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Pan flute physics We are trying to create a custom pan flute. With some searching, we found a formula relating the length of a closed cylinder with the frequency at which it resonates, namely L = c / 4f, with c being the speed of sound. Based on this, we calculated that a cylinder to produce 880 Hz, or A5, would need to be about 97.4 mm long. But, having printed such a cylinder with a 3D printer, when blown, it produces a frequency of about 840 Hz. We have checked and re-checked the dimensions, and the plastic does not appear to be porous at all. Why are we getting a different frequency than predicted by the theory?
| Your formula ignored the "end correction" to the length of the pipe. The standing wave in the air does not end exactly at the end of the pipe, but at a distance approximately 0.6D outside it, where D is the pipe diameter.
If your pipe diameter was about 7mm, that would explain the difference between 880Hz and 840Hz.
Note that the "exact" length correction will also depend on the change to the air flow pattern around the pipe caused by your face, when you are blowing the instrument, so you will always have to make some minor adjustments by trial and error.
See https://en.wikipedia.org/wiki/End_correction.
| {
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Calculating new velocities of $n$-dimensional particles after collision I am working on a particle simulation where there is no gravitational force exerted on particles, they simply travel through space and, upon collision, change trajectories accordingly. There isn't a set number of dimensions the simulation can be run on, with the minimum being 2.
Each particle has three attributes: radius (each particle is an $n$-ball of $n$ dimensions corresponding to the simulation's dimensions), coordinates (a tuple describing its position), and velocity (another tuple describing how quickly it moves in any given direction, determining where it will be on the next iteration).
With the coordinates one can determine exactly the angle at which it strikes another particle, and that should be enough to determine the new velocities of both particles, or so I think. However, I'm not sure how to do this for an unknown amount of dimensions. I would presume a sigma function could describe this behaviour well, but I'm not sure how to write it. How might I go about doing so?
Any necessary information can be added upon request.
| The exchange in momentum (impulse $J$) happens along a single direction. With the simple model of just spheres you can calculate the direction from the positions of the particles
$$ \boldsymbol{n} = \frac{ \boldsymbol{r}_j - \boldsymbol{r}_i }{ \| \boldsymbol{r}_j - \boldsymbol{r}_i \| } $$
where $\| \boldsymbol{r}_j-\boldsymbol{r}_i \| = \sqrt{ (\boldsymbol{r}_j-\boldsymbol{r}_i) \cdot (\boldsymbol{r}_j-\boldsymbol{r}_i) } $. This depends on the dot product $\cdot$ which is defined for any dimension vectors.
So the change in velocity is $$ \begin{aligned} \Delta \boldsymbol{v}_i &= - \tfrac{J}{m_i} \boldsymbol{n} & \Delta \boldsymbol{v}_j &= + \tfrac{J}{m_j} \boldsymbol{n} \end{aligned} $$
and the value of $J$ is determined by the type of collision
$$ \begin{array}{c|c}
\text{elastic} & \text{plastic}\\
\hline
J = 2 \frac{\boldsymbol{n} \cdot ( \boldsymbol{v}_i - \boldsymbol{v}_j )}{ \tfrac{1}{m_i} + \tfrac{1}{m_j}} & J = \frac{\boldsymbol{n} \cdot ( \boldsymbol{v}_i - \boldsymbol{v}_j )}{ \tfrac{1}{m_i} + \tfrac{1}{m_j}}
\end{array}$$
Note that symbols in boldface are vectors and normal letters are scalar values.
| {
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Loop-correction for non-commutative quartic theory
What is the meaning of the second, third and fourth graph? The image is from arXiv:hep-th/9912072.
| *
*The Feynman rules in non-commutative field theory are classified using 't Hooft double-line notation. In particular, the legs in a vertex has a cyclic order, cf. eq. (2.5) in Ref. 1.
*Interestingly, non-planar graphs are suppressed due to extra phase factors. (This is somewhat similar to the planar limit/large-$N_c$-expansion in $SU(N_c)$ Yang-Mills theory, although the non-planar suppression is there caused because non-planar diagrams have fewer traces.)
*Returning to OP's question, the second, third and fourth graph in Fig. 7 are non-planar graphs (again due to that the cyclic order of legs in vertices matters).
References:
*
*S. Minwalla, M. Van Raamsdonk & N. Seiberg, Non-commutative Perturbative Dynamics, arXiv:hep-th/9912072.
| {
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Why does blending hot food in the Nutribullet blender create a "pressure seal"? I have noticed that when blending cold foods in the Nutribullet, it is easy to unscrew the blender once the blending process is finished. When I blend something hot (i.e, boiled tomatoes, to create a tomato cream soup), I noticed it is very difficult to unscrew the blender, as if an "air-seal" had been created during the blending process. Upon opening the blender, a "frizzle noise" can be heard.
Is the difficulty all to do with the screw-on part expanding in the heat? (But that heat should also expand the plastic container, so I doubt that's all there is too it?). Is there another physical process taking place that creates some sort of vacuum seal?
| At the time that you fasten the lid, the vapor pressure of the hot liquid in the head space is equal to the equilibrium vapor pressure of the liquid at the high temperature of the liquid in the container. But, during the mixing, the liquid cools a little, and its vapor pressure decreases. So some of the vapor in the head space condenses, and the partial pressure of the volatile species drops. This creates a partial vacuum in the mixer chamber which is often referred to as a vacuum seal. This is the same thing that happen in home canning of foods, and in commercial canning operations (like soups). When you puncture the lid with a can opener, you hear the hissing of air entering the head space to equalize the pressures.
| {
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Derivation of curl of magnetic field in Griffiths Can someone please derive how $$\frac{d}{dx} f(x-x') = -\frac{d}{dx'} f(x-x')~?$$
In Griffiths electrodynamics, this is directly mentioned. I'm really confused, can someone elaborate!
| This follows directly from the chain rule:
$$\frac{\partial}{\partial x}[f(x-x')] = f'(x-x')\frac{\partial}{\partial x}[x-x'] = f'(x-x')$$
whereas $$\frac{\partial}{\partial x'}[f(x-x')] = f'(x-x')\frac{\partial}{\partial x'}[x-x'] = -f'(x-x').$$ (Here I take $f'(x-x')$ to mean that (total) derivative of $f$ with respect to its single independent variable.)
Thus, we see that the two expressions are simply the negations of each other. More complex versions of this can similarly be derived for other vector calculus operators, such as $$\nabla_x f(x-x') = -\nabla_{x'} f(x-x'),$$ where $\nabla_x$ denotes the gradient with respect to $x$. I think this is also explained in Griffiths somewhere, but hopefully this explanation suffices.
Hope this helps.
| {
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What is gravity's relationship with atmospheric pressure? I'm asking for clarification here. If Earth had the same atmospheric mass per square unit of ground but the Earth had suddenly gained mass so it had twice the gravity at the surface, would the Earth now have twice the atmospheric pressure just because of the doubling of gravity? I know pressure is defined as force per area, but I'm not sure if air pressure works the same way. When I look up gravity and air pressure on the Internet it just has information on air pressure with height, but I'm not looking for that.
| The phenomenon that determines atmospheric pressure is very similar to Pascal's Law for incompressible liquids. But air isn't an incompressible liquid: as a gas its density is quite dependent on pressure and temperature, for instance.
A doubling of $g$ would nonetheless double also the atmospheric pressure.
I know pressure is defined as force per area, but I'm not sure if air
pressure works the same way.
The definition of pressure isn't dependent on the type of fluid.
| {
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Why every system tends to be more stable? An entire topic of inorganic chemistry, i.e. chemical bonding, which is also one of the most important topics, is based on the idea of stability. But whenever I ask why every system tends to get stable, I never get a satisfactory answer. People say that's how nature "wants" to be. How does a system know that it wants to get stable?
| 1- Because that's the definition of stability: you can't escape it
In the words of Peter above :
In this sense a system tends to become stable because stable states
are the only states a system can remain in.
2 - Not all systems are "stable" some are oscillatory (like our star system or a single atom). But you may call these stable as well depending on how you look at it.
3 - In the chemistry context, evolution towards stability is described by the 2nd law of thermodynamics which is a consequence of statistical mechanics which relies on the law of large numbers that basically says that things tends to average out in the long run.
| {
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Why is stress defined in the way as it is? Stress is like pressure and it doesn't matter in which direction the force acts (given it is perpendicular to the surface).
I read in my book that if we have a rope which is being pulled on both sides by a force $F$ , then the stress at any cross section of the rope is defined as
$\sigma = \frac{F}{area}$
.
But my question is that since the rope is pulled from both the ends, the molecules of the considered cross section is being pulled by other molecules from both sides of the cross sections.
So the stress is due to both the forces.
So shouldn't stress be defined as
$$\sigma =\frac{2F}{area}$$
Edit : The answer of Bob D forces me to add an edit. By $\sigma = \frac{2F}{A}$ , I meant to say
$\sigma =\frac{|F_{left}|}{A}+\frac{|F_{right}|}{A} = \frac{2F}{A}$ .
Here $|F_{left}|$ and $|F_{right}|$ are the forces applied by molecules on the left and right of the considered cross section on the cross section.
Hope it is clear now.
| Firstly, by its definition, the restoration force developed per unit area inside the body when subjected to the deforming force is called stress.
Secondly, by Newton's third law of motion every action has an equal and opposite reaction and both act on two different bodies.
Now let's take your rope or whatever body it is in consideration
We apply a pair of deforming forces of magnitude F on both ends of the rope, to make sure the body is in rest. Now,
Let's consider a cross-section of the body at a general distance x from left end (sorry, I forgot to show it in the above diagram). There would be some tensile force developed inside the body. If the rope is kept at rest by balanced force F, we can say that T=F.
Now let's consider the equilibrium of any of the parts of the rope.
The tensile force will act as a restoring force (because you know, your rope didn't just snap). Also, applying Newton's third law of motion, we can see that development of tensile restoring force in one of the parts would result in in development of tensile force in its anti-parallel direction, which would be present on the other part. (I hope this line has cleared something for you). Thus, by applying the definition of stress on any of the part we have, $\sigma=\frac{T}{A}\\ \Rightarrow\sigma=\frac{F}{A}$
Where ofcourse, A= cross-sectional area of the rope.
| {
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Nature of Work done by friction I have always been told that work done by friction can, at most, be zero, but never positive.
But consider two blocks placed one on top of the other, such that their surfaces in contact are rough. If we give the block on the top a certain horizontal velocity, then in crude words, we can say that friction will try to slow down the block on the top and speed up the other one, thus opposing relative motion. Then in this case, wouldn't the work done by the friction on the block at the bottom, be positive? Please correct me if I have gone wrong.
| *
*Friction is a force that apposes the motion. It does negative work.
*The second block is subjected to reaction force due to friction and as such is not friction force.
*Just some terminology.
| {
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Why is it easier to handle a cup upside down on the finger tip? If I try to handle a tumbler or cup on my fingertip (as shown in fig), it is quite hard to do so (and the cup falls most often).
And when I did the same experiment but this time the cup is upside down (as shown in fig), it was quite stable and I could handle it easily.
In both the cases, the normal force as well as the weight of that cup is the same but in first case it falls down and in the other it is stable.
I guess that it is falling because of some torque but why is there no torque when it is upside down.
What is the reason behind this?
| Maybe because when the cup is the right way up, it’s centre-of-mass is above the point on your finger meaning that as your finger tries to balance the cup any small motion will generate a torque about this c.o.m making it harder to balance.
When the cup is upside down, you have your finger on or going through the c.o.m and so any small motion by your finger will not generate torque about the c.o.m making it much easier to balance.
| {
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How is flow velocity defined in Navier-Stokes equations? I know Navier-Stokes equations rely on the continuum assumption. In this context, how is the flow velocity mathematically defined? Is it merely a spatial average of the micrscopic particles velicities inside the Representative Volume Element? Or is it a mass-weighted average, so that it results in the centre-of-mass velocity of the RVE? Or is it a even different kind of average?
Any link to a detailed description of this topic would also be appreciated
| First of all, I would not necessarily call it the continuum approximation. I think that physically, "coarse grained" is a better word. Obviously, there is no system for which $\vec{v}(\vec{x})$ has meaning at arbitrarily short distances.
Also note that fluid dynamics is about the motion of conserved charges in a system close to thermal equilibrium. Quantities like $T(\vec{x}),\mu(\vec{x})$ and $\vec{v}(x)$ refer
to suitable definitions of thermodynamic variables that can be used to express the conserved currents in such a system using "constitutive relations". Since the system is not in perfect equilibrium, there is some ambiguity in defining what we mean by these quantities. This ambiguity is unavoidable -- all we can ask is that the predictions of fluid dynamics do not depend on these ambiguities order by order in an expansion in gradients of $T,\mu,\vec{v}$.
Having said this, there is an essentially universally agreed upon definition of the velocity of a non-relativistic fluid (the relativistic case is more tricky). Take the total momentum in a volume element (a well defined object in any microscopic theory), and divide by the mass
$$
\vec{v}_{cell}=\vec{P}_{cell}/M_{cell}
$$
This leads to the constitutive relation
$$
\vec{\pi} = \rho\vec{v}
$$
for the momentum density of the fluid. In Navier-Stokes theory, this is the definition of $\vec{v}$, so it receives no correction in $\nabla_i v_j$. Contrast with the energy current, which receives corrections involving $\nabla_i T$.
| {
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Relativistic Jets in Black Holes It is understood till now that nothing is faster than speed of light and we also know that even light cannot come out of black hole(that’s why the name black). So does that mean that the Relativistic jets which (assuming I am true) come out of the black hole are faster than the speed of light. And if possible can you also please explain the phenomenon of these Relativistic jets and how they seem to appear on both the sides of black hole like going in from one side and coming out of another....
| Relativistic jets do not come out of the black hole. They originate from near the black hole but not, of course, from inside the event horizon.
The relativistic jets consist of streams of charged particles that are accelerated close to the black hole and travel, often in both directions, along the presumed rotation axis of the black hole.
The launching mechanism for these jets is still a substantial topic of research. The most likely ideas are that ionised, accreting material becomes entrained in a strong magnetic field that becomes compressed near the black hole. The field becomes toroidally twisted due to the rapid rotation around the black hole and this sets up a very strong magnetic field gradient along the rotation axis, which accelerates the charged particles in that direction - an example mechanism is the Blandford-Znajek effect. The ultimate energy source here is the rotational energy of the black hole.
Jets would be expected to be launched in both directions along the rotation axis. However, it is more usually the case that the jet in one direction is more prominent. This is due the Doppler boosting phenomenon - radiation emitted by particles with a component of their velocity towards the observer will be significantly boosted. Conversely if a jet is moving away from us in the line of sight then the radiation seen by us will be considerably weakened.
An excellent popular account of the mechanism can be found here.
| {
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Hamiltonian formalism of General Relativity Textbook I've been reading Wald's book on General Relativity and in appendix $E_{2}$ it discusses the Hamiltonian formalism of General Relativity.I would like to understand it more, can you recommend me a textbook about ADM formalism?
P.S.
I have already read Eric Poisson book too.
| Besides the original literature DanielC enumerated,
*
*Bojowald's Canonical Gravity and Applications is specifically about the Hamiltonian formalism;
*D'Inverno's Introducing Einstein's Relativity has a fairly good introduction to the Hamiltonian formalism, as well, but simpler...more friendly for students (I'd imagine).
| {
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What quantum gates are needed to get the state $|01\rangle+|10\rangle$ from $|00\rangle$? I was wondering if I start with two qubits in the state
$$|00\rangle$$
If it's possible to apply gates to get it to the state
$$\frac{|01\rangle + |10\rangle}{\sqrt{2}}$$
I have tried applying the Hadamard Gate, Controlled X etc, But I couldn't make this state. So I'm curious if it's possible and I am just missing something very obvious.
| Let's consider a space of four states $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$. The question is whether exist any operator for which,
$$
\left(\begin{array}{cccc}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44} \\
\end{array}\right)
\left(
\begin{array}{c}
1 \\ 0 \\ 0 \\ 0
\end{array}
\right) = \frac{1}{\sqrt{2}}
\left(
\begin{array}{c}
0 \\ 1 \\ 0 \\ 1
\end{array}
\right)
$$
Of course one can invent such an operator, e.g.
$$
\hat{a} = \frac{1}{\sqrt{2}}\left(\begin{array}{cccc}
0 & a_{12} & a_{13} & a_{14} \\
1 & a_{22} & a_{23} & a_{24} \\
0 & a_{32} & a_{33} & a_{34} \\
1 & a_{42} & a_{43} & a_{44} \\
\end{array}\right),
$$
but I do not know if it corresponds to any known quantum gate. Remaining $a_{ij}$ can be arbitrary.
| {
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Is it possible for a wavefunction to partially collpase? What would happen in this situation: You have an ideal empty space with only two objects: a photon emitter, and a small observer. The emitter emits a photon in a random direction, with every direction being equally likely.
Given that the probability of observing the photon travels at the speed of light, what happens at the point when the observer is able to observe the photon? I have come up with two possibilities:
*
*The wavefunction is collapsed everywhere: the possibility of measuring the photon forces the photon to collapse to some random point in the spherical shell of probable locations
*The wavefunction only collapses if it collapses onto the observer. Otherwise, the wavefunction will be modified to exclude passing through that point
Is one of these possibilities correct? Or are both wrong? I am completing an introductory quantum mechanics course, and trying to build intuition.
Here is a picture about how I am visualizing the setup:
|
The wavefunction only collapses if it collapses onto the observer. Otherwise, the wavefunction will be modified to exclude passing through that point
That's correct, except that what you called "modified" is also a collapse. The case where the photon is not detected, but the wavefunction nonetheless is modified/is updated/collapses to a state reflecting its non-detection, is called interaction-free measurement.
The wavefunction is collapsed everywhere: the possibility of measuring the photon forces the photon to collapse to some random point in the spherical shell of probable locations
This would happen only if you had a detector everywhere on the sphere. "Complete" position measurements where there's a detector everywhere are the norm in introductory quantum mechanics courses, but they never happen in the real world.
| {
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Will the velocity be constant in a vertically oriented tube?
The tube here is placed into a water stream moving with a velocity $v$
My query is will the velocity be constant throughout the pipe or only in some places?
I have conflicting opinions on this, the equation of continuity would say the velocity is constant through the pipe but Bernoulli's equation would say it slows down the higher it went.
Where have I flawed conceptually here?
The person who solved the question said the velocity at the blue points will be the same but in the portion above the level of water, it will not. This just confused me even more.
Link to the question although not necessary.
| The equation of continuity and Bernoulli's equation should always be satisfied (at least in ideal condition i.e. no viscosity and ...)
What you are forgetting is the pressure in Bernoulli's equation. The pressures at the beginning of blue line and at the end of the blue line are not the same. That explains how the velocity in the pipe can be the same everywhere. Let's write down the equation for the beginning of the blue line (point A) and the end of the blue line (point B)
$P_A + \frac{1}{2}mv^2 + \rho g h_A = P_B + \frac{1}{2}mv^2 + \rho g h_B$
$\frac{1}{2}mv^2$ cancels and we get,
$P_A + \rho g h_A = P_B + \rho g h_B$
$P_A - P_B = \rho g (h_B- h_A)$
because $h_B > h_A$ the LHS is positive. Thus, we can conclude that pressure at point A is larger that the pressure at the points B. This should make sense because the water moves from A to B so there should be some force from A toward B.
| {
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How can photons destructively interfere? This is a concept I don't fully understand. If I have two photons each with frequency $\nu$, then they each have an energy of $E = h\nu$. If they get matched with an inverted phase, then the summed wave will be null due to destructive interference. Then where does the energy go? It cannot radiate, since that would produce an extra E-M wave, right?
| In the present day physics standard model photons are elementary particles , on par with the other particles in the table. This means they are point particles, of fixed (in this case zero) mass with spin 1 and $E=hν$ . The $ν$ is the frequency that the classical light will have, as it is composed of zillions of photons This can can be seen experimentally, how classical interference appears because the beam is composed out of a large number of same energy photons.
camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.
Single photons leave a point consistent with the particle nature. It is the accumulation of photons that shows the classical interference pattern.
So two photons will not interfere in any way, except if one is studying photon photon scattering, which is very improbable for low energy photons. For high energy photons, gamma rays, a lot of particle antiparticle pairs can be created and there are plans of gamma colliders.
So there is no problem with the individual photons, they do not interfere. It is the wavefunction of the set up ( in the case above "photon scattering through two given slits given distance apart") that carries the frequency information of the photon, and can thus appear in the probability distribution. This should not be surprising as it is a quantized maxwell equation that gives the photon wavefunctions.
| {
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General plane motion and freely floating rigid body
Consider a rigid rectangular plate of length $l$, width $w$ and thickness $t$ which is at rest and is floating freely in space (no gravity). The center of the plate is at $O_L$ with respect to global coordinate frame $O_G$. The initial pose (position and orientation) $\mathbf{T}$ of the rigid body is assumed to be known and is given by a $3\times 3$ Rotationmatrix and a $3\times 1$ translation vector. Also as shown in the figure, there are $n$ points on the rigid body whose position is known. On each of these points forces are applied which is also known. After time interval $\Delta t$ the pose of the rigid body is given by $\mathbf{T'}$.
Is the information provided above sufficient to find the new pose $\mathbf{T'}$? If not, what information is missing and how do I proceed to find the new value of $\mathbf{T'}$?.
Any comments and suggestions are welcome : )
EDIT
In simple words what I wish to find is a solution (if possible) that says something like: shift the plate by so and so amount in $x$, $y$ and $z$ direction and then rotate by so and so amount about $x,y$ and $z$ axis respectively so that the plate lands at $\mathbf{T'}$.
Please note that the Forces remain constant during the short time interval $\Delta t$.
| you have to solve those equations
\begin{align*}
&m\,\boldsymbol{\ddot{R}}=\boldsymbol{S}(\boldsymbol\varphi)\,\sum_i\,\boldsymbol{F}_i\\
&\boldsymbol\Theta\,\boldsymbol{\dot{\omega}}+\boldsymbol\omega\times\,\left(\boldsymbol\Theta\,\boldsymbol\omega\right)
=\sum_i \left(\boldsymbol{r}_i\times \boldsymbol{F}_i\right)\\
&\boldsymbol{\dot\varphi}=\boldsymbol{A}\,\boldsymbol\omega
\end{align*}
with the initial conditions
\begin{align*}
&\boldsymbol{R}(0)= \boldsymbol{R}_0\\
&\boldsymbol{\dot{R}}(0)= \boldsymbol{0}\\
&\boldsymbol{\varphi}(0)=\boldsymbol{\varphi}_0\\
&\boldsymbol\omega(0)=\boldsymbol{0}
\end{align*}
where
*
*$\boldsymbol{S}$ Rotation matrix between body system and inertial system
*$\boldsymbol{R}$ Center of mass position vector
*$\boldsymbol{\omega}$ Angular velocity
*$\boldsymbol{\varphi}=\left[\alpha~,\beta~,\gamma\right]^T$ the Euler angles
*$\boldsymbol\Theta$ Intertia tensor
\begin{align*}
\boldsymbol\Theta= \left[ \begin {array}{ccc} \frac{m}{12}\, \left( {w}^{2}+{t}^{2} \right) &0&0
\\ 0&\frac{m}{12} \left( {l}^{2}+{t}^{2} \right) &0
\\ 0&0&\frac{m}{12} \left( {l}^{2}+{w}^{2} \right)
\end {array} \right]
\end{align*}
from the solution of the differential equations you obtain the position of the center of mass $~\boldsymbol{R}(t)~$ and the body rotation matrix $~\boldsymbol{S}(t)$
Edit
how to obtain the matrix $~\boldsymbol{A}$
you start with the rotation matrix for example:
\begin{align*}
&\boldsymbol S=\left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left(
\alpha \right) &-\sin \left( \alpha \right) \\ 0&
\sin \left( \alpha \right) &\cos \left( \alpha \right) \end {array}
\right]\,
\left[ \begin {array}{ccc} \cos \left( \beta \right) &0&\sin \left(
\beta \right) \\ 0&1&0\\ -\sin
\left( \beta \right) &0&\cos \left( \beta \right) \end {array}
\right]\,
\left[ \begin {array}{ccc} \cos \left( \gamma \right) &-\sin \left(
\gamma \right) &0\\ \sin \left( \gamma \right) &\cos
\left( \gamma \right) &0\\ 0&0&1\end {array}
\right]\\\\
&\text{with}\\
&\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}}
\\ \omega_{{z}}&0&-\omega_{{x}}\\
-\omega_{{y}}&\omega_{{x}}&0\end {array} \right]
=\boldsymbol{S}^T\,\frac{d}{dt}\,\boldsymbol{S}\\
&\Rightarrow\\
&\begin{bmatrix}
\omega_x \\
\omega_y \\
\omega_z \\
\end{bmatrix}=\underbrace{\left[ \begin {array}{ccc} \cos \left( \beta \right) \cos \left( {
\gamma} \right) &\sin \left( {\gamma} \right) &0\\ -
\cos \left( \beta \right) \sin \left( {\gamma} \right) &\cos \left( {
\gamma} \right) &0\\ \sin \left( \beta \right) &0&1
\end {array} \right]
}_{\boldsymbol{J}_R}\,\begin{bmatrix}
\dot{\alpha} \\
\dot{\beta} \\
\dot{\gamma}\\
\end{bmatrix}\\
&\boldsymbol{A}=\left[\boldsymbol{J}_R\right]^{-1}=
\left[ \begin {array}{ccc} {\frac {\cos \left( \gamma \right) }{\cos
\left( \beta \right) }}&-{\frac {\sin \left( \gamma \right) }{\cos
\left( \beta \right) }}&0\\ \sin \left( \gamma
\right) &\cos \left( \gamma \right) &0\\ -{\frac {
\sin \left( \beta \right) \cos \left( \gamma \right) }{\cos \left(
\beta \right) }}&{\frac {\sin \left( \beta \right) \sin \left( \gamma
\right) }{\cos \left( \beta \right) }}&1\end {array} \right]
\end{align*}
The initial conditions $~\boldsymbol{\varphi}_0=\left[\alpha_0~,\beta_0~,\gamma_0\right]$
with:
\begin{align*}
& \boldsymbol{S}_{t=0}=\left[ \begin {array}{ccc} m_{{1,1}}&m_{{1,2}}&m_{{1,3}}
\\ m_{{2,1}}&m_{{2,2}}&m_{{2,3}}
\\ m_{{3,1}}&m_{{3,2}}&m_{{3,3}}\end {array}
\right]\\\\
&\text{with}~\boldsymbol S= \boldsymbol{S}_{t=0}\\
&\Rightarrow\\
&\tan \left( \alpha_{{0}} \right) =-{\frac {m_{{2,3}}}{m_{{3,3}}}}\\
&\tan \left( \gamma_{{0}} \right) =-{\frac {m_{{1,2}}}{m_{{1,1}}}}\\
&\sin \left( \beta_{{0}} \right) =m_{{1,3}}
\end{align*}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/601472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Entropy reducing power generator An 8.3 EER 1200 W air conditioner will move 10000 BTU of heat per hour. 1 BTU is 1055 J of heat. Per second this is moving 2930 W of heat energy. It is possible to get them even more efficient e.g. 12 EER.
This means 4130 W of heating on hot side, and 2930 W of cooling on cold side.
The heating side of the air conditioner is used to boil water to power a steam turbine. The cooling side is used to produce coolant.
A condensing steam turbine is approx 40% efficient, so the 4130 W of heat will produce 1652 W of electricity, and 2478 W of heat.
The 2478 W of heat is cancelled by the coolant, with 452 W of cooling left over.
The 1200 W of power needed by the air conditioner is cancelled by the turbine, with 452 W of electricity left over.
The excess cold can be released into the environment or used some other way, and the excess power put into the power grid.
I was thinking about this while playing the computer game Oxygen not Included, and wondered if such a machine could exist in reality?
| No, such a machine cannot exist in reality. One critical thing that you neglected in your analysis is the temperature. Every heat engine or heat pump has a hot side and a cold side. The efficiency depends strongly on the temperature difference. The heat pump operates with a much lower heat difference than the steam engine. If you run it at the larger difference as you described then it will be nowhere near as efficient as you quoted.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Generalization of first law of thermodynamics What is the mathematical statement for the first law of thermodynamics, accounting for kinetic energy, potential energy, internal energy, work, heat and most importantly taking into consideration the work-energy theorem?
Also, is $∆U=∆Q-∆W$ only valid for systems whose center of mass is at rest in an inertial frame, or is it also valid for other systems?If so, please specify. Thanks
| The law of conservation of energy states that the total energy of an isolated system is constant; For a thermodynamic process (closed system) without transfer of matter, the first law is often formulated
$$dU=\delta Q+\delta W$$
or
$$dU=TdS-PdV$$
In the case of a closed system in which the particles of the system are of different types and, because chemical reactions may occur, their respective numbers are not necessarily constant, the fundamental thermodynamic relation for $dU$ becomes:
$$dU=TdS-PdV+\sum_i\mu_i dN_i$$
. If the system has more external mechanical variables than just the volume that can change, the fundamental thermodynamic relation further generalizes to:
$$dU=TdS-\sum_i X_idx_i+\sum_i\mu_i dN_i$$
Here the $X_i$ are the generalized forces corresponding to the external variables $x_i$.
The first law of thermodynamics is equivalent to the law of conservation of energy: energy cannot be created or destroyed; the total amount of energy in the Universe is fixed. Still, a modification is needed as you talk about non-inertial frames. As the normal energy conservation laws need modification as you go to the non-inertial frame. For example, the generalized force gets transform when you get into non-inertial frames.
Thus you need to account for the fact that you are in a non-inertial frame. Other than that it's energy conservation and thus valid everywhere.
| {
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Why can't dark matter lose energy by gravitational waves and collapse into itself? Because of lack of electromagnetic induction, dark matter can't lose its gravitational potential energy. That is preventing it from collapsing like an ordinary matter cloud in space.
But why can't dark matter lose energy by gravitational waves and collapse into itself?
| Well, we do not know what it is. On top of that IF it is made of particles, the density is very very low. So, it needs a lot of time to collapses.
| {
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Why does air pressure decrease with altitude? I am looking to find the reason: why air pressure decreases with altitude? Has it to do with the fact that gravitational force is less at higher altitude due to the greater distance between the masses? Does earth’s spin cause a centrifugal force? Are the molecules at higher altitude pushing onto the molecules of air at lower altitudes thus increasing their pressure? Is the earths air pressure higher at the poles than at the equator?
|
Has it to do with the fact that gravitational force is less at higher
altitude due to the greater distance between the masses?
The gravitational force does decrease as you go higher up, but that's not the reason. The pressure would still be greater at the bottom even in some weird physics where gravity got stronger further from the surface.
Does earth’s
spin cause a centrifugal force?
It does, but again, that's not part of the reason.
Are the molecules at higher altitude
pushing onto the molecules of air at lower altitudes thus increasing
their pressure?
Yes. That is exactly the answer.
Is the earths air pressure higher at the poles than at
the equator?
No. Even if the effective gravity is different, air at sea level will flow from where there is more pressure to where there is less until it balances out. Of course, pressure changes due to weather but over time I believe seal level pressure is the same around the world.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/602020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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When did liquid first appear? When and where did matter in liquid state first appear in our universe?
Liquid has never been a global state of the universe, but it certainly appears nowadays locally in some places.
As far as I know, liquid currently only exists on planets, but I can well be mistaken.
I can imagine a couple of different possibilities for first appearance:
*
*After the first generation of supernovae there were heavier elements outside stars and they could form planets with suitable composition to have liquids.
*Perhaps there were early gas planets of suitable size around the time that stars started forming that produced suitable conditions for liquid hydrogen inside. The outer layers would be gaseous, but perhaps pressure could produce liquid inside.
*If the previous scenario is impossible without an external source of heat, maybe during the first generation of stars it would have been easier.
Or perhaps it was something else.
I am not looking for a transient situation which could be vaguely be described as a liquid for a fraction of a second.
Instead, I am looking for something in or close enough to equilibrium so that the liquid stays around for at least a minute and there is at least a liter of it.
Feel free to adjust the parameters, but I hope my intention is clear enough.
| Have a look at a paper 1312.0613 by A. Loeb. He considers possible existence of rocky planets with liquid water as early as $100<(1+z)<137$ (about 10-17 million years since Big Bang).
From the footnote 2:
After the first stars formed, the subsequent delay in producing heavy elements from the first supernovae could have been as short as a few Myr. The supernova ejecta could have produced high-metallicity islands that were not fully mixed with the surrounding primordial gas, leading to efficient formation of rocky planets within them.
So the mechanism for liquid formation would be OP's item 1.
| {
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Dirac fields: Do particle and antiparticle creation operators act differently on the vacuum? Given a Dirac field $$\Psi(x):=\int\frac{d^4k}{(2\pi)^4}\delta\left(p_0-\omega(\mathbf{k})\right)\sum_s\left(a_s(k)u_s(k)e^{-ikx}+b^\dagger_s(k)v_s(k)e^{ikx}\right)$$
with the creation operators $a^\dagger_s(k),b^\dagger_s(k)$ for particles and antiparticles respectively, how do these operators act on the vacuum?
In particular, is it true that $|k\rangle=a^\dagger_s(k)|0\rangle=b^\dagger_s(k)|0\rangle$?
| Ah I think I understand your question now and I think this is a simple notational issue. The single particle states for the particles and antiparticles should be denoted differently, i.e. trying to be as close to your notation would give something like
$$|k,s\rangle \equiv a^\dagger_s(k)|0\rangle \ \ \ \ , \ \ \ \ |\tilde{k},\tilde{s}\rangle \equiv b^\dagger_s(k)|0\rangle \ .$$
And all the usual commutation relations are the same. Perhaps more standard notation would be $|1_{k}\rangle \equiv a^\dagger_s(k)|0\rangle$ and $|\bar{1}_{k}\rangle \equiv b^\dagger_s(k)|0\rangle $, but I'm not totally sure what's most common.
| {
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Why only reversible adiabatic processes are called isentropic and not irreversible ones? For both reversible and irreversible adiabatic processes, $dQ =0$,and by the definition of entropy $dS=dQ/T$, it should imply that entropy is constant for both. Why it is not so?
| In general, you can write that for all processes $\frac{\delta q}{T} \le dS$ (Clausius) but in the case of a reversible process you have equality $\frac{\delta q}{T}|_{rev} = dS$. If the process is adiabatic then by definition $\delta q =0$ hence for the reversible case you have the equality $dS=0$, i.e., an isentropic process, but for an irreversible process you can only say that $0<dS$ which is usually verbalized by saying that the entropy can only increase (never decreases) in an adiabatic process that is not reversible.
| {
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Is it the gravitational field created by an object which exerts force on another object or the interactions between their fields that exerts force? We are taught that Gravitational force exerted by an object is a two-step process:
*
*The object creates a field around it.
*The field exerts a force on bodies present in the field.
Now, since we know that an object any object of certain mass creates its own field, is it correct or incorrect to say that 'Field-field interactions exerts force on the bodies'. Can a similar doubt be raised in case of Electric fields created by charges?
| Gravity is neither a force or a field. It is the effect of matter, or energy, on the curvature of spacetime, which affects the matter or energy imbedded in that spacetime. The single least understood component of physics as it is currently understood. Many paradoxical aspects of gravity exist.
Most notably, gravity at the quantum scale. It does not dovetail into either special or general relativity.
A breakthrough is needed, and is highly sought after.
| {
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Last step derivation of hamiltonian of particle in electromagnetic field My textbook quantum mechanics does an analogues derivation of the hamiltonian as given here, but I'm struggling to understand the last step: The final obtained hamiltonian is (in my textbook's case)
$$H = \frac{1}{2}m\dot{\vec{x}} + eV$$
with V the electromagnetic potential, my professor (and the link) then claims that this is the same as:
$$H = \frac{1}{2m}(\vec{p} - \frac{e}{c}\vec{A})^2 + eV$$
but when I expand the bottom equation I get:
$$\frac{1}{2}m\dot{\vec{x}} + eV -\frac{e}{2mc}(\vec{A}\cdot\vec{p} + \vec{p}\cdot\vec{A}) + \frac{1}{2m}\left(\frac{e}{c}\right)^2A^2$$
How do the two terms on the right side cancel out?
Thanks in advance.
| The mistake you're making is in assuming that $$p_x = m \dot{x}.$$ This is not true. $m \dot{x}$ is known as the kinetic momentum, while $p_x$ is known as the conjugate momentum (to the coordinate $x$). In most (all?) problems in Classical Mechanics these two concepts are equal, but they aren't the same in general.
The conjugate momentum is obtained from the Lagrangian, using the relation $$p_x = \frac{\partial \mathcal{L}}{\partial \dot{x}}.$$
In most "usual" problems in Classical Mechanics, the only dependence the Lagrangian has on velocity is in the kinetic energy term, and thus $$\frac{\partial \mathcal{L}}{\partial \dot{x}} = \frac{\partial}{\partial \dot{x}}\left(\frac{1}{2}m\dot{x}^2\right) = m \dot{x},$$
but in electromagnetism, the Lagrangian is given by: $$\mathcal{L}_\text{em} = \frac{1}{2} m \dot{x}^2 - qV + q \dot{x}A_x,$$ and so $$p_x = \frac{\partial \mathcal{L}_\text{em}}{\partial \dot{x}} = m \dot{x} + q A_x.$$
As a result, $m \dot{x} \neq p_x$, but rather $$m \dot{x} = p_x - qA_x.$$ Using this in formula for the Hamiltonian: $$\mathcal{H} = \frac{1}{2}m \dot{x}^2 + qV = \frac{(p_x - qA_x)^2}{2 m} + q V.$$
| {
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Did I make an electric circuit with my cat? This is definitely the dumbest question I've asked, but I'm curious. My cat was resting her head on my left hand while I pet her with my right. Suddenly, my left hand began receiving regular static electric shocks. Was I pulling electrons from her fur with my right hand, thus causing them to flow back through my left?
If it really is an example of a circuit, I was thinking it's a pretty hilarious way to explain circuits to students.
| I think you did steal a bunch of electrons from your cat. Hair typically loses electrons very quickly. For more information look up triboelectric series.
| {
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Is this constraint equation written right here?
This is a constraint equation he has written here.
This is a trick to solve constraint equations faster.
Now he has made a equation that a2 ( acceleration of 2kg block) and a1(acceleration of 1kg block).
a2 = 2a1.
I have done earlier constant equation by other method.I think it should be wrong since the directions of both acceleration should be different.So it should be -a2 = 2a1.
But am not if I am right.He has written these statements by saying that work done = tension force * their displacement (of blocks).
| Your question will not be understood by people who don't know this trick you should provide a link or give a short derivation. That being said,
$\sum T.a$ Is the summation of the dot product of the tension force and acceleration you should know that the dot product is the product of their magnitudes and the angle between them.
So the equation he arrived at is the relation between the magnitudes of $a_1$ and $a_2$
| {
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Scalar product of fields in Schwarzschild space-time The scalar product of fields in curved space-time is defined by (Birrel, Davies)
$$\left(\phi_{1}, \phi_{2}\right)\equiv-\mathrm{i} \int_{\Sigma} \phi_{1}(x) \overset{\leftrightarrow} {\partial_\mu}\phi_{2}^{*}(x)\left[-g_{\Sigma}(x)\right]^{\frac{1}{2}} \mathrm{~d} \Sigma^{\mu}$$
where $\mathrm{d} \Sigma^{\mu}=n^{\mu} \mathrm{d} \Sigma$, with $n^\mu$ a future-directed unit vector orthogonal to the spacelike hypersurface $\Sigma$ and $\mathrm{d}\Sigma$ is the volume element in $\Sigma$. If $\Sigma$ is taken to be a Cachy surface in the globally hyperbolic spacetime, then the scalar product is independent of $\Sigma$.
I don't know how to apply this formula to a particular example. For instance, consider a Schwarzschild spacetime. How will this formula look like when I choose $\Sigma$ to be the past light-like infinity (of course, one can choose from many coordinate systems)?
| First of all that is not a scalar product but it is a symplectic form.
Yes you can use the lightlike infinity provided
(a) no information escapes through the timelike infinity,
(b) you have rewritten that integral in the language of differential forms ,
(c) the field is massless (otherwise it vanishes too fast before reaching the past light infinity).
I used lots of times that mathematical technology in the past.
See for instance this couple of papers https://arxiv.org/abs/gr-qc/0512049 and
https://arxiv.org/abs/gr-qc/0610143 I wrote in the past (around Eq.(47) in the former there is an explicit discussion about the expression of the symplectic form in terms of differential forms) and the rigorous construction of the Unruh state in the Kruskal-Schwarzshild manifold I obtained in collaboration with two colleagues https://arxiv.org/abs/0907.1034. Also this brief monography https://doi.org/10.1007/978-3-319-64343-4 may be useful.
In these papers you can find the explicit version, in terms of forms, of that symplectic form in a manner that can be used for light like 3-surfaces.
| {
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Why does wind blowing across a bird bath cause the water to freeze even though the ambient air temp is 39 - 40 degrees F? Frequently, after cold frontal passage, a strong NW wind blows across the open marsh and through our back yard. With ambient temps still well above freezing, the surface of the water in our concrete bird bath begins to freeze. Venturi? Bernoulli? Just heat transfer ("wind chill")?
| Surely evaporative cooling is a significant factor. The wind carries away water molecules that have evaporated from the water, so not as many return to the water (through collisions with air molecules) than without the wind. Therefore the net evaporation rate is enhanced, and so is the evaporative cooling.
The cooling occurs because only water molecules with considerably more than the mean kinetic energy can escape from the water surface, so lowering the mean kinetic of the remaining water molecules, and hence the temperature.
This effect can be demonstrated with a fan blowing air over a shallow bowl of water. The fall in water temperature is not huge, but if the water temperature and the air temperature are only a few degrees above freezing in the first place, the water might just be made to freeze.
| {
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What happens to phase after wavefunction collapse? Suppose an initial quantum state $\psi = a_1\phi_1 + a_2\phi_2 + ... + a_n\phi_n$, where $\phi_i$ is the eigenfunction with eigenvalue $\lambda_i$ of some measurement operator. Post-measurement, we will find the system in state $\phi_i$ with probability $|a_i|^2$.
What happens to the phase post-measurement? The principle that immediate subsequent measurements should always return the same value would be satisfied no matter the resulting phase. We might find the system in any state $b\phi_i$, so long as $|b|^2=1$. I am sure the postulates of quantum mechanics specify something about this, but I haven't managed to find any text that addresses it. What should $b$ be?
| In quantum mechanics, states are represented by rays in Hilbert space, or more accurately, the space of states is projective Hilbert space - for example, for a finite dimensional system, the space is $H_n / \sim \ \cong \mathbb{C}P^{n-1}$, where for $u, v \in H_n$, $u \sim v$ if $u = \alpha w$ for some non-zero complex number $\alpha$.
Now usually we prefer to work with the plain Hilbert space rather than the projective one, choosing to impose the quotient whenever useful - simply because we have many more useful tools at our disposal while working with Hilbert spaces.
However, you must always remember that the actual space of states is the projective Hilbert space, which means that the statement "We might find the system in any state $b\phi_i$ as long as $|b|^2 = 1$" is meaningless, because there aren't separate states $b\phi_i$ - neither is it that all these states are the "same" - the real reason is that there is only one state $\phi_i$ in projective Hilbert space.
| {
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What would happen if a 10-kg cube of iron, at a temperature close to 0 kelvin, suddenly appeared in your living room? What would be the effect of placing an object that cold in an environment that warm? Would the room just get a little colder? Would it kill everyone in the room like some kind of cold bomb? What would happen?
Don't think about how the cube got there, or the air which it would displace.
| There are about 30 m3 of air in a room, and the density of air is about 1.2 kg/m3. So the mass of air in the room would be about 36 kg. The heat capacity of air is about 1 kJ/kg-K. So the mass times heat capacity of the air is about 36 kJ/K. The mass times heat capacity of the 10 kg of iron is about 4.5 kJ/K. So, if the room started out at 300 K, the final equilibrium temperature (neglecting the walls) of the room would be about $$\frac{(36\ kJ/K)(300\ K)}{36\ kJ/K + 4.5\ kJ/K}=267\ K$$
| {
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Understanding dependent/independent variables in physics How does one determine the independent and dependent variables?
What do the terms mean?
Can they be derived from a formula?
For example I saw in a textbook $F = k\Delta l$, Hooke's Law, that $F$ is the independent variable. Is this because $\mathbf {F} $ is the subject, therefore it is independent?
|
How does one determine the independent and dependent variables?
It's totally relative. In $F = k\Delta l$ all three variables can be considered dependent or independent, depending on your purpose.
E.g. in $\Delta l=\frac{F}{k}$, $\Delta l$ would now be considered the dependent variable.
Now suppose you studied a set of different springs, so that:
$$k=\frac{F}{\Delta l}$$
$k$ is 'normally' the proportionality constant (or factor) but in that study it would be the dependent variable.
| {
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Does Work become state function in an Isothermal Process and what are other processes in which it happens? In a reversible isothermal process and for an ideal gas we know from the definition of Helmholtz free Energy
$dF= -SdT -PdV$.
And as temperature doesn't change for an isothermal process, $dT$ must be zero. So dF can be written negative of change in Helmholtz free Energy. Since $F$ is a state function and $dF$ a perfect differential, work also should be.
Also, does work become state function for adiabatic processes also? Please throw light on it.
| Here is an additional analysis that is consistent with my previous answer:
For a process in a closed system, the first law of thermodynamics tells us that $$\Delta U=Q-W$$Now, if we define an isothermal process (either reversible or irreversible) as one in which the temperature of the system in its initial and final thermodynamic equilibrium states is T, and that, during the process, all heat transfer takes place (at the interface) between the system and surroundings at temperature T, then from the Clausius Inequality, we have $$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is the entropy generated during the process as a result of irreversibility (a positive definite quantity). If we combine these two equations, we have $$\Delta U=T\Delta S-T\sigma-W$$or$$W=-\Delta F-T\sigma$$From this it follows that for any isothermal process path, the work done by the system on the surroundings is less than (irreversible) or equal to (reversible) the decrease in the Helmholtz free energy.
Here is a specific example: If we have one mole of an ideal gas at $P_1$ and $V_1$ and we suddenly drop the external pressure on the gas to $P_2$, and then let it equilibrate, what is the change in F and how much work is done on the surroundings. Well, the change in F is just $$\Delta F=-\int_{V_1}^{V_2}{\frac{RT}{V}dV}=-RT\ln{\left(\frac{V_2}{V_1}\right)}=-RT\ln{\left(\frac{P_1}{P_2}\right)}$$The work done on the surroundings is $$W=P_2(V_2-V_1)=P_2V_2\left(1-\frac{V_1}{V_2}\right)=RT\left(1-\frac{P_2}{P_1}\right)$$Mathematically, the decrease in F, given by $RT\ln{\left(\frac{P_1}{P_2}\right)}$ is always greater than the work W, given by $RT\left(1-\frac{P_2}{P_1}\right)$, irrespective of the pressure ratio (even for compression).
| {
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How do we know it is the Centripetal Force we have to calculate? A helicopter rotor blade is 7.80 m long and has a mass of 110 kg. (a) What force is exerted on the bolt attaching the blade to the rotor axle when the rotor is turning at 320 rev/min? (Hint: For this calculation the blade can be considered to be a point mass at the center of mass. Why?
| There are actually three important components of the bolt force: vertical, to keep the blade from falling, horizontal-tangential which pushes or pulls on the blade perpendicular to its length, and horizontal, to keep the blade from flying away from the rotor axle.
In order to calculate the vertical force you would need to know the shape and size of the fulcrum that keeps the blade level. Since that isn't given to you, you can't calculate a number. One could make a statement like "if the bolt is $x$ meters from the end of the support structure to which the blade is attached" you could express the vertical force in terms of $x$. I doubt that is what the question intends for you. Also, we don't know anything about the lift exerted on the blade, nor the weight of the helicopter.
To calculate the horizontal-tangential you would need to know air resistance and whether the blade is increasing, decreasing, or constant in rotational speed. There's not enough information to calculate this.
Your only remaining choice is the horizontal force. Because the motion of the blade is circular, that force must be providing the centripetal acceleration for the blade to move on that non-linear path.
So $$\sum \vec{F} = m\vec{a_c} = m\omega^2 r (-\hat{r}) $$
| {
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How is a tensor operator defined in terms of commutators? If $J_i$ represent the angular momentum operators, then a scalar operator $S$ (rank-0 tensor) is defined as an operator which satisfies $$[S,J_i]=0$$ for $i=1,2,3$.
$A_i$ is a vector (rank-1 tensor) operator, if it satisfies $$[J_i,A_j]=i\hbar\epsilon_{ijk}A_k$$
How does a rank-2 tensor operator defined in terms of commutators?
| Given
\begin{align}
\hat L_\pm \vert \ell m\rangle &= \sqrt{(\ell\mp m)(\ell\pm m+1)}\vert \ell,m\pm 1\rangle\, ,\\
\hat L_0 \vert \ell m\rangle &= m \vert \ell m\rangle
\end{align}
then by definition $\hat T^{(\ell)}_m$ commutes as
\begin{align}
[\hat L_\pm, \hat T^{(\ell)}_m]&=\sqrt{(\ell\mp m)(\ell\pm m+1)}\,\hat T^{(\ell)}_{m\pm 1}\, ,\\
[\hat L_0, \hat T^{(\ell)}_m]&=m\,\hat T^{(\ell)}_{m}\, ,
\end{align}
valid for any $\ell,m$.
| {
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How impulse is fundamentally different than momentum? Are impulse and momentum really different quantities?
Is there a conservation law of Impulse too just like momentum conservation.
And lastly, will impulse be conserved for the cases where body (say a rod)is hinged. I know that momentum conservation can't be applied to cases where body is hinged.
| Momentum is the base quantity. Impulse is the change in momentum from an initial value $\mathbf{p}_i$ to a final value $\mathbf{p}_f$:
$$\mathbf{I} = \Delta \mathbf{p} = \mathbf{p}_f - \mathbf{p}_i$$
However, and hence, they both have the same dimensions and units (force times time, e.g. newton-seconds and prefixed derivatives in SI).
| {
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Do pseudovectors also transform differently to vectors under spatial dilation, not just reflections and parity? It is frequently expressed online that the only difference between vectors and pseudovectors is a change in sign with reflection/parity transformations etc.
For instance the pseudovector angular momentum $\vec L=\vec r\wedge \vec p$ keeps the same direction under parity inversion while the vectors $\vec r$ and $\vec p$ change sign.
However is it not also true that under spatial dilation transformations, the pseudovector angular momentum would stretch by $\kappa^2$ (if $\vec r\rightarrow\kappa \vec r$ and $\vec p=m \frac{d \vec r}{dt} \rightarrow m \frac{d (\kappa \vec r)}{dt}=\kappa \vec p$) while the vectors $\vec r$ and $\vec p$ only stretch by $\kappa$? So it isn't just the difference under parity transformations etc?
| Well, one possibility is that this is just a matter of where attention is focused. Since, excluding the weak force, all physical laws are invariant under parity, it's interesting to note that pseudovectors transform differently than vectors under that operation. However, real physics isn't, as far as we know, invariant under spatial dilations, and so for most people this isn't an interesting (approximate) symmetry and likely not included in the mental set of operations one could apply to vectors or pseudovectors. Perhaps researchers of conformal field theory have a different point of view.
A second point is that $\vec{L} = \vec{r} \times \vec{p}$ is just one equation for one pseudovector. By dimentional anslysis, the angular momenentum is proportional to the length scale squared, and so this double-dilation could be viewed as just its dimensional properties. Pseudovectors are, more technically, elements of the exterior algebra of a vector space, whose basis elements are the wedge products of basis elements of the regular vector space. Since physical space has dimensions, so do its basis elements, and thus the basis elements of the exterior algebra are dimensionfull as well.
| {
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Why do we need instantaneous speed? I am new to this topic and was just wondering about the use of instantaneous speed. I mean, we use to calculate the speed of car let us say at 5 sec. So we take the distance travelled in 4.9 to 5.0 seconds and divide it by time. We get instantaneous speed. We could simply as well have had taken distance travelled from 0 to 5 seconds and then divide it by time. So what is the use of instantaneous speed then?
| Because instantaneous speed affects physics. Imagine a wall $10~\textrm m$ in front of you. You walk towards it smoothly over a timeframe of, say, $20~\textrm s$, and without getting slower, you walk into the wall. You'll feel a slight bonk, but nothing serious is going to happen. Now imagine the same 20 seconds going differently: You wait for 17 seconds, then you sprint towards the wall at full speed.
Both scenarios will give you the same average speed over the 20 seconds, but you better be wearing a helmet for the second one. The difference lies in the fact that the instantaneous speed at the end of the 20 second interval is different. It's a quantity that affects things. So it makes sense to talk about it.
| {
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Is displacement current the rate of change of induced polarization? I know that the displacement current is proportional to the rate of local polarization change. Yet, I'm not sure if it is technically correct to attribute the displacement current to the rate of change in the induced polarization, as polarization itself is kind of the induced displacement of charges due to some external field.
My gut feeling says that it's not wrong. However, it's confusing as "induction" is really reserved for another phenomenon. But I could be (very) wrong. What do you think?
| No, you can have a displacement current in a vacuum, where there is no polarization possible.
| {
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Application of the Cartan Structure Equations seems to imply the Einstein-Palatini action is zero? The Einstein-Palatini action can be written as
$$ S = M_{pl}^2\int\varepsilon_{abcd}\left(e^a\wedge e^b\wedge R^{cd}\right), $$
where $e^a={e^a}_\mu\text{dx}^\mu$ is the basis one-form and $R^{ab}=\frac{1}{2}{R^{ab}}_{\mu\nu}\text{dx}^\mu\wedge\text{dx}^\nu$ is the Riemann curvature two-form. The Cartan Structure Equations for the torsion-less and metric-compatible connection of GR are
$$ R^{ab} = D\omega^{ab} = d\omega^{ab} + {\omega^a}_c \omega^{bc}, \quad 0 = De^a = de^a + {\omega^a}_be^b, $$
where $\omega^{ab}={\omega^{ab}}_\mu\text{dx}^\mu$ is the (antisymmetric) spin connection one-form.
Now, my confusion comes from the fact that if I apply the first structure equation, integrate by parts (neglecting boundary terms), and apply the structure second equation, the whole action seems to vanish.
\begin{align}
S &= M_{pl}^2\int\varepsilon_{abcd}\left(e^a\wedge e^b\wedge D\omega^{cd}\right) \\
&= M_{pl}^2\int\varepsilon_{abcd}\left(-D(e^a\wedge e^b)\wedge\omega^{cd}\right) \\
&= M_{pl}^2\int\varepsilon_{abcd}\left(-De^a\wedge e^b\wedge\omega^{cd} + e^a\wedge De^b\wedge\omega^{cd}\right) \\
&= M_{pl}^2\int\varepsilon_{abcd}\left(0+0\right)=0
\end{align}
This obviously doesn't seem correct, so is there an error in my understanding somewhere? Is it incorrect to use the structure equations and integrate by parts in the action in this way?
| This part of yours is incorrect:
$$
\begin{align}
S &\sim \int\varepsilon_{abcd}\left(e^a\wedge e^b\wedge D\omega^{cd}\right) \\
&= \int\varepsilon_{abcd}\left(-De^a\wedge e^b\wedge\omega^{cd} + e^a\wedge De^b\wedge\omega^{cd}\right)
\end{align}
$$
The correct derivation is:
$$
\begin{align}
S &\sim \int\varepsilon_{abcd}\left(e^a\wedge e^b\wedge D\omega^{cd}\right) \\
&= \int\varepsilon_{abcd}\left(-(de^a + \frac{1}{2} {\omega^a}_k\wedge e^k)\wedge e^b\wedge\omega^{cd} + e^a\wedge (de^b+ \frac{1}{2} {\omega^b}_k\wedge e^k)\wedge\omega^{cd}\right) \\
\end{align}
$$
Obviously:
$$
de^a + \frac{1}{2} {\omega^a}_k\wedge e^k \neq De^a= de^a + {\omega^a}_k\wedge e^k
$$
Added note after discussion with @JeffK (the person who originally asks the question), since @JeffK opted to stand by the wrong calculation.
Moving covariant derivative $D$ between $\omega$ and $e$ is a bit tricky, because of the unique definition of $D\omega$.
One can arrive at the Einstein action by applying the zero torsion condition
$$
0 = De^a = de^a + {\omega^a}_be^b
$$
to the Einstein-Palatini action, via expressing $\omega$ as a function of $e$, effectively eliminating explicit dependence of the Einstein action on $\omega$.
The Einstein action is apparently not zero. I hope @JeffK can see the light and would avoid making similar mistakes in future study/research.
| {
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Is there a phase shift $\pi$ radians when a pressure wave is reflected upon a medium having less acoustic impedance? In my text book it is written that a sound wave modeled as pressure fluctuations does not undergo a phase shift of $\pi$ radians upon reflection as there will be a pressure antinode at the interface and the pressure needs to be continuous at the interference. I wonder if a phase shift $\pi$ radians occurs when a pressure wave is reflected upon a medium having less acoustic impedance, as this would mean there is a pressure node at the interference?
| Short answer: Yes, for a pressure wave reflected from a surface with less acoustic impedance than the original medium the phase shift would be $\pi$ radians, neglecting absorptive processes.
Longer answer:
I have some equations, so let me provide the full context. Consider a plane pressure wave propagating in a homogeneous fluid and is normally incident upon a flat interface to another domain. Let the characteristic acoustic impedance of the original domain be denoted as $Z_1$ and for the second domain as $Z_2$. The reflection coefficient may then be written as
$$
R = \frac{Z_2-Z_1}{Z_2+Z_1}.
$$
In general $Z_1$ and $Z_2$ may be complex quantities (frequency domain; they would be convolution operations in the time domain), but if we neglect absorptive processes they are real. Thus, for $Z_2>Z_1$ (e.g., rigid surface) the reflection coefficient is positive, and the phase change is 0 radians. If $Z_2<Z_1$ the reflection coefficient is negative, which is a phase change of $\pi$ radians. Obviously, complex impedances lead to reflection coefficients that are not restricted to 0 or $\pi$ radians phase shifts, and in general are not those simple cases.
| {
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Where does the law of conservation of momentum apply? Take the scenario of a snowball hitting a tree and stopping. Initially, the snowball had momentum but now neither the snowball nor tree have momentum, so momentum is lost (thus the law of conservation of momentum is violated?). Or since the tree has such a large mass, is the velocity of the tree is so small that it's hardly noticeable?
If the explanation is the latter, this wouldn't hold for a fixed object of smaller mass. So in that case, how would the law of conservation of momentum hold?
| The snowball will likely break apart and some of that momentum will be transferred to its subsequently far-flung constituent far flung parts. It will probably deform, transferring some of that momentum into the new shape. Also, some of that kinetic energy will be converted to heat, both into the tree and into the material of the snowball itself. Then there is the sound produced when the snowball hits, taking some of the energy of momentum away with it.
| {
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Angular Momentum of a rigid body w.r.t to an axis about which it is not rotating suppose we need to calculate $\overrightarrow{L}$ about an axis, but the rigid body is not rotating about this axis. Can we define the $\overrightarrow{L}_{axis}$ still? I think we should be able to since $\overrightarrow{L}$ (for an infinitesimal point on the rigid body) is just $dm(\overrightarrow{r}\times\overrightarrow{v})$
If it is possible then is there a way to determine angular momentum in such cases more systematic approach (with proof) for
*
*Special case where we have to only deal with axis parallel to axis of rotation
*The general case i.e. the axis of rotation is tilted/skewed w.r.t to axis of rotation.
| Yes. If the angular momentum of a body about the origin is $\vec{L}=\vec{r} \times \vec{p}$, the angular momentum about an axis in the direction $\hat{n}$ is given by-
$$L_{axis}=|\hat{n}.\vec{L}|$$ You can see that for yourself pretty easily. This is simply the projection of the angular momentum of the body about that axis. The perpendicular component of the angular momentum will vanish in the axis frame.
This answers all the remaining questions as well.
| {
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When a microwave oven stops, why are unpopped kernels very hot and popped kernels not hot? A commonplace empirical observation is that when a microwave oven stops, unpopped kernels are very hot (it's physically painful to touch them) and popped kernels are not.
Is there an elementary (or not) exposition of the physics involved?
| Popcorn pops in a microwave oven due to the microwaves interacting with the moisture in the popcorn kernel raising its internal temperature and pressure. Once the pressure increases enough the kernel pops and the moisture escapes and cools. The moisture in the un popped kernel remains hot.
Hope this helps.
| {
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Why isn't the magnetic field density zero here? In my lecture notes for magnetostatics, my professor has this explanation of why H is not necessarily $0$ that I dont understand.
$$\nabla \times \bf{H} = \bf{J} \\ \bf{J}=0 \Rightarrow \nabla \times \bf{H} = 0 \not\Rightarrow H = 0 \\ \nabla \cdot H \neq 0 $$
"H is only fully defined by its curl and divergence."
I thought that the divergence of B is always zero and since B and H are related by only a constant divergence of H should also be $0$ so I am not sure why the last expression is there. Please help me understand his explanation.
| The field $\vec{H}$ in a medium is defined as
$$\vec{H}(r,t) = \frac{1}{\mu_0}\vec{B}(r,t)- \vec{M}(r,t)$$
where $M(r,t)$ is the magnetization field which depends on the medium properties. Maxwell's law states that $\vec{\nabla}.\vec{B}=0$, i.e., magnetic monopole does not exist. But that does not imply that there is not magnetic polarization inside the medium. The presence of the $\vec{M}$ field may give rise to the nonzero divergence of $\vec{H}$.
However, if we are talking about the vacuum or a medium that follows $\vec{H}(r,t) = \frac{1}{\mu}\vec{B}(r,t)$ where $\mu$ is just a constant independent of spatial coordinates (usually linear media), then from Maxwell's law it follows that $\vec{\nabla}.\vec{H}=0$.
| {
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Partial derivative of thermodynamics properties I know that in finding the partial derivative of certain thermodynamics property such as $H=H(P,V)$, we can hold the other variable as constant. But what will happen if the relation have more than two variables? For example, if a certain thermodynamics property of pure substance is given by $\Gamma$ where,
$$ \Gamma=S-\frac{U}{T}-\frac{PV}{T}$$
If I need to to find the value of $\left(\frac{\partial\Gamma}{\partial P}\right)_T$, when I perform a partial derivative can I also consider other thermodynamics properties (S and U) as constant and the equation becomes
$$\left(\frac{\partial\Gamma}{\partial P}\right)_T=0-0-\frac{V}{T}$$
| When you evaluate that partial derivative, you do not need to consider other properties, U and S, as constant. You have $$d(T\ \Gamma)=TdS+SdT-dU+VdP+PdV$$But, $$dU=TdS-PdV$$Therefore, $$d(T\ \Gamma)=\Gamma dT + Td\Gamma=SdT-VdP$$So, $$\left(\frac{\partial \Gamma}{\partial P}\right)_T=-\frac{V}{T}$$
| {
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Does all EM radiation consist of photons? Since university I've always accepted without question, that electromagnetic radiation consists of photons; but is it true?
I believe it was Einstein who demonstrated that light comes in small packets, 'photons', and since light is EM radiation, it is tempting to conslude that all such radiation consists of photons. On the other hand, most (all?) light originates from discrete processes in atoms, so perhaps that is why it comes in discrete packets?
To take an extreme example, if I charge something, say a metal sphere, and then 'wave it about', I will generate EM radiation - will that radiation consist of photons?
| Photons are what light is and they can have frequencies anywhere from radio to gamma. Billions of coherent photons resemble a wave.
| {
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What does it mean that a neutron has a 'negative' magnetic moment? Most questions about this ask why, or how, a neutron has a magnetic moment at all, or why it is negative....
But I am curious as to what it means, physically or experimentally, for a magnetic moment to be 'negative'....
I am reading that a neutron's 'angular-momentum' spin is pointing in the opposite direction of its 'magnetic-moment' spin, but I thought that the quantum spin of a particle WAS its magnetic moment spin....
At any rate, what experiment(s) showed that neutrons have a 'negative' magnetic moment spin? Maybe reading about that will help...
Edit: P.S.: I still don't understand how a neutron's negative magnetic moment is actually manifested... Perhaps an antiproton or electron, being negatively charged, is antiparallel, or whatever, but a neutron is neutral....
| Suppose that a spinning body has equal amounts of positive and negative charge --- so its total charge is zero --- and the positive charge is near the rotation axis, while the negative charge is further away from the axis. Then the magnetic moment will point in the opposite direction to the spin. Presumably this is a classical model of the charge distribution in a neutron.
| {
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Which force exerted a torque? Suppose a horizontal disc fixed in the center with a vertical shaft passing perpendicular to the plane, is rotating at some angular speed and there is an insect sitting initially at center. The insect then starts moving radially outwards. From conservation of angular momentum of insect+disc system, we can state that the angular speed of disc reduces as the moment of inertia of the insect with respect to the axis increases. My question here is that, if we see only the disc without the insect as the system(or as a single rotating object), the angular speed of it changed and hence, some external torque has definitely acted upon the disc. What is the force that produced this torque?
I tried to think about it -
If we say that the insect is moving along $x$ axis and the axis of rotation is the $z$ axis,
the weight of the insect should not cause any torque because the cross product of distance and insect's weight lies in $y$ axis while we need a torque about z axis. Therefore, there must be a force along the y axis, what can be such a force? Can anyone help me out in analyzing the motion and dynamics out here?
|
Therefore, there must be a force along the y axis, what can be such a force?
Friction is that force. It speeds the bug as it moves to tangentially faster moving parts of the ring. Resulting in an equivalent and opposite force slowing the ring down. In the disc's frame friction opposes the fictitious Coriolis force.
Tangentially faster? Isn't the ring slowing down too? Here is why the angular momentum of the bug always increases even though the ring slows down.
Let the total angular momentum of the system be 'L' which it =$I_{system}\omega_{system}$
so the angular velocity of the bug as a function of distance from the axis(r):$$ \frac{L}{\frac{MR^2}{2}+Mr^2}$$
Assuming ring and bug to have the same mass for ease of calculation.
The angular momentum of the bug will be$$ \frac{L}{\frac{MR^2}{2}+Mr^2}Mr^2$$$$ \frac{Lr^2}{\frac{R^2}{2}+r^2}$$
Let the disc have a radius of 5 and the total angular momentum 10 all in its SI units.
Graphing our result:
The blue line represents r=5 which is the radius of the disc and it's where the bug stops, note how the angular momentum constantly increases. Now a Graph with how the angular momentum of the disc varies(green):
| {
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How can we discern so many different simultaneous sounds, when we can only hear one frequency at a time? As I understand it, the eardrum works like any other kind of speaker in that it has a diaphragm which vibrates to encode incoming motion into something the inner ear translate to sound. It's just a drum that moves back and forth, so it can only move at one rate or frequency at any given time.
But humans have very discerning ears and can simultaneously tell what instruments are playing at the same time in a song, what the notes in the chord of one of those instruments is, even the background noise from the radiator. All of this we can pick apart at the same time despite that all of these things are making different frequencies.
I know that all of these vibrations in the air get added up in a Fourier Series and that is what the ear receives, one wave that is a combination of all of these different waves. But that still means the ear is only moving at one frequency at any given time and, in my mind, that suggests that we should only be able to hear one sound at any given time, and most of the time it would sound like some garbled square wave of 30 different frequencies.
How can we hear all these different frequencies when we can only sense one frequency?
| As you mentioned, all the different waves get added together in a Fourier series.
The hair cells in the inner ear are essentially performing a Fourier analysis of this combined wave, splitting it back into its component frequencies. The amplitudes of each frequency are then sent to the brain, which performs higher level analysis to recognize specific types of sounds (and also determines the direction of the sound sources by comparing the timing from each ear).
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"source": "stackexchange",
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Calculating the direction of friction on an inclined plane/Determining the fate of a round body on an inclined plane
Question:
If a round object (say ring or a sphere) is kept on a rough inclined plane of inclination $\theta$ and coefficient of friction $\mu$ is such that it exactly balances the component of weight of that object, that is, $mg \sin \theta=\mu mg \cos \theta$, then it's acceleration would be $0$, right?
So, will the object move downwards on that plane? We see that there will be a net torque of friction about the centre of mass, so it has a rotating tendency (in the counter-clockwise direction), so will it still move downwards despite that there is no force in the downward direction that would give it translational motion along the incline?
Or, could it be that friction acts downwards initially adding to it's component of weight and gives it acceleration and then starts acting again in the backward/upward direction to give it a pure rolling motion? However, this would mean that it has a clockwise torque which seems absurd for an object kept solely on it's own.
Note that it's not a homework question. It's asking for conceptual clarity.
| This is the mistake you are making:
$mg \sin \theta=\mu mg \cos \theta$
Friction will never be equal to $\mu mg \cos \theta$. Why?
Friction will be static and will only try to prevent the slipping between the point of contact and the ground. The net acceleration of point of contact of ground $\ne g\sin\theta$ due to the torque produce by friction in anticlockwise direction. (like you said)
Also this para :
Or, could it be that friction acts downwards initially adding to it's component of weight and gives it acceleration and then starts acting again in the backward/upward direction to give it a pure rolling motion? However, this would mean that it has a clockwise torque which seems absurd for an object kept solely on it's own.
This is wrong too.
| {
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Wavelength of reflected light vs transmitted light In the basic setup of the experiment below, the transmitted light is used to infer the absorbed wavelengths. The explanation of the experiment states that a high transmittance reading for a specific wavelength means that wavelenght is not absorbed.Does this mean that the wavelength of the transmitted light and the reflected right are equal?
| Why are you asking about reflected light levels? Are you worried that the flask and/or the solution have spectrally-varying reflectances? If so, then the standard way to deal with that is to run a reference calibration with the flask and whatever solution you intend to use, but none of the test material itself. Treat the transmitted power curve as a normalizer for the data you collect thru the test specimen.
| {
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Can sound be used for propulsion? I'm no physicist so this might seem absurd.
I Remember watching a cartoon as a kid where the character uses a powerful speaker to propel his cart and I was wondering if this was actually possible.
Being a highschooler I am aware to propel forward you shoot something backward.
So maybe in the case of a speaker it could "shoot out" sound waves?
| Sure! Sound waves carry energy and momentum. So you could use a speaker like a very weak rocket. Totally impractical, though.
| {
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Rotation matrices in Schwinger's oscillator model of angular momentum I am Section 3.9 in Sakurai's Modern QM, 3rd ed (which is Section 3.8 in 2nd ed.) I am trying to obtain the given form for $\hat D(R)|jm\rangle$:
I employ $\hat D^{-1}\hat D=1$ and ignore the denominator to write
\begin{align}
\hat D(R)|jm\rangle&= \hat D \bigg[\big(\hat a^\dagger_+\big)^{j+m}\big(\hat a^\dagger_-\big)^{j-m}|0,0\rangle\bigg]\\
&= \hat D \bigg[1^{j+m}\times\big(\hat a^\dagger_+\big)^{j+m}\times1^{j+m}\times\big(\hat a^\dagger_-\big)^{j-m}\times1^{j-m}|0,0\rangle\bigg]\\
&= \hat D \bigg[\big[\hat D^{-1} \hat D \big]^{j+m}\big(\hat a^\dagger_+\big)^{j+m}\big[\hat D^{-1} \hat D \big]^{j+m}\big(\hat a^\dagger_-\big)^{j-m}\big[\hat D^{-1} \hat D \big]^{j-m}|0,0\rangle\bigg]\\
&= \underbrace{\big(\hat D^{-1} \big)^{j+m-1}}_{*} \big(\hat D \,\hat a^\dagger_+\,\hat D^{-1} \big)^{j+m}\underbrace{\big(\hat D \big)^{2m}}_{*}\big(\hat D \,\hat a^\dagger_-\,\hat D^{-1} \big)^{j-m}\underbrace{\big(\hat D \big)^{j-m}}_{*}|0,0\rangle \\
\end{align}
Among the three indicated $*$ terms, I have one extra factor of $\hat D$ so that I will obtain the expression given in Sakurai. However, I need to show that $\hat D$ commutes with $\hat a_\pm^\dagger$ or else that it commutes with $\hat D \,\hat a^\dagger_\pm\,\hat D^{-1}$. What would be the easiest way to show this? I think it will be unnecessarily involved to find an expression for $\hat J_y$ in terms of the Schwinger oscillator operators.
| You may use the following formula:
$$(ABA^{-1})^{m} (ACA^{-1})^{n}=(ABA^{-1})(ABA^{-1})...(ABA^{-1})(ACA^{-1})(ACA^{-1})...(ACA^{-1})=AB^{m}C^{n}A^{-1}$$
where $A,B,C$ are operators and $m,n$ are some positive integers.
| {
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Physical meaning of the inner product of quantum states I would like to ask a simple question about quantum mechanics. Let $|a\rangle$ and $|b\rangle$ two states in which a quantum system can be. Then, I can consider the following inner product:
$$
\langle a|b \rangle
$$
Does this quantity have any physical meaning?
I thank you in advance.
| Physically, the quantity $\langle a | b\rangle$ is the probability amplitude that a measurement will find the system in state $|a\rangle$, given that it was originally prepared in state $|b\rangle$.
To get from this to the probability, you need to take the mod-square, i.e. $|\langle a | b\rangle|^2$.
| {
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Does alpha decay have anything to do with weak interaction? In alpha decay, an atomic nucleus emits an alpha particle (identical to a helium-4 nucleus). This happens in large nuclei because the nuclear force keeping the nucleus together is outweighed by the electromagnetic force.
Makes sense to me, but isn't the weak interaction supposed to be "responsible for all radioactive decay of atoms"? Alpha decay is considered as such, so is the weak interaction involved in any way?
| The basic radioactive decay modes are alpha decay (the emission of a helium nucleus), beta decay (the emission of electrons/positrons and neutrinos), and gamma decay (emission of photons). These decays are governed by the strong, weak, and electromagnetic force, respectively.
Obviously, there are many details. There are other forms of radioactivity (neutron emitters, proton emitters, .. ), and all interactions contribute to the precise rate of alpha, beta, and gamma decays. For example, since alphas are charged, the alpha decay rate is governed by the Coulomb barrier.
| {
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Would we need Alternating Current if superconducting wires existed? The major advantage of Alternating Current is that it can be transmitted to large distances without significant losses, which is not possible in Direct Current.
Had economical superconducting wires existed, DC could be transmitted to any distance without any loss, and DC is much safer compared to AC.
So, I want to know, do we need AC if long distance transmission is no longer a problem because of superconducting wires?
Would DC be better in that case, or we would still need AC?
| Ohmic power line losses occur in both DC and AC systems and are always proportional to the RMS value of the current, so your assertion that AC transmission has insignificant losses compared to DC is not correct.
The major advantage of AC power transmission is that the transmitting voltage can be transformed down as needed locally at any point along the line with a simple transformer. This is not possible with DC, which is why it is rarely used for long-distance power transmission. In addition, three-phase AC power transmission directly enables three-phase electric motor technology, which is overwhelmingly preferred in industrial use.
Note that AC power transmission is not inherently more dangerous than DC, and that lossless superconducting lines for long-distance power transmission could be used in either DC or 3-phase AC mode.
| {
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Finding an exact value for energy in perturbation theory
Supose a particle of mass $m$ and electric charge $q$, subject to harmonic potential in 1D, is placed in an area with electric field $\vec E = E \hat u_x$. Determine the exact change in its energy spectrum caused by interacting with this field.
I started by writing down the Hamiltonean operator as:
$\hat H = \frac{1}{2m}\hat p_x^2 + \frac{1}{2}m\omega^2 \hat x^2 + qE \hat x$, where the last term represents the perturbation caused by interacting with the electric field. (We can treat it as a perturbation since usually $q \ll 1$.)
Using this, for any state $|n \rangle$, the first order correction to the energy is always going to be zero, since, using the ladder operators: $\hat a_{\pm} = \sqrt{\frac{m\omega}{2\hbar}}(\hat x \pm \frac{i}{m\omega} \hat p)$,
$$\epsilon_1 = \langle n | \hat W |n\rangle = qE\cdot 2\sqrt{\frac{2\hbar}{m\omega}} \langle n | \hat a_+ + \hat a_-|n\rangle = C (\sqrt{n+1}\langle n |n+1 \rangle + \sqrt{n}\langle n |n-1 \rangle) =0,$$
because the states are orthogonal to each other. Therefore, the first order correction is zero. However, this still leads to an approximate answer:
$$E(n) \approx \epsilon_0 (n) + q\epsilon_1 + O(q^2) = \epsilon_0 (n) + O(q^2), $$
so I don't understand how to get to an exact value for the change in energy, specially since in my Quantum Mechanics class we didn't cover higher order corrections.
Is there another way to approach the problem that I'm not seeing?
| If you want an exact solution, you can't use perturbation theory.
Try a change of variables $x\rightarrow y$ so that:
$$ \frac 1 2 m \omega x^2 + qE x = \frac 1 2 m\omega (y-a)^2 - b^2 $$
where $a$ is an offset to the (classical) ground state, and $-b^2$ is a global energy shift.
Classically, that energy shift is stored in the spring, and quantum mechanically, it's in the field that causes $V(x)$.
| {
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Reason of saturation region in MOSFET I can't wrap my head around the circumstance that, somewhen when I increase $ V_{DS} $ the current $ I_{DS} $ will remain constantly. Why is that?
https://www.electronics-tutorials.ws/amplifier/mosfet-amplifier.html
| For concreteness, consider an n-channel enhancement mode MOSFET whose source terminal is connected to the body. If $V_{DS}=0$ and $V_{GS}>V_{th}$, a conducting channel is opened between the drain and source. When $V_{DS}$ is increased, current flows across the channel.
However, increasing $V_{DS}$ has another effect - it changes the charge distribution on the gate. When the source voltage is higher than the drain voltage, the positive charge which was initially (more or less) evenly distributed across the gate gets pushed toward the drain side. This has the effect of making the conducting channel asymmetric - wider near the drain and narrower near the source.
If $V_{DS}$ exceeds $V_{GS}-V_{th}$, then the channel will be closed near the source region. This is referred to the pinch-off regime. Past this point, increasing $V_{DS}$ causes a proportional increase in resistance because the channel closes off even further away from the source region. As a result, the current does not appreciably increase as $V_{DS}$ increases. This is why the current saturates.
The current does increase with changes in $V_{GS}$, however, and the increase can be quite dramatic. That's why amplifiers operate in this regime - a signal on the $V_{GS}$ terminal produces a current signal $I_{DS}$.
| {
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Stability analysis basics I would like to see a rigorous treatment of stability analysis.
For example, a lot of high-school level texts give examples like this: https://courses.lumenlearning.com/physics/chapter/9-3-stability/
But they describe stability in very qualitative terms like "Hence, the chicken is in very stable equilibrium, since a relatively large displacement is needed to render it unstable. The body of the chicken is supported from above by the hips and acts as a pendulum between the hips. Therefore, the chicken is stable for front-to-back displacements as well as for side-to-side displacements."
What is the quantitative theory behind this?
If I give you an arbitrary object mesh, it's center of mass, and its position and orientation, can you give me a formula to determine if it is stable or not?
What if I give you $n$ objects?
| Today I'm in the mood of trying to be an artist :
Chicken is stable only partially, otherwise why it needs nails ? Humans uses somatosensory system in brains to control stability and balance in a stationary or moving positions. Chicken has small nervous system, thus nails for compensation.
| {
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What is the meaning of $F=ma$? Does it mean that an object with mass $m$ will have an acceleration $a$ if an external force $F$ is applied to it? I know this is a very simple question, but I am just learning physics. I am seeing the basics of how a block on a horizontal frictionless surface gets accelerated by a force F in any direction.
However, I ask myself if F=ma is only used on the object the force is being applied to, not the source of the force (which could be a finger, another block, or anything that pushes)
Could one use F=ma in order to calculate the force an object of mass M would produce on its own if it already had an acceleration.
Thank you for you helpful responses in advance
| The answers here are already great, but here's a small suggestion I can add: when learning Newton's laws, I think it's easier to start off by thinking of Newton's second law as
$$
a = \frac{F}{m}
$$
This is mathematically the same as $F = ma$, of course, but I've noticed that students tend to think that an equation $X = Y$ means that $Y$ causes $X$ -- likely because they are use to plugging in values on the r.h.s. to get the value on the l.h.s. I think it's this that's behind your question about whether $F=ma$ means a mass with acceleration can generate a force. This isn't the right causal order, though; it's the force that causes the acceleration. It's crucial to remember this and be familiar with it when solving problems, and thinking of the law as $a = F/m$ might help with that.
| {
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Hypothetically, why can't we wrap copper wires around car axles and turn them into electromagnets to help charge the batteries? We already have a magnetic core, why can't we use it to recharge the batteries? The only problems I see with it are potentially wiping magnetic data, but doesn't the electromagnet have to be revolving around the damageable device?
| Not stated in your question is the type of vehicle we are discussing - I am assuming this is about vehicles powered by internal combustion engines.
Attaching some sort of electricity generating device to the car's final drive to recover kinetic energy during braking comes down to cost/benefit.
Every piece of equipment you add to a car increases its mass, which reduces its overall efficiency - more fuel consumed to haul it all around.
Car engines are equipped with alternators to provide electrical power whenever the engine is running, in part because the car doesn't have to be moving in order to power lights and accessories, charge the battery for the next start, and so forth.
The amount of power moving around the electrical system and the amount of energy stored in an automotive battery is relatively small compared to the kinetic energy of a moving vehicle and power required to get one up to highway speed in an acceptable amount of time.
An energy recovery system would need to be rather sophisticated to capture a useful amount of energy from a decelerating car - the slower it is moving, the less efficiently it will work.
Electric cars use regenerative braking, it works, and is worthwhile because the motors used to drive the vehicle are operated in reverse; the same strategies which allow the motors to efficiently accelerate the vehicle over a wide range of speeds allow for efficient energy recovery when decelerating. It's a matter of re-using equipment that is already there.
In summary, the existing alternator meets the need and is well-matched; attempting to recover kinetic energy during braking as electrical energy would be too poorly matched to the requirements of the electrical system to be of practical benefit for the cost incurred.
| {
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Particle Physics: Decomposition of a Helicity Spinor I would have a general question: If we consider the decay of the $W^{-}$ boson into $l^{-}\nu_{\bar{l}}$, how can we calculate the polarization of the $l^{-}$?For example, Mark Thomson has on page 299, Eq. (11.17), the following decomposition of a right-handed helicity spinor $u_{\uparrow}$:
$$u_{\uparrow} = \frac{1}{2}\left( 1 + \frac{p}{E + m}\right)u_{\text{R}} + \frac{1}{2}\left( 1 - \frac{p}{E + m}\right)u_{\text{L}} \qquad [1], $$
where $u_{\text{L}}$ and $u_{\text{R}}$ denote chiral states.
Question:
Is there a similar decomposition for a left-handed helicity spinor $u_{\downarrow}$ as in Eq. [1]? I coulnd't find it in the Thomson.
EDIT:
Following Cosmas Zachos' comment, here is where I am stuck on proving [1] on my own. I think I might manage to prove for myself a representation for $u_{\downarrow}$ once I understand [1]. So: One line before Eq. (6.38) in Thomson, he has the following Eq.: $$u_{\uparrow}\left( p, \theta, \phi \right) = \frac{1}{2}\left( 1 + \kappa\right)N\begin{pmatrix} \cos\frac{\theta}{2} \\ \sin\frac{\theta}{2}e^{i\varphi} \\ \cos\frac{\theta}{2} \\ \sin\frac{\theta}{2}e^{i\varphi}\end{pmatrix} + \frac{1}{2}\left( 1-\kappa\right)N\underbrace{\begin{pmatrix} \cos\frac{\theta}{2} \\ \sin\frac{\theta}{2}e^{i\varphi} \\ -\cos\frac{\theta}{2} \\ -\sin\frac{\theta}{2}e^{i\varphi} \end{pmatrix}}_{\left(\star\right)}\qquad [2],$$ and then Eq. (6.38) (he also wrote somewhere that $s \equiv \sin\frac{\theta}{2}$ and $c\equiv \cos\frac{\theta}{2}$):
$$u_{\uparrow}\left( p, \theta, \phi \right) \propto \frac{1}{2}\left(1+\kappa\right)u_{\text{R}} + \frac{1}{2}\left( 1-\kappa\right)u_{\text{L}}.$$
I do not understand how $\left( \star \right)$ is supposed to be proportional to $u_{\text{L}}$. According to Thomson, "the above spinors all can be multiplied by an overall complex phase with no change in any physical predictions", page 108. According to Eq. (6.32), $$u_L = N\underbrace{\begin{pmatrix} -\sin\frac{\theta}{2} \\ \cos\frac{\theta}{2}e^{i\varphi} \\ \sin\frac{\theta}{2} \\ -\cos\frac{\theta}{2}e^{i\varphi} \end{pmatrix}}_{\left(\star\star\right)}.$$
Comparing $\left(\star\right)$ to $\left( \star\star\right)$ for the first component for now, and taking into account that we are allowed to habe a global phase factor of $e^{i\xi}$, I get:
$$e^{i\xi} \cdot \cos\frac{\theta}{2} = -\sin\frac{\theta}{2}$$
For me, this Equation is never satisfied, regardless of what I choose for $e^{i\xi}$ ...
| Consider a particle moving in the $\hat z$ direction, for simplicity. Define $\kappa= p/(E+m)$ and note it collapses to 1 for m =0. In this frame,
$$
u_ \uparrow =\sqrt{E+m} \begin{pmatrix} 1\\ 0\\ \kappa\\ 0\end{pmatrix}, \qquad u_ \downarrow =\sqrt{E+m} \begin{pmatrix} 0\\ 1\\ 0\\ -\kappa \end{pmatrix}.\tag{4.65}
$$
Up to normalizations, the eigenstates of
$$
\gamma_5 = \begin{pmatrix} 0& 1\!\!1 \\ 1\!\!1 &0 \end{pmatrix}
$$
have eigenvalues $\pm 1$, respectively. We can resolve the identity $I=P_R+P_L$,
$$
P_R= \frac{1}{2} \begin{pmatrix} 1\!\!1 & 1\!\!1 \\ 1\!\!1 &1\!\!1 \end{pmatrix} \qquad P_L= \frac{1}{2} \begin{pmatrix} 1\!\!1 & -1\!\!1 \\- 1\!\!1 &1\!\!1 \tag{6.34}\end{pmatrix}
$$
so that the R,L projections of the above helicity eigenstates are
$$
u_ \uparrow =P_R u_ \uparrow +P_L u_ \uparrow \propto \left ( \tfrac{1+\kappa}{2}\right ) \begin{pmatrix} 1\\ 0\\ 1\\ 0\end{pmatrix}+ \left ( \tfrac{1-\kappa}{2}\right ) \begin{pmatrix} 1\\ 0\\ -1\\ 0\end{pmatrix} \tag{6.38}$$
and finally
$$ u_ \downarrow = P_R u_ \downarrow +P_L u_ \downarrow \propto
\left ( \tfrac{1-\kappa}{2}\right ) \begin{pmatrix} 0\\ 1\\ 0\\ 1\end{pmatrix}+ \left ( \tfrac{1+\kappa}{2}\right ) \begin{pmatrix} 0\\ 1\\ 0\\ -1\end{pmatrix} ,
$$
your resolution of the negative helicity spinor sought. Note its right and left chiral components are different than those of the above positive helicity given, even though they have the same eigenvalues, (and mismatched with your text's (6.32).... Probably sloppy notation).
Note that at $\kappa\to 1$, $u_\uparrow$ collapses to its right chiral projection, and $u_\downarrow$ to its own (different) left-chiral projection; which is why negative helicity is improperly/confusingly sometimes called "left", and positive helicity is misidentified as "right", to remind you of its connection to chirality seen here.
Apologies I failed imagining your text might lack consistent notation.
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Centrifugal Force Dilemma While learning Classical Mechanics, I am confused on nature and application of centrifugal force.
In my textbook, it is written that centrifugal force is a pseudo force that, depends on reference frame, but I can't understand that if it is pseudo force then why we feel something pushing us outwards during a tight turn in vehicle.
Also, I am very confused that when we will apply centrifugal force, since centrifugal will cancel centripetal force so how will the object move in circle in absence of radial acceleration? (My understanding)
Please clarify my confusion and tell any flaws in my understanding if any,
| It is also called fiticious force, or d'Alembert force, or inertial force. I prefer the term inertial force, because we do feel it, not an imagination.
We apply centrifugal force only when we are in a rotational frame, for example we stand on earth, there is a centrifugal force due to the spinning of the earth. The centrifugal force cancles the centripetal force, and makes us on earth without a relative rotation w.r.t earth.
There are other inertial forces generated from derivating a rotational motion on an inertial frame besides centrifugal force, including Coriolis force which making typhoon to turn in a certain direction, $-m \vec{\omega} \times \vec{v}$, and for accelerating rotation frame $- m \frac{d\vec{\omega}}{dt} \times \vec{r}$.
| {
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Does soap clean grease because it lowers surface tension? Or are these things independent?
Have you noticed how much easier it is to clean dirty, greasy hands or plates once you add soap to the water? By reducing the surface tension of water, soaps and detergents allow it to mix with oil and grease more easily. Warm water melts the oils and fats so that it is easier for the soap to work – that is why warm, soapy water is so effective.
Is this quote correct? I thought soap worked by connecting directly to grease by its hydrophobic parts, not by allowing water to mix with oil and grease due to reduced surface tension.
| Soap has one end of its molecule which is soluble in water, while the other end is soluble in oily substances. This makes soap act like a glue that sticks oil molecules to water molecules, allowing the water to carry away the oil. In this way, soapy water will remove oily stains from fabric.
In addition, soap reduces the surface tension of the water, allowing it to enter small crevices more easily, which makes it more effective as a cleaning agent. So both effects are important, but the thing that gets grease off surfaces is the actions of the two different ends of the soap molecule.
| {
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Why do we need to introduce thermodynamic potentials? Each thermodynamic potential (Enthalpy, Helmholtz free energy, Gibbs free energy) is the same as the 1st law of thermodynamics. Then, why do we need them? Why did people introduce them in the first place?
| Two central questions in thermodynamics are
*
*What are the conditions for spontaneous reaction?
*What are the conditions for equilibrium?
Constrained by First law of thermodynamics (Conservation of Energy), and hinted by second law of thermodynamics (Increasing entropy), under different conditions, these questions are answered by different types of thermodynamic potentials.
Consider any dynamic reaction, for generalised constant externally applied force and displacement, the work input to the system is $\text{đ}$W $\leq$J$\cdot$ dx, also from second law $\text{đ}$Q $\leq$ TdS; now use first law dU=$\text{đ}$Q+$\text{đ}$W $\leq$ TdS +J$\cdot$dx
Rearragement gives: dU-TdS-JdX $\leq$0. This inequality indicate when system is not in equilibrium, the quantity U-TS-JX tends to minimise, that's the answer for spontaneous reaction, ie: when the quantity is not minimal, spontaneous reaction take place to minimise this quantity. Similarly, for equilibrium, once that quantity is minimise, equilibrium is achieved. This quantity is called availability in thermal physics.
Return to the question why people introduce various thermodynamic potentials, actually these thermodynamic potentials are special case of availability.
For instance Helmholtz free energy,
TdS $\geq$ dQ
Substitution gives:
dU=TdS-pdV= $\text{đ}$ Q+ $\text{đ}$W
Rearragement gives
dU+pdV=TdS $\geq$ dQ
For $dT=0$ and $dV=0$, this gives:
d(U-TS)=dF $\leq$ 0
Which means for a system with fixed volume at constant temperature, the condition for equilibrium is to minimise Helmholtz free energy.
| {
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"timestamp": "2023-03-29T00:00:00",
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Quantum Harmonic Oscillator Virial theorem is not holding I'm asked to calculate the average Kinetic and Potential Energies for a given state of a quantum harmonic oscillator. The state is:
$$
\psi(x,0) = \left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{\frac{-2m\omega}{\hbar}x^2}
$$
The thing is, calculating
$\langle T\rangle=\int_{-\infty}^{\infty}\psi(x)(-i\hbar)^2\frac{d^2}{dx}\psi dx=\left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{2}\int_{-\infty}^{\infty}e^{\frac{-4m\omega}{h}x^2}dx-\left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{2}\int_{-\infty}^{\infty}x^2e^{\frac{-4m\omega}{h}x^2}dx=\hbar\omega$
Where I used that the momentum operator is $p=-i\hbar\frac{d}{dx}$
$\langle V\rangle=\dfrac{m\omega^2}{2}\left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{2}\int_{-\infty}^{\infty}x^2e^{\frac{-4m\omega}{h}x^2}dx=\dfrac{\hbar\omega}{16}$
But then the Virial Theorem is not satisfied. I've read the virial theorem holds for any bound state and all states in a Quantum Harmonic Oscillator are bound. Can someone point out where I am going wrong?
Thank you
| You can have Gaussian fields that are not eigenstates, but then they are not time independent -- and time independence is the essential element of the virial theorem. For example, the harmonic oscillator time-dependent Schrödinger equation
$$
i\frac{\partial \psi}{\partial t} = -\frac 12 \frac {\partial^2 \psi}{\partial x^2} +\frac 12 \omega^2 x^2 \psi
$$
has a time-dependent solution
$$
\psi(x,t)= \left(\frac{\omega}{\pi}\right)^{1/4}\frac 1{\sqrt{e^{i \omega t} +R e^{-i\omega t}}}\exp\left\{ - \frac \omega 2 \left(\frac{1-R\,e^{-2i\omega t}}{1+R\,e^{-2i\omega t}}\right)x^2\right\},
$$
where the parameter $|R|<1$. Only if $R=0$ are its $x$ and $p$ distributions time independent. If $R\ne 0$ the gaussian "breathes" in and out. Your wavefunction is a snapshot of this one at some particular time.
Below is a visualisation of $|\psi(x,t)|^2$ (taking $\omega=1$) for different values of $R$, showing how the Gaussian "breathes". As you can see, as $R\to 0$, the probability distribution tends to not change as much.
| {
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Magnetization and Polarization in an electromagnetic field theory I am currently reading through a paper by Hughes and Ramamurthy (ref: https://arxiv.org/abs/1508.01205), which describes the electromagnetic response of a line-node semimetal by the action
$$S[A,B] = \frac{e}{16 \pi^2}\int d^4x \; \epsilon^{\mu \nu \rho \sigma} B_{\mu \nu} F_{\rho \sigma},$$ where $F_{\rho \sigma} = \partial_\sigma A_\rho - \partial_\rho A_\sigma$ is the usual electromagnetic field, and $B_{\mu \nu}$ is a two-form. At the top of the second page of this paper, the authors go on to say that from the form of the action, you can see that $B_{\mu\nu}$ is related to the magnetization and polarization as
$$e B_{0 i} = 4 \pi^2 M_i$$
and $$e B_{ij} = 4 \pi^2 \epsilon^{ijk} P_k.$$
This statement is not obvious to me, how does one determine this relation between the magnetization/polarization and the two-form $B$?
| See here. For a more detailed source on the Lagrangian formulation of electrodynamics in matter, there is a wonderful book "Classical Field Theory" by David Soper.
| {
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Why does a wooden spoon creates bubbles when put in hot oil? This question might be a bit weird, but I just asked myself why a wooden spoon creates bubbles when put in oil at about 170°C. My idea is, that the water in the spoon reacts with the Oil, but why does this just starts to happen when the Oil reaches approx 170°C? Why does this not happen at 100 or 120°C ?
| I doubt whether this has to do with the content of the spoon, as old and new spoons behave in the same way. I suspect the reason is that oil is a mixture of different oils, and that oils with smaller molecules have a lower boiling temperature. The oil may contain a small proportion with a boiling temp of about 170°C. The spoon has a rough surface on which bubbles are more likely to form. You could test this by seeing whether any other objects with a similar rough surface have the same effect.
| {
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How warm are radioactive metals? I read that radium is warm to the touch -- is that because of actual heat or is that because, for example, the radiation it emits creates the sensation of warmth? How high of a temperature can a radioactive element or isotope actually have?
| Subcritical chunks of enriched uranium and plutonium are naturally warm to the touch, because of the thermal energy released as they spontaneously fission. Plutonium has a higher spontaneous fission rate and this effect is stronger for Pu.
If you place two such subcritical chunks near one another and slide them together gradually, at some point the neutrons released by one will induce extra fissions in the other, and the resulting energy release will make their facing sides hot enough to glow red.
Such (incredibly unsafe!) demonstrations and experiments were done at research laboratories like Los Alamos during the Manhattan Project. Two people died at Los Alamos due to radiation exposure when they slipped up while moving the chunks around. Don't do this at home!
| {
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What does particles with asterisk superscript * mean, such as $Z^*$ or $W^*$, in decay production? I can always find that in some articles the production of higgs decay written as "$\rm H\to {ZZ}^*\to 4l$", "$\rm H\to ZZ\to 4l$", "$\rm H\to Z\gamma^*\to\ldots$” What dose it mean when some particles with a superscript *? Such as “$\rm Z^*, W^*, \gamma^*, \ldots$". What’s the difference between them and "$\rm Z, W, \gamma$"?
| In this context, the asterisk refers to a virtual particle, one that is off shell.
A $\rm Z$ boson, for instance, has a mass of $\sim 91~\rm GeV$, while a Higgs boson has a mass of $\sim 125~\rm GeV$. A Higgs boson is less massive than two $\rm Z$ bosons, and so the decay $\rm H\to ZZ$ is not possible unless one of the particles is off shell.
| {
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Why is energy lost here? Let's say a $1 \ \text{kg}$ block is moving.
With a speed of $1 \ \text{m/s}$ so its kinetic energy is $\frac{1}{2} \ \text{J}$. Now let's gently place a block of mass $3 \ \text{kg}$. Now as linear momentum is conserved due to lack of external forces on the system the blocks move together with velocity $1/4 \ \text{m/s}$ but the energy is now $\frac{1}{8} \ \text{J}$ which is lesser than it used to be.
Where has the energy gone?
| Let the initial body be $A$ and let it move along $X$ axis. Let the other one be $B$
The reason why we get loss of energy is because we are looking at half the picture. Ideally if body $B$ was to fall on to body $A$, then it should bounce back from conservation of momentum (no matter how small the speed of placement was). But some stickiness to the surface is preventing this. So the energy of motion of block $B$ was absorbed by this attractive sticking force. The full story of what happens to the energy and why we generally call this loss a heat loss is there below.
As some answers mention, we cannot add stuff to the system. So a different form of this question is more reasonable.
Imagine a body $A$ moving along $X$ axis over a table (no gravity no friction). Let another mass $B$ be slowly approaching $A$, collides with $A$ and sticks to $A$. This is equivalent to placing the body gently.
where the energy goes
If no stickiness existed, Body $B$ bounces off from $A$ trying to move away but the sticky force pulls it back with all molecules in glue pulling back.
Now a when the molecules of the body and glue attract each other, both molecules would gain speed. the mutual attraction transfers the energy slowly from $B$ into the glue molecules.
This higher speeds would increase the vibration of atoms in the glue, which is equivalent to a rise in temperature. That is why we also call this heat loss
the main take away is that - as feynman says - there is no non-conservative forces. Energy is always conserved but we are lazy to calculate all the energy.
Also to mention this is generally known as a perfectly inelastic collision.
| {
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Non-Analytic Equations and Chaos Could anyone please tell me an example of an equation with no analytic solution(s) that is not a chaotic one? And what is the physical meaning of having analytic solution? For instance, the three body problem does not have in general analytic solution and it leads to chaos. But I don't know if this is a general statement. I have absolutely no idea. Could anyone explain me, please?
| Like the following?
$$ \sin (x) = \lambda x \tag{1} $$
for $\lambda < 1$
Or do you want an ODE? You did not specify in the question.
| {
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How to calculate the heat that leaves the furnace through one opening? I recently read in the newspapers that one man died through the flame that left the iron furnace. I wasn't sure about this. Iron is melting at 1500 °C. The furnace they used is 8 cubic meters, he was standing one meter from the furnace. The opening is 1 squared meter. By the story, he was outdoors when malfunction door opened and made him burns from which he died. The temperature in our city that day was 35 °C. I wasn't sure about the story is this heat really enough to cause the burns that can kill you. I tried to calculate the heat that left with this formula (Q=m×c×ΔT) but I have no idea of the air heat after the accident, so I can't calculate ΔT.
I'm sure I'm missing something, could someone point me in the right direction?
| I think there was a sudden burst of flame out of an opening in the furnace that impinged on his body, not from radiative heat transfer from the furnace. This can happen very quickly allowing no time to move away. This can occur due to a combination of extra oxygen combined with a localized higher fuel source, and/or sudden movement (collapse) of combustible fuel.
| {
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Is projectile motion an approximation? Doesn't the acceleration vector points towards the center of the Earth and not just downwards along an axis vector. I know that the acceleration vector's essentially acting downwards for small vertical and horizontal displacements but if the parametrization of projectile motion doesn't trace out a parabola, what is the shape of projectile motion?
| Assume Galilean relativity and Newtonian gravity. Neglect the drag due to the atmosphere. The gravitational field of the Earth is the same as the one produced by a point particle in its center (with the same mass, the usual $1/r^2$ gravitational force field). Now, you may know that a test particle in this $1/r^2$ force field of the Earth can have different orbits (closed or open, depending on the initial velocity and the initial position).
Leave out open orbits, which means that you are shooting the projectile at infinity. All other orbits are ellipses. However, the Earth is not a point and has a finite radius: some of those ellipses (starting at the Earth surface) will intersect at later times the Earth surface again.
Why do you have parabolas in the "simple" setting you are describing? Because you can always approximate locally an ellipse with a parabola. See e.g. Can a very small portion of an ellipse be a parabola?
Edit: yes, in a spatially constant and uniform gravitational field, a parabola is exact. Other answers can be found in these "duplicate" posts: Elliptical Trajectory, or Parabolic? and Why does the Earth follow an elliptical trajectory rather than a parabolic one?
| {
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Is it even theoretically possible for a perfect clock to exist? I have heard that even atomic clocks lose a second every billion years or so. That raises the question, is it even theoretically possible for a perfect clock to exist, one that never gains or loses time?
| To say that something is perfect, accurate or absolute, is in itself a lie if we consider the realm of our cosmos being governed by the laws of physics - be it classical, modern, or others.
The most convenient way to say is that nothing is absolute in this real world ( ha! I know that the speed of light is constant in vacuum. But, yes, the truthfulness is just unachievable ). Even space and time is not just, because we know the instances where space-time is just not as accurate as it seems to be.
The theories and their related phenomena, of length contraction, time dilation, red/blue shift in macro physical words and gravitational lensing, gravitational waves and much else more in the quantum level, suggests and portrays a world that we live in as something of a ever changing and ever revealing Pandora's box.
| {
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Branching ratio of two particles decaying to the same channel When looking at the numbers for the $WW$ decay's branching ratios, I have the impression that a factor $2$ is taken into account only when the $W$s decay to different channels.
For example:
$Br(WW \to qql\nu) = 2* Br(W \to qq) \cdot Br(W \to l\nu)$
$= 2 * 0.685 * 0.316 = 0.432$
but
$Br(WW \to l\nu l\nu) = Br(W \to l\nu) \cdot Br(W \to l\nu)$
$= 0.316 * 0.316 = 0.099$
where the numbers are taken from here (a 20 years old reference, so the best numbers might be slightly different now)
Is this the case ? If it is, why is so ?
| You have two $W$.
For the final state $qql\nu$, one $W$ decays to $qq$, the other to $l\nu$. But the decay to $qq$ could come either from the first W or the second W, thus you have the following possibilities :
First possibility
-first $W$ decays to $qq$, thus the second $W$ decays to $l\nu$.
-second $W$ decays to $l\nu$, thus the first $W$ decays to $qq$.
So you need to consider the two possibilities. Since the two possibilities have exactly the same branching ratio, it is equivalent to have twice the product of branching ratio for final state $qq$ and $l\nu$.
For the second final state $l\nu l\nu$, there is a unique solution, because one cannot distinguish first $W\rightarrow l \nu$ and second $W\rightarrow l\nu$, thus there is no factor 2.
If the $W$ could be distinguished (which is not the case), for example, if the first particle would be $W'$ and the second $W$, then there would be the factor 2.
Whenever the two $W$ are of same nature, there is no factor 2.
| {
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Why do we need Newtonian gravitational field (vector)? Let's ignore GR(scalar) and I am wondering why do we need to model Newtonian gravitational field using vectors? I can understand electromagnetism because of Lorentz force (right hand rule) but what about gravitational field it just the difference in strength at each point in space! Could there be some problems that can only be solved using vector field for gravity? Maybe I should use temperature (scalar) as a better example to compare Newtonian gravitational field ;D
| The way gravity comes into the framework of Newtonian mechanics is as a force, i.e., it has a direction from the get-go. So, it has to be a vector. More directly, as mentioned elsewhere, the gravitational field at the North pole and the South pole are roughly of the same magnitude but they are still different vis-à-vis their direction.
Of course, since gravitational force is a conservative force, one can also describe it using a potential formulation where the gravitational potential is simply a scalar which varies from point to point only in its magnitude. However, the physically observable aspect of this scalar gravitational potential is the force that it exerts on a particle. This force would depend on the gradient of the scalar potential, not the value of the scalar potential. This gradient is, of course, a vector, namely, the gravitational field.
| {
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Least count of cesium clock and maximum possible significant figures for time We know that a second is defined as being equal to the time duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the fundamental unperturbed ground-state of the caesium-133 atom(form wikipedia).Hence least count of this clock will be $\frac{1}{9192 631770}$ seconds
From my understanding of least count and significant figures I conclude that it is not possible to measure time duration less than $\frac{1}{9192 631770}$ seconds accurately. Is my conclusion correct?
Is there any workaround for this?
| The number $N=9\, 192\, 631\, 770$ is used to define a second. Do not mistake this as an accuracy or precision. These are completely different concepts from the definition. Thus it is of course possible to measure things with an accuracy lower than $1/N$.
| {
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Can beats be produced by two waves moving in opposite direction? I've always seen beats to be produced when two waves are said to be moving in the same direction with different frequency.
Can beats be produced by addition of waves moving in opposite direction ?
| Two waves moving in opposite directions that have the same amplitude and frequency produce a standing wave. If one of those waves has a slightly different frequency than the other, it should produce a "standing wave" that slowly moves. As the anti-nodes and nodes move past an observer's ear, they would hear an increase in amplitude followed by a decrease in amplitude, and this variation in sound amplitude would be interpreted as beats.
Generating such an effect would be "tricky". A sound generator would have to broadcast a pure tone and face a sound reflector such that a standing wave was created. Then, the sound reflector would have to be put in motion, moving either directly towards the sound generator or directly away from it such that the Doppler effect slightly shifted the frequency of the reflected sound. If the sound reflector was very efficient (close to 100% reflectivity) and a small microphone was placed in the path of the reflected sound waves, it should be possible to record the effect.
| {
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Rotation matrices and reference frames Lets say I have 2 3x3 rotation matrices, we will call them A and B. I am told to find frame B relative to A. how would I go about doing this basic operation? would I do it similar to position (with taking B's coordinates and subtracting A's points from them) or is it different? (or am I just an idiot and doing position with reference frames wrong?)
| If you have a vector with co-ordinates $v_A$ relative to frame $A$, how do you find its co-ordinates relative to frame $B$ ?
First you multiply $v_A$ by $\mathbb A$ to rotate it back to the original frame, then you multiply by $\mathbb B^{-1}$ to rotate to frame $B$. Now you just need to find one matrix that combines these two operations.
| {
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Why does mist on glasses disappear when you go back outside in freezing weather? This evening after a walk in the cold weather, about -5 degree Celsius, my glasses were misting up when I got back into the warm house, about +20 degree Celsius.
However, shortly after I went outside in the cold again, and the mist on my glasses disappeared. This puzzles me, since I guess the outside air to have nearly 100% relative humidity, since it is cooling down in the evening.
How come that the mist on my glasses can disappear in the freezing weather?
| The $absolute$ humidity is much lower outside. The lenses of your glasses have some thermal inertia, and while you were outside they got pretty cold. When you step inside, where the absolute humidity is higher, a thin layer of air near the lenses cools to below the dew point, and condensation occurs.
Then when you go back outside, the droplets evaporate. They would eventually even if the glasses were at ambient temperature, but they’re warmer than that—they warmed up a bit during the brief period you were inside, and they get heat from your head. So now the thin layer of air around the lenses is warmer than ambient, leading to low relative humidity and rapid evaporation.
| {
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Superposition of eigenstates in statistical mechanics Consider the simplest case in quantum statistical mechanics, where we find the density of states in the case of a cuboidal 3 dimensional box. In the derivation we take only those states which are product seperable into wavefunctions along the three directions i.e. can be denoted by three quantum numbers $(n_1, n_2, n_3)$ henceforth written as $|n_1,n_2,n_3\rangle$ . However I feel that even states which are not product seperable should be considered. For example a particle in the system could be in the state $\frac{|1,0,0\rangle+|0,1,0\rangle}{\sqrt{2}}$. This will alter the counting of number of states. Why are such states excluded?
| The question that we are concerned with is the following, given a point in the $k$ space $(k_{1},k_{2},k_{3})$ and an infinitesimal volume centred at this point, how many states can be found within this volume? The superposed state $$\frac{1}{\sqrt{2}}\left(|1,0,0 \rangle + |0,1,0\rangle\right)$$ is found in two different volumes at a finite separation from each other in the $k$ space and is therefore, of no relevance.
| {
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Is a wave function a ket? I just started with Dirac notation, and I am a bit clueless to say the least. I can see Schrödinger's equation is given in terms of kets. Would I be correct to assume if I were given a wavefunction, say $\Psi(x)=A\exp(-ikx)$, would I be able to just use the notation $\lvert \Psi\rangle =A\exp(-ikx)$?
| It depends on the equation, that is: $V(x)$. If it's free particle ($V(x)=0$), with $\hbar=1$, then:
$$ \psi(x) = Ae^{-ikx}$$
is proportional to the momentum eigenstate:
$$ |k\rangle$$
Then:
$$ \hat p |k\rangle = \hbar k|k\rangle = k|k\rangle$$
is easier to deal with then:
$$ i\frac{d}{dx}\psi(x)=i\frac{d}{dx}Ae^{-ikx} = -i^2kAe^{-ikx}=k\psi(x)$$
and likewise for:
$$ \langle k'|k\rangle = \delta(k'-k)$$
| {
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What distinguished physical and pseudo-forces? Why are some forces are considered pseudo-forces while some are considered real or physical forces?
The definition of pseudo-forces that I know of is that they exist in noninertial reference frames but don't exist in inertial reference frames. Among other arguments is that they don't have an identifiable source.
However, a person certainly "feels" the pseudo-force such as in a car that is rounding a corner. How can it be proved, then, that the centrifugal force is not a real force? Why isn't what is "real" determined by the non-inertial reference frame?
| Because it is different in different reference frames. You would expect a universal force to be the same regardless of how you look at it.
Also, it is not easy to figure out the source of the force.* You feel like being pushed by something you can't identify, which appears to break with Newton's 3rd law. That is an indicator (although not certain proof) of it possibly being an effect we feel due to something else than forces.
Finally, you won't feel the centrifugal force if you close your eyes.** This indicates that it might be an illusion, a mind trick, due to the surroundings changing.
*Think centrifugal forces, Euler forces and the push you feel when standing in a braking bus, and then try to look for the source.
** Sure, in the turning car or braking bus you feel the seat underneath you pulling sideways or backwards. You feel the car door or bus rail slamming into you, pushing you. But try jumping just before the turn or the braking - you won't feel anything and with your eyes closed you won't notice any changes at all. Not until you touch something again, that is.
| {
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Infinite acceleration without loss of energy in a vacuum, gravity-less void? Suppose we have a void that is free of any other objects or forces, even gravity (impossible I know but hypothetically). If one were to propel an object forward, would that object be able to accelerate infinitely without requiring further addition of energy?
My line of thinking may be totally off, but an object that had energy applied to it loses the energy after work is done right? And when it comes to movement that initial acceleration energy dissipates when it encounters resistance through friction or gravity in a vacuum.
But if one were to remove all external forces aside from the initial burst of energy, that energy would have nowhere else to go, other than to propel the object forward faster and faster?
It’s been awhile since I’ve done calculus and physics so forgive me if I’m missing something ha.
| First of all, energy is conserved. An object could not, without external forces, propel itself faster and faster as this would mean an ever-increasing kinetic energy. The only way to achieve this would be to apply, non-stop, a force that accelerates it. If you do work on a ball and accelerate it, the ball would travel and the same speed once you stop doing work on it. That kinetic energy wouldn't have anywhere to go and so it remains constant, as per conservation of energy (and because kinetic energy is the only energy).
| {
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In electromagnetic theory, do photons obey the inverse-square law? The energy ($E$) of a photon depends on its frequency ($f$):
$$E = hf = hc/\lambda$$
Notes:
*
*$E$ = energy of the photon (in joules)
*$h$ = the Plank constant ($6.63 \times 10^{-34}$ joules)
*$f$ = frequency in hertz, Hz
*$c$ = speed of light
*$\lambda$ = wavelength
This seems to yield an invariant value of $E$.
How can that be reconciled with the inverse-square law, whereby the intensity (field strength) of a light-source reduces with increasing distance from the light-source?
| The confusion comes from mixing the classical and QM theories of light. In reality, the classical theory is beautifully built up by the herd of photons as described in QM.
https://motls.blogspot.com/2011/11/how-classical-fields-particles-emerge.html
In your example, as you move away from the source, the same number of photons is distributed over a larger area.
Therefore, if you move away from a light source, the same number of photons is distributed over a larger area, thus decreasing the intensity. So it is not an effect of an individual photon, but of the distribution of all the photons emitted.
How does a photon lose intensity as it passes through space?
What this tells us is that the number of photons per unit area (as you say intensity) decreases as you go away from the source. Of course, because the EM wave spreads spherically from the point source, and as it does, the same EM wave has to cover a larger surface area (the expanding sphere if you will). Since the unit area is constant, and the energy content of the EM wave is constant (in your example), the number of photons per unit area has to decrease as the surface area increases.
| {
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Why is the magnetic field $B$ a pseudo-vector? Physically speaking, "pseudo-vectors" are vectors $v\in \mathbb{R}^3$ which transform as $ v'= (\det {R})v$ if the "system were to transform as $R\in O(3)$". However, what does this mean mathematically? And in particular, why is the magnetic field $B$ a pseudo-vector?
I would imagine that by "vectors", we actually mean smooth differential forms with the isomorphism $v\mapsto \sum v_i dx_i$, and by "transforming the system as $R\in O(3)$", I would imagine that it means we are applying a pullback on 1-forms corresponding to the map $x\mapsto Rx$. Assuming that $B =*dA$ where $*$ is the hodge-star operator, how would $\det R$ be factored into this transform?
| I thought about this a little further, and I think I can explain that $B$ is a pseudo-vector (without using too much math jargon).
Let $R\in O(3)$ and let $\omega:\mathbb{R^3} \to \mathbb{R^3}$ be a smooth vector function such that $A$ or $B$. Let me define transforming the vector as $R$ as group action on $\omega$ defined by
$$
(R\cdot \omega)(x) = R\omega (R^{-1}x)
$$
That is, to get the new vector function $\omega'=R\cdot \omega$ at some postion $x$, we need to transform the system back as $R^{-1}x$, find the corresponding vector of $\omega$ at that value, and then rotate/reflect that vector based on $R$. One way to see that this is the canonical way of transforming the system is to think about the differential forms
$$
\omega_i(x)dx_i
$$
If we take $y=Rx$ (i.e. system transforms as $R$), then the differential form becomes
$$
(R\cdot\omega)_i(y) dy_i
$$
Then using the fact that $B=\nabla \times A$, we see that
$$
R\cdot B = (\det {R}) (\nabla\times(R\cdot A))
$$
This is where the determinant comes from. In particular, if for some special system $R\cdot A =A$ for some reflection $R$, then $R\cdot B = -B$.
| {
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D'Alembertian of a delta-function of a space-time interval (i.e. on the light-cone) How one differentiates a delta-function of a space-time interval? Namely,
$$[\partial_t^2 - \partial_x^2 - \partial_y^2 - \partial_z^2] \, \delta(t^2-x^2-y^2-z^2) \, .$$
Somewhere I saw that the result was $$4\pi\delta^{4}(x).$$ However, have no idea on how to obtain it.
| The action of any distributional "function" such as a delta function is really only determined when you integrate it with a test function; and two distributions are equal if and only if their actions agree on all test functions. So let's see what this distribution does when we integrate it with a smooth test function $f(t, \vec{r})$.
We start with the integral
$$
I[f] = \iiiint d^4 x \, \left[ f(t, \vec{r}) \Box \delta(\Delta s^2) \right]
$$
where $\Box$ stands for the wave operator under the metric convention $(+ - - \, -)$. Integrating by parts twice and assuming that $f$ has compact support (as all good test functions should), this becomes
$$
I[f] = \iiiint dt d^3\vec{r} \, \left[ (\Box f) \delta(t^2 - x^2 - y^2 - z^2) \right]
$$
We can then use the rules concerning composition of the delta-function with a function to rewrite this as
$$
I[f] = \iiiint dt d^3\vec{r} \, (\Box f) \left[ \frac{\delta(t - |\vec{r}|)}{2 |t|} + \frac{\delta(t + |\vec{r}|)}{2 |t|} \right].
$$
Finally, we can integrate over $t$ to obtain
$$
I[f] = \iiint d^3 \vec{r} \, \frac{ \Box f(|\vec{r}|, \vec{r}) + \Box f(-|\vec{r}|, \vec{r})}{2 |\vec{r}|}.
$$
The notation here for $\Box f(\pm|\vec{r}|, \vec{r})$ is awkward, but it means "$\Box f$ evaluated at $t = \pm|\vec{r}|$."
This action will differ from the action of the distribution $4 \pi \delta^4(x^\mu)$. In particular, we have
$$
J[f] = \iiiint d^4 x \, f(t,\vec{r}) \left[ 4 \pi \delta^4(x^\mu) \right] = 4 \pi f(0, \vec{0}).
$$
$I$ and $J$ are only equal if their results agree on all test functions $f$. But if we choose a test function that vanishes at the origin and for which $\Box f$ is positive on some portion of the light cone, then $I$ and $J$ will yield a positive result and a zero result respectively. Thus, the distributions are not equal.
It is possible that the actions of these two distributions will be equal on test functions $f$ that have special properties (for example, if $f$ is the Green's function for the wave operator or something like that.) This may be all that is necessary in a particular context. But the two distributions are not equal in a strict sense.
| {
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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?
| The overshoot behavior you noticed is called cogging and occurs when the magnet arrangement in the motor "catches" the rotating magnetic core of the motor during shutdown and jerks it back to one of the local strong spots in the field.
You can demonstrate this yourself by carefully rotating the fan blade around with your finger when the motor is off. You will notice there are certain rotation angles where the fan wants to come to rest and others which it wants to avoid. If the motor passes one of the preferred spots but fails to rotate far enough to "climb the hill" and snap forward into the next cog spot, the motor will very briefly rotate backwards a fraction of a turn and go "boing-oing-oing-oing" as it settles into that cog position.
| {
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Will a changing $E$ field induce a current in a loop similar to a changing $B$ field? An induced current in a wire loop that is caused by a changing B field is a common EM question. However, I couldn't find examples online where the B field was substituted for a changing E Field.
The following question was given on a test and the goal was to find the current flow caused by a varying B Field first, then a varying E Field. My answer is illustrated below.
While it was simple to deduce the direction of the current with a changing B field (clockwise), when the E field was subbed in below, my answer was completely different. Instead, I ended up with an induced B field that was counterclockwise on the outside of the loop and clockwise on the inside of the loop.
| I think that the idea of the question is to note not only that there is a changing field, but also that the time derivative is constant.
So, your conclusion is OK in the first part. In the second part $B$ is constant around the loop, so the charges are not affected and there is no current.
The electric field acts inside the loop, not in the wire as far I understand.
| {
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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?
| In error analysis, the errors multiply/add up (basically effect) only if all these sources of error are independent. That need not be the case always. And further using different instruments like moving form Cloud Chamber detector to Semiconductor detectors have increased the resolution my 100s of magnitudes, the reason being some materials are more sensitive than other, and you can always create new materials!
| {
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Amplitude spectral density vs power spectral density I'm reading the Wikipedia article on spectral density. It is said:
Sometimes one encounters an amplitude spectral density (ASD), which is the square root of the PSD; the ASD of a voltage signal has units of V Hz−1/2.[6] This is useful when the shape of the spectrum is rather constant, since variations in the ASD will then be proportional to variations in the signal's voltage level itself. But it is mathematically preferred to use the PSD, since only in that case is the area under the curve meaningful in terms of actual power over all frequency or over a specified bandwidth.
Could somebody elaborate what this means? Why is it that only in the case of power spectral density (PSD) the area under the curve is meaningful?
| The units are the key, so let's consider an example. Suppose we are measuring a voltage. Thus, the unit of the $PSD$ is $V^2/ Hz$, and the integral over the frequency range
$$
\int_{f_1}^{f_2}PSD\; df
$$
has the unit $V^2$. Hence, the result of the integration has the unit $[signal]^2$. In signal processing the square of the signal is called "power" -- note that this differs from the definition we use in physics ($\rm{Watt} \ne \rm{Voltage}^2$). Also note that the square root of this integral is the RMS value of the signal within the frequency range $[f_1, f_2]$.
In contrast, if we integrate the ASD across the frequency range
$$
\int_{f_1}^{f_2}ASD\; df
$$
the result has the unit $V \cdot \sqrt{Hz}$. The unit of this result does not have a simple interpretation or relation to the measured signal. However, I would not call it a meaningless result.
| {
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Is galactic gravitational lensing self-magnification big enough to contribute noticeably to the galaxy rotation curve problem? Gravitational lenses magnifies the appearance of themselves, so in the case of a galaxy it looks bigger than it is. Thus the outer stars in a galaxy seems to have a higher tangential velocity.
For example, the Sun deflects light passing from minus infinity to us 1.75 seconds of arc. Say that half of that value is the amount it deflects its own light at its edge, then the Sun looks 200 km wider.
Does this effect noticeably contribute to the galaxy rotation curve (dark energy/matter) problem?
| For a spherically symmetric potential, the apparent radius of an object seen from far away is
$$R_{\infty} = R \left(1 - \frac{2GM}{Rc^2}\right)^{-1/2}\, $$
where $M$ is the total mass-energy and $R$ is the radius in Schwarzschild coordinates.
For a big spiral galaxy (like the Milky Way), and ignoring any contribution from dark matter, then approximate numbers would be $M \sim 5\times 10^{10} M_{\odot}$ and $R \sim 20$ kpc.
Using the equation above, $R_{\infty} = 20.0000002$ kpc.
If one includes dark matter, then you might lose one of the zeroes.
So no, the effect you ask about has no significant influence on the measurement of galactic sizes or velocities.
| {
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Hamiltonian Mechanics without a Lagrangian Let's say I want to develop Hamiltonian mechanics from scratch without going through Lagrangian mechanics and Legendre transformations. How would I go about doing that? What I am struggling with is a definition of conjugate momentum. It is usually defined as a derivative of Lagrangian, and there does not seem to be a natural way of defining it in terms of trajectory. Is there an alternative definition that does not involve a Lagrangian?
| You can just define some Hamiltonian $H(\vec{q},\vec{p})$ and the equations of motion are Hamilton's equations
\begin{eqnarray}
\frac{{\rm d} q_i}{{\rm d} t} &=& \frac{\partial H}{\partial p^i} \\
\frac{{\rm d} p^i}{{\rm d} t} &=& - \frac{\partial H}{\partial q_i}
\end{eqnarray}
If you integrate the equations you'll get $q_i(t)$.
You might ask how to write down a Hamiltonian to represent a given system. If the system is fairly simple you can just start with a standard form like $H=\frac{\vec{p}^2}{2m} + V(\vec{q})$. For more complicated systems, it depends on the problem which formalism is the easiest to derive the equations of motion.
Finally I'll just mention that Hamiltonian mechanics is quite deep and there are some very beautiful geometric ways to formulate the theory as a symplectic manifold in a coordinate-independent way. If you have the mathematical background, this is an even more elegant way to formulate the equations; you just need to specify some two-form on the manifold that will generate the dynamics.
| {
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Why does changing $X\to iX$ in the definition of the CNOT make an important difference? How to find a ket $|\psi \rangle$ that illustrates how changing X to iX in the definition of the CNOT gate makes an important difference because of what happens when CNOT is applied to $|\psi \rangle$?
The definition of the CNOT gate is:
$$CNOT = |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes X $$
What is the physical meaning of this?
| Assuming your question is "What $| \psi_i \rangle $ would have different resulting $| \psi_f \rangle $ having undergone a $C-X$ gate vs. a $C-iX$ gate?"
Consider $| \psi_i \rangle = \frac{1}{\sqrt2} \Big( |00 \rangle + |11 \rangle \Big)$
$$C-X | \psi_i \rangle = \frac{1}{\sqrt2} \Big( |00 \rangle + |10 \rangle \Big)$$
$$C-iX | \psi_i \rangle = \frac{1}{\sqrt2} \Big( |00 \rangle + i|10 \rangle \Big)$$
Obviously these two resulting states are different since there is a relative phase. To make the difference more clear you can try to consider measurements in the Y-basis. Just as a note a $C-U$ gate is a "Controlled $U$ gate", and often times you will find $CNOT$ written as $C-X$ to be consistent with $C-H$ (Controlled Hadamard), $C-Z$ (Controlled Z gate), etc.
The Physical meaning of this is that adding a global phase to a single Qubit Gate on a multi Qubit system results in a change in the multi Qubit system. Notice how if we only had a single Qubit, there would be no way of knowing the difference between $X$ and $iX$, because it would be a global phase; however in a 2 Qubit system, that phase shows itself as a relative phase which indeed is measurable.
| {
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Pseudo-forces in revolving frame in contrast with rotating ones A person in a rotating coordinate system, to do Newtonian mechanics has to use pseudo forces such as Coriolis and centrifugal.
We use such forces for rotating coordinate systems,ones that rotate about a fixed axis.
A person on the surface of earth is not in a rotating but a revolving coordinate system, that revolves around the axis of earth ( neglecting rotation around sun) shown as $E$,that goes around the centre of earth.
How do we describe motion in such a revolving frame?
What pseudo forces should I add to analyze motion using newton's laws?
When I look at my textbook this difference isn't touched upon and perhaps the motion is same as would've been for a purely rotating one.
I'd be glad if someone pointed out , I don't have a teacher to go to.
| The method I'm familiar with is a little strange. It uses the tools of general relativity. I did this awhile ago, but I don't have that notebook anymore. I tried to find it computed somewhere online. (I found it. but ignore the t->t' transformation)
$F^i =\frac{d^{2}{x^i}}{dt^2}+ {\Gamma^i}_{rs} \frac{dx^r}{dt} \frac{dx^r}{dt}$
It's all in this little formula. You just have to mechanically unpack it. By the Einstein Summation Convention, you sum over repeated indices (r and s) and $i=1,2,$ or $3$ where $1$ represents $x$ , $2$ represents y, and $3$ represents z.
For example,
$F^y =\sum_r \sum_s \frac{d^{2}{y}}{dt^2}+ {\Gamma^y}_{rs} \frac{dx^r}{dt} \frac{dx^r}{dt}$
https://mathworld.wolfram.com/EinsteinSummation.html
$\Gamma$ is defined in terms of the metric $g_{ij}$.
https://mathworld.wolfram.com/ChristoffelSymboloftheSecondKind.html
https://mathworld.wolfram.com/MetricTensor.html
$g_{ij}$ transforms like a second order tensor. It's easy but tedious to transform from the ordinary flat space metric to the rotating metric. Put rotating metric into the formula for $\Gamma$. The whole problem reduces to finding every $\Gamma$.
The coordinate transformation is
$x'=cos(\omega t) x + sin(\omega t) y$
$y'=-sin(\omega t) x + cos(\omega t) y$
$z'=z$
And I recall using the identity $\cos^2 (\theta) + \sin^2 (\theta)=1$.
| {
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Reflection on moving mirrors Say I have an endless mirror, in a x y plane, at y=1.
Situation 1: the mirror is stationary and when we send light vertically from the origin, the light reflects back and returns to the origin.
Situation 2: let the mirror move horizontally at a constant velocity. Would the light reflected return to the origin?
| A nice way to answer this question is to do the calculation in the rest frame of the mirror. Light propagating in the $y$ direction in the lab frame will propagate in some other direction in the rest frame of the mirror. It will then reflect off in the ordinary way in that frame (angle of reflection equals angle of incidence). Finally, one transforms back to the lab frame to see what direction the reflected beam of light is travelling in.
Here is a calculation using 4-vectors and Lorentz transformation.
Initial 4-wave-vector of beam propagating in +ve $y$ direction, as expressed in lab frame:
$$
K = \left( \begin{array}{c} k \\ 0 \\ k \\ 0 \end{array} \right)
$$
Lorentz transformation for relative motion of the frames in the $x$ direction:
$$
\Lambda = \left( \begin{array}{cccc}
\gamma & -\beta \gamma & 0 & 0 \\
-\beta \gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right)
$$
where $\beta$ is the speed of the mirror in the $x$ direction.
Hence the 4-wave-vector of the upward propagating beam, in the rest frame of the mirror, is
$$
K' = \Lambda K = \left( \begin{array}{c}
\gamma k \\ -\beta \gamma k \\ k \\ 0 \end{array} \right)
$$
(And for a check let's confirm that this is null: $K' \cdot K' =
-\gamma^2 k^2 + \beta^2 \gamma^2 k^2 + k^2 = k^2(1 - \gamma^2(1-\beta^2)) = 0$.)
Now the reflection process in the rest frame of the mirror simply reverses the vertical ($y$) component of $K'$ so the final wave vector (in mirror frame) is
$$
K'_f = \left( \begin{array}{c}
\gamma k \\ -\beta \gamma k \\ -k \\ 0 \end{array} \right)
$$
and therefore the final wave vector in the lab frame is
$$
K_f = \Lambda^{-1} K'_f =
\left( \begin{array}{cccc}
\gamma & \beta \gamma & 0 & 0 \\
\beta \gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right)
\left( \begin{array}{c}
\gamma \\ -\beta \gamma \\ -1 \\ 0 \end{array} \right) k
= \left( \begin{array}{c} 1 \\ 0 \\ -1 \\ 0 \end{array} \right) k.
$$
Thus we find that the light comes straight back down again: the horizontal motion of the mirror does not matter.
Note that motion of the mirror in some other direction, not parallel to its surface, would change the reflection angle for non-normally incident light.
A comment
When I started on this answer I began to write reasons why, by symmetry, the motion of
the mirror could not matter. Those arguments are probably ok, but it occurred to me
that mirror reflection off a dielectric stack, for example, involves propagation inside
the stack and it is well known that the motion of the medium affects the motion of light inside the medium. So then the symmetry argument was not quite so self-evident as it may first appear. But I think the above calculation via two changes of frame is water-tight.
At least, I find I am more confident of this answer than I would be trying to argue that a moving medium doesn't affect the reflection at normal incidence in the lab.
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