L,t) \Longrightarrow Ce^{ik_2L} + De^{-ik_2L} = Ee^{ik_1L}$$
4 -
$$\left[ \frac{\partial{y_2}}{\partial{x}} \right]_{x=L} = \left[ \frac{\partial{y_3}}{\partial{x}} \right]_{x=L} \Longrightarrow k_2(Ce^{ik_2L} - De^{-ik_2L}) = k_1Ee^{ik_1L}$$
Then, solving that 4 unknown equation system in terms of A, we obtain the following results:
$$B=A_r=\frac{i\left(\frac{k_1^2-k_2^2}{k_1k_2}\right)\sin{k_2L}}{2\cos{k_2L}-i\left( \frac{k_1^2+k_2^2}{k_1k_2} \right)\sin{k_2L}}A$$
$$E=A_t=\frac{2Ae^{-ik_1L}}{2\cos{k_2L}-i\left( \frac{k_1^2+k_2^2}{k_1k_2} \right)\sin{k_2L}}$$
So far I think these results are fine. However, when I try to do the second part of the exercise and I try to calculate the R and T coefficients, I don't know what to do with the imaginary parts of both expressions because the R and T coefficients are only real.
Please, could anyone tell me which would be the next step to obtain R and T?
Thanks in advance!
| For example, the transmission coefficient $T = |\frac{A_t}{A}|^2$, since
$$ \frac{A_t}{A} =\frac{e^{-ik_1L}}{\cos{k_2L}-i\left( \frac{k_1^2+k_2^2}{2k_1k_2} \right)\sin{k_2L}}$$
Multiply this with its complex conjugate:
$$
T = \frac{A_t}{A} \frac{A^*_t}{A} = \frac{e^{-ik_1L}}{\cos{k_2L}-i\left( \frac{k_1^2+k_2^2}{2k_1k_2} \right)\sin{k_2L}} \times \frac{e^{+ik_1L}}{\cos{k_2L}+i\left( \frac{k_1^2+k_2^2}{2k_1k_2} \right)\sin{k_2L}} \\
= \frac{1}{\cos^2{k_2L}+\left( \frac{k_1^2+k_2^2}{2k_1k_2} \right)^2\sin^2{k_2L}}\\
=\frac{4 k_1^2 k_2^2 }{4 k_1^2 k_2^2 \left(1-\sin^2{k_2L}\right)+\left( k_1^2+k_2^2\right)^2\sin^2{k_2L}}\\
=\frac{4 k_1^2 k_2^2 }{4 k_1^2 k_2^2+\left( k_1^2-k_2^2\right)^2\sin^2{k_2L}}
$$
| {
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Is there a way to measure the amount power of visible light against a surface? If you were to hypothetically shine a single beam of light that is specific in its power source and output efficiency against a wall across a specific measurable distance. Is there a possible way to measure the amount of visible observable light against the surface when it finally reaches said surface over a specific pre-defined distance?
Any help or what type of subject matter this applies to would be appreciated. Trying to learn about such topics. I assume it has to do with the speed of light. Any formula or documentation that might be helpful and appreciated as well.
The light and the object it shines against are not moving and static in position.
Thanks
| I'm assuming you're looking for experimental techniques. You can use a photodiode to measure power. For selecting visible light you could use one or more cut-off filters. There also exists monochromatic filters that select specific wavelengths, if you're interested.
| {
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Flat space limit of the $AdS$ metric: Very basic question Suppose I am given the following global coordinates in empty $AdS_n$:
$$ds^2 = \alpha^2\left(-\cosh^2\rho \, d\tau^2 + \, d\rho^2 + \sinh^2\rho \, d\Omega_{n-2}^2\right)$$
where the length scale $\alpha$ is related to the cosmological constant as $\Lambda = \frac{-(n - 1)(n - 2)}{2\alpha^2}$. I had two very basic questions:
*
*In the limit $\alpha \to \infty$, the cosmological constant tends to 0. Is it correct to conclude that we thus obtain a flat space limit of $AdS_n$?
*In that case, how do we obtain the standard Minkowski metric from the above metric using $\alpha \to \infty$?
| Far far away from the boundary, AdS looks flat. We can see this by noting that AdS can be thought of as a spacetime with a "background potential" $V(\rho) \sim \rho^2/\alpha^2$ so the region in the neighborhood of $\rho=0$ is flat. As we move farther away from this point, the potential of AdS becomes stronger and the effects of curvature become more and more important.
OK, so in order to take the flat space limit, we must take a limit which allows us to zoom into this region. We do this by setting $r=\rho \alpha$, $t=\tau\alpha$ and then take $\alpha \to \infty$ keeping $r$ and $t$ fixed. In this limit we are simultaneously taking $\rho \to 0$ and $\tau \to 0$ which is the region we wish to zoom into. Note that $\tau$ has to be similarly rescaled so that we are sitting far away from ALL boundaries of AdS, not just the timelike boundary.
Doing this, we find
\begin{align}
ds^2 &= \alpha^2 \left[ - \cosh^2 \rho d\tau^2 + d\rho^2 + \sinh^2 \rho d\Omega_{n-2}^2 \right] \\
&= \alpha^2 \left[ - \alpha^{-2} \cosh^2 (r/\alpha) dt^2 + \alpha^{-2} dr^2 + \sinh^2 (r/\alpha) d\Omega_{n-2}^2 \right] \\
&\stackrel{\alpha\to\infty}{\to} - dt^2 + dr^2 + r^2 d\Omega_{n-2}^2.
\end{align}
| {
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Jumping into String Theory I am a third year student in physics and mathematics and taking a course in which each student need to prepare and give one lecture about some topic related to physics. As I saw String Theory in the topics list I thought it could be a good opportunity to start learn it.
I have background in quantum mechanics, statistical physics, analytical mechanics, group theory, and I read the book "quantum field theory for the gifted amateur" until chapter 12.
*
*What topics should I learn before string theory?
*I would appreciate any recommendations for books or other material about string theory.
| It's really wonderful to read a young physicist interested in string theory.
For the matter of the lecture I recommend to follow Witten's essay What Every Physicist Should Know About String Theory.
The following videos are about the enormous impact that string theory has produced for theoretical physics as a whole:
-What Every Physicist Should Know About String Theory
-Fundamental Lessons from String Theory with Cumrun Vafa
For a brief summary on string theory textbooks read the following blog post entry:
*
*String texbooks
*My answer to the following PSE question: Texbook recommendation for superstring theory
However I absolutely sure that you should start with the introductory textbook A first course in string theory by Barton Zwiebach.
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A light problem: What happens when light completely destructively interferes? So here's the setup. We have a spherical source. It emits a pulse of light in all directions with some wavelength $\lambda$. It reflects off of a spherical mirror that is centered around this source.
Now, when the light comes back, it bounces off of the source again (or some percentage, whatever). The source emits another pulse of light at the same time with exactly the same energy as the light that bounced off, but shifted back $\lambda$/$2$, so all the crests line up with all the troughs and the light completely destructively interferes.
I can't see a way out of this. I've spent energy - but where did it go?
| Think of an analogous but simpler problem. You have a string attached to a wall in one side and you are handling it on the other side (see this video, your hand is the spherical source of light and the wall is the spherical mirror). If you left your hand without moving when the reflection comes back (as in the video) the pulse will bounce again. If you move your hand with the same phase as the reflected pulse you will increase its energy by constructive interference. If you move your hand in anti-phase then the pulse will transmit its energy to your hand and it will disappear because you created the exact opposite wave and canceled it. You can think of this case as moving your hand in such a way as the string "believes" that there is more string to go so the energy keeps flowing from the string into your body. Once it is into your body you do whatever you want with it.
| {
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Why can't a system reach equilibrium with non-interacting particles? I am new to Statistical Mechanics, and have just started reading from this book Tony Genault Statistical Physics
where he writes the following (paraphrased for convenience)-
Consider a system of $N$ weakly interacting particles. If the energy of one particle is $\epsilon$, the total energy of the system is-
$$U=\sum_{l=1}^{N} \epsilon(l)$$
Any such expression implies that the
interaction energies between particles are much smaller than these (self) energies ε. Actually any thermodynamic system must have some interaction between its particles,
otherwise it would never reach equilibrium. The requirement rather is for the
interaction to be small enough for the above equation to be valid, hence ‘weakly interacting’ rather
than ‘non-interacting’ particles.
And hence my question- Why does non interacting not work?
| Non-interacting in this context means the particles cannot collide and exchange kinetic energy. Without the ability to share energy during collisions, the system of particles has no opportunity to distribute the individual energies of the particles between all those particles and hence cannot eventually achieve a maxwell-boltzmann energy distribution, the peak value of which defines the temperature of the ensemble.
| {
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Where from is the distance calculated in Newton's law of universal gravitation? In the equation for universal gravitation $(1)$ between two objects, where from is $r$ calculated? From the surface, from the center? Also, are the objects assumed to be particles in this equation or could the be multiple particles like molecules?
$$F=G \frac{m_{1} m_{2}}{r^{2}}\tag{1}$$
| Strictly speaking, this equation for the force due to gravity only holds between point-like objects. In the case of a point-like object, the notion of "distance" between them is simply the distance between their positions.
For extended objects, things get a little more complicated. Really, if you had an extended object (a sphere, cylinder, amorphous blob) you would need to imagine breaking the object into many many tiny pieces of mass, compute the force due to gravity caused by each small piece, and add them all up (as vectors! not all the forces will point in the same direction, see for example tidal force).
The only exception to this is the gravity due to a sphere...in this case we can use the distance from the center of the sphere to compute the force due to gravity. But this is a very special result that comes about from the "breaking apart" that I described above...it just so happens that things work out the same as if we had just used the center of the sphere. For example, see here.
| {
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Is conduction band discrete or continuous? My question is very simple. I just want to know that is conduction band discrete or continuous?
|
A useful way to visualize the difference between conductors, insulators and semiconductors is to plot the available energies for electrons in the materials. Instead of having discrete energies as in the case of free atoms, the available energy states form bands.
Read on the link.
You ask:
My question is very simple. I just want to know that is conduction band discrete or continuous
The band theory is a quantum mechanical model, such that the difference between energy levels in the band is very small, mathematically discrete but experimentally continuous. That is why it is called a "band".
| {
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Newtonian physics and equivalence principle: a doubt on acceleration and gravity First of all, the famous Einstein's elevator experiment is quite clear in my head, both of versions.
But now, consider the following:
Suppose then you wake up inside a car that is traveling in perfect straight path in a autoban (but you don't know that). The car have a constant velocity $v$ and is a self-driving car with totally dark-glass windows. You don't have any information about the outside world. After a time $t$ travelling in the straight path, the car enters in a curve. You then fells an acceleration (exactly with $9,8 m/s^2$) accelerating you.
Now, in my opinion, the person inside the car cannot say that the centrifugal acceleration is different from artificial constant gravitational field. The equivalence principle states something similar, since a person inside a elevator in a gravitational field is equivalent to a person inside a elevator which is accelerated with $9,8 m/s^2$. Furthermore, we can construct a ring-like structure to produce, via circular motion, a artificial gravitational field.
So, can I say that any accelerated frame, due to equivalence principle, is equivalent to a gravitational field?
| Locally, yes. The geodesic equation of GR and differential geometry contains $\Gamma$. When you move to rotating frame these $\Gamma$s are nonzero. The geodesic equation doesn't care if the terms in $\Gamma$ come from a real gravitational field or a non-inertial reference frame.
| {
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Is it possible to lift an object from rest with constant power? This is inspired by the following question.
Consider some object which I want to lift from rest with a constant power throughout the whole process; the power I apply when lifting the object from rest is the same power I apply to keep lifting it. The force $F$ and the speed $v$ may change, but power may not.
The power is $P=Fv$. If we want constant power, then $dP/dt=0$.
First, differentiate wrt time, $$\dfrac{dP}{dt} = F \dfrac{dv}{dt}+v \dfrac{dF}{dt}.$$
Set equal to zero, this guarantees constant power, which implies:
$$F \dfrac{dv}{dt} = -v\dfrac{dF}{dt}.$$
From there, I use $F=\dfrac{d(mv)}{dt}\implies F = m\dfrac{dv}{dt}$ and get $$\left( \dfrac{dv}{dt} \right)^2 + v\dfrac{dv}{dt} = 0.$$
How may I solve this non-linear ODE?
EDIT: Initially, I got to the wrong conclusion that $F(v-v_0) = -v(F-F_0)$ is a solution. That's what the answer by Vilvanesh addresses.
| Your discussion is valid till $\displaystyle{F\frac{dv}{dt}=-v\frac{dF}{dt}}$.
I think the only way to reach $F(v-v_o)=v(F-F_o)$, is by assuming $F$ is constant while integrating on the left side and by assuming $v$ as constant while integrating on the right side. This is inconsistent because on the left side $F$ is being treated as a constant and on the right side it is treated as a variable. Same is applicable for $v$.
| {
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What is a resonance element? I'm looking at a horn antenna and why its a good addition to a waveguide. I've read that the horn can operate at a wide range of frequencies since it has no resonant element. But I can't find a good description of what a resonant element is in this application.
| In this context, a resonant element would be a chunk of the system that possesses a substantial quantity of impedance (either capacitance or inductance), different from that of the system as a whole. So if the horn had a local swelling or constriction somewhere along its length, the impedance of that segment would not match that of the horn inlet or outlet, and reflections would occur.
By the way it is not strictly true that a horn has no resonant elements in it. It has both inductance and capacitance contained in each little element of its length- however, by smoothly tapering the horn the impedance change from the throat to the exit of the horn is made gradual and the horn thereby becomes a transformer which matches the impedance of the waveguide to that of free space.
| {
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How does the growing up of trees works with respect to energy conservation? We know that the energy (and hence mass) can neither be created nor destroyed. In the light of above statement, I would like to understand views on growing up of fauna and flora on earth.
A small seed is planted in earth. The only visible input for the seed is amount of manure, fertilizer and water added to it. Agreed, it receives a good amount of energy from sun and other minerals from the earth.
The output: The tree grows up amounting into hundred of tons of wood mass, thousands of flowers and fruits.
Considering that this has been happening for billions of year, Is it fair to assume that the above output is produced from the above set of inputs only and there has been no increase of mass in the process.
| Of course it is. Mass doesn't magically appear out of nowhere.
In the case of a tree, most of the mass is taken out of the air (the carbon from the carbon dioxide) and water (the oxygen and hydrogen).
| {
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What are so-called opposite colors? Colors are said to be electromagnetic frequencies.
How is this compatible with the notion of color opponency, or opposite colors, since those frequencies form a single dimension?
| The notion of opposite colors has nothing to do with the electromagnetic spectrum. Human eyes have three different photoreceptors, named cones (I won't enter into subtleties such as rod spectrum sensitivity here) with each a certain response curve to the electromagnetic spectrum.
When those receptors are excited by light in their detection range, they inhibit their nervous discharge. Here the inhibition is irrelevant because neurons can convert that into an excitation again. Anyway, we are at neurons communication level now.
Neurons form circuits that tend to generate categories from rather continuous values, and these networks are the ones producing colors as categories instead of continuous values.
Black and white are opponents, and this is because there is a map of these values that do not circle. I.e., you can go from white to gray to black but this does not loop.
In the case of trichromacity, this is a ring. You go from red to orange to yellow to green to teal to blue to violet to magenta to red again.
Obviously this circling from red to blue through magenta is not physical in terms of EM frequencies.
This circle of colors has diametrically opposed colors. Those are the opposite colors you are refering to.
The opposite colors emerge from the existence of a circular organization of those categories.
Addentum: to be more precise, there is also opponency at a very low level in the visual system, e g. between blue and the sum of red and green (i.e. yellowish). This creates a special category for yellow as if it was a single photoreceptor. This is why you can have a blueish green but not a blueish yellow, as blue and yellow truly are incompatible at an early stage. But this opponency is not necessary for the circle of colors to appear at later stages. In the same way, there are other opponencies at mid level, between bluish violet and lime and teal vs orangeish. Again this is not the complete story.
| {
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If electrons can be created and destroyed, then why can't charges be created or destroyed? I read on Wikipedia that electrons can be created through beta decay of radioactive isotopes and in high-energy collisions, for instance when cosmic rays enter the atmosphere. Also, that they can be destroyed using pair annihilation.
We also know that charge is a physical property which can be associated with electrons.
My question is why can't charges be created or destroyed if electrons can?
| Well most of what needs to be said already has been said. But there is something else that I think would interest you.
Charge in a system is only conserved when a very specific symmetry in the system in not broken. Noether's theorem links symmetry to conservation laws. If you can find a way to break the symmetry that governs conservation of charge, then that conservation law will not hold.
An example of this would be introducing an external force in a system, thus breaking the symmetry that govern conservation of linear momentum.
Electrons can be created and destroyed because the process does not violate any of the given system's conservation laws. If you can find a situation in which the symmetry governing conservation of charge is broken, you shouldn't have much problem creating or destroying charge.
| {
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The definition of Quasi-static process? A quasi-static process is often defined as a process "that occurs infinitely slowly such that equilibrium holds at all times."(Harvard, Matthew Schwartz, statistical mechanics Spring 2019). My question is a simple but possibly subtle one which I haven't seen mentioned anywhere.
Simply put, does the system need to maintain equilibrium with the surroundings at all times during the process in order for the process to be classified as a quasi-static process or is the lesser requirement of simply having the system maintain internal equilibrium with itself good enough to classify the process as quasi-static?
As an example, suppose we have a fixed volume system immersed in a heat bath (the surroundings). The system is at temperature $T_{sys}$ while the heat bath is at fixed $T_{bath}$ with $T_{sys}<<T_{bath}$. The walls/boundary of the system are virtually adiabatic but not totally (i.e they posses a very low thermal conductivity) and so heat can and will seep into the system across a large temperature gradient but this will happen very slowly (perhaps even infinitely slowly). After a very long time, the systems temperature will equal the heat baths temperature. Does this count as a quasi-static process? Throughout the process, the system had a well-defined internal equilibrium however it was never in equilibrium with the surroundings and so I am not sure whether it counts as quasi-static or not.
Any help on this issue would be most appreciated!
| I would consider the process you described as quasi static. In fact, in that particular process, both the system and the surroundings experience reversible changes. However, for the low thermal conductivity medium in-between, the process is not reversible, and entropy is generated within this medium. This generated entropy is transferred to the system. So the increase in entropy of the system is greater than the decrease in entropy of the surroundings, and the net result is an increase in entropy of the "universe."
| {
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How do I find the approximate surface area of a chicken? I'm working on building a chicken army and I'm trying to find out how much metal or kevlar (still deciding) I need to make armor for the chickens. this measurement does not need to be exact I'm just trying to get an estimate for how much I will need. You will be spared when my chickens take over the world if you give me a working answer.
| Archimedes' principle can also be used to measure rate of change of volume as the chicken is lowered into liquid. Sedate the chicken before doing this. Plot a curve of volume vs depth. This tells you the area of a cross section of the chicken at the surface of the liquid at each depth. Each cross-sectional part of the chicken can be approximated pretty well as a circular disk whose top surface has the corresponding area. Calculate the area of the chicken as the sum of all the circumferences times the thickness of the cross sections. Of course the leg cross sections will need to be treated as pairs of disks; and you'll need to treat the wings separately. Just make "gloves" that fit the wings, and measure the area of fabric required to make the gloves. You should end up with a surprisingly accurate estimate.
A possibly more accurate way would be to wrap a string around the chicken at different heights to get the length of the perimeter of whatever shape the chicken is at each height. Plot those perimeters as a curve, with the height axis along the x-axis. Calculate the area under the curve, as height times the spacing between the heights. This would handle any convex cross sections, and give you the area of a soap bubble covering the chicken parts. Again, wings would present an easily solvable problem.
And finally, if you plan to put the chicken in stew, first wring its neck, then wrap its body and wings separately with shrink-wrap plastic you can get from the butcher. Use a heat gun to shrink the plastic tight against the chicken. With a razor blade, make a long cut so you can peel the shrink wrap off the chicken. That has the area of the chicken, and even records the shapes of the parts you need to make for the armor.
| {
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Assumption while deriving Time Independent Schrödinger equation While deriving the Time Independent Schrödinger equation we assume that the wave function is composed of the two separate functions of time and space. And since we do not have any information regarding the wave function, then how is it correct to assume this?
P.S- I am still a beginner in Quantum mechanics so it would be great if you will elaborate your answer.
| The Time Dependent Schrödinger Equation (TDE) is a linear, second order partial differential equation (PDE) with variables time $t$ and position $\mathbf{r}$.
When trying to obtain the Time Independent Schrödinger Equation (TISE) we apply a well-known method known as separation of variables (SoV). This method is not inherent to quantum mechanics, wave functions or the Schrödinger Equation, it just works for that type of PDE.
Using the method of SoV we assume that the wave function can be written as:
$$\psi(\mathbf{r},t)=\Psi(\mathbf{r})\varphi(t)$$
Insertion of this assumption into the TDSE then yields two ordinary differential equations, one for $\Psi(\mathbf{r})$ and one for $\varphi(t)$.
Similar SoV methods are used for the classical wave function, the heat equation and the diffusion equation (among others).
| {
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Is no acceleration a cause or consequence of no net force? If a body is moving with constant velocity, or is at rest, then the net force on it must be $0$. If the net force on a body is $0$, then it must be moving with constant velocity or must be at rest.
Is $0$ net force a consequence of being at rest or moving with constant velocity or is moving at constant velocity or being at rest a consequence of $0$ net force?
| This is a question about philosophy not physics. Here is the answer. If I kick a ball it accelerates. It is not because of the ball accelerating that I kick it, for what would be the point of football (EU English) if it was the other way around?
| {
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Forces and point of application Let us consider a rod and two forces acting on it that are not concurrent, that is, to say that their points of application are not identical, but they intersect somewhere outside the rod. Now, if we theoretically find the direction of the resultant $P$ from the point where both the lines of action of forces are intersecting, will it act in the same direction as the real resultant $R$ (that is, the the force or resultant acting on the rod caused by the two forces as mentioned before)?
Is there any theorem or proof in which this matter is explained?
Will the direction and magnitude of $F$ and $R$ be as in the image?
| Yes. You can slide any force along its line of action (parallel line through the point of application) and it does not change the system.
In your case $R$ is going to be along the line of action where it intersects the rod. As shown below:
Note that $F$ and $R$ must have the same magnitude and direction.
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Raising a canal's water level using anchored tug boats? You've all read about the enormous container ship stuck in the Suez Canal:
source
They're planning to attempt to unstick it today when the tide is high, as the 50cm rise in the water level should help significantly.
That, plus all the tugboats running around the scene of the stranding, led me to wonder if anchored tugboats could be used to significantly raise the water level in the canal. If you anchor two tugboats in a canal, facing away from each other and pulling as hard as they can against their rodes, it should raise the water level in between them. But by how much?
Assumptions:
*
*We are in the Suez Canal, which is 313m wide at the surface, 121m wide at the bottom, and 24m deep
*Typical tugboats have a bollard pull capacity of about 450 kilonewtons
*The tugboats are far enough apart that we don't have to worry about local flow effects
*The water level on either side of the pair of tugs is constant
| If there was only one tugboat, then the effect would be to pump water along the canal. But the opposing tug prevents that; the force of the pushed water from one balances against the push of the water from the other.
Since the tugs are isolated, we can treat the push of each tug as a generalized push against the water. Model this as a surface spanning the canal and pushing against the water with the force of the tug. The pressure applied to the surface will be balanced by the higher water level on the pushed side.
The cross-section of the canal is 5200 square meters, so a 450 kn-capacity tug would be equivalent to a pressure across the surface of 87 newtons per square meter. This is balanced by the additional weight of the higher water level on the other side of the surface. 87 newtons is the force applied by 8.8 kg of water, so there's an additional 8.8kg of water on top of each square meter of surface, which corresponds to 0.88cm of water. So, the pair of tugs would raise the water between them by 0.88cm.
To achieve the 50cm rise that is expected to come from the high tide, we'd need 114 typical tug boats, 57 on either side, steaming away. The world's strongest tug has a bollard pull of 4,680 kN, so we'd only need 12 of them, six on either side. Easy peasy.
| {
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What is the significance of the sign of the velocity for a particle executing SHM? So while deriving equation for the velocity of particle executing SHM at any point, I noticed a difference in the result depending on what wave (sine or cosine) you chose.
For $x=A\cos\omega t$:
$\quad \ \,v=-\omega\sqrt{A^2-x^2}$
For $x=A\sin\omega t$:
$\quad \ \,v=\omega\sqrt{A^2-x^2}$
Can anyone explain to me why the difference is there and what it means, since both equation are basically the same with only a phase difference?
| It is just related to two different initial conditions for the system.
Assume the system is a mass attached to a spring.
If $$x=A\cos(\omega t)$$ then this represents holding the mass with the spring extended in the +x direction at t=0. When you let go, the velocity is in the negative x direction.
This agrees with the calculated velocity you get by taking the derivative of the above expression: $$v=-\omega A\sin(\omega t)$$
The second expression represents the mass passing through x = 0 at t=0 and moving in the +x direction (as time increases, x grows positive), which is what the calculated velocity tells you also: $$v=\omega A\cos(\omega t)$$
| {
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Distribution of charge when 2 metallic spheres are connected It is given in my textbook that when 2 charged conducting sphere of different radius are connected by finite wire the redistribution of charges takes place such that the potential just outside of both spheres become equal.
But why potential is the necessary condition?
Like if net electric field just outside one sphere is 0 then even if there is some potential difference charge will not flow ,so why doesn't electeic field is necessary condition ??
And what should be the relation between the charges of two sphere after equillibrium is established when they are just touched?
| When you touch two spheres, you can consider them as one system, in other words, one big conductor. Now, if a conductor has different potentials on either side, then current (charges) flows through it from higher potential to lower potential.
This happens as a system always wants to be in the lowest energy state, hence it will try and minimize potential energy by redistribution of charges.
If we talk about net electric field, then it can be zero at only one point between the spheres, it is non zero everywhere else.
Assuming distance between the spheres as d, we get:
$\frac{kq{1}q{2}}{r^{2}}=\frac{kq{1}q{2}}{\left(d-r\right)^{2}}$
where r is the distance from one sphere, upon solving we find only 1 value of r for which the net field is zero.
Since net field is zero at only one place, the charges can still move under the influence of non zero electric field elsewhere.
Consider the equation $E=\frac{dV}{dr}$ written in different format:
$V=\int_{ }^{ }E\cdot dr$
now, V is the integral of E•dr, hence it is the area under the curve of electric field vs distance. In such a situation E can be negative and positive, depending on direction, which means that the net area under the curve can be zero. This doesn't imply that E is 0 everywhere, it just means the positive area cancels with negative area
| {
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Derivation of the radial Hartree-Fock equations Can anyone help deriving the radial Hartree-Fock equations of atoms from the Hartree-Fock equations
$$(h_i + J_i - K_i) \psi_i = E_i \psi_i$$
where each orbital $\psi_i$ involves three space coordinates? I'm interested in the restricted Hartree-Fock formulation.
| If found the following formula
$$
\frac{1}{\vert \mathbf{x}-\mathbf{x}'\vert} =
\sum_{k,q} \frac{r_<^k}{r_>^{k+1}} \frac{4\pi}{2k+1} Y_{kq}(\Omega) Y_{kq}^*(\Omega')
$$
in this page. That's all what I need for helium.
| {
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How to know if the error is in a law or in uncertainty of the measurement? I read these words in a (great) answer to this question:
There are errors that come from measuring the quantities and errors that come from the inaccuracy of the laws themselves
But how do we know that the errors are in the measuring or in the law about which we make measurements?
| Measurement errors or experimental errors can be reduced by, for example, using more accurate and more sensitive equipment; making multiple measurements and taking an average; thinking about possible sources of noise and trying to compensate for or reduce these. If you do all these things and there is still a difference between actual and expected results - and if this difference is larger than can be accounted for by the remaining sources of noise or error - then the problem is with the law or theoretical model that produced the expected results.
Usually, the "null hypothesis" is that the law or theoretical model is correct, so discrepancies in results are due to experimental error. A scientist will typically only consider modifying a law or model after all the alternative explanations of discrepancies have been ruled out.
| {
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What is the physical importance of topological quantum field theory? Apart from the fascinating mathematics of TQFTs, is there any reason that can convince a theoretical physicist to invest time and energy in it?
What are/would be the implications of TQFTs?
I mean is there at least any philosophical attitude behind it?
| In condensed matter physics, topological quantum field theories provide an effective description of (many, but not all) gapped phases of matter at low energies and long distances. A phase of matter is gapped if it costs a finite amount of energy to create any excitation above the ground state.
Examples of gapped phases of matter that admit a low-energy TQFT description include quantum Hall phases, which are described by Chern-Simons theories, and superconductors with dynamical electromagnetic fields, which are described by $BF$ theories.
Examples of gapped phases that may not have a TQFT description are so-called fracton or gapped non-liquid phases.
| {
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Will the cosmic microwave background radiation eventually not be microwaves? I was reading through this answer about the cosmic microwave background radiation, which implied that we receive it as microwaves because it's been shifted to that wavelength by the expanding universe. Did I understand that correctly?
If so, wouldn't that mean that the rate of expansion would change the wavelengths perceived? Based on our understanding, has the CMB ever not been microwaves? Will it eventually stop being microwaves? If/when it is visible light, would we be able to see it?
| Yes you read the other answer correctly. The expansion does change the wavelength and not just the perceived wavelength. Each individual photon gets stretched as its traveling through a stretching spacetime. The CMB was released back when the entire universe was filled with a hot opaque plasma which emitted light as a black body radiation in a similar way as stars do. There was originally a lot of visible light. It will eventually cool off further than microwaves if the universe expansion continues which at this point most theories predict it will.
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In a solar cell when an electron is freed due to light in the depletion layer, why does it move to N-type layer even though it is negatively charged? Even though in a solar cell the N-type layer is negatively charged why do the electrons from the depletion layer get attracted to it?
| I think the main misunderstanding is that the n-layer is neutrally charged overall so it does not attract anything.
Only in the depletion region is there an electric field. The electric fields acts to sweep electrons towards the n-layer and holes toward the p-layer.
The depletion region partially overlaps both n and p doped layers. The part in the n-side is positively charged because the region is depleted of electrons which were once bound to the n-type dopants. This makes the n-side of the depletion region positive which attracts electrons towards the n-layer.
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Friction force on a rotating car I can't understand how the frictional force is the reason for both the tangential and the centripetal acceleration in a car that is rotating. I found some useful answers on how it provide one of them, but I can't visualize how it provides both at the same time.
| a) Tangential acceleration:
By definition, frictional force opposes the direction of motion. For the part of the car's tire that is in contact with the ground, the friction acting is static friction unless the car is skidding (in the case of skidding, the point in contact with the ground moves along the surface). Because this point on the rim of the tire remains motionless, you can see that the frictional force opposes the direction of velocity and is tangential.
b) Centripetal accelaration:
If frictional force did not exist, the wheels of the car would not turn. Through the propulsion of the engine, the car would simply slide. So, when the car wheels turn (as a result of frictional force), a point chosen along the rim of the wheel displays circular motion, which means that there must be a centripetal acceleration.
| {
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Series combination of springs When a spring mass system is connected vertically with two massless springs in series whose spring constants are $k_1$ and $k_2$ to a block of mass $m$ we know that equal forces act on both the springs. Let that force during oscillations be $F$.
When we calculate effective spring constant $k_s$, why don't we say the net force acting on the system is $2F$?
Finding net force acting on the above system:
When the block is attached,the system attains equilibrium position through displacements $x'_1$ and $x'_2$.
At equilibrium:
$2F'=mg$(Where $F'$ is magnitude of spring force initially by each spring)
So, $k_1x'_1+k_2x'_2=mg$ (equation 1)
When the system is pulled down it makes oscillations,now:
Total elongation be $x$
Elongation in spring 1 be $x_1$ and elongation in spring 2 be $x_2$.
Total spring force $= -k_1x'_1-k_2x'_2-k_1x_1-k_2x_2$
Total forces acting on the system $= -k_1x'_1-k_2x'_2-k_1x_1-k_2x_2+mg = -mg-k_1x_1-k_2x_2+mg$
(from equation 1)
So, total force $= -k_1x_1-k_2x_2 = F_1+F_2=2F$(as we know that both forces are equal)
So net force acting on the system is $2F$
The way I calculated effective spring constant is:
$x=x_1+x_2$
$2F/k_s = F/k_1 + F/k_2$
$2/k
_s = 1/k_1 +1/k_2$
But that is not a correct equation.
What's wrong in taking net force acting on system as $2F$.
| ok so the first condition for the series spring system is :
The spring force in the entire system is the same i.e the tension in the springs is the same
therefore
$k_1x_1=k_2x_2$
$k_e(x_1+x_2)=k_1x_1$
where $k_e$ is the equivalent spring constant
so if the system is vertical let gravitation force, mg, be $F$
hence
$x_1=F/k_1$ ,$x_2=F/k_2$ and $x_1+x_2=F/k_e$
$x_1+x_2=F/k_1+F/k_2=F/k_e$
$\Rightarrow 1/k_e=1/k_1+1/k_2$
Now let's deal with your question,
So I recommend You look into the free body diagram again. The tension in the entire system is equal to $F$
If u are facing troubles you can also try relating this with electricity resistor circuits.
In a series connection, the current is the same through the resistors similarly in a series spring connection the tension is the same throughout. And in the case of a parallel connection the current splits in the branches but with each branch having the same potential difference, similarly, in a parallel spring system the two springs experience different tensions but extend/ contract equally
For further reference:
I have some trouble with springs in series
and
https://physics.stackexchange.com/questions/311111/why-springs-in-series-experience-equal-force#:~:text=In%20series%20circuits%3B%20the%20current,force%20must%20be%20the%20
| {
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Angular momentum commutation relations The operator $L^2$ commutes with each of the operators $L_x$, $L_y$ and $L_z$, yet $L_x$, $L_y$ and $L_z$ do not commute with each other.
From linear algebra, we know that if two hermitian operators commute, they admit complete sets of common/simultaneous eigenfunctions. The way I understand this statement is that the eigenfunctions of both operators are the same. So, if that were the case, that would mean that $L_x$ has the same eigenfunctions as $L^2$. The same goes for $L_y$ and $L_z$. That would mean that $L_x$, $L_y$ and $L_z$ all have the same eigenfunctions, which doesn't seem to be true since they do not commute with each other. How is this resolved?
| This is possible precisely because $L^2$ is degenerate: for an eigenvalue $l(l + 1)$, it has an eigenspace of dimension $2l + 1$ (i.e., it has this many linearly independent eigenstates). The choice of basis in a degenerate eigenspace is not unique -- thus explaining how it can be that the eigenstates $L^2$ shares with $L_x$ are not the same as the ones $L^2$ shares with $L_y$.
It is possible to simultaneously diagonalize $L^2$ and $L_x$, or $L^2$ and $L_y$, or $L^2$ and $L_z$. But the most you can do for $L^2$, $L_x$, $L_y$, and $L_z$ together is simultaneous block diagonalization: $L_x$, $L_y$, and $L_z$ are nonzero only in each $(2l + 1) \times (2l + 1)$ block corresponding to each $L^2$ eigenvalue. And $L^2$ commutes with all of them because it is proportional to the identity matrix in each block.
| {
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What is light, a wave or a particle or A wave-particle? What is light?
And how do we know that light is an electromagnetic wave?
I asked my teacher and he said that when you place a compass in light's path, the needle of the compass rotates. Which I think is not a valid answer and thats not what actually happens when we place a compass in path of a light.
| Light is a range of frequency of electromagnetic waves which our eyes can detect.
Light consists of photons, which is a weightless particle. Light is a medium through which energy can be released.
| {
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Reconstructing wavefunction from the density matrix Say I have a state, $$| \Psi \rangle = \frac{1}{\sqrt 2} \left( | 0 \rangle + \exp( \text{i} \phi ) | 1 \rangle \right) = c_{0} | 0 \rangle + c_{1} | 1 \rangle.$$
Now I construct the density matrix (DM), $$\hat \rho = | \Psi \rangle \langle \Psi | = \frac{1}{2} \left( | 0 \rangle \langle 0 | + \exp( - \text{i} \phi )| 0 \rangle \langle 1 | + \exp( \text{i} \phi ) | 1 \rangle \langle 0 | + | 1 \rangle \langle 1 | \right).$$
So from the DM $\hat \rho$, I can read off $|c_{0}|^{2}$, $|c_{1}|^{2}$, $c_{0}c_{1}^{*}$, and $c_{0}^{*}c_{1}$. Basically $3$ equations and $4$ unknowns.
Is there a way to reconstruct $| \Psi \rangle$ uniquely from the DM, $\hat \rho$?
| On solving, we find:
$$\frac{1}{2}\begin{pmatrix}
1 & e^{-i\phi}\\
e^{i\phi} & 1
\end{pmatrix}= \begin{pmatrix}
|c_0|^2 & c_0c^*_1\\
c_1c_0^* & |c_1|^2
\end{pmatrix} $$
$$\Rightarrow |c_0|=|c_1|=\frac{1}{\sqrt{2}}$$
$$c_0c^*_1=\frac{1}{2}e^{-i\phi}\Rightarrow c_0=e^{-i\phi}c_1$$
$$|\psi\rangle =c_0|0\rangle +c_1|1\rangle =c_0\left(|0\rangle+\frac{c_1}{c_0}|1\rangle \right)=\frac{1}{\sqrt{2}}e^{i\chi}(|0\rangle +e^{i\phi}|1\rangle )$$
So the wave function would be unique up to phase factor $\chi$.
| {
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Is dark energy (like) normal energy? I know that the term 'dark' is used because a) dark matter does not interact with light b) we know so little about them. Dark matter I guess could be just an unexplained type of particle we don't know about. But what about dark energy?
When we say 'dark energy', are we saying that it's a different concept altogether than energy? Is it saying that there is a third energy type to potential and kinetic? Or is it just 'normal' energy that we cannot explain the origin of? Or do we just not know?
| What we know is that the mass-energy density of the universe must be much higher than expected from baryonic matter/dark matter in order to fit the accelerated expansion observed. It could be a particle. It could some gravitational feature we don't understand... The thing is, what we currently think is that the density of dark energy is constant over time. That's pretty crazy by itself, but it's even crazier if you try to figure out how to do that with some kind of particle.
The most reasonable way to get a density that doesn't change over time, is to say that it's a property of space; if you get more space, you get more "space energy". Dark matter is likely to be a particle, since it "clumps" around galaxies. But dark energy seems to be basically everywhere.
| {
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What is the exponential (or geometric) rule (or law) for uranium enrichment? Uranium ore starts at about .72% U-235...
At ~20% U-235, it is considered to be about '90% of the way' to weapons-grade uranium, which is about ~90% U-235...
Because uranium enrichment in centrifuges follows a geometric (or exponential) law...
I have read about this repeatedly when hearing about Iran's enrichment program....
Does anybody know what the 'rule' or 'equation' is for uranium enrichment...
(I am not trying to build a bomb, I swear....)
Edit: P.S.:
In the Work equation $W_extract = -T R ln(x)$ , what are T, R, and x?
I can find that equation nowhere else....
| The exponential law for enrichment is that doubling the percentage of U-235 requires the same amount of effort regardless how much there already is. Getting from 20% to 40% is just as hard as getting from 1% to 2%.
Though obviously, this only works well when the percentage is below half. Getting from 95% to 99% is a lot of work. In this regime, the law goes the other way, i.e. halving the remaining amount of U-238 requires the same amount of work, regardless how much remains.
| {
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What is granularity in particle physics? What is granularity in particle physics? It is used throughout "The ATLAS Experiment at the CERN Large Hadron Collider" http://nordberg.web.cern.ch/PAPERS/JINST08.pdf without being defined anywhere.
An example use: "high detector granularity is needed to handle the particle fluxes and to reduce the influence of overlapping events"
Resolution would simply be referred to as "resolution" so it can't be it.
| In this instrumentation review:
Next follow the detectors, whose key parameters are efficiency, speed, granularity and resolution.
The term is used and not defined further because it is descriptive of the "number of sensitive to measurement ΔV(ΔxΔyΔz) per unit volume : how many "grains" and what a "grain" is will depend on the material and design composing the specific detector.
See fig 32. That is one "grain" in the silicon detector.
| {
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According to general relativity planets and Sun bend the spacetime (explaining gravity), but does this hold true for smaller objects? According to general relativity planets and the sun bend spacetime, and that is the explanation of gravity. However, does this hold true for smaller objects, like toys, pens, etc.? Do they also bend spacetime?
| Yes, small objects such as pens, paper, tennis balls etc, all bend spacetime.
Not only does spacetime bend, warp, and stretch, but it also flows. You can see that idea in this video which is a really good visualization of gravity.
Another way to think about gravity is that gravity is a difference in proper time. When an object falls, it falls because there is a gradient difference of proper time. You can see that idea in this video.
I hope that helps.
| {
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Muon $g-2$ experiment: is there any theory to explain the results? The nature of the experiment has been discussed here, but my main question is this: is there any theory that has predicted the results of this experiment or are we completely clueless about what's happening? In other words, have we come up with a new hypothetical interaction that could explain the results?
| Yes there is a theory that explains the results ... the Standard Model. In other words, the claim is that the Standard Model already is consistent with the experimental data, and the original "prediction" was calculated wrong.
Check out the paper published in Nature together with the muon g-2 results, or the writeup at popular level.
| {
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Why do images not appear inverted when looking directly through a pinhole camera? I understand that the way light takes through a pinhole creates an inverted image on a surface behind the pinhole. I remember this effect from school experiments, it's also described in this wikipedia article. I punctured a piece of paper and looked through it (instead of watching the reflection), the image appeared as normal to me. Why is that? Why doesn't the scene appear upside down when looking through the hole?
| I assume that what you are describing is peering through a pinhole such that the field-of-view you get is smaller than normal. That is, I assume you describe a looking through the pinhole that comes with "tunnel-vision".
So, I assume that what you are describing is peering through a pinhole such that for the light to traverse the distance from the pinhole to the point where the light transits into the eye is like traversing the length of a short tube.
The distance from your eye to the pinhole is very short. The closer your eye is to the pinhole, the larger the field-of-view you obtain (but some reduction of field-of-view is inevitable).
So: think of peering through the pinhole directly as peering through a tube. When you are peering through a tube you are not surprised that your field-of-view is smaller, and you are not surprised that what you see is still the right side up.
| {
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Why is $ \text{div}(\vec{r}/r^3) = 0 $? Let's begin with some context. I was reading Griffiths's introduction to electrodynamics.
This makes sense given all the content one can find online in order to understand visually the divergence.
for example: https://youtu.be/rB83DpBJQsE or https://en.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/divergence-and-curl-articles/a/intuition-for-divergence-formula
So the problem arises when one tries to understand conceptually (or visually) the fact that $$ \nabla\cdot\frac{\hat{r}}{r^2} = 0, \qquad r\neq 0. $$
I've done the calculation where:
$$\nabla \cdot \frac{\hat{r}}{r^2}=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{1}{r^2})=\frac{1}{r^2}\frac{\partial}{\partial r}(1)=0.$$
That's fine but I find it impossible to understand it visually. Take for example this picture.
$ \vec{r}/r^3 = 0 $ " />
Having in mind the intuitive explanation of the divergence, like the one showed in the links, is easy to see that the flow through a small region in the image above, not containing the origin, is not equal to zero. Furthermore, it is obvious that the divergence is not zero because it doesn't look anything like the next image whose divergence is zero.
(In the image above the vector field isn't changing in space.)
Finally, I know that using the divergence theorem one can show mathematically that:
$$\int \nabla\cdot\vec{r}/r^3 d\tau=\int \frac{\vec{r}}{r^3}\cdot d\vec{a} .$$
And if we consider the surface of a sphere of any radius, centered at the origin, one can show that $$\int \frac{\vec{r}}{r^3}\cdot d\vec{a}=4\pi.$$
Again, that's okay but doesn't help me to understand visually why $$ \nabla\cdot\frac{\vec{r}}{r^3} = 0~? $$
| As @Uyttendaele comments(1)
\begin{equation}
\boxed{\:\:
\boldsymbol{\nabla}\boldsymbol{\cdot}\left(\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\,\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)\boldsymbol{=}4\pi\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right)\:\:\:\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}}
\tag{A-01}\label{A-01}
\end{equation}
see my answer here Divergence of $\frac{ \hat {\bf r}}{r^2} \equiv \frac{{\bf r}}{r^3}$, what is the 'paradox'?
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$
(1)
$''\texttt{Introduction to Electrodynamics}''\texttt{ D.J.Griffiths, Edition 3 or 4, Sect.1.5.3, eq.(1.100)}$
| {
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What is momentum? Momentum tells you the mass of the object and how fast it is going right? So if I have a 2 kg ball moving at 2 m/s, then the ball has 4 kg⋅m/s of momentum. My question is why do we multiply mass and velocity to get momentum. (From the example above) Why cant we just say the ball is 2 kg moving at a speed of 2 m/s and that is momentum. Why do we have to multiply it?
| In the context of classical mechanics, and especially rigid body motions momentum can have the following interpretation. The statements below might seems a bit circular, but they work.
Momentum is the quantity needed to completely remove all movement from a rigid body or a point mass.
Specifically, momentum is a vector quantity applied along an infinite line in space (the axis of percussion). An impulse (a short lived force) of equal and opposite magnitude and direction, but along the same line applied on a moving body will instantaneously stop all motion(s). There are specially cases than more than one impulses needed to be applied for this to happen, and in those cases momentum is defined from the net impulse acting on the body.
In this context the units of momentum are Newton-second [$\mathrm{N\,s}$] representing a force action (force over time).
The resulting change in velocity $\Delta \vec{v}$ due to an impulse $\vec{J}$ is $$ \Delta \vec{v} = \frac{1}{m} \vec{J}$$ This is true for a point mass, or for the center of mass of a rigid body. The mechanics of how an impulse $\vec{J} = \int \vec{F} \;{\rm d}t$ is defined does not matter here.
You can think of this backwards also. Take a stationary object and apply an impulse equivalent to the desired momentum, in magnitude, direction and location and the body with move after the application in said way.
Example(s):
*
*A point mass of $m=2\,\text{kg}$ is moving with velocity (vector) $\vec{v} = \pmatrix{5 & 0 & 0}\;\text{m/s}$
An impulse of $\vec{J}=\pmatrix{-10 & 0 & 0}\;\text{Ns}$ causes the following step change in velocity $$ \Delta \vec{v} = \tfrac{1}{m} \vec{J} = \pmatrix{-5 & 0 & 0} \; \text{m/s} $$
The final velocity is thus $$\vec{v}^\star = \vec{v} + \Delta \vec{v} = \pmatrix{0 & 0 & 0}$$ If the final velocity is zero, it means the impulse applied must have been equal and opposite to the momentum of the body. So the momentum of the body was $\vec{p} = -m \Delta{v} = - \vec{J} = \pmatrix{10 & 0 & 0}\;\text{Ns}$
*A solid disk (rigid body) of mass $m=2\;\text{kg}$ is rolling on a horizontal plane with the velocity of the center $\vec{v} = \pmatrix{5 & 0 & 0}\;\text{m/s}$. The radius of the disk is $R=1 \;\text{m}$ mass moment of inertia of the disk is $I = \tfrac{m}{2} R^2 = 1\;\mathrm{kg\, m^2}$. The rotation of the disk is $\vec{\omega} = \pmatrix{0 & 0 & -5}\,\text{rad/s}$. The motion response of the disk that rolls must maintain that $\Delta \vec{v} = \Delta \vec{\omega} \times \pmatrix{0 & R/2 & 0}$.
An impulse of $\vec{J} = \pmatrix{-10 & 0 & 0}\;\text{Ns}$ applied at a distance $b = \frac{I}{m R}= \tfrac{1}{2}R = 0.5\;\text{m}$ above the center of mass has the following affect in the motion
$$\begin{aligned}
\Delta \vec{v} & = \frac{1}{m} \vec{J} & \Delta \vec{\omega} & = I^{-1} ( \vec{b} \times \vec{J} ) \\
& = \pmatrix{-5 & 0 & 0}\;\text{m/s} & & = \pmatrix{0 & 0 & 5} \; \text{rad/s}
\end{aligned} $$
with the final motion all zero for both rotation and translation.
So the momentum of the rolling disk is $$\vec{p} =-m \Delta \vec{v} = -\vec{J} = \pmatrix{10 & 0 & 0}\;\text{Ns}$$ and the percussion axis being horizontal through a point $$\vec{b} = \pmatrix{0 & 0.5 & 0}\;\text{m}$$ above the center of mass.
| {
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Is the distance between all points on the event horizon zero? These are the Kruskal-Szekeres coordinates of Schwarzschild spacetime:
It is isomorphic to the split-complex plane. But on the split-complex plane the distance between all points on the null diagonals, as measured by the norm of the difference between two spacetime points, is zero.
Does this imply that the distance between all points on the event horizon is also zero?
| If this were a 1+1 dimensional spacetime then you'd be correct, but there are two other spatial dimensions not shown on the chart. Each point on the chart is really a sphere whose radius is a function of $X^2-T^2$.
The event horizon is a 3-cylinder (direct product of a 2-sphere and a line), but with a degenerate metric along the length of the cylinder. If you put $(θ,\phi,z)$ coordinates on it in the obvious way, then the metric is $ds^2 = r_s^2 (dθ^2 + \sin^2 θ\,d\phi^2) \; [+\;0dz^2]$. The distance between points at the same position on the horizon at different "times" is zero, but between different points it's nonzero and spacelike.
Note that these coordinates are not the Schwarzschild coordinates restricted to $r=r_s$. Schwarzschild coordinates are singular at $r=r_s$ and don't cover the horizon. But these coordinates are Eddington-Finkelstein infalling coordinates restricted to $r=r_s$, and with $t$ renamed to $z$.
| {
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Why does there being no finite dimensional unitary representations of the Lorentz group lead to QFT? In Schwarz's Introduction to QFT and the standard model they say that because there are no finite-dimensional unitary representations of the Lorentz group we must have fields and not particles.
However, for a single particle quantum state the basis of the Hilbert space is still infinite and so this would suggest we would be able to find a lorentz invariant single particle theory.
| There are a number of reasons why the single particle theory of QM can't be extended into a quantum theory of fields and this is another one. Generally speaking we have to move into a picture where particles can be created and annihilated. This is the multi-particle picture.
Freeman Dyson discusses some of the reasons in his Advanced QM.
The reason why there are no finite dimensional unitary irreps of the Lorentz group is because the Lorentz group is not compact. All non-compact groups have only infinite-dimensional unitary irreps, assuming that they have any.
| {
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How to inject the maximum acoustic power into a block of granite? I know that we can use transducers that are glued to a surface to achieve this. If I want, for example, to have 200 watts of actual acoustic power in the audible range in the granite, is a transducer the easiest cheapest way? Are there other methods?
| *
*Use an industrial size vibrator motor bolted to the granite. The motor shaft has an eccentric mass on it which will vibrate at the rotational frequency of the motor. Not enough power? Use more than one.
*Use a plate compactor if the process can be conducted outdoors.
*Use a jackhammer, or its smaller cousin, an electric demolition hammer. If you don't want a hole in the granite, cut off the pointy end of the bit and weld a thick flat plate to spread the impact over a larger area.
I have no idea if any of these methods will deliver 200 watts of power to the granite, but the power will be in the acoustic range and the cost is low enough to try it and take measurements. Not to mention the fun factor.
| {
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Trouble showing from first principles the equation describing the motion of a water drop through a cloud If we had a water drop, initially of mass $m_0$ and speed $v_0$, falling through a cloud for a time
T and during this time its mass increases at a constant rate $\alpha m_0$, through accretion of
water vapour initially at rest.
How would we show from first principles that, neglecting air resistance, the equation describing
the motion of the water drop through the cloud is : $$d[(1 + αt)v]/dt = g(1 + αt)$$
I 'm trying to apply the technique we get from the rocket equation where we consider the change in momentum $\Delta P$, and consider the momentum before and after - but all I run into is $$ -m_0g \delta{t} = m_0(1+\alpha \delta{t})\delta{v} - \alpha \delta{t}v $$ which I seemingly can't manipulate any further.
Any pointers? Thanks
| Starting with the second law of dynamics:
$$ \frac{dp}{dt} = mg$$
We have to consider that here the mass is variable.
$$p(t) = m(t)v(t) = m_0(1 + \alpha t)v $$
Then:
$$ \frac{dm}{dt}v + \frac{dv}{dt}m = mg$$
Since the rate of change of the mass is $\frac{dm}{dt} = m_0\alpha$
$$ m_0 \alpha v + m_0(1 + \alpha t)\frac{dv}{dt} = m_0(1 + \alpha t)v$$
Finally we have:
$$ \frac{d[(1 + \alpha t)v]}{dt} = (1 + \alpha t)g$$
| {
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Perception of simultaneous events I have a two-fold question about the light-cone structure of spacetime, specifically about space-like separated events.
As far as I understand it, any two events that happen at the same time in a given reference frame are space-like separated. If so, any two simultaneous events occurring on my arm and leg are space-like separated.
What confuses me is this:
If I am not mistaken, all the events that we perceive are in our past light-cones. This is because we only perceive events that have emitted light that has reached us, and thus has causally affected us. If so, what happens when I look at my arm and leg? It seems to me that for any two simultaneous events A and B, where A occurs on my arm while B occurs on my leg, A and B must be time-like (since I have perceived them) and not space-like separated. In a nutshell, if simultaneous events are space-like separated, while I only perceive time-like separated events, how can I perceive simultaneous events?
And second, could someone recommend me an article or a book explaining how the light-cone structure relates to ordinary perception? I struggle to connect the light cone structure to real-life events, so some kind of graph or an explanation of this would be useful.
| You are correct to think that all the events we perceive are in our past light cone.
We perceive, more or less simultaneous events, since for all intents and purposes, for us, light is more or less instantaneous.
To actually see how Einstein goes about constructing a synchronisation of clocks - that is simultaneity - you could try his original paper on special relativity, On the Electrodynamics of Moving Bodies. It is, unlike many modern papers, short and uses only school level algebra and quite straight forward to follow.
You can also try the first chapter of d'Inervo's book on general relativity to see how simultaniety is to be understood. He explains it on the basis of Bondi's k-calculus, and which has nothing to do with calculus. Here k stands for the radial Doppler factor.
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Which types of strain tensor are positive definite? I am taking a look at different types of strain tensor. Specifically, I am thinking about if the infinitesimal strain tensor
\begin{align*}
\epsilon_{ij} = \frac{1}{2} (\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i})
\end{align*}
is positive-definite. I have Google searched some resources, and one of them says it is positive-definite. However, I think that it is not always positive-definite, as in the one-dimensional trivial case, if $\partial u/\partial x$ is negative, then it will not be positive-definite. Other sources say that other strain tensors, like the Lagrangian strain tensor, are positive-definite. I am not sure which types of strain tensor are positive-definite and also the implications if so. (I am thinking about the strain surface being an ellipsoid or not.) Any ideas will be greatly appreciated!
| I think the material you found is just wrong. A trivial counter example is zero displacement = zero strain, which is not positive definite.
Nor do I see where such a property would be particularly useful when calculating anything.
It's a different story when talking about the other tensors we deal with (such as the elasticity tensor), where it definitely is a important property.
Looking at reputable resources (wikipedia, https://www.continuummechanics.org/) i see no mention of such claims when they present different strain measures.
| {
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Euler Equation in General Relativity For a perfect fluid the stress-energy tensor is $T^{\mu \nu} = (\rho + p)u^{\mu}u^{\nu} + pg^{\mu \nu}$, thus the equations of motion read:
$$0 = T^{\mu \nu}_{\; \; \; ;\nu} = (\rho + p)_{;\nu}u^{\mu}u^{\nu} + (\rho + p)(u^{\mu}_{\; ;\nu}u^{\nu} + u^{\mu}u^{\nu}_{\; ;\nu} ) + p_{;\nu}g^{\mu \nu}$$
Now I want to project this equation using $P_{\alpha \mu} = u_{\alpha} u_{\mu} + g_{\alpha \mu}$. According to the book's calculations the following term: $$P_{\alpha \mu}(\rho + p)_{;\nu}u^{\mu}u^{\nu}$$ vanishes and I don't understand why.
How does this term vanish?
| Just, to make it more clear (as mentioned in the comment), first prove this
$${P_{\alpha \mu }}{u^\mu } = ({u_\alpha }{u_\mu } + {g_{\alpha \mu }}){u^\mu } = {u_\alpha }\underbrace {{u_\mu }{u^\mu }}_{ = - 1} + \underbrace {{g_{\alpha \mu }}{u^\mu }}_{ = {u_\alpha }} = 0,$$
and hence, you get this
$${P_{\alpha \mu}}(\rho + P){ _{;\nu }}{u^\mu }{u^\nu } =\underbrace{{P_{\alpha \mu }}{u^\mu }}_{ = 0}(\rho + P){_{; \nu }}{u^\nu } = 0.$$
| {
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Derivation of proper time with respect to time I’m confused on how the author got this answer.
He started with this:
$$\frac{d(\text{proper time})}{dt}=\frac{\sqrt{dt^2-dx^2}}{dt}
=\sqrt{1-v^2}$$
I don’t get how this is solved, specifically where the $1$ came from in $1-v^2$.
| The author is doing typical "physics mathematics", so it may be considered to be a little sloppy to a pure mathematician. ;)
The $1$ comes from $\frac{dt}{dt} = 1$. Perhaps it will make more sense if we do the derivation in a more formal way.
It's common to denote proper time using, $\tau$, the Greek letter tau. Note that the author is using natural units where $c$, the speed of light, equals $1$.
Let
$$(\Delta\tau)^2 = (\Delta t)^2 - (\Delta x)^2$$
Dividing through by $(\Delta t)^2$,
$$\left(\frac{\Delta\tau}{\Delta t}\right)^2 = \left(\frac{\Delta t}{\Delta t}\right)^2 - \left(\frac{\Delta x}{\Delta t}\right)^2$$
or
$$\left(\frac{\Delta\tau}{\Delta t}\right)^2 = 1 - \left(\frac{\Delta x}{\Delta t}\right)^2$$
Taking limits as $\Delta t \to 0$,
$$\left(\frac{d\tau}{dt}\right)^2 = 1 - \left(\frac{dx}{dt}\right)^2$$
But $\frac{dx}{dt}=v$, so
$$\left(\frac{d\tau}{dt}\right)^2 = 1 - v^2$$
Taking square roots,
$$\frac{d\tau}{dt} = \sqrt{1 - v^2}$$
which is the desired result.
Note that
$$\gamma = \frac{1}{\sqrt{1 - v^2}}$$
is the Lorentz factor.
| {
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Isotopy class of spacetime We know that spacetime is an orientable manifold:
Can spacetime be non-orientable?
But supposing that spacetime is an orientable closed 2D surface, one might envision a variety of non-equivalent solutions in the following sense:
Given a 2D strip, by one rotation(twist), one can create a Moebius strip (it's non-orientable so discarded), but by another rotation (360 degrees) one finds an orientable 2D surface.
Suppose one can repeat this for arbitrary many times(integer multiples of 360 degrees), then one has a countable set of possible orientable spacetimes
Is there any way to determine which spacetime relates to ours(2D), given the fact that Einstein's Field equations are pretty much open-minded regarding the topology of spacetime?
Can I find any physical observable in QFTs on such spacetime that is related to the number of turns in general?
If not, can one hypothetically say that the real spacetime is a superposition of all these possibilities?
Is it possible to extend the idea of twist to 3D hypersurfaces?
|
Is there any way to determine which spacetime relates to ours(2D)
No: all these spaces are (globally) homeomorphic. They are not isotopic, but that is a property of embedded manifolds, not of all of spacetime, and general relativity depends only on the intrinsic geometry of spacetime.
Can I find any physical observable in QFTs on such spacetime that is related to the number of turns in general?
For the same reason as the previous, not as such. The tangent bundle and all derived tensor bundles and principal bundles will be the same.
Is it possible to extend the idea of twist to 3D hypersurfaces?
A very direct generalization would be too start out with a solid cube, and identify two opposite faces via one of their four orientation preserving isometries. This should give you three different spaces. Embedding this in $\mathbb R^3$ (though not isometrically) you can twist it as many times as you want, but it will be homeomorphic to one of the three. Likewise for other solid prisms or the solid cylinder (the latter having infinitely many non-equivalent twists).
| {
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Covariant derivative of the spin connection I wish to compute $$[\nabla_{\mu}, \nabla_{\nu}]e^{\lambda}_{~~a}. $$
To do so, I make use of $\nabla_{\nu}e^{\lambda}_{~~a} = \omega_{a~~~\nu}^{~~b}e^{\lambda}_{~~b}$, so that I may write
$$\nabla_{\mu}\nabla_{\nu}e^{\lambda}_{~~~a} = \nabla_{\mu}(\omega_{a~~~\nu}^{~~~b})e^{\lambda}_{~~b}+\omega_{a~~\nu}^{~~b}~\omega_{b~~~\mu}^{~~c}e^{\lambda}_{~~c}$$In the end, I intend to anti-symmterize in $\mu, \nu$ to get the desired object. Therefore, I would like to know what is the covariant derivative of the spin-connection $\omega$ in order to finish my computation. Is $\omega$ a scalar, a vector or what? How do you decide? Can someone help?
| Note that from the relation $e^\lambda{}_{a;\nu} = \omega_a{}^b{}_\nu e^\lambda{}_b$ you give you can deduce by contracting with $e_{\lambda c}$
$$e^\lambda{}_{a;\nu}e_{\lambda c} = \omega_{ac\nu}$$
Note, however, that I am using the definition of the covariant derivative that takes tetrad indices $a,b,c$ as mere labels and thus the covariant derivative of the tetrad leg vector field is
$$e^\lambda{}_{a;\nu} = e^\lambda{}_{a,\nu} + \Gamma^\lambda{}_{\kappa\nu} e^\kappa{}_{a}\,.$$
That is, the covariant derivative is covariant with respect to coordinate transforms but not wrt vielbein transforms.
In other words, $\omega_{ab\nu}$ transforms as a tensor in the $\nu$ index and as such it will have the corresponding Christoffel connection term in the covariant derivative. This should help you derive the commutator (relation between the Riemann tensor and $\omega$) as desired.
| {
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Why does silver paper work as Faraday cage for x-rays and does not shield just a simple phone call? I believe the topic of the present question fully explains the issue I would you to talk with me.
It is well known that the higher the electromagnetic frequency, the more difficult is its shielding.
I have seen many x-rays devices at university covered just with silver paper in order to prevent that any radiation outlay could affect the experimentalists.
So I tried to isolate my phone with a silver paper, but still I could call it with another phone.
Why do radio waves overtake that shielding while x-rays do not?
| I personnally wrapped my phone with silver paper (aluminium foil to be precise) and the phone was shielded. However, the phone has to be completely covered in foil for it to work, a few mm hole will leak enough for the experiement to fail.
| {
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What are the weaknesses, if any, of the relational interpretation of quantum mechanics? Carlo Rovelli's 1996 relational interpretation of quantum mechanics (RQM) seems to solve many of the quandaries of traditional theories, including the Copenhagen interpretation (what privileges the observer? Why does he/she instigate wavefunction collapse); the many-worlds interpretation (an infinity of unobservable universes); and spontaneous collapse theories such as that of GRW.
However, there seems to have been limited academic discussion of the theory. Is this due to a shortcoming of the theory, or has there simply not been enough time for it to disseminate?
| It's fair to say that the discussion about RQM has begun to gain momentum in recent years. I have the impression that it has increased in popularity among physicists and philosophers of physics.
But to make contact with the title of your question, there certainly are substantial open problems. Inter alia, some critics say that it leads to solipsism, and that it suffers from a sort of preferred-basis problem. It's also slightly unclear how the story extends to more general interaction settings (say, continous interactions). If you're interested in diving a littler deeper into some of these issues: here and here are some recent critiques, which Rovelli has replied to here and here, respectively.
| {
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How does a copper ring levitates above a AC supply primary coil? Doesn't AC current gives decreasing current and so decreasing magnetic fields or flux. This should then attract the copper ring above primary coil. When magnitude of current increases in AC then due to Lentz law as we know ring will get repelled. But what about the decrease of current. Dosen't current decreases like it increases in AC. What is the reason of that lift. Even when anyone blocks ring at a place it still try to lift. Why??? I am very confused on that topic. I am trying to find it hard from everywhere and thinking myself too. But it seems like how?why? Why only lift above?
| I was wrong in my first answer. It is slightly more subtle than I thought.
Perhaps it has to do with the shape of the exciting field and its oscillation frequency.
If the field diminishes quickly (spatially) with the distance to the exciting coil (e.g. if the field diverges strongly), the second coil could be in the air during the attracting phase, so that it might not see said "attractive phase" as strongly as he felt the repulsive phase.
If the frequency of the oscillator, the shape of the field, and the mass of the second coil (the ring), are right, and the attractive phase always coincides with the ring being in the air, then it could work.
So, if I have got it right this time, it means that it is not a purely electromagnetic problem, but that is also has a mechanical aspect.
One can picture it naively as the exciting field "joggling" with the second coil (the way a ping-pong player bounced a ball with a paddle).
| {
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How can parallel rays meet at infinity? I found that in every book (till my 12th) it is written that, in concave mirror, when object is at focus, then reflected rays will be parallel and they meet at infinity to form a real image.
But, as we know, parallel rays never meet. Then, does this mean that all books are wrong ? If not, then why?
| When physicists say something "goes to infinity", what they mean is "as you take the limit, this value gets bigger and bigger without any bound, and will eventually exceed any number you choose".
In the standard system of real numbers (which is used for most things in classical physics), infinity isn't actually a number; it's more like a notational shorthand. So a more technically accurate way to say this would be:
As the object gets closer to the focus, the image (where the rays meet) gets farther and farther away, without any bound. You can make the image be as far away as you want, by bringing the object close enough. When the object is exactly at the focus, the rays are parallel, and thus never meet.
"The rays meet at infinity" is just shorthand for this.
EDIT: As Don Thousand points out in the comments, situations like this are sometimes handled in projective space, where infinity has a concretely defined meaning, with the projective real numbers (or projectively extended real numbers), where infinity is in fact a number. But in my experience, introductory texts tend to avoid this in favor of the Euclidean space and real numbers students are used to.
| {
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Phase difference approximation I'm sitting and trying to solve the equation of the phase difference given by:
$\Delta \phi = k (\sqrt{a^2+d^2} -d) \approx \frac{ka^2}{2d}$
Where $a$ is the size of an aperture and $b$ is the distance of the point at the aperture's center as shown in the figure below.
I'm not a math expert here, so I wondering if anyone can explain the approximation that has been done above here.
Reference:
Applications of Classical Physics by Roger D. Blandford and Kip S. Thorne - Chapter 8 - Diffraction
| After fixing the typo in your equation you can use the binomial theorem
$$
\sqrt{1+x}= 1+\frac 12 x+\ldots
$$
to get
$$
k(\sqrt{a^2+d^2}-d)= k\left(d\left(1+\frac{a^2}{d^2}\right)^{1/2}-d\right)\\
= k\left(d\left(1+\frac 12 \frac{a^2}{d^2}+\ldots\right)-d\right)\\
=k\frac{a^2}{2d}+\ldots
$$
| {
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Energy of Hydrogen Atom (Electron vs Proton) In many textbooks, energy changes of the hydrogen atom are attributed to the electron transitioning between energy levels. However, the energy itself is that of the whole system (proton+electron) so how can we attribute its changes to the electron? what's preventing us from attributing these changes to the proton??
| THe protons are reside inside the nucleus . Hence they arent the ones who can transitions between different orbits . The thought of protons leaving the nucleus is just a vague thought . That would just mean that we can make ions by loss and gain of protons too which we all know is not possible
| {
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In the theory of special relativity speed is relative so who decides which observer’s time moves slower? If for example we have 2 people one moving in speed v relative to the other, according to special relativity the time passing for the moving person is slower than for the stationary person. However from the moving person’s prospective he is stationary and his friend is moving so time should move faster for him. What’s going on?
| They are both right.
But, then, what happens if they meet? Well, in order to meet, you have to accelerate. And if one is accelerating and the other is not, the symmetry is broken.
If they both accelerate equally, they'll end up agreeing anyhow.
Under relativity, events can be separated by spacelike or timelike curves. Any two world lines will agree that things on timelike curves occur in a specific order, and for any two things seperated by spacelike curves, you'll find wordlines that consider them one to be before the other, vice versa, or "at the same time".
That is, barring exotic geometry. Relativity ends up depending on the geometry of your universe; if the universe is flat and boring, the above statements work. If your universe's geometry contains wormholes, or wraps around, things get more interesting.
| {
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Cosmology - Confusion About Visualising the Universe as the Surface of a 3-Sphere Consider the FRW metric for the Universe in the form found in many standard cosmology textbooks:
$$ds^2 = -dt^2 + a(t)^2\left(\frac{dr^2}{1-Kr^2}+r^2(d\theta^2 + \sin^2\theta d\phi^2)\right)$$
I am confused as to what $r$, $\theta$ and $\phi$ represent in this formula. For example, some texts introduce this topic by considering a 2-sphere as opposed to the 3-sphere that is described when $K=1$ in the formula above. For the 2-sphere we have the spatial line element:
$$dl^2 = \frac{dr^2}{1-r^2}+r^2d\theta^2,$$
where $r$ is the distance of a point on the surface of the 2-sphere from the $z$ axis, and $\theta$ is the angle that $r$ makes with the positive $x$-axis. It occurs to me that we have essentially used cylindrical coordinates to describe points on the surface of the sphere in this case. We may then associate $r$ with $\sin \chi$, where $\chi$ is the angle that the position vector of a point on the sphere makes with the positive $z$ axis.
My trouble comes when we then extend this argument to the 3-sphere. What exactly do the parameters now represent? To illustrate my problem: suppose we wish to calculate the volume of a sphere of radius $R_0$ that exists on the surface of the 3-sphere ($K = 1$ Universe). How would we do that using this metric? The volume element would be easy enough to write down, but in order to perform the integration we would need to know what limits to place on $r$, $\theta$ and $\phi$. This is an impossible task if one does not understand the physical significance of the parameters in this more general case.
| A common elementary coordinatization of the $2$-sphere of radius $R$ uses two angles $(\theta,\phi)$ as (co)latitude and longitude. If we embed the sphere into $\mathbb R^3$, then the points $(x,y,z)$ on the surface take the form
$$\pmatrix{x\\y\\z} = \pmatrix{R \sin(\theta)\cos(\phi)\\R\sin(\theta)\sin(\phi)\\ R\cos(\theta)}$$
If we restrict $\theta \in (0,\pi)$ and $\phi\in (0,2\pi)$, then this constitutes a coordinate chart which covers all of the $2$-sphere except for the poles and the line $\phi=0$ which connects them. In this chart, the metric takes the form
$$\mathrm ds^2 = R^2 (\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2)$$
An alternative approach is the following. Rather than using the polar angle $\theta$ as a coordinate, we can choose to use the distance from the $z$-axis, given by $R\sin(\theta)$, which we will call $r$. Equivalently, $r$ is the circumference of the circle centered at the north pole divided by $2\pi$. Note that we can only cover the northern (or southern) hemisphere of the sphere in this chart, but that's okay. Embedding the $2$-sphere in $\mathbb R^3$, we would have
$$\pmatrix{x\\y\\z} = \pmatrix{r \cos(\phi)\\r\sin(\phi) \\ \sqrt{R^2-r^2}}$$
In this chart, the metric takes the form
$$\mathrm ds^2 = \frac{1}{1-kr} \mathrm dr^2 + r^2 \mathrm d\phi^2,\qquad k\equiv \frac{1}{R}$$
This should look familiar.
The extensions to the $3$-sphere are straightforward. The "spherical coordinate" embedding uses three angles $\psi,\theta,\phi$ and takes the form
$$\pmatrix{x\\y\\z\\w} = \pmatrix{R\sin(\psi)\sin(\theta)\cos(\phi)\\R\sin(\psi)\sin(\theta)\sin(\phi)\\R\sin(\psi)\cos(\theta)\\R\cos(\psi)}$$
Defining $r\equiv R\sin(\psi)$, this becomes
$$\pmatrix{x\\y\\z\\w}=\pmatrix{r\sin(\theta)\cos(\phi)\\ r\sin(\theta)\sin(\phi) \\ r\cos(\theta)\\ \sqrt{R^2-r^2}}$$
and in this chart, the metric takes the form
$$\mathrm ds^2 = \frac{1}{1-kr} \mathrm dr^2 + r^2(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2)$$
where once again $k\equiv 1/R$.
My trouble comes when we then extend this argument to the 3-sphere. What exactly do the parameters now represent?
In the 2D case, the set of points equidistant from the origin constitutes a (generalized) circle (a $1$-sphere); the coordinate $r$ is the circumference of that circle divided by $2\pi$. It is not the radius of the circle, despite the deceptive name.
Generalizing to the 3D case, the set of points equidistant from the origin constitutes a $2$-sphere; the coordinate $r$ now represents the surface area of that sphere divided by $4\pi$. This is the same interpretation as the "radial" Swarzschild coordinate, for example.
Once you've chosen some $r$, you've restricted your attention to a $2$-sphere of points which sit at the same distance from the coordinate origin. The angles $\theta$ and $\phi$ specify a point on this $2$-sphere precisely as they usually do in elementary spherical coordinates.
To illustrate my problem: suppose we wish to calculate the volume of a sphere of radius $R_0$ that exists on the surface of the 3-sphere (K=1 Universe). How would we do that using this metric?
The surface area of a sphere of radius $r$ is simply $4\pi r^2$, which follows immediately from the interpretation provided above. The volume of the ball of radius $R_0$ is then straightforwardly
$$V = \int_0^{R_0} 4\pi r^2 \frac{1}{1-kr} \mathrm dr = 4\pi \frac{kR_0(kR_0+2)+2\log(1-kR_0)}{2k^3}$$
which reduces to $\frac{4\pi R_0^3}{3}$ in the limit as $kR_0 \rightarrow 0$, as expected.
| {
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Calculating the Wigner transform of operators Recently I started to study the formulation of quantum mechanics in the phase space. So I was introduced to the concept of Wigner function and Weyl transform.
I learned that if F is an operator, then I can represent it by an integral as follows:
\begin{equation}
F = \int_{-\infty}^{+\infty}\frac{dpdq}{2\pi\hslash}f(p,q)\Delta(p,q)
\end{equation}
Where $f(p,q)$ is the Wigner transform given by:
\begin{equation}
f(p,q) = \int_{-\infty}^{+\infty}e^{\frac{i}{\hslash}qu}\langle p+\frac{u}{2}|F|p-\frac{u}{2}\rangle du
\end{equation}
and $\Delta(p,q)$:
\begin{equation}
\Delta(p,q) = \int_{-\infty}^{-\infty} e^{\frac{i}{\hslash}pv}|q+\frac{v}{2}\rangle \langle q-\frac{v}{2}|dv
\end{equation}
all of the above expressions were derived using the completeness relations as follows:
\begin{equation}
F = \int_{-\infty}^{+\infty}dp'dp''dq'dq''|q''\rangle\langle q''|p''\rangle \langle p''|F|p'\rangle\langle p'|q'\rangle \langle q'|\end{equation}
and the following variable change was also taken
\begin{equation}
2p =p'+p'',
2q = q'+q'',
u = p''-p',
v = q''-q'
\end{equation}
Could someone show me what these calculations would look like for a numerical example of some observable operator. I tried to calculate for one of the pauli matrices, but I was stuck in the middle of the calculations.
My learning becomes more consistent when I see practical examples, if anyone can help me with this problem, I will be very grateful.
| Pauli matrices are mere constant matrices acting on 2d spinors, not functions of x or p, so you may be barking up the wrong tree.
I assume you or your text have evaluated the free particle hamiltonian,
$$
h(p,q) = \frac{1}{2m}\int_{-\infty}^{+\infty}e^{\frac{i}{\hslash}qu}\langle p+\frac{u}{2}|\hat p^2|p-\frac{u}{2}\rangle du\\
\frac{1}{2m}\int_{-\infty}^{+\infty}e^{\frac{i}{\hslash}qu}(p-u/2)^2\langle p+\frac{u}{2}| p-\frac{u}{2}\rangle du\\ = \frac{1}{2m}\int_{-\infty}^{+\infty}e^{\frac{i}{\hslash}qu}(p-u/2)^2 ~\delta(u) ~ du =\frac{p^2}{2m}~~ .
$$
The operators need not be observables. Try the hermitean parity operator,
$$
P=\int \!\! dp ~~|-p\rangle\langle p| ~~\leadsto
$$
$$ \Pi (q,p)=
\int\!\!du dp' e^{iqu/\hbar} ~~\langle p+u/2 | -p'\rangle \langle p'|p-u/2\rangle \\
=\int\!\!du dp' ~ e^{iqu/\hbar} ~~ \delta( p+u/2 +p') \delta( p-u/2 -p') \\
=\int\!\!du ~ e^{iqu/\hbar} ~~ \delta(2 p ) =\frac{h}{2}\delta(q) \delta(p)~.
$$
| {
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Are there any known or theoretical substances that don't experience creep? Creep is very slow elasticity that most materials have, that cause pipes to sag over decades. Are there any materials, in theory or in practice, that are creepless?
| Creep in many engineering materials requires dislocation travel. Since dislocation travel is assisted by diffusion, and since diffusion processes in metals kick in exponentially at temperatures above about ~1/2 the melt temperature (in degrees absolute), creep can be prevented by choosing alloys with the highest possible melting points, called superalloys (typically iron with lots of chromium, nickel, vanadium, etc.) and by not exposing the material to high temperatures.
This also means that creep can be inhibited by choosing a material that has no dislocations in it. Since grain boundaries create dislocations, a material that has no grain boundaries will be creep resistant. Single-crystal, directionally-cast gas turbine blades are made this way to prevent creep.
Creep is also caused by progressive dislocation pileup at grain boundaries which allows the grains themselves temporarily to decouple themselves and rotate slightly in response to creep stress. This can be inhibited in iron-based alloys by adding in certain alloying constituents, called carbide formers, which selectively segregate out at grain boundaries to stabilize them.
Even in the presence of active dislocations, creep can be prevented by "pinning" the dislocations so they cannot move much. This is done by adding materials to the metal which precipitate out of solid solution upon cooling from the melt or during heat-treatment as nanoscale particles within the crystalline matrix of the metal (precipitation hardening).
You can also prevent creep in polycrystalline materials by choosing one that has no dislocation movement mechanisms in it that are active at the temperature of interest. This means ceramic materials will be very resistant to creep even at relatively high temperatures.
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Kinetic Energy and Moment of Inertia In this video, at around 12:00, it is said that spinning about the axis with the smallest moment of inertia gives the most kinetic energy. But, isn’t rotational kinetic energy equal to $(1/2)(I)(ω)^2$ . Thus, shouldn’t kinetic energy increase with increase in moment of inertia?
| It is because the angular momentum is conserved while the kinetic energy is not.
So: $$I_1\omega_1=I_2\omega_2$$
When $I_2$ decreases, $\omega_2$ increases. $\omega$ is squared in the expression of kinetic energy and hence the net kinetic energy increases.
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Why are the lenses of a Keplerian Telescope positioned the sum of their focal points apart? I am currently designing an extremely simple Keplerian telescope and am confused as to why many explanations say that the objective lens and eyepiece lens are positioned a distance of the sum of their focal points away (fobjective+feyepiece). As pictured in the figure below, when the two lenses have positioned a distance of the sum of their focal points away, the light rays that ultimately hit the eye will be parallel; meaning that they will never converge and therefore the eye will never be able to piece together an image. Or the other way I think about it is that the object distance of the eyepiece lens is at its own focal point, which as we know, won't form an image.
From my belief, if the objective lens is positioned just where fobjective is inside feyepiece, this will create an extremely magnified image and will be the perfect alignment of a telescope. So should I align the focal points as just described above, or should they be aligned as shown in the picture and described by many books? If so, why?
| Parallel rays entering a relaxed eye with normal vision will be focused by the cornea and lens of the eye to form an image on the retina. As far as the eye is concerned, the parallel rays coming from a Keplerian telescope are entirely equivalent to the (nearly) parallel rays coming from a distant object. If one relaxes one's eye (and one's far point is sufficiently far away), one will be able to see the image through the Keplerian telescope clearly.
The real power (pun intended) of the Keplerian telescope is that it increases the angle these rays make with the optical axis. This means that an object that would have subtended (for example) 1 arcminute and just barely have been resolvable by the naked eye might now subtend 20 arcminutes and be readily resolvable.
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Why 2nd Shell can have 8 electrons? I recently watched this video:https://youtu.be/INYZy6_HaQE
and understood why 1st orbital can have only 2 electrons: According to Pauli's exclusion principle, two electrons cannot have the same quantum states.In the first energy level,all other quantum states are same except for the spin. And spin can have two values $+\frac{1}{2}$and $-\frac{1}{2}$.So,there can be two electrons in the first shell.
I need a similar explanation for 2nd shell
In other words, why the 2nd shell can have 8 electrons?
I know that there are other quantum states. I am requesting to explain how many values each quantum state can take.For example spin can have two values.Likewise,how many values can the angular momentum can have?
How all these lead to the 8 electrons in the second shell?
(Don't make it simple,you can use maths.Iam really on to this)
| There is no such thing as a "second orbital". There is a second shell which is "made of" four orbitals which each can hold 2 electrons – this is where the 8 comes from.
So there is no contradiction: Each atomic orbital can hold two electrons with opposite spin (according to the Pauli exclusion principle), but there are simply more orbitals in higher shells. The orbitals in one shell have the same principal quantum number $n$, but they differ in the other quantum numbers – the azimuthal quantum number and magnetic quantum number.
For more information, see Atomic orbital and Electron shell.
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Does the albedo of a photovoltaic cell correlate with its load? Exactly what the title says. Since solar panels convert (to my understanding) visible light into electricity, will a solar panel under load appear visually darker compared to a solar panel under no load?
Better phrasing: Does the albedo of a photovoltaic cell correlate with its load?
| For "regular" (silicon-based) solar panels, I believe the answer is no. Silicon is an indirect band gap material. Recombination rarely releases photons, so an increase in conduction band electrons does not make the material glow. Likewise, an increase in valence band electrons does not make the material darker.
Now if you had something like a GaAs panel, I'd be less certain. I suspect that it would be hard to see it visibly, but it's very possible that the material would emit significantly more recombination photons when under no load than when the cell is nearly short circuited.
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Why electric field is scaled by gamma? Two opposite charges are in a spaceship and are attracted by the electric field $E_s$
But for an observer on earth the Electric force is $$E_e=\gamma E_s$$
Normally the forces are scaled down by $\gamma$ in the earth frame and here also the total force is scaled down.
But why the Electric component of force is scaled up?
Is it because the Electric fields are now closer together because of length contraction?
But i think this is not the answer because if that was the case,then gravitational fields and hence Forces would have scaled up.But this is not the case in nature.
Can you provide a derivation or something which explains how the Electric field is scaled up?
|
Normally the forces are scaled down by γ in the earth frame and here
also the total force is scaled down. But why the Electric component of
force is scaled up?
The actual derivation is based on Lorentz transformation equations but one intuitive way to visualise this is to visulalise the electric field lines. In a charge at rest, the field lines are pointing outward uniformly in all directions.
But, in a charge that is moving , the field lines get "scrunched up" in the transverse direction to the field of motion, so that the electric field strength in that direction increases
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Nonlinear extension of Lorentz Group The Lorentz group is defined to be the set of linear transformations that leave $ds^2 = -dt^2 + |d\vec{x}|^2$ invariant. The Poincaré group contains the Lorentz group, but now we allow transformations of the form $t' = t+a$ where $a$ is some constant.
Is there a name for the group of transformations that leave $ds^2 = -dt^2 + |d\vec{x}|^2$ invariant without assuming linearity? The answers to this question here claims that any nonlinear transformation will have some "privileged" point, and hence we disregard them as nonphysical. However, I am curious nonetheless if there is a name for this group and any interest in it.
| NOTE ADDED IN PROOF The linearity for Minkowski metric preserving diffeomorphisms is treated here:
Interval preserving transformations are linear in special relativity
However, one may go a step further and ask a better "foundation of SR question": what are the linear/non-linear transformations that preserve wave-fronts, and do they form a group? (it is known that in the foundations of SR light signals are treated as spherical waves propagating at speed "c").
There are two separate issues here:
1. What is the most general form of space-time transformations which leave the D'Alembert equation "conserved":
$$\Box f(x_0,x_1,x_2,x_3) =0 \Rightarrow \Box'f(x'_0,x'_1,x'_2,x'_3) =0) ~\tag{1}$$
2. Do the transformations at point 1., if non-linear, form a group?
Question 2. excludes some (if not all non-linear) solutions to point 1. because non-linearity conflicts with the group property of unique inverse transformation.
Think of (as example) $\text{T:} ~ x'=\sin x^2 + a, y'=y, ~ z'=z, ~ t'= \sqrt{t^2 + 75}$. This may be a solution to problem 1., but you cannot invert it, therefore is not something the OP may seek.
If we remove from the OP the condition of being a group, then question 1. is really interesting in itself and a solution has been offered by Weyl (quoted by V.A. Fock in Appendix A of his relativity book: The theory of space, time and gravitation, Pergamon Press, 1959, +---).
Answer to 1.
The form of the most general transformation $x'_i= f_i(x_0, x_1, x_2, x_3), ~ (i=0,1,2,3)$ satisfying $(1)$ is either:
$$ x'_i =\frac{x_i - \alpha_i \sum_{k=0}^{3} e_k x_k^2}{1-2 \sum_{k=0}^3 e_k \alpha_k x_k + \sum_{k=0}^{3} e_k \alpha_k^2 \sum_{l=0}^{3} e_l x_l^2 } \tag{2}$$
or
$$ x'_i = a_i + \sum_{k=0}^{3} e_k a_{ik} x_k, \tag{3} $$
with $a_{ik}$ satisfying
$$ \sum_{i=0}^{3} e_i a_{ik}a_{il} = e_k \delta_{kl} \tag{4}$$
with $a, e, \alpha$ constants with respect to $x_i$.
$(3) + (4)$ are called Lorentz transformations and they do form a group, while the (2) are called Möbius (spherical) transformations and unfortunately do no form a group.
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What does it mean for the gravitational force to be "between" two bodies? What is the meaning of the word "between" in the law that the force between two masses at separation $r$ is given by $\frac{GM_1M_2}{r^2}$? I am confused about how can a force be in-between, either it is on body A or on body B, or on both.
Suppose body A exerts force $F$ on Body B, so according to Newton's 3rd law of motion B should also exert a force on A.
Let's consider this case for gravitational force between two bodies. If body A exerts force $g$ on Body B, then B body should also exert a force $g$ on A, but B is also exerting the gravitational force $X$ on A, hence A will also exert force $X$ on B.
So, how are two forces acting?
I have given the representation in this diagram.
| The meaning of the word "between" in this case is the same as the meaning of between in the sentence:
The love between two people
Of course it is understood that love does not exist in air. One person loves another. Air has no brain and thus has neither emotions nor feelings.
The word between in the sentence above means that person A loves person B and person B loves person A.
Thus the phrase: "the force between two bodies" means body A exerts a force on body B (this force is on body B) and body B exerts a force on body A (this force is on body A).
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In this ray diagram, a plane mirror seems to form a real image In this ray diagram the image formed seems to be real with the given position of the eye. I have learnt that plane mirrors cannot form real images at any circumstance. But at this one it does. Please explain the answer like I'm 5 and how you deduced what you propose.
| Some diverging rays have been left out as shown below!
A cone of diverging rays which appear to come from $B_1$ are reflected off the mirror and the eye focuses those rays onto the retina to form a real image of the bottom of the feet.
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Which force came first in Newtonian gravity? Force A on B or force B on A? Can a body $X$ apply a gravitational force on a body $Y$ without itself being acted on by a gravitational force from $Y$? Just per intuition, we deduce from Newton's third law that if $A$ applies force on $B$ then $B$ applies force on $A$. But in the case of gravitation, they both apply force on each other. So which force came first as an action, and which came later as a reaction?
| Both forces appear at the same time.
No, a force cannot be applied by one body on another body without the first body feeling a reaction force. This is due to Newton's third law, as you mention.
For the action/reaction force pair we can't say that one "came first". Both forces of the pair must be present simultaneously. Calling one the action and the other the reaction is only done as an aid for the intuition; it has no physical significance. They both appear exactly simultaneously and one did not "come first".
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Velocity is relative, which means acceleration is relative, which further implies that forces are relative as well So how would we know whether a force truly exists or not. I can be accelerating a car my 5 meters per second squared but another car accelerating with the same acceleration would think that my car is at rest relative to them. So is there any force on the car? Or are forces just relative and their existence just depends on our reference frame?
| In Galilean relativity, physics is unchanged for frames that are related to uniform boost. That is, we introduce an equivalence relation $\sim$ among frames and define $\mathcal{A} \sim \mathcal{B}$ if frame $\mathcal{B}$ moves with constant velocity vector as seen by the frame $\mathcal{A}$. Then Newton's first law says that there exists a particular equivalence class such that inertial motion appears as uniform linear motion. Then there comes the second law.
In fact, we can think in this way: $\vec{a}$ is Galilean invariant, and $\vec{F}=m\vec{a}$ is the simplest physical law that conforms with Galilean relativity! And this is because force, a physical quantity, should not depend on observers. Indeed, "(Physical quantity)$ = $(Geometrical invariant)" is a recurring leitmotif throughout the history of physics.
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Is Weyl transformation part of diffeomorphism? Does a gravitational anomaly capture also the anomaly due to Weyl transformation?
*
*Weyl transformation is a local rescaling of the metric tensor
$$
g_{ab}\rightarrow e^{-2\omega(x)}g_{ab}
$$
*Diffeomorphism maps to a theory under arbitrary differentiable coordinate transformations
(Diffeomorphism is an isomorphism of smooth manifolds. It is an invertible function that maps one differentiable manifold to another such that both the function and its inverse are smooth.)
Question 1: Is Weyl transformation part of diffeomorphism?
It seems that the answer would be yes,
*
*if this $e^{-2\omega(x)}$ is arbitrary differentiable and
*if the starting manifold with a $g_{ab}$ is differentiable.
Question 2: Because the gravitational anomaly is also known as diffeomorphism anomaly, related to the diffeomorphism of manifold. Is this correct to say that the gravitational anomaly capture also the anomaly due to Weyl transformation?
p.s. I asked more additional details in a previous post Weyl transformation vs diffeomorphism; conformal invariant vs general in/covariant, but I got no answer. So let us zoom into a specific case. I hope someone can give a definite correct answer this time.
| No to both. A Weyl transformation will send a metric to another metric that will not be the image of the first under diffeomorphism. Some diffeomorphisms have the effect of a Weyl transformation on the metric; those are called conformal maps and are quite special (finite in number outside of dimension 2 and 1).
| {
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Does the Oort cloud act as a kind of shield for the Solar System? Does the Oort cloud act as a kind of shield for the Solar System? When an interstellar object impacts the cloud, does its momentum get absorbed substantially?
| The Oort cloud is extremely diffuse, and while there are numerous small bodies, they're spread out over a very large region of space. Assuming that the outer Oort Cloud ranges from 20,000 AU to 50,000 AU and contains $\sim10^{12}$ bodies on the scale of a kilometer or greater, I find an average number density $n$ of 0.002 objects per cubic astronomical unit. Put another way, there would be two of these objects contained within a cube with sides 10 AU across.
We can define an optical depth of sorts, $\tau$, as
$$\tau=n\sigma l$$
with $\sigma$ the cross-section of the two objects and $l$ the thickness of the cloud. Assuming that $\sigma$ is on the order of maybe a few square kilometers (probably an overestimate) and that $l$ is 30,000 light-years, we find that the optical depth of the outer Oort cloud is $\tau\approx10^{-14}$ (a bit lower, but I've rounded up). The probability of an incoming body colliding with an Oort Cloud object is then
$$P=1-e^{-\tau}\approx0$$
The upshot is that no, the Oort Cloud doesn't act as a shield, as it's extremely unlikely that an interstellar interloper would be stopped.
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Do partial derivatives of different coordinate systems commute? Consider an arbitrary set of coordinates $x^\mu$ and another set of coordinates $y^{\mu}$, which is a (lorentzian) transformation from $x^\mu$ given by $y^\mu = f(x^\mu)$.
So I want to know whether $\frac{\partial}{\partial x^\alpha}\frac{\partial}{\partial y^\beta} = \frac{\partial}{\partial y^\beta}\frac{\partial}{\partial x^\alpha}$ holds true or false?
| It depends on the transformation at hand, but in general the answer is no. It boils down to whether $x^\alpha$ can change while $y^\beta$ is kept constant.
Denote by $J_{\nu}^\mu(x) = \frac{\partial y^\mu}{\partial x^\nu} $ the Jacobian of the change of coordinates.
By the chain rule
$$ \frac{\partial}{\partial x^\alpha} = J_{\alpha}^\nu(y)\frac{\partial}{\partial y^\nu} $$
It is not difficult to see that it makes a difference whether this expression is acted upon by $\partial_{y^\beta}$ from the left or the right. The former generates an additional term
$$ \frac{\partial J_{\alpha}^\nu(y)}{\partial y^\beta} \frac{\partial}{\partial y^\nu} $$ which needn't vanish.
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Why do I feel electric shock even in the presence of wood? I feel mild electric shock when my laptop with aluminum body is kept on a wooden desk is charging, while both my leg rests on the the leg of the table . Why is that ? Why isn't wood behaving like an insulator here ?
| If the chassis of your laptop is "hot,*" then you would feel a tingle when you touch it even if your body was perfectly insulated from the Earth. That's because the "hot" wire in home electricity supply is an alternating voltage with respect to Earth, and your body and the Earth act like the plates of a capacitor. Each time the "hot" voltage changes sign,** a tiny amount of electric charge flows in to or out from your hand where it touches the laptop. I say "tiny," but it's enough for you to feel it. (Don't ask me how I know!)
* If the chassis of your laptop is "hot," (a.k.a., "live") then that's a serious fault in your home wiring, and you should have an electrician fix it before somebody is injured or killed. If you accidentally touched the laptop with one hand and, for example, a water pipe with the other hand, that could be lethal.
** 50 or 60 times per second, depending on where you live.
| {
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Why doesn't $v_T = \omega r$ involve the direction of its variables? We derived $v_T = \omega r$ by the following procedure, and it's said that $v_T = \omega r$ "is a relation between the magnitudes of the tangential linear velocity and the angular velocity". Why doesn't the formula involve the direction of its variables?
We first define
| Suppose we have a rigid object rotating at some angular velocity $\vec{\omega}$ about some fixed axis. It can be shown that in general, $\vec{v} = \vec{\omega} \times \vec{r}$, where $\vec{r}$ is the instantaneous position of the object. This more general equation does take the directions of the quantities into account via their cross product.
However, it is common (particularly in introductory physics texts) to consider only the case where $\vec{\omega}$ and $\vec{r}$ are at right angles to each other. In that case, this equation simplifies to the version you have. Recall that the magnitude of a cross product is $|\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin \theta$. This means that if $\vec{r}$ and $\vec{\omega}$ are perpendicular to each other, then we have $\theta = 90^\circ$ and
$$
v = |\vec{v}| = |\vec{\omega} \times \vec{r}| = |\vec{\omega}||\vec{r}| \sin \theta = \omega r.
$$
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