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Is there a nice physical interpretation of this formula? As a trivial example in our vector analysis class, we did the following computation.
Let $\overrightarrow{\omega} = (\omega_1, \omega_2, \omega_3)$ be the angular velocity and $\overrightarrow{r} =(x,y,z)$ be the position. Then we have a vector field $\overrightarrow{R} = \overrightarrow{v} = \overrightarrow{\omega} \times \overrightarrow{r}$.
We quickly calculated the rotor and got:
$\text{rot} \overrightarrow{v} = 2 \overrightarrow{\omega}$.
The calculation is trivial of course, but I can't see any physical meaning behind this. But the equation is so simple that there must be some neat way to interpret this! Does anyone know of a nice intuitive explanation of why this equality holds?
| I think the comments of @J.G. and @Cleonis add up to a complete answer. In particular see the answer of Ian to that maths stack exchange question: the factor of 2 is geometric in origin. The geometric definition of the curl, or rot, at a point P is the limit of the loop integral around an arbitrary loop enclosing P divided by the area. If we subdivide the area into many triangles apex at P and base on the loop, then the contribution of each triangle to the loop integral is $(\omega h) b$ where $h$ is the height of the triangle and $b$ is the base, and the contribution to the area integral is $bh/2$. Thus the factor of 2 comes directly from the geometry.
| {
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"Earth spinning faster will break GPS and atomic clocks".. Umm, how? Since it's just come out that Earth is apparently spinning about a milliscecond and a half faster.. I see a bunch of "articles" claiming this will.. somehow.. break atomic clocks and make GPS useless.. (Google "earth spinning faster break atomic clocks" and go to crazy-town..)
Umm, how and what exactly do they think will happen? A couple things I saw (I read two articles, then quit trying to play whack-a-mole to find one that might actually explain the basis of this claim) did admit that we already have to correct for relativity.. And we already have positive leap seconds added every so often, and my understanding is that various ephemeris data has to be updated every so often.. So would this be be really any different?
I can see that if the speed kept increasing by something like.. say two or three seconds per day, or the Earth started wobbling around that could cause problems.. But we're not talking about that (at least until next year)
| None of the articles (e.g. this one) I came across say anything about breaking atomic clocks. The problem is simply a logistical one - if we have to implement a negative leap-second, then we will be doing something which most of the software which implements ordinary leap-seconds has not been explicitly designed to do. And if you've ever written any large-scale computer code, you are probably aware that a tiny change can wreak massive havoc if that change hasn't been thoroughly tested.
| {
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Why does the graviton polarization satisfy $\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k},\lambda') = 2 \delta_{\lambda\lambda'}$? I am reading the paper ``Graviton Mode Function in Inflationary Cosmology'' by Ng (link here). The graviton $h_{ij}$ is here expanded (in the TT gauge) where
$$
h_{ij}(x) \sim \epsilon_{ij}(\mathbf{k},\lambda) h_{\mathbf{k}}(\lambda,x)
$$
and in equation 9 it is said that
$$
\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k},\lambda') = 2 \delta_{\lambda\lambda'} \ .
$$
Where does this come from? And how is the relation adjusted for differing momenta $\mathbf{k} \neq \mathbf{k}'$, ie. can one write down a relation for $\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k}',\lambda) = \ldots$?
EDIT: Why does one need this condition? Is it so the Lagrangian is properly normalized when written in terms of $h_{\mathbf{k}}$?
| I can't comment with certainty on $\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k},\lambda') = 2 \delta_{\lambda\lambda'}$, but I suspect that it's because gravitons actually have only two independent polarizations, so this is a way to go from polarization labels ($\lambda$) to real space ones ($i,j$).
I can, however, say for certain that the polarizations at different $\mathbf{k}$ are unrelated, by necessity. To the extent that they are related is a question of the net polarization you'll observe over a certain bandwidth. I want to say it's related to coherence, but that's more related to the phase at different wavelengths.
| {
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Why flapping rudder produce net thrust if one half-stroke produce thrust and second half-stroke drag? In small sailing boat like optimist is well know technique when there is no wind, rudder pupming which push boat forward.You just need push-pull rudder stick left to right with fast movement.
Rudder works complety under the hull, so there is no pressure interaction between stern and rudder.
Forward half-stroke is when rudder rotate from centerline to left or right
(from 2 to 1 or from 2 to 3).
Why stiff rudder(not felxibile like flippers) produce net thrust if forward half-stroke produce drag?
(Or maybe forward half stroke produce thrust as well? I dont think so..)
Please explain your answer with pressures at rudder sides for two condition;
*
*boat speed zero
*boat is moving
Avoid Newton 3 law.
| I suspect it has nothing to do with regions of higher or lower pressure (if those even exist). When you pump the rudder you are pushing water backward and by Newton's Third Law that water exerts an equal and opposite force on the boat, pushing it forward.
| {
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What is the average recessional velocity of an object in the universe? I’m trying to get a better grasp on cosmological horizons and have a question regarding recessional velocity. In particular:
What is the average recessional velocity of a cosmological object (irrespective of distance) an observer would expect to observe if measured at some time in the future?
Here is my attempt:
If the current cosmic event horizon $D_{eh}$ is the maximum proper distance from which light emitted now can ever reach the observer in the future, then to calculate the average recessional velocity $\bar v_{rec}$ of an observable that could potentially be detected by the observer at some time between now $t_0$ and $t=\infty$, one can calculate the recessional velocity of the current event horizon $v_{eh}$ and divide by two—i.e., as recessional speed is proportional to distance, this is the average between $v_{eh}$ and the observer’s recessional velocity, $v_{ob}=0$. Therefore,
\begin{align}
\bar v_{rec} = \frac{1}{2}v_{eh}(t_0) = \frac{1}{2}H_0 D_{eh}.
\end{align}
If the Hubble constant $H_0$ is about $70$ km/s/Mpc, and assuming a standard $\Lambda$CDM-model, $D_{eh}$ is about 5 Gpc, then
\begin{align}
\bar v_{rec} \approx 0.58c.
\end{align}
Does this reasoning make sense? Thanks for your help.
| If I understand your assumption that all of the matter objects are similar, and all on a large scale have similar densities, then what determines one's velocity from us is determined by the equation:
$$V = D * H_0.
$$
$V$ is velocity from us, $D$ is distance from us, and $H_0$ is the Hubble constant with the reciprocal value: $1/H_0 = 14.4 \, \text {Gyrs}$.
So, what needs to be calculated is the average of distance A of points within a sphere from its center. The radius of the sphere R will be
$$R = c * 14.4 \, \text {Gyrs}. $$
(Note: c = speed of light.)
I think you that should have the first try to calculating A. If you can't solve for A, then I will offer some more help.
The reason for this value of R (also known as the event horizon) is that the distance R is where anything further away travels away from the center faster than the speed of light.
| {
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What is virtual photon concept in classical electrodynamics? If we observe a charged particle like an electron passing us at some high speed $u$, then as $u \to c$ the field we observe looks like a superposition of plane waves normal to the trajectory of the electron. The field can be Fourier transformed, and the modes associated with virtual photons. See for example the discussion in chapter 19 of Classical Electricity and Magnetism by Panofsky and Phillips.
Is this virtual photon we talk about in classical electrodynamics the same as virtual photon that is the the force carrier in quantum electrodynamics?
| It's the same photon but, in this context, it's part of a "back of the envelope" type calculation based on energy quantization ($E=hf=\hbar \omega$) and the assumption that each sufficiently small interval of the energy spectrum $U_\omega\mathrm d\omega$ consists of a number of "equivalent photons" $N_\omega$. It's probably a good approximation but it isn't the modern quantum field theory approach where the Number operator is defined in terms of the Hamiltonian/Energy operator and creation and annihilation operators.
The section on the "virtual photon" is in chapter 18. Section 3 "The Virtual Photon Concept"
The energy of the electromagnetic wave is (in terms of its Fourier components) is calculated.
$$U= \int_0^\infty U_\omega\mathrm d\omega$$
$$U_\omega
=\frac{2}{\pi} \frac{e^2}{4\pi \epsilon_0 u} \ln\left(\frac{\gamma u}{\omega b_{min}}\right),$$
where $u$ is the velocity.
The number of "equivalent photons" $N_\omega$ is defined by $U_\omega\mathrm d\omega = \hbar \omega N_\omega\mathrm d\omega$ such that
$$N_\omega\mathrm d\omega
= \frac{2\alpha}{\pi} \ln(E/A\hbar\omega) \frac{\mathrm d\omega}{\omega}$$
and
$$N_\omega\mathrm
= \frac{2\alpha}{\pi} \ln(E/A\hbar\omega) \frac{\mathrm 1}{\omega}.$$
$\alpha$ is the fine structure constant, $A$ is a numerical constant and $E$ is the energy of the particle.
Hence the spectrum of “equivalent photons” varies approximately as
$l/\omega$. The number of equivalent photons per electron is small, namely,
of the order of 1/137. Equation (18-38) is very useful in relating the
probability of processes induced by electrons, or by other particles
which act essentially only through their electromagnetic field, to the
probability of processes induced by electromagnetic radiation.
| {
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Measured Data Question regarding Galaxies with or without Dark Matter and Supermassive black holes My understanding is that all Ultra Diffuse Galaxies (UDGs) have no Dark Matter nor do they have Supermassive Black Holes. It is also my understanding that all other galaxies have both Dark Matter and Supermassive Black Holes. Or am I wrong:
Is anyone able to name or site a reference to a galaxy that has Dark Matter, but has no Supermassive Black Hole at its center?
Is anyone able to name or site a reference to a galaxy with no Dark Matter, but that does have a significant Supermassive Black Hole at its center?
| *
*Ultra-Diffuse Galaxies actually have a wide range of dark matter content. Some indeed seem to have little or no dark matter, but others are quite dominated by dark matter. For example, Dragonfly 44 in the Coma Cluster has an estimated dark-matter halo mass similar to that of the Milky Way, even though the total mass of its stars is about 100 times smaller than the Milky Way's.
*Not all galaxies have SMBHs. M33 (aka the Triangulum Galaxy), for example, has a very small upper limit of $\sim 1500$ solar masses on any possible central black hole (Gebhardt et al. (2001)), well below the usual definition of an SMBH (i.e., $> 10^{6}$ solar masses).
*We currently know nothing about any possible SMBHs in UDGs. In general, these galaxies are simply too faint for any of the usual methods for measuring SMBHs to work. This is especially true given that their low stellar masses and stellar velocity dispersions imply that any central BHs that did exist and which followed the standard relations between SMBH mass and galaxy properties would have BH masses $< 10^{5}$ or $< 10^{4}$ solar masses, making them even harder to detect.
| {
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Computing the maximum force a rod can bear Suppose I had a rod of diameter $d$ composed of some material with tensile strength $T$. If I then exterted a pulling force $F$ on the ends of the bar, how do I compute the force $F$ for which the rod will break apart? Is there some general equation that I can use to compute this?
| Breaking stress is the maximum force that can be applied on a cross sectional area of a material in such a way that the material is unable to withstand any additional amount of stress before breaking.
Breaking stress is calculated with the formula:
Breaking Stress = Force / Area (in your case Area= pi*d^2/4)
Breaking stress may also be known as ultimate tensile stress or breaking strength.
This is a picture from wikipedia which explains this for more info you can visit them
https://en.wikipedia.org/wiki/Stress%E2%80%93strain_curve
| {
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What's a good simple model for wind attacking drone? I'm trying to simulate a drone on https://rapier.rs/. My idea is to do an X shaped object, with a force $P$ on each end of the X, always perpendicular to the drone plane.
I want to simulate wind, which I think would be good if it were a $W$ vector force acting on each of the propellers. Or should I just use a force acting on the middle?
Is there a better way?
The goal is to construct a https://en.wikipedia.org/wiki/PID_controller to control the drone, so I don't need a very complicated wind model.
| Assuming all of the propellers rotate the same direction, when the wind hits them it will create more lift and drag on the side of the propeller that is moving toward the wind. The right propellers will have more lift on the outside blade, and the left propellers will have more lift on the inside blade (for example)
Looking at the torque on the drone about its center of gravity - the outside blades have disproportionately large torque effects due to their increased lever arm, so the asymmetric outside lift will create a rolling action around the wind vector, and the asymmetric outside drag will create a yawing action around the up axis.
That is to say - a simple model looking only at a single force acting on the drone will not capture the full dynamics. Nor will simple forces acting on each propeller.
| {
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Besides traveling at the speed of light, how can we be sure that it is possible to have energy and momentum without mass? How can we be sure that it is possible to have energy and momentum without mass? If something were to continually lose energy, would it not also lose a corresponding amount of mass? I understand that photons are predicted to have no mass because they travel at the speed of light, but is there another reason this is believed?
| How about this: even in classical electrodynamics (no quantum, no relativity) electromagnetic waves carry momentum. Waves can't have mass because they are not objects. And electromagnetic waves are disturbances in a medium that isn't made of stuff with mass.
And it has to be that way for conservation of momentum to be compatible with electrodynamics. Recall the basic picture of radiation: accelerating charges produce electromagnetic fields, which travel at a finite speed until they meet other charges and cause them to accelerate.
If the waves don't carry momentum then momentum appears to disappear from the first charge, then reappear in the second charge later. Conservation of momentum can only be true at every stage in this process if the wave itself has momentum.
So even with just classical electrodynamics you have to extend the concept of momentum to things beyond particles with mass. That means it should not be surprising to find other surprising kinds of particles with momentum.
| {
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Acceleration of wedge and mass locked on axes Consider this diagram:
Mass m1 can only move in the x direction and mass m2 in the y direction. Find the acceleration of both m1 and m2, and the reaction of the wedge. Negligible Friction.
I tried attempting to solve this problem by finding the normal force on m1 from m2, and what its horizontal movement component was, however I couldn't quite understand it. How would I solve this?
| Using this algorithm you can solve almost all of high school mechanics(except rotational mechanics):
*
*I will ignore the wheel at the end of the rod since friction is negligible and therefore no rolling.
*Draw the free body diagram for the experiment.
*Find the constraint equation(In this case, the wedge and the rod should always stay in contact, i.e. the acceleration of rod and wedge along the line perpendicular to tangent at point of contact are equal).
*The accerleration of the rod and the wedge can now be calculated by simple artihmetic.
| {
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Is the equal sign in "$0 ^\circ \mathrm{C} = 273.15\, \mathrm{K}$" fair? I'm in doubt whether the equal sign in an expression like "$0 ^\circ \mathrm{C} = 273.15 \,\mathrm{K}$" is fair because, normally, if $A=B$, then, say, $2A=2B$, which is hardly applicable to "$0 ^\circ \mathrm{C} = 273.15 \,\mathrm{K}$".
So is it okay to use the "$=$" if we cannot actually perform multiplication or division by the same number on both sides? Which sign would suit better here if the "$=$" does not work? Thanks a lot in advance!
|
I'm in doubt whether the equal sign in an expression like "$0 {\ ^\circ\mathrm{C}} = 273.15 {\ \mathrm{K}}$" is fair because, normally, if $A = B$, then, say, $2 A = 2 B$, which is hardly applicable to "$0 {\ ^\circ\mathrm{C}} = 273.15 {\ \mathrm{K}}$".
I think that the equals sign is totally appropriate, and the rule "if $A = B$, then $2 A = 2 B$" is perfectly applicable to this equation.
The only tricky part is that, unlike most unit symbols, the unit symbol $^\circ\mathrm{C}$ can't be treated as though it were a quantity that is multiplied by the number next to it. It must be treated as an operator which is written on the right-hand side of the quantity that it operates on.
As a result of this, the rule for multiplying a temperature in degrees Celsius by a scalar is strange. Specifically, the rule is that
$$n (t {\ ^\circ\mathrm{C}}) = (n t + (n - 1) 273.15) {\ ^\circ\mathrm{C}}.$$
But in any case, by applying this rule, we can see that
$$
\begin{align}
2 (0 {\ ^\circ\mathrm{C}}) &= (2 \cdot 0 + (2 - 1) 273.15) {\ ^\circ\mathrm{C}}\\
&= (2 \cdot 0 + 1 \cdot 273.15) {\ ^\circ\mathrm{C}}\\
&= (0 + 273.15) {\ ^\circ\mathrm{C}}\\
&= 273.15 {\ ^\circ\mathrm{C}}\\
&= (273.15 + 273.15) {\ \mathrm{K}}\\
&= (2 \cdot 273.15) {\ \mathrm{K}}\\
&= 2 (273.15 {\ \mathrm{K}}).\\
\end{align}
$$
| {
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Wave Function Collapse and the Dirac Delta Function When the wave function of a quantum system collapses, the probability of finding it at some specific point is given depends on $||\Psi||^2$:
$$
\int_{\mathbb{R}^3}{d^3 \mathbf x \; ||\Psi||^2} = 1
$$
Could this modulus square, the instant you measure, be thought as the Dirac Delta Function, because all the probability condensates to a single point, and its integral over all $\mathbb{R}^3$ gives 1.
$$
||\Psi||^2 = \delta(\mathbf x)\\
\int_{\mathbb{R}^3}{d^3 \mathbf x \; \delta(\mathbf x)} = 1
$$
If yes, what are the initial conditions the wave equation must have the instant after being collapsed.
The first one shall be this:
$$
\Psi(\mathbf x, t) \\
||\Psi(\mathbf x, 0)||^2 = \delta(\mathbf x)
$$
No? How would you plug this condition onto the Schrödinger Equation?
| I suppose that you think that when a measurement is done then probability of a finding particle at a given point is certain, thus it can be equated with delta function which has value of one at a point and zero otherwise. Also this is after measurement, so wavefunction is collapsed.
In the case been asked, once a measurement has taken, state of a particle is changed. And how could particle is find at given position with certainity, when its probability of to be find there is a fraction smaller than one. So a wavefunction can't be equated with delta function even after collapse. Well, delta function is used to represent cause and not response, because such a confined response is not physical. A wavefunction is response to operator, in given conditions.
| {
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What causes light passing through a hole to change direction? On diagrams showing light passing through a hole, the wave of light appears to change direction when it emerges from the hole.
What causes that change of direction? Is it maybe the walls of the hole imparting a pulling force or the sudden absence of light next to the emerging beam causes the light to spread?
Or maybe light does this all the time and we only notice when we put a wall with a hole in the way.
Please explain this to like I'm a five year old.
| Light is a wave , at the edge of the hole the oscillation spreads in all directions but the highest intensity is forward.
| {
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Why electron do not eject out even when there is photon of less threshold energy with increasing in time When photon having certain energy less than threshold energy strikes on the electron of metallic plate. Electron do not eject out. But my question is when photon are falling continuously then electron must gain the energy to eject out.as electron have the energy transferred from earlier photon and next photon will transfer his energy too.so combination of these energy must be greater than the required energy for moving out from his path.
I know it's a silly question but i have doubt so asking it...
| It's not that simple. Electron can only absorb such photons which exactly hits electron quantum jump levels, for example check Lyman, Balmer series, etc. Otherwise non-compatible photons "are ignored" by electrons. Unless incident light is very strong, i.e. you shine with an intensive laser light on metal, even laser wavelength does not pass metal work function,- in this case non-linear effect can happen, such as multi-photon absorption. It is explained that intensive laser light weakens atom potential barrier, so electrons can escape atom by tunneling ionization process. But non-linear tunneling ionization frequency is covered by such law :
$$ {{\omega }_{t}}=\frac{eE}{\sqrt{2m_e{ {\mathcal E} }_{i}}} $$
where $E$ is amplitude of incident electric field, ${ {\mathcal E} }_{i}$ - ionization potential. So answer is that electrons can eject out of metal even photons does not pass ionization energy barrier, but... just in case strong electric field applied when non-linear effects can begin to happen.
| {
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Why does the opposing force differ in when falling on concrete vs on water in spite of Newton's third law? If a person jumps from the first floor of a building and lands on a concrete surface, they will suffer serious injury because of Newton's third law.
If the same person jumps the same distance and lands in swimming pool filled with water, however, then there will not be any serious injury.
The person in both cases lands with same amount of force. Why doesn't water offer the same amount of force in return as concrete?
|
But the person in both case lands with same amount of force. Then why
doesn't water offer the same amount of force in return as concrete
does?
The person does not land with the same amount of impact force. The average impact force that the concrete exerts on the person is greater than the average impact force the water exerts on the person because the person's stopping distance is much less for the concrete.
This can be seen by applying the work energy principle, which states that the net work done on an object equals its change in kinetic energy, along with the definition of work.
The work done on the body by the concrete or water where $F_{ave}$ is the average impact force and $d$ is the stopping distance
$$W=F_{ave}d$$
Ignoring the change in gravitational potential energy after impact, this work equals the change in kinetic energy of the object that is brought to a stop, or
$$F_{ave}d=-\frac{1}{2}mv^2$$
Where $v$ is the velocity of the person just prior to impact and the final velocity is zero when brought to a stop. So
$$F_{ave}=-\frac{1}{2d}mv^2$$
Since the concrete gives very little compared to the water, the stopping distance $d$ for the concrete is much less than the water, meaning the average impact force (and the damage it does) is much greater for the concrete.
Hope this helps.
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If Aristoteles was right and heavier objects falled faster towards the ground how would be Newton's Laws of Motion described? It seems like it would be like:
a(m)=km
and may be a(m1,m2)=K(m1-m2)
Am I doing any sense?
Btw I'm no negationist, nor I'm trying to create a negationist movement here, I just wonder how physics would be If we lived on different physical environment in order to understand better the physics we have now.
| When solving Newtons equation for a free falling object the mass cancels out on both sides, which implies that two objects of the different mass fall with the same acceleration
$$
m a = F = - m g \quad \Rightarrow \quad a = -g
$$
For the acceleration to be mass dependent you would need to change either side of the equation. For example, you could say that the laws of gravity are different, say $F = -m^2 g$, which leads to $a = -m g$, i.e. the acceleration increases linearly with its mass. You could also change the other side of the equation ($F = m a$), i.e. how force is related to acceleration. In this case you are changing the Newtonian laws of motion entirely.
As pointed out in the earlier answer, the implications of these equations being different are of course absurd.
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Why is it easier to raise AC current to high voltage than DC? In my country (and maybe all around the world I don't know) once electricity has been generated, it is then raised to 200k Volts for transportation.
I know this is to reduce the loss. Given $P=U.I$ and $P=I^2.R$, raising U will lower I and so limit the loss by joule effect.
From what I've read, one of the reason electricity is transported in AC is because this is easier/cheaper to raise AC to 200k Volts than if it was in DC.
Why?
| As with most engineering decisions, it all boils down to "what's the best trade off between conflicting constraints and requirements".
*
*AC allows very simple conversion between different voltages using passive transformers
*AC conversion scales nicely across a large range of voltages and powers
*AC is the historical standard and all appliances expect roughly 110V or 220V AC, even if most of them immediately convert this to something else.
*DC is more energy efficient, especially over long distances
As a result of this the majority of electrical power is transmitted using AC but there are plenty existing system that use DC. Some of the larger ones (in both terms of distance and power) are in China and Brazil. For a description of the technology, see https://en.wikipedia.org/wiki/High-voltage_direct_current.
It's a simple matter of choosing the right tool for the specific job.
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Computation of $\nabla_a\nabla_b X^d$ All my notation follows Chrusciel's book "Elements of General Relativity". For a vector field $Y$ and connection $\nabla$, we define
$$\nabla_a Y^b:=dx^b(\nabla_a Y)$$
where $\nabla_a:=\nabla_{\partial_a}$. So in English, this would be taking the b-th component of the vector field $\nabla_a Y$, NOT taking the covariant derivative of the coefficient function $Y^b$ in the direction of $a$. Apologies if this is standard, but I haven't encountered such notation before. Continuing, we also define Christoffel symbols as
$$\nabla_a \partial_b:=\Gamma^c_{ab}\partial_c$$
My question concerns the following lines in proof of Proposition 1.4.1 on page 29:
\begin{align*}
\nabla_a\nabla_b X^d&=\partial_a(\nabla_b X^d)+ \Gamma^d_{ae}\nabla_bX^c - \Gamma^e_{ab}\nabla_e X^d\\
&=\partial_a\partial_bX^d+\partial_a\Gamma^d_{be}X^e+\Gamma^d_{be}\partial_aX^e+\Gamma^d_{ac}\partial_bX^c+\Gamma^d_{ac}\Gamma^c_{be}X^e-\Gamma_{ab}^e\nabla_eX^d
\end{align*}
I don't understand the first equality. Let me try computing in a way more familar to me. If we take $X=X^i\partial_i$, and define coefficeints $Y^i$ by $\nabla_b X=Y^i\partial_i$, we compute
\begin{align*}
\nabla_a\nabla_b X&=\partial_aY^i\partial_i+Y^j\nabla_a\partial_j\\
&=\partial_a Y^i\partial_i+Y^j\Gamma^i_{aj}\partial_i\\
&=(\partial_a Y^i+Y^j\Gamma^i_{aj})\partial_i
\end{align*}
A similar computation determines $Y^i=\partial_b X^i+X^k\Gamma^i_{bk}$, so we subsitute to get
\begin{align*}
\nabla_a\nabla_bX&=\left[\partial_a(\partial_bX^i+X^k\Gamma^i_{bk})+(\partial_bX^j+X^k\Gamma^j_{bk})\Gamma^i_{aj}\right]\partial_i\\
&=\left[\partial_a\partial_bX^i+\partial_aX^k\Gamma^i_{bk}+X^k\partial_a\Gamma^i_{bk}+\partial_bX^j\Gamma^i_{aj}+X^k\Gamma^j_{bk}\Gamma^i_{aj}\right]\partial_i
\end{align*}
Note that the two answers do NOT match! In Chrusciel's computation, there's an extra number of terms, namely
$$-\Gamma^e_{ab}\nabla_e X^d,$$
but the rest of the terms match up. Could someone help me find the computation in my mistake, explain where these extra negative terms come from, or explain the first equality in Chrusciel's proof?
| Your mistake is a subtle one, and is related to the notational ambiguity you mention at the beginning of your question. In abstract index notation, the expression $X^b$ is to be understood to mean the vector field $X$, with the superscript $b$ not being a numerical label (i.e. the $b^{th}$ component) but rather as a formal symbol to remind us that $X$ has one "slot" to fill (i.e. it is a function of a single covector).
Therefore, in the expression $\nabla_a\nabla_b X^d$, we should identify $X^d$ as a $(1,0)$-tensor, $\nabla_b X^d$ as a $(1,1)$-tensor, and $\nabla_a \nabla_b X^d$ as a $(1,2)$-tensor. In other words, $\nabla_a \nabla_b X^d \equiv \nabla \nabla X$ in coordinate-free notation.
The action of $\nabla$ on a $(1,1)$-tensor $T$ is given in component form$^\dagger$ by
$$(\nabla T)^i_{jk} = \partial_j T^i_{\ \ k} + \Gamma^i_{j\ell}T^\ell_{\ \ k} \color{red}{- \Gamma^\ell_{jk} T^i_{\ \ \ell}}$$
which is the origin of the mysterious extra term.
$^\dagger$Note that this is not abstract index notation - the indices in this final expression take numerical values, and the quantities being labeled are the components of tensors (and the Christoffel symbols).
| {
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Negative energy solutions in Dirac equation for hydrogen atom Solving Dirac's equation for the hydrogen atom $^1$,
$$(\boldsymbol{\alpha}\cdot\boldsymbol{p}+\beta\cdot m)\Psi=E\Psi$$
after some mathematical machinery we find the condition for energy eigenvalues
$$\varepsilon^2=\frac{1}{1+\left(\frac{\alpha}{1+\gamma+n'}\right)^2}\tag{1}$$
where
$$\varepsilon:=\frac{E}{m}\tag{A}$$
$$\lambda=j+\frac{1}{2}\qquad\gamma=-1+\sqrt{\left(j+\frac{1}{2}\right)^2-\alpha^2}\tag{B}$$
and $n'$ is a quantum number.
Taking the square root in equation $(1)$ and using $(A)$ and $(B)$ only the positive energy solution is retained$^2$
$$E_{n'j}=\frac{m}{\sqrt{1+\left(\frac{\alpha}{\sqrt{\left(j+\frac{1}{2}\right)^2-\alpha^2}+n'}\right)^2}}.\tag{2}$$
Equation $(2)$ yields the correct non-relativistic result in the first terms of McLaurin expansion for $\alpha$. Is there any other argument to reject the negative energy solution of $(2)$ i.e.
$$E_{n'j}=\frac{-m}{\sqrt{1+\left(\frac{\alpha}{\sqrt{\left(j+\frac{1}{2}\right)^2-\alpha^2}+n'}\right)^2}}.\tag{2}$$
other than yielding the correct non-relativistic limit?
The electron here is not a free particle, so I can't see any obvious reason to discard the negative energy as we would if it were.
$^1$J.J.Sakurai, J. Napolitano, Modern Quantum Mechanics. Third edition. Chapter 8, section 8.4.
$^2$See $(8.149)$ and $(8.150)$ of footnote 1.
| It is easy to see that the negative energy solutions would be unphysical.
First of all, the proton is positively charged and the negative energy solutions would be a positron rather than an electron, and there should not be a positron bound to a proton.
If you are more mathematically minded, you could try to solve the original equations without the squaring step, and then there will not be the unphysical negative energy solutions coming in.
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Why spontaneous emission? From what I have understood Einstein somehow deduced the A and B coefficient for spontaneous respectively stimulated emission to match the observed blackbody radiation/Planck spectrum. How did he come to this conclusion that there should be emissions that are spontaneous? What causes that all emissions aren’t stimulated but it is necessary to invoke spontaneous emissions? Of course I know that it is not the case and most of the photons around us were caused by spontaneous emissions, but I don’t understand the fundamental/mathematical reason for it. How do we know that there aren’t enough „photons/phonons“ there of the right frequency to be causing stimulated emission for all emissions? Why is it therefore necessary to invoke spontaneous emission to be able to describe the black body radiation?
| If stimulated emission was the only radiative emission process then objects in thermal equilibrium would be transparent.
Any absorption of a light beam would be exactly balanced by stimulated emission that is in phase and in the same direction as the incident radiation. The absorption coefficient, which is the true absorption minus the stimulated emission would be zero.
A blackbody must absorb all radiation incident upon it, and the ratio of emission to absorption coefficients should equal the Planck function. If the absorption coefficient is zero then clearly that isn't possible.
| {
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Spectral theorem in QFT I know that the spectral theorem holds for unbounded normal linear operators on infinite dimensional Hilbert-spaces. We usually employ it in Quantum mechanics to explain the role of self-adjoint operators.
However, I'm not sure wether the theorem also applies to the observables of QFT, the reason being (for an interactive QFT) that we don't even know how the Hilbert space of the QFT looks like, or that the fields in QFT are operator valued distributions, and not operators. Hence the question: Is there a version of the spectral theorem that still holds in QFT (possibly with some restrictions to the used operators).
| I am not sure to understand the nature of the problem.
The spectral theorem, as it is a mathematical fact, holds also in QFT. It does not matter if we do not know how the Hilbert space is made, it is sufficient to know that it is a Hilbert space and that the used operator is selfadjoint. Regarding operator valued distributions $\phi$, the spectral theorem applies to (usually the closures of) the images of these distributions $\phi(f)$ when they are selfadjoint operators.
If the theorem did not hold, then we would conclude that the space of states is not Hilbert or the operator is not selfadjoint (more generally normal).
| {
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Delta v of a trans-Mars injection (TMI) Why does it only take about 600 m/s more than Earth's escape velocity to have an encounter with Mars while it takes much more Delta v (about 3 km/s) from a solar orbit (same as Earth orbit) to have an encounter with Mars?
What explains this difference?
Is it the Oberth effect or something else that explains this?
| Yes, it seems related to the Oberth effect. Starting from the solar orbit, we need $\Delta v$ to enter an orbit to Mars. By energy conservation the velocity of the spaceship at launch from Earth has to be
$$v_0 = \sqrt{v_{\text{esc}}^2+\Delta v^2} \ ,$$
where $v_{\text{esc}}$ is the escape velocity. For $v_{\text{esc}} \gg \Delta v$, we find that the $\Delta v^{\prime}$ on top of the escape velocity is
$$\Delta v^{\prime} \simeq \frac{1}{2}\frac{\Delta v^2}{v_{\text{esc}}} \ .$$
This formula relates your Delta-v's in your question. We observe that $\Delta v^{\prime}$ becomes smaller the bigger the escape velocity from Earth. Bigger escape velocity means that you're deeper in the gravitational well. By the Oberth effect, a small Delta-v deep in the gravitational well has a bigger effect outside of the well.
In fact, irrespectively of the above approximation, we always have
$$\Delta v^{\prime} = \sqrt{v_{\text{esc}}^2+\Delta v^2}-v_{\text{esc}} \leqslant \Delta v \ .$$
| {
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Why are sunspots cooler than the other regions? The Sunspots appear darker than the other regions because they are cooler; and I know that according the Babcock model, the Sunspots are places where the tangled magnetic fields burst out through the Sun's surface. How does these bursting magnetic field lower the Sun's surface temperature?
I found a good explaination:
Collins, G. W. (2021). Sunspots. In K. H. Nemeh & J. L. Longe (Eds.), The Gale Encyclopedia of Science (6th ed., Vol. 7, pp. 4326-4329). Gale. https://link.gale.com/apps/doc/CX8124402391/CIC?u=utoronto_main&sid=bookmark-CIC&xid=d939e1ba
| As I discuss in https://physics.stackexchange.com/a/415248/59023 and https://physics.stackexchange.com/a/708183/59023, a sunspot is a pressure-balance structure. The total pressure in a collisionally mediated plasma includes the thermal and magnetic pressures. Since the regions involved in generating sunspots arise from the enhanced magnetic fields of flux ropes rising to the photospheric surface, we know that the magnetic pressure is high. We assume the thermal pressure is given by:
$$
P = n \ k_{B} \ T \tag{0}
$$
where $n$ is the number density [number per unit volume], $k_{B}$ is the Boltzmann constant, and $T$ is the temperature.
Since this is a collisionally mediated system, the density inside and outside these structures will roughly remain the same but the temperature can vary. To remain in pressure balance (i.e., total pressure outside equals total pressure inside), the temperature must then drop to accomodate the higher magnetic pressure.
| {
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Is the non-simply connected version of AdS space a maximally symmetric spacetime? A common construction of anti-de Sitter space is the following:
*
*Start with the flat five-dimensional manifold with metric $ds_5^2 = -du^2 - dv^2 + dx^2 + dy^2 + dz^2$.
*Consider the hyperboloid submanifold given by $-u^2 - v^2 + x^2 + y^2 + z^2 = -\alpha^2$.
*Define the hyperbolic coordinates $(t, \rho, \theta, \phi)$ on the hyperboloid. The coordinate $t$ is periodic with period $2 \pi$. The submanifold geometry inherited from the ambient space yields the metric $ds^2 = \alpha^2 \left( -\cosh^2(\rho)\, dt^2 + d\rho^2 + \sinh^2(\rho)\, d\Omega_2^2 \right)$.
*Consider the universal cover of this hyperboloid, which has the same metric as above but the new coordinate $t$ ranges over the whole real line. This is AdS space.
My questions are about the original hyperboloid constructed in steps 1-3, before taking the universal cover.
Q1. What is the global topology of this spacetime? Is it $S^1 \times \mathbb{R}^3$?
Q2. Is this spacetime maximally symmetric? It seems like it should be, since (I think) it has the same Killing vector fields as AdS. But I've sometimes seen the claim that "AdS is the unique maximally symmetric spacetime with constant negative scalar curvature." Is that true, or is this hyperboloid a counterexample?
| Related answers: this one by me and this one by Slereah, which has a classification of maximally symmetric spaces with one timelike dimension (citing Spaces of Constant Curvature: Sixth Edition by Joseph A. Wolf).
Your hyperboloid (which Wolf calls $\mathbb H^3_1$) is topologically $S^1\times\mathbb R^3$. The maximally symmetric spacetimes of that signature and curvature are the covers of $\mathbb H^3_1 / \mathbb Z_2$, so $\mathbb H^3_1$ and $\text{AdS}_4$ are maximally symmetric. $\text{AdS}_4$ is unique if you also require simple connectedness.
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Displacement Vector and resultant vector I am struggling with the concept of displacement. From my understanding displacement can be found for 1D motion along the x-axis as $\Delta x= x_{f}-x_{i}$. For example someone walks $1\,\mathrm{m}$ to the left then $1\,\mathrm{m}$ to the right then $2\,\mathrm{m}$ to the left and finally $3\,\mathrm{m}$ to the right. I know that I can find the net displacement by adding them all up as $-1+1-2+3$ and get $1\;\mathrm{m}$ to the right as my resultant displacement. My question is why can't I use $\Delta x$ and use the $3$ as my final position and $0$ as my initial position? I know that it clearly doesn't give me the correct answer but why does that not work?
| Responding to your question and follow up comment, displacement is indeed final position minus initial position. But in your example, the numbers you are listing are in fact displacements, not positions.
Let us number your positions as follows:
\begin{array} {r|r} \hline x_0 & 0 \\ x_1 & -1 \\ x_2 & 0 \\ x_3 & -2 \\ x_4 & 1 \\ \end{array}
Those are the positions after each step.
The displacements after each step are given by $\Delta x = x_{current} - x_{previous}$, or in other words:
\begin{array} {r|r} \hline \Delta x_1 & -1 \\ \Delta x_2 & 1 \\ \Delta x_3 & -2 \\ \Delta x_4 & 3 \\ \end{array}
Which are the numbers you presented. You can easily verify that the current position is the cumulative sum of all previous displacements, and that any given displacement is the difference between the current and previous positions. The total displacement is $x_4-x_0 = 1$, but subtracting the individual displacements like $\Delta x_4 - \Delta x_1$ does not give anything meaningful.
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What makes a photon a photon? As i understand photons are excitation of the electromagnetic field. Therefore charged particles are affected by this excitation. But what if we have (highly theoretically) a particle that has the exact same properties like a photon (spin 1, no electric charge, no color charge, no mass etc.) but is an excitation of an other field. Is this even (again, theoretically) possible? Is a particle with the properties of an photon always the excitation of EM fields? And would this photon-like particle interact with charged particles (because they are not belong to EM fields)?
| I think the existing answers are already excellent. I wish merely to add that this kind of question is one which is asked in theoretical physics all the time. Whenever we have some aspect of physics which seems to show something not accounted for in the Standard Model (dark matter is an example) then one thing to try is to suggest some new field. The apparatus of field theory allows one to 'cook up' a field with whatever properties you think worth trying (e.g. zero mass, no coupling to electric charge, but with energy and consequently gravitation, and possibly other couplings). If the field is in all respects like the electromagnetic field but without coupling to charge, then clearly it is not the electromagnetic field. But it has to couple to something or it will have no impact on the physical world.
Existing experiments rule out many such fields right away, however. If your field couples to $X$ then it will contribute to decays of particles with $X$. Measurements of those decays put quite strict limits on the possibility of further fields.
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What work does a microwave oven do? I learned that when energy is transfered it either produces work or it becomes thermal energy (heat).
Work implies a force that acts on an object producing changes in its position.
I'm learning these concepts from Khan Academy, and in this article they say:
A hot cup of coffee has a measurable amount of thermal energy which it
acquired via work done by a microwave oven, which in turn took
electrical energy from the electrical grid.
What work did the microwave oven if all the energy transferred became heat?
What is heat actually? What is so special about it?
Does heat imply movement of atomic particles and hence the microwave oven sucessfully did work on them?
Thanks for your patience.
| *
*Work in thermodynamics is usually associated with motion of the boundaries of the system. The standard example is a gas in a cylinder fitted with a piston. To move the piston to either compress or expand the gas involves work. But there are more ways to transfer work to a thermodynamic system and one example is #4 below.
*In the microwave example we have a series of energy conversions. Electrical work, by which we refer to moving electrons, is converted into MW radiation and then it is absorbed by the coffee as internal energy. In general, electrical work can be transformed into various energy forms, heat (via resistance), motion (via a motor), radiation, etc.
*Heat is energy that is transferred between systems at different temperatures. The molecules of a hot body move faster than those of a colder body. When the two bodies come into thermal contact, collisions transfer energy in such a way that the molecules of the hotter body slow down by passing some of their energy to the colder body, whose molecules now move a bit faster.
*Does heat imply movement of atomic particles and hence the microwave oven successfully did work on them? Yes, in the same sense that rubbing our hands does work on our palms that is manifested as higher temperature.
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Deriving $\langle H\rangle$ from average momentum and position for a LHO Assume that we know the values of $\langle x\rangle$ and $\langle p\rangle$ for a LHO, that is in a random superposition of zeroth and first state. Derive $\langle H\rangle$.
So I tried solving this problem with writing $H=\hbar\omega (a^*a+1/2)$. We know that $a^*a= (\text{Re} a)^2+(\text{Im} a)^2$ and we can derive those from $\langle p\rangle\propto\text{Im}\langle a\rangle$. However if I average this expression for creation and annihilation operator, I end up needing $\langle p^2\rangle$ and not $\langle p\rangle^2$, which I have (same for position).
How should I solve it?
| generally you can't derive $\langle H \rangle$ only from $\langle x \rangle$ and $\langle p \rangle$, because you end up needing the expectation values of $\langle x^2 \rangle$ and $\langle p^2 \rangle$, as you wrote. However, here you are given another crucial piece of information: that the state is in a superposition of the ground (zeroth) state and the first excited state.
One thing to remember is that the expectation value of the position and the momentum is zero for any eigenstate of the quantum harmonic oscillator: $\langle n | x | n\rangle = 0 = \langle n | p | n \rangle$ for any $n$. You can see that from the fact that both $x$ and $p$ are linear superpositions of $a$ and $a^{\dagger}$ and naturally $\langle n | a |n \rangle = 0 = \langle n | a^{\dagger} | n\rangle$, as these operators take an eigenstate $n$ to another eigenstate.
This means that all non-zero contributions to $\langle x \rangle$ and $\langle p \rangle$ must come from the cross terms between the ground and first excited states.
Write the state in general as $|\psi\rangle = a |0\rangle + b |1\rangle$, and get the expectation values of $\langle x \rangle$ and $\langle p \rangle$ from $a$ and $b$. Add to that the fact that $|a|^2+|b|^2=1$ and you'll get three equations for the four variables $a$ and $b$ (they are complex so each one has both real and imaginary part you need to find out). However the total phase is arbitrary so you can fix it at will, and then you have three equations for three variables, which you can easily solve and find out $\langle H \rangle = \frac{\hbar \omega}{2} |a|^2 + \frac{3\hbar\omega}{2} |b|^2$
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Work done by an eccentric force on a rigid body I am hoping someone can explain in simple terms where I might be going wrong in my logic below.
We have a thin rigid rod (see diagram A) where a couple is acting and rotates the rod about its COM by an angle θ .
I am assuming that the work done by the couple is F.r.θ
Now consider diagram C where an eccentric force F is acting on the rod at point r from its COM and continues to do so for a distance d. I understand that an eccentric force will cause a translation of its COM and a couple rotation about its COM.
Let's assume that as the eccentric force F acts over distance d that the rod is rotated an angle θ (see diagram B).
I am assuming that the total work done on the rod by the eccentric force is F.d . But the rod has also been rotated to the same orientation in space as in diagram A which implies angular work has also been done by the couple caused by the eccentric force.
Therefore, doesn't this mean that the total work done on the rod is greater than F.d ?
| The work applied is still Force times distance.
The angular travel of the body reduces the travel of the body’s COM. Additionally, the travel of the COM reduces the angular travel.
If we were using this work to accelerate the bar (give it kinetic energy) some would go to the rotational motion and some would go to the translational motion. The amount given to each would depend on the bar’s moment of inertia and mass of course.
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Does the earth’s rotational angular velocity change? This is what is written in The Feynman Lectures on Physics, Vol. 1 (ch.5)
We now believe that, for various reasons, some days are longer than others, some days are shorter, and on the average the period of the earth becomes a little longer as the centuries pass.
Why should some days be longer than the others? There is no “gravitational” source of external torque acting on the earth, so why does its rotational angular velocity change?
|
There is no “gravitational” source of external torque acting on the earth
Yes, there is. The tides are caused by the Moon's gravity. That energy has to come from somewhere. The drag caused by the tides is slowly changing the angular momentum of the Earth, and the tides from the Sun also doing so, albeit even more slowly.
Moreover, the Earth's moment of inertia is not constant. For instance, when two faults smash into each other and push up a chain of mountains, that increases the moment of inertia. Whenever something moves towards the equator, that also increases the Earth's moment of inertia. Etc. These changes in the moment of inertia can cause tiny fluctuations in the Earth's angular velocity.
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Why Is Capacitance Not Measured in Coulombs? I understand that the simplest equation used to describe capacitance is $C = \frac{Q}{V}$. While I understand this doesn't provide a very intuitive explanation, and a more apt equation would be one that relates charge to area of the plates and distance between them, I'm having trouble understanding it in general. Capacitance seems to be describing, well, the capacity of two plates to store charge (I understand that the electric field produced between them is generally the focus more so than the actual charge). Shouldn't it just be measured in units of charge such as coulombs? I'm sure this is due to a lack of more fundamental understanding of electric potential and potential difference but I'm really not getting it.
| Capacitance is not a measure of how much charge that is stored on the plates.
It is a measure of how much charge that is stored per volt of sustained voltage.
Just like how, say, pressure is not the amount of force exerted, but the amount of force per area exerted. You mustn't ignore the qualifier in such definitions.
Thus, naturally, the units of capacitance are not merely Coulomb but Coulomb per volt. (This can be rewritten to Coulomb-squared-per-Joule, if you wish, since a volt is a name for Joule-per-Coulomb, a sometimes more used SI combination.) This unit combination is then given a name: Farad. The other definition of capacitance that you refer to is in terms of geometric parametres along with the permittivity - that combination ends up with equivalent units, although it is not obvious.
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Assigning initial conditions for Schrodinger's equation I am self-teaching myself quantum mechanics, and my understanding so far is as follows. In the most general case, we would like to find a wave function $\varphi(x,t) \in \mathcal{H}$, where $\mathcal{H}$ is some appropriate Hilbert space. This wave function encodes the desired information about the system, with measurable quantities being obtained by applying certain operators. If we know the total energy of the system we may construct a Hamiltonian $H$. If this Hamiltonian is time-independent, we may solve the time-independent Schrodinger equation
$$\hat{H}\varphi_i = E_i \varphi_i$$
which amounts to finding the eigenvectors $\varphi_i$ and eigenvalues $E_i$. Then by the spectral theorem we may compute the wave function as a linear combination of these eigenvectors and eigenvalues:
$$\varphi = \sum_i E_i \varphi_i.$$
Now if the Hamiltonian is not independent of time, we use the full and so called time-dependent Schrodinger equation:
$$i\hbar\frac{d}{dt}\varphi = \hat{H}\varphi.$$
My confusion is mainly with the time-dependent case. Since the wave function is unknown a priori, how do we assign an initial condition to it? Also, is my understanding of everything correct?
| Given a state $\Phi(x,0)$ at $t=0$ you can always expand in in eigenstates of $H$:
$$
\Phi(x,0)=\sum_k c_k\varphi_k(x)
$$
where the coefficients $c_k$ (they are not $E_k$) are obtained from
\begin{align}
c_k=\int dx \varphi_k^*(x) \Phi(x,0)\, .
\end{align}
The time evolution of $\varphi_k(x)$ is known:
\begin{align}
\Phi_k(x,t)=e^{-i E_k t/\hbar}\varphi_k(x)
\end{align}
so then it’s simply a matter of rewriting
$$
\Phi(x,t)=\sum_k c_k e^{-i E_k t/\hbar}\varphi_k(x)\, . \tag{1}
$$
Since (1) satisfies the BC $\Phi(x,0)=\varphi(x)$, you have the complete time evolution.
| {
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"timestamp": "2023-03-29T00:00:00",
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What happens if I rewrite the acceleration in work formula this way? Work is known as
\begin{equation}
W=\vec{F}.\vec{q},
\end{equation}
thus for a gravitational force $\vec{F}_g=m\vec{g}$ we have
\begin{equation}
W=mgh.
\end{equation}
My derivation is as follows:
\begin{equation}
dW=\vec{F}_g\cdot d\vec{q}
\end{equation}
\begin{equation}
\implies W=\int_0^h\vec{F}\cdot d\vec{q}=\int_0^hm\vec{g}\cdot d\vec{q}=mg\int_0^h dq=mgh.
\end{equation}
Now consider a spring force
\begin{equation}
\vec{F}_s=-k\vec{q}.
\end{equation}
Inserting this into the first equation gives
\begin{equation}
W=-k\int_0^x \vec{q}\cdot d\vec{q}=-k\frac{x^2}{2}.
\end{equation}
Using
\begin{equation}
v^2-v_0^2=-2gq
\end{equation}
I rewrite the acceleration in $F_g$:
$$F_g=m\frac{v^2-v_0^2}{2q}$$
and insert it into the equation of work:
$$W=\int m\frac{v^2-v_0^2}{2q}dq=m\frac{v^2-v_0^2}{2}\int\frac{dq}{q}=m\frac{v^2-v_0^2}{2}\ln|q|\neq mgh.$$
I want to know which part is wrong and why. Thank you in advance.
| In the integral
$$
\int m \frac{v^2-v_0^2}{2q}dq
$$
$v$ is a function of $q$, since $v$ changes over the range of integration. So it cannot be treated like a constant and cannot be pulled out of the integral as you have done.
In fact, as you showed, $v^2 = v_0^2 + 2 q g$, and so the integrand simplifies to $mg$ and the integral is just $W = \int mg \, dq$.
| {
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Energy from spacetime expansion? Are there any processes or mechanisms in the universe involving spacetime expansion where energy is produced somehow out of this expansion?
I ask this in part due to this article by Sean Carroll 1 which it says at some point:
(...) In general relativity spacetime can give energy to matter, or absorb it from matter
Therefore, could energy be produced somehow out of the Universe's expansion, continuing to be produced as long as it keeps expanding?
| General relativity, in an arbitrary spacetime, has local conservation of energy but not global conservation of energy. On practical scales, spacetime can be modeled as asymptotically flat. In an asymptotically flat spacetime, one can define a conserved measure of energy. So if you're going to extract energy from cosmological expansion, the expectation is that your apparatus is going to have to be cosmic in size.
in general relativity spacetime can give energy to matter, or absorb it from matter
An example of this would be that a gravitational wave detector can, in principle, gain energy from the gravitational waves it detects. That's different from harvesting energy from cosmological expansion.
| {
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Why does particle leave circular motion after string slacks? If a particle is attached to a string and made to move in a vertical circle with initial velocity of $\sqrt{4gl}$ $m/s$ where l is the length of string, at some angle (approx $131°$ with the initial position), the string slacks and the particle leaves the circular path and undergoes projectile motion. Why does this phenomenon occur even though the component of weight can provide centripetal acceleration?
If we throw the particle at $\sqrt{2gl}$ $m/s$ , the velocity and tension both become 0 when the string is in horizontal direction. How can we know whether the particle will now oscillate or leave the circular path?
| Answering the second part first: While the force of gravity provides exactly the required centripetal force at the highest point, the force of gravity does not change direction as required for the centripetal force. The centripetal force has a horizontal component everywhere except for the highest and lowest points, and that horizontal component is required to force the weight into a horizontal oscillation. Because of the lacking horizontal force after the string has been released, the weight does not slow down horizontally, widening its trajectory into an inverted parabola.
Concerning the first part, note that the force of gravity is not in the direction of the string during the entire upswing. So, while gravity does supply increasing amounts of centripetal force (only after the point of vertical motion has been passed), it also supplies a braking force that slows the circular motion down. A the point where the string slacks, the circular motion becomes so slow that requires less centripetal force than gravity provides, and gravity is still slowing the weight down. As such, the radius of the actual curve grows every smaller as the weight becomes even slower. Until the weight moves horizontally, gravity switches from braking upward motion to accelerating downward motion, and the radius rises again so that the weight follows an inverted parabola trajectory.
| {
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Is the normal force the reaction force? There is a box on the surface of earth. The earth exerts a force to the box (black arrow). The box exerts a reaction force to the earth(brown arrow). But this reaction force is exerted to the earth not the box, so where is the normal force? If the reaction is the normal force then why they put the arrow starting in the box upwards and not starting in earth? (second diagram).
| The best way to visualize and solve force problems like this is with a Free Body Diagram, in which you isolate each body and draw all forces acting upon it. That way it is easy to see what forces are acting where.
In this case, the block has two forces acting on it: gravity, and the normal upward force.
The ground also has two forces acting on it: the normal force downward from the block, and then the "offscreen" upward force from whatever is supporting it.
| {
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Explain Heisenberg's uncertainty principle There was one homework question that asks what Heisenberg uncertainty tell us about the energy of an electron in an infinite square well when the length of the well decreases. The correct answer is that the energy decreases when length increases. I know that the energy should decrease by the formula for energy eigenstate, but I feel like this has nothing to do with Heisenberg's uncertainty principle. Uncertainty principle only tells us how accurate is the measurement. Can someone explain how is the uncertainty in energy related to the actual energy of the electron?
| The expected answer is direct. A particle in a stationary state in the box has zero averaged momentum, just because the particle stays there stationarily. Hence $(\Delta P)^2 = \langle P^2\rangle$ which is proportional to the averaged energy of the particle. However, this value is also the eigenvalue of the energy since the state has definite energy by hypothesis. If we decrease the size of the box, due to the Heisenberg inequality, then $\Delta P$ and thus the energy must increase.
However, in my view, this answer, though popular, is wrong as it stands (see my comment below however). That is because the momentum observable does not exist in the box with vanishing boundary conditions (so the validity of the H principle is disputable). Also the energy observable in the box is different of the energy observable in the whole real line where the Heisenberg principle is valid (is a theorem).
ADDENDUM. To make more acceptable the argument from the physical side, we can proceed as follows. We can say that the infinite well is nothing but a very steep well, defined along the whole real line. The energy levels are the ones computed with the ideal well just approximately. On the complete real line we have no problems in defining the momentum operator and approximately, in our physical context, the energy is only kinetic (proportional to $P^2$). From this perspective, within the assumed approximations, the reason why the energy increases when the well width decreases is in fact the Heisenberg inequality. However this line of reasoning is very difficult to follow from a mathematical perspective.
| {
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Rule swing with spring experiment: how can I modify it?
Basically I want to replicate this experiment (https://youtu.be/GqPGbHq2fxU). It's a ruler oscillating with one fixed end and one end attached to a spring. In my previous experiment, I used a short spirng (didn't measure the spring constant which I regret a lot). The oscillation was so fast, period of which was around $0.10 \ s$. I had to record it and count the number of frames to determine the period. What's more, the oscillation was very unstable (dk whether this is the right word), meaning that it was swinging in every direction.
So I want to know if there is any way I can modify my experiment and make it as smooth as in the video?
I don't have a ruler with holes in it, so I stuck a piece of tape on it and used a thread through the tape, to hang the ruler to the clamp. I suspect it contributed to the instability, but can't find any other ways to hang it. Are there methods I can hang it with little friction?
If you want to have further discussion about the experiment, feel free to comment!
I would really appreciate your help! Have a nice day :)
| Why not bore a hole or two in the ruler? Or use a cheap wooden strip, you don't need a ruler since you can measure the distances. If you don't use the hole, it may always swing in more directions. Use for the spring: attach first a weight comparable to the ruler or strip and measure the frequency, if it is slow enough it is slow enough for your experiment, otherwise your spring is too strong.
| {
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Work Integral and its derivation The work integral is something I saw long time ago and in completely understood it.
\begin{align}
W_{12} & =\int F(x)dx=m\int^{t_2}_{t_1}adx=m\int\left(\frac{dv}{dt}\right)dx=m\int\left(\frac{dv}{dx}\right)\left(\frac{dx}{dt}\right)dx\\
&=m\int\left(\frac{dx}{dt}\right)dv=\frac12\left(mv_2^2-mv_1^2\right)
\end{align}
which is clear as day.
But then i saw another version and i cannot follow it:
So my question is:
How did $$m\int\frac{d}{dt}[\dot{x}(t)]\dot{x}(t)dt$$ become $$\frac{m}{2}\int\frac{d}{dt}[\dot{x}(t)]^2dt$$
Where does the $\frac{1}{2}$ come from?
And how did that expression become the change in kinetic energy equation?
I'm sure this is a simple question but i havnt been able to find a solution online so im asking here. Thanks for the help
| Simple use of chain rule:
\begin{equation*}
\frac{d \dot{x}^2}{dt}=2\dot{x}\frac{d\dot{x}}{dt}
\end{equation*}
| {
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How to teach yourself physics systematically？ I'm just a university student and I am not physics majors, but I am interested in physics. I plan to self-taught physics. However I don't know how to teach yourself physics systematically. I want to know what math I need to learn, what is the order of learning, which textbooks are recommended？
| At my university (MIPT) there are the subjects and plans for them:
1st semester:
-General physics: mechanics - https://pastebin.com/UgHnVtaY
-Introduction to Mathematical Analysis - https://pastebin.com/fkbPuC36
-Analytical geometry - https://pastebin.com/MKyniqtS
2nd semester:
-Multidimensional analysis, integrals and series - https://pastebin.com/B7wxbMnp
-Linear algebra - https://pastebin.com/H3ZBp3uT
-General Physics: thermodynamics and molecular physics - https://pastebin.com/C9YHeEnC
I think it will be enough for you now, if you can handle it then come here again. Any complex math from list above is not necessary for general physics in first semesters but linear algebra and special functions for example will be necessary in quantum mechanics in 3rd year and later if you'll decide to study physics seriously. I am sorry that I can't suggest any literature but I am studying in another language so my materials will be useless for you
| {
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I'm having trouble understanding the intuition behind why $a(x) = v\frac{\mathrm{d}v}{\mathrm{d}x}$ I was shown
\begin{align}
a(x) &= \frac{\mathrm{d}v}{\mathrm{d}t}\\
&= \frac{\mathrm{d}v}{\mathrm{d}x}\underbrace{\frac{\mathrm{d}x}{\mathrm{d}t}}_{v}\\
&= v\frac{\mathrm{d}v}{\mathrm{d}x}
\end{align}
However, this feels somewhat unintuitive, and somewhat questionable mathematics-wise. Perhaps it's the best way to explain it, but I was hoping for a more intuitive understanding of this formula.
| For a simple function like $x=t^2$ you can show that the chain rule works.
$x=t^2 \to \dot x = v = 2t \to \ddot x = \dot v = a = 2$
$t=x^{1/2} \to v=2x^{1/2} \to \frac{dv}{dx} = x^{-1/2}= 1/t$
$\frac {dv}{dx} \cdot v = 1/t \cdot 2t = 2 = a$
Looking at the slopes of the graphs you can imagine that as time progresses the increasing gradient of the one on the left multiplied by the decreasing gradient of the one in the middle could produce a constant value for the graph on the right.
| {
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Change of velocity without doing work Body of mass $1\rm\, kg$ is moving towards left side at velocity of $-2\rm\,m/s$ and is slowing down due to the constant external force until it stops. After that, it's speeding up towards the right side due to the same force until it reaches velocity of $2\rm\,m/s$. How much work has the force done on the body?
Correct answer is 0 Joules and we get that if we use formula $E_k=\frac 1 2 mv_f^2 - \frac 1 2 mv_i^2$. But that makes no sense to me. Basically, an object was stoped and accelerated without work. How is that possible?
It would have more sense for me if it was 8 Joules because $E_k= \frac 1 2 mv^2 = \frac{1\rm\, kg\cdot 16\rm\, m^2/s^2}{2} = 8\rm\, J$.
| Suppose we apply a uniform force $F$, which is of course positive. Now let's say that the total work $W$ is the sum of the work when the body moves to the left side $W_-$ and when the body moves to the right side $W_+$:
$$W=W_-+W_+$$
By definition of work,
$$W=W_-+W_+= F\cdot \Delta x_-+ F\cdot \Delta x_+= F\cdot (\Delta x_-+\Delta x_+)$$
As can be seen from the graph $v(t)$, if the body starts at $x=x_{0}$, it ends at $x=x_{0}$ as well.
$$W=F\cdot (\Delta x_-+\Delta x_+)=F\cdot ((x_{min}-x_{0})+(x_{0}-x_{min}))=0$$
where we used the definition of the displacement $\Delta x=x_{final}-x_{initial}$. Therefore, the work is 0, indeed.
The fundamental idea is that at first the force points to the other direction of the displacement (negative work), and then points to the same direction as the displacement (positive work). At some point in time, the total energy of the system is smaller than the initial, but in the end it is the same. The work was done, but the net work is zero.
| {
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What is the correct direction of turbulence energy cascade? I have learned from a fluid mechanics textbook [1] that the turbulence energy is cascaded from the largest eddy to the smallest eddy and is then dissipated by the molecular viscous effect. But recently I was reading Chapter 3 of a thermodynamics textbook [2], Prof. A Bejan claimed that such a classic Richardson picture is wrong and the turbulence is ALWAYS cascaded from the smaller scale to the larger scale, which is supported by Prof. C H Gibson [3]. While I found the terminology "inverse cascade" from this thread and it seems that there is already a lot of research on this topic, e.g. [4]. Hence, I got confused if Prof. A Bejan and Prof. C H Gibson are talking about this "inverse cascade" phenomenon and what is the correct direction of turbulence energy cascade?
[1] Pope, Stephen B., Turbulent flows. Cambridge university press, 2000.
[2] Bejan, Adrian. Entropy generation minimization: the method of thermodynamic optimization of finite-size systems and finite-time processes. CRC press, 2013.
[3] https://thejournalofcosmology.com/APSPittsGibson.pdf
[4] Chen, Shiyi, Robert E. Ecke, Gregory L. Eyink, Michael Rivera, Minping Wan, and Zuoli Xiao. "Physical mechanism of the two-dimensional inverse energy cascade." Physical review letters 96, no. 8 (2006): 084502.
| Depends on the flow! In homogeneous isotropic turbulence, energy may go in both directions, though there is a clear preference for it going to the smallest eddies (called forward cascade), in line with the theories of Kolmogorov and many others. When energy goes the other way, it is called backscatter.
In this paper (arXiv link), they found that forward cascade happens twice as often as backscatter.
Read more here.
| {
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Does "non-interacting" (fermions) really mean "no interactions other than Pauli exclusion"? When one speaks of non-interacting elections (or other ferimons), doesn't one technically mean non-interacting but with the exception of Pauli exclusion? I wonder if it is appropriate to view Pauli exclusion as essentially an infinitely strong short ranged interaction, the same as the condition one imposes to create a model of "hard core bosons".
| Not really. The concept of 'non-interacting fermions' refers to the fact that a many-body Hamiltonian quadratic in fermion operators can be reduced to a single-particle Hamiltonian, where the wavefunction and energy of a single fermion is unaffected by the others. To see this, consider Hamiltonian
$$
H = \sum_{ab} H_{ab} \psi^\dagger_a \psi_b,
$$ then instead of analyzing the entire Hilbert space ($2^N$ dimensional, with $N$ flavors of fermions), one can diagonalize the matrix $H_{ab}$ ($N$ dimensional) to get $$
H = \sum_{\alpha} E_\alpha \gamma^\dagger_\alpha \gamma_\alpha,
$$ with $E_\alpha$ the eigenvalues. The spectrum is very simple now. Occupying a fermion $\gamma_a^\dagger$ will increase the energy of the state by $E_a$, regardless of whether other flavors $\gamma_\beta$ are occupied or unoccupied. This is what we mean by non-interacting; the presence/absence of a fermion operator $\gamma_\alpha$ will change the energy by $E_\alpha$: there is no effect of other fermions on this fact. Pauli exclusion is built into the fermions, and the property that the many-body Hamiltonian reduces to a single particle (noninteracting) Hamiltonian is unaffected by this property. You could make the same argument for bosons, with the minor change that you can have more than one boson per flavor.
When we refer to interactions, we mean terms of the form such as
$$ \sum_{ab} V_{ab} \psi^\dagger_a \psi_a \psi^\dagger_b \psi_b,$$
which ruin this single-particle decomposition.
| {
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Waves and linear dependence on space and time Any function that depends on space and time through the combination $\vec{k}\cdot\vec{r}-\omega t $, namely a function $$f(\vec{r},t)=g(\vec{k}\cdot\vec{r}-\omega t)$$ where g is an arbitrary function of a single real variable, represents a perturbation that propagates in the direction of $\vec{k}$ with a velocity $v=\omega/k$. For sure this tipe of function can be considered a wave, if the broadest definition of wave is adopted: "wave = moving perturbation".
But do all types of wave have such space-time dependence? Is this the only way to implement a function that describes a moving perturbation?
|
But do all types of wave have such space-time dependence?
Certainly not. General form of wave equation is :
$$ \Box u=0, ~~~(1)$$
where,
$$ \Box ={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-\nabla ^{2} ~~~(2)$$ is
a d'Alembert operator
.
So any function $u(\vec r,t)$ which satisfies eq. 1, has wave properties. Such functions should be many.
| {
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Equivalence between small distance and high energy I see in a lot of particle physics literature statements along the lines of: 'This is valid for high energies (small distances)'. Exactly what do we mean by small distances in this case? From QFT, QM and other physics courses, the connection between small distance and high energy makes intuitive sense, but I am wondering if there is a concrete connection that authors have in mind?
| Connections between energy and distance tend to involve the factor
$$
\hbar c= \rm 197\ eV\ nm=197\ MeV\ fm
$$
and some additional factors from doing algebra. The algebraic factors tend to be values like $\frac13$ or $\sqrt 6$, but it is unusual to do a page of first-principles algebra and come up a purely numerical factor of a thousand.
The most common example is a Yukawa-type force mediated by a massive particle, which has potential energy
$$
V=\alpha\hbar c\frac{e^{-r/r_0}}{r},
\qquad\text{where }r_0=\frac{\hbar c}{m c^2}
$$
Electromagnetism, where the photon is massless, is a Yukawa interaction in the limit $r_0\to\infty$, with $\alpha_\text{e.m.}≈10^{-2}$ the fine structure constant. In a meson-mediated model of nucleon-nucleon interactions, the longest-range component comes from the pion, whose mass $mc^2=140\rm\ MeV$ corresponds to a distance $r_0=1.4\ \rm fm$. For nucleons many femtometers apart, the pion-mediated attraction is exponentially suppressed. At shorter distances, you have to include heavier mesons in your calculations. At very high energies, you even have “forces” which mediated by the giga-eV weak bosons.
| {
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Electromagnetic radiation reflected normally causing interference to itself Is it possible to have a body reflect almost perfectly any radiation falling normally with the insident ray so that the reflected ray interferes with the incident and hopefully reduce or cancel it out ?
| "Interference" is the linear superposition of waves, i.e. it is really the absence of interaction of the wave with itself. As a result the energy in waves can not disappear and waves can not actually "cancel" out. They form minima and maxima that can be located in fixed positions in space ("standing waves") instead. We can therefor find "dark" regions with a diminished amplitude, but they are always near "bright" regions (often called interference stripes, rings etc.) where the amplitude is increased.
If we want to reduce the total energy of a wave, then we need an absorber material that removes energy from the wave. One can, however, increase the efficiency of imperfect absorbers by placing them in the regions of maximal intensity or, alternatively, one can reduce the absorption of such materials by using them in standing wave minima.
| {
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Is there a general version of Bell's inequality, like the general version of uncertainty principle? By the general version of uncertainty principle, I mean the result involving general operators $A$ and $B$, which says that the products of standard deviations is equal to the sum of the expected value of the commutator plus the expected value of the anti-commutator (the result is not exactly this. I forgot the exact expression).
The Bell inequality is proved for spin measurements. Is there a general version of the inequality? Maybe something involving the commutator? I want to understand the key reason behind the weird correlations in Quantum mechanics. I think it might have to do with non-commutative observables, just like the uncertainty principle has to do with non commutativity
| Probably the most generic statement about correlation that is still useful in this context is Tsirelson's bound for the CHSH inequality:
The setup are four observables $A_0,A_1,B_0,B_1$ with possible outcomes $\pm 1$ and $[A_i,B_j] = 0$ (but not, crucially, $[A_0,A_1] = 0$ or $[B_0,B_1] = 0$). This isn't as restrictive as it might seem because you can convert any observable with discrete spectrum into a family of such observables by just taking the projectors onto the eigenspaces of that observable and adding minus the projector onto the orthgonal complement to each projector, and likewise you can convert any experiment where you want to measure some quantity into a series of binary questions whether that quantity is in inside or outside some interval and then assign +1 to "inside" and -1 to "outside".
Then Tsirelson's bound says that the correlations of these observables are bounded as
$$ \sum_i \sum_j \langle A_i B_j\rangle \leq c$$
where $c = 2$ if $[A_0,A_1] = 0$ and $[B_0,B_1] = 0$ and $c = 2\sqrt{2}$ for the general case. The vanishing commutator corresponds to a classical local realist theory. Hence the proof of Tsirelson's bound shows us that the reason local realist theories cannot reproduce quantum mechanics indeed is that the commutativity of local realist observables places stricter bounds on correlation functions than general quantum theory.
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Does a random number generator have real entropy? In thermodynamics, entropy is defined for gases. Of course, my laptop is not a gas. However, it contains a random number generator and I have seen the word ‘entropy’ being used in this context. Is this the same entropy? How can this entropy be linked to the definitions from thermodynamics?
UPDATE
I think my question is different from this question. That question is about information content, for example a book. However, this question is about the entropy of a random number generator. Those seem to be different because the contents of a book are fixed while the output of a random number generator is not yet fixed.
| Are you asking if running an algorithm on your laptop generates entropy? Yes, it does. It is an irreversible process. The electronics requires electrical energy and converts it into heat. Is that the minimal entropy generation of the algorithm? No. It's many orders of magnitude more than that with current technology. However, there is a fundamental limit to the power requirements of computers and it is given by the energy needed to "flip a bit" in the computer's memory. Since a bit can only hold information if the threshold energy for a flip is several times kT (otherwise it will flip randomly due to thermal noise), every time we perform such a change of memory content we have to expend that much energy and that energy becomes heat.
| {
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Calculating heat removed by cooling system using output fluid temperature I am trying to calculate the amount of energy removed by a cooling system for some medical research. I'm a little out of my depth with the physics calculations.
I have fluid flowing through a variable temperature object (an organ being heated) at a known unchanging flow rate. The fluid input temperature is constant. A thermocouple on output side measures the temperature of the fluid leaving the object at high frequency.
So I have the mass of water, the change in temperature of the water (and obviously it's specific heat capacity). I do not have the average temperature of the water at the end of the experiment.
I think I can calculate the amount of energy removed by taking the area under the curve of the time temperature graph but I'm not sure what the units on the x-axis should be. Would it be mass of water?
Would taking the average temperature of the output water be equally accurate given the flow rate is unchanging? The temperature rises and falls several times.
Thanks so much
| The following variables apply:
$Q$ = the amount of heat transferred out of the organ, BTU/s
$m$ = the mass flow rate of the water, lb/s
$C_p$ = the specific heat of water, BTU/lb-degF
$T_i$ = the temperature of the water entering the organ, deg F
$T_f$ = the temperature of the water leaving the organ, deg F
The heat transfer out of the organ over a short time interval can be calculated by the equation $Q=mC_p(T_f-T_i)$. By summing up all of the short term heat transfers, you can track the total heat transfer as a function of time.
Note that this method does not require the area under a plot, it does not require the average $T_f$, and it does not require higher level math (e.g., integrals) if you record $T_f$ at a high frequency (e.g., one or a very few seconds per measurement). In addition, if you are using metric units, the same equation will apply but all of the units will be different.
| {
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Thermodynamics : an empty container surrounded by an atmospheric gas I have a small exercices of thermodynamics if anyone can help me :
We start with an empty container of volume $V$. The walls of this container are adiabatic and will not change over time surrounded by a gas of pressure $P_0$ and of temperature $T_0$.
*
*Define and give the characteristics of the system we'll study
*Write the internal energy inside the box using the characteristic of the exterior gas when the equilibrium is reached
*What is the internal energy if the box is closed when the mechanical equilibrium is reached but not the thermal one ?
I know the system is the volume of the box + the matter inside the atmosphere which will go inside the box.
Idk how to characterize it and how I can write just the internal energy and not its variations.
Edit : by applying the first principle in open system, supposing that work and thermal energy are equal to 0 :
$u_2 = P_0(v+v_0) - P_f \times v$
with $m$ the mass of the matter of the system, $v = \frac{V}{m}$, $v_0$ the volume occupied by the gas outside the box at initial time.
I think this relation is wrong because next question ask for internal energy when the box is closed when you reach mechanical equilibrium but not thermal one (my relation doesn't need temperature at all ??!!).
| There are two ways of doing this. One is using the closed system version of the 1st law, and the other is using the open system version. I'll show you how it is done using the open system version. Let n be the number of moles of gas which eventually enter the container. Applying the open system version to the situation yields: $$\Delta U=nu=h_0n$$where $\Delta U$ is the change in internal energy within the container during the process, and the enthalpy per mole entering the container during the change is $$h_0=u_0+P_0v_0=u_0+RT_0$$If we combine these two equations, we obtain $$u=u_0+RT_0$$or$$u-u_0=C_v(T-T_0)=RT_0$$The rest is easy.
I leave it to you to show how the closed system version of the 1st law delivers this same result. In this case, you would choose as your closed system all the gas that eventually ends up in the container.
| {
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Associativity of covariant derivatives I'm having trouble proving that covariant differentiation is an associative operation.
Essentially I'll have to show
$$\nabla_\mu( \nabla_\nu \nabla_\sigma) = (\nabla_\mu\nabla_\nu) \nabla_\sigma. $$
But is it enough to show that both LHS and RHS yield the same result when acted up on a scalar or a contravariant vector?.
Will this hold for any general tensor?
Is there any other method to show this ?
| If you see the covariant derivation as a bundle-map between the relevant vector bundles of tensors, associativity is automatically guaranteed by the usual associativity of composition of functions.
| {
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Weight in Interplanetary Space How is weight zero in interplanetary
space? The Moon is orbiting the Earth because of the gravitational pull of earth. Then gravity must exist in interplanetary space too. So any body in space must also have an acceleration due to gravity ($g$) but $g$ must actually be 0 for weight to be zero.
Can anyone please help me with this?
| It's not that gravity doesn't exist in interplanetary space - gravity has no "maximum range", it exists everywhere - but that you don't feel it. Imagine a weighing machine attached to the floor of the spacecraft and you standing on it. If the reading is $0$, then you are weightless, although you are still being acted on by gravitational forces.
If you stand on a weighing machine and jump off a building with the weighing machine, then you feel weightless because you and the machine are both falling at the same speed. Something similar applies to astronauts in the International Space Station.
| {
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Can we physically explain superconductor mean field order parameter (gap function) as Cooper pair wave function? We usually define mean field order parameter (gap function) in BCS theory
$$
\Delta(r_1,r_2)_{s,s'} = \langle GS | \hat{\psi}_s(r_1)\hat{\psi}_{s'}(r_2) |GS\rangle,
$$
where $\hat{\psi}_s(r_1)$ is field operator of electron on position $r_1$ with spin $s$ and $|GS\rangle$ is ground state of a superconductor.
Can we understand this quantity as a wavefucntion of a Cooper pair just as solution of two electrons interacting with each other via an attractive potential?
| I don't know my answer make sense or not what i understand about that order parameters given by
wave function mod square it represents the copper pair forming probability density which is zero above some critical temperature and critical fields.
Now the formation copper pair is little bit tricker but this is how i imagine with classical pictures. Imagine +ve charge uniform density lattice . Now imagine two electron moving parallel to each other one of them slightly behind other. Since electron is negatively charge it will attract near by +ve core resulting a compression like mesh we call that creation of phonon represents as 'psi+' opperator acted on vacuum. Due to that a higher dense +ve charge density occurs resulting attractive force on electron slightly behind it. Now the oscillation is in sound velocity range while electron moving much faster than sound . Even electron not their the phonon created will be gonna continue their resulting attracting in other electron thats electron phonon electron interaction make the Cooper pair possible
| {
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Experimentally Measuring the Velocity of Water coming out of an Orifice I plan on doing an investigation into Torricelli's Law, where I will be looking at one of the following:
*
*How the cross-sectional area of an orifice affects the velocity of water coming out of it (constant height).
*How the height of an orifice affects the velocity of water coming out of it (constant orifice area).
However, I was unsure about how to accurately measure the velocity of water coming out of the orifice. Videos on YouTube only suggest one method, which is using the horizontal and vertical displacements of the water stream to calculate velocity. However, when I've done this experimentally I've found an about $15\%$ error compared to expected values. The process is also not very exact per se, i.e. it is hard to judge the exact marking of a ruler that the stream lands on.
Therefore, I was wondering if there were any accurate means to measure the velocity of water coming out of an orifice, using equipment typically found in a school laboratory.
| I did some work on this once and came across a range of references on this topic. I can't remember if they specifically cover what you ask for here, but might be in one of these:
Paulo Murilo Castro de Oliveira, Antonio Delfino, Eden Viera Costa and Carlos Alberto Faria Leite,“Pin-hole water flow from cylindrical bottles”, Phys. Educ. 35 110 (March 2000)
Rod Cross, “Filling or draining a water bottle with two holes”, Phys. Educ. 51 045014 (July 2016).
Stephen M. Durbin, “Combined demonstration of non-viscous and viscous flow”, Am. J. Phys. 87(47), 305 (March 2019).
Laura Pavesi, “Investigating Torricell’s Law (and More) with a 19th-Century Bottle”, Phys. Teach.57, 106 (January 2019).
| {
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What is the definition of the phase of a particles (points) on a travelling wave? If we have a travelling wave of equation $y=5\sin(\omega t-\pi x +0.5)\:\rm m$. I know that the phase is 0.5 but how can I visualize this and what is the definition of phase difference between 2 points along the same wave that are $n$ meters apart in the same medium?
| The phase of an oscillation is the 'position' (usually expressed as an angle) that it has reached in its cycle. For example the phase of your wave at time $t$ and distance $x$ from some fixed point is $(\omega t -\pi[\text m^{-1}] x+0.5)$. It needs also to be stated whether the function is sine or cosine.
In your example, 0.5 is the phase constant.
The phase difference, $\Delta \phi$, at two points distance $\Delta x$ apart along the direction of propagation is the difference in these phase angles at any given time. Thus, $\Delta\phi=2\pi \frac {\Delta x}\lambda$. In your example, $\lambda = 2$m, so $\Delta \phi = 2\pi \frac {\Delta x}{2\text m}=\pi \ \text m^{-1} \Delta x$.
| {
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Why does a sensitive thermometer absorb little heat? In an experiment to measure the specific heat capacity of water I'm trying to make it as accurate as possible. And somewhere I read that a sensitive thermometer absorbs little heat. By "sensitive" I am referring to the amount of change in thermometric property for a unit change in temperature.
| A thermometer is "sensitive" when it absorbs as little heat as possible from the item it is trying to measure. Any heat absorbed by the thermometer will lower the temperature of the item, and hence cause a false reading.
This problem occurs whenever you are trying to measure anything. Your measuring device should always aim for the smallest possible disturbance of its environment, as any disturbance will affect the measurement. For example, a voltmeter should draw as little current as possible, to avoid changing the measured voltage; an ideal ammeter has zero resistance, to avoid changing the measured current.
| {
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Why is an equilateral triangle not a 2d unit cell? An equilateral triangle obeys the crystallographic restriction theorem, but it is not a part of 2d crystal structure. What symmetry does it lack? Why can't it be a Bravais lattice?
| It is a possible 2d lattice structure, it just so happens that in the case of equilateral triangles we can observe them to tile together into hexagons. A hexagonal lattice is sometimes called a triangular lattice. https://en.wikipedia.org/wiki/Hexagonal_lattice
We can see clearly how the lattice vectors of a hexagon can form the lattice vector of an equilateral in the following figure
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Local $SU(2)$ symmetry breaking and unitary gauge In a $SU(2)$ gauge field theory with scalar field $\phi$ in the fundamental representation of the $SU(2)$ group with lagrangian $$\mathcal{L} = -\frac{1}{2}TrF_{\mu\nu}F^{\mu\nu} + (D_{\mu}\phi)^\dagger(D^{\mu}\phi) + \mu^2\phi^\dagger\phi - \frac{1}{2}\lambda(\phi^\dagger\phi)^2,$$ we can pick a vev $\phi_{0} = \frac{1}{\sqrt{2}}(0 \;\; v)^T$, with $v^2 = \frac{2\mu^2}{\lambda}$, and break the symmetry doing $\phi = \varphi + \phi_0$.
Then, the new lagrangian has a couple mixing terms $ig\partial_{\mu}\varphi^{\dagger}A^{\mu}_{a}t^a\phi_0$ + h.c., which can be set to zero by choosing the unitary gauge. This implies that the field $\varphi$ satisfies the conditions $\varphi^{\dagger}t^a\phi_0 = 0$, where $t^a$ are the generators of $SU(2)$ picked so that $t_a = \frac{1}{2}\sigma_a$, where $\sigma_a$ are the Pauli matrices.
The unitary gauge imposes constraints on the field $\varphi$. However, if we consider $\varphi^{\dagger}t^1\phi_0 = 0$ and $\varphi^{\dagger}t^3\phi_0 = 0$ and take $\varphi^{\dagger} = (\varphi^*_1 \;\; \varphi^*_2)$, I get that
$$
\begin{pmatrix} \varphi^*_1 & \varphi^*_2 \end{pmatrix}
\frac{1}{2}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}
\frac{1}{\sqrt2}
\begin{pmatrix} 0 \\ v \end{pmatrix} = \frac{1}{2\sqrt2}\varphi^*_1v = 0
\Rightarrow \varphi^*_1 = 0.
$$
Similarly,
$$
\begin{pmatrix} \varphi^*_1 & \varphi^*_2 \end{pmatrix}
\frac{1}{2}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}
\frac{1}{\sqrt2}
\begin{pmatrix} 0 \\ v \end{pmatrix} = -\frac{1}{2\sqrt2}\varphi^*_2v = 0
\Rightarrow \varphi^*_2 = 0.
$$
I'm confused because this would imply that
$\varphi = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.
| Your text should be illustrating the H-E-Brout phenomenon by parameterizing the Higgs doublet in the chiral parameterization, as opposed to the linear σ-model one you are using. They are, of course, equivalent, and it behooves you to catch your mistake in translating back to yours. (Observe $( \varphi_1^* \varphi_1+\varphi_2^*\varphi_2) \phi_0\neq 0$; they are not independent!)
The standard SM (Gürsey) parameterization is
$$ \phi = UH\equiv e^{it^a \xi^a/v} \frac{1}{\sqrt2}
\begin{pmatrix} 0 \\ v+h(x) \end{pmatrix},
$$
where the three ξs are the Goldstone bosons in the adjoint of SU(2), but the h, the dross higgs, is an SU(2) singlet. You must appreciate that $\phi$ has four components/d.o.f., and one of them, the "radial" one, does not transform under SU(2). Note if you think of the ξs as transformation parameters, no non-vanishing ξ can leave H and the SSB vacuum invariant!
It is then straightforward to see that, in any gauge,
$$
D_\mu\phi= (\partial_\mu -igt^a A^a_\mu) UH= U(U^{-1}\partial_\mu U +\partial_\mu -igt^a ~U^{-1}A^a_\mu U) H.
$$
The unitarity gauge is defined as the one in which all three goldstons are gauge-transformed to zero, i.e. U=1, and their gradients are absorbed by the gauge fields: in that gauge,
$$
\leadsto \qquad (\partial_\mu -igt^a {\cal A}^a_\mu) \frac{1}{\sqrt2}
\begin{pmatrix} 0 \\ v+h(x) \end{pmatrix}.
$$
No goldstons, but the Higgs is still there! You may now read off the $\cal A_\mu^a$ mass terms from the lower diagonal entry of their $(D\phi)^\dagger \cdot D\phi$ matrix, etc...
(In the unitary gauge, $h/\sqrt{2}=\Re \varphi_2$. It turns out the h.c. term in your expression for mixing terms is not an innocent bystander at all! It is needed to cancel the Higgs component in your second expression which should not vanish individually, unlike what you guessed…)
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Should a system be “uniform” to qualify for it to be in steady state? I am wondering about whether a system need to be “uniform” to qualify for it to be in steady state or can not uniform systems also act as a steady system.
Can someone please clarify this.
Thanks in advance!
Any help would be appreciated.
| Steady state just means whichever state is in question doesn't change. If you pick uniformity as the state in question then it's a tautology that uniformity is required for a steady state.
If you pick virtually any other state then uniformity is not required for steady state. Suppose you pick the number of elements as the state. As long as that number stays constant then the system is in a steady state.
| {
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Given a system, how to decide whether a closed orbit is homoclinic, not periodic, solely based on its phase portrait? Background and definitions:
*
*A system is conservative if it has at least one conserved quantity.
*In a phase portrait of a nonlinear conservative system, trajectories that start and end at the same fixed point are homoclinic. (A homoclinic orbit is not periodic because it takes infinite time so that the trajectory reach the fixed point after leaving it.)
Assume that we have the phase portrait associated with a system whose energy conservation is not known. If we knew the system were conservative, closed orbits in its phase portrait were homoclinic. But without this assumption, (since phase portrait does not convey information about time) how can one decide whether a closed orbit is periodic or homoclinic?
| One misconception here is to assume that
If we knew the system were conservative, closed orbits in its phase portrait were homoclinic.
Actually a typical conservative system's closed orbits are almost always quasiperiodic or periodic. For instance, the only homoclinic point in the phase portrait below is the x-shaped crossing at the origin:
All the other curves are periodic orbits. (Source of the image.)
FWIW, homoclinic orbits tend to be non-smooth at the periodic point, since it sits at the transversal intersection of invariant manifolds.
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Cleaning silt off magnets I gone done a foolish thing; I got my set of spherical (⌀ 5 mm) neodymium magnets covered in of (what is presumably iron based) ferromagnetic silt/sand particles.
With some difficulty, I've been able to remove most of the sand sized particles (by grabbing them with some needle-nose pliers). But, the silt sized particles are much harder to remove — as the magnets are very strong, and the particles are very small.
How can I effectively removed these small ferromagnetic particles from the magnets?
Ideally I'd like to keep the silt, but my priority is to clean the magnets.
Obviously I don't want to permanently demagnetise the magnets.
| I have found that using an adhesive can work well to remove small pieces of dust like this. I have used packing tape to pull the particles off of small neodymium magnets.
I found this method on the K&J magnetics FAQ page, link here.
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Surface tension per unit length Can shear stress be expressed as surface tension per unit length? How do I interpret it physically?
| Surface tension has the dimension of a force per unit length, stress has the dimension of a force per unit surface.
You can imagine the surface tension as the integral of stress over a surface with one infinitely small dimension,
$\Delta \mathbf{F} = \displaystyle \int_{\Delta S} \mathbf{t_n} dS = \displaystyle \int_{\Delta S} \underbrace{\mathbf{t_n} dh}_{\boldsymbol{\gamma}} \ d\ell = \displaystyle \int_{\Delta \ell} \boldsymbol{\gamma} d \ell$,
being:
*
*$\mathbf{t_n} = \mathbf{\hat{n}} \cdot \mathbb{T}$ the stress vector acting on a surface with unit normal vector $\mathbf{\hat{n}}$, where the stress tensor is $\mathbb{T}$
*$\boldsymbol{\gamma} = \mathbf{t_n} \ dh$ the force acting per unit length at the interface surface.
You can imagine the stress vector due to surface tension written using a delta in space $\mathbf{t_n} = \boldsymbol{\gamma} \delta(\mathbf{r} - \mathbf{r}_S)$, whose contribution is non-zero only at the points $\mathbf{r}_S$ of the interface, whose integral on the surface thickness gives
$\displaystyle \int_{\Delta S} \mathbf{t_n} dS = \displaystyle \int_{\Delta S} \boldsymbol{\gamma} \delta(\mathbf{r} - \mathbf{r}_S) dS = \displaystyle \int_{\Delta \ell} \boldsymbol{\gamma} d \ell$,
to derive the results given above, a bit more rigorously.
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How can acceleration in special relativity be uniform? Acceleration is defied as the rate of change in velocity, implying that $v(t) = at + v_0$. Say that an object is accelerating at $5 m/s^2$ with respect to an inertial frame in special relativity. Using the definition of acceleration alone, we now have $v(t) = 5t + v_0$. This means that given enough time, $v(t) > c$ which is supposed to be impossible in relativity. The solution is to say that uniform acceleration in special relativity is hyperbolic, so while it may approach c it never actually gets there. The problem with this is how can that be said to be uniform acceleration? If something is accelerating at $5m/s^2$ in order for it to stay less than $c$ it has to eventually decelerate even more and more to asymptotically approach $c$.
How is hyperbolic acceleration uniform?
| I’ll just formalize my previous comments. Let me restrict to 2D flat spacetime with a certain inertial frame $t,x$ ($c=1$ and the metric signature is $(+,—)$ like in particle physics). Then hyperbolic motion of proper acceleration $a$ can be parametrized by its proper time $\tau$ given (up to a space-time translation):
$$
t = \frac{\sinh(a\tau)}{a} \\
x = \frac{\cosh(a\tau)}{a} \\
$$
As you pointed out, the acceleration viewed from the original frame is not uniform:
$$
x = \frac{\sqrt{1+(at)^2}}{a} \\
\frac{dx}{dt} = \frac{at}{\sqrt{1+(at)^2}}\\
\frac{d^2x}{dt^2} = -a\frac{1}{\sqrt{1+(at)^2}^3}
$$
the velocity reaches $1$ asymptotically, so you have a non uniform deceleration reaching $0$ asymptotically.
Note however that acceleration $d^2x/dt^2$ in the frame coincides exactly with the proper acceleration $a$ at $t=0$ ie when the frame coincides with the rest frame of the particle.
This is true in general. At any event of the world-line, I can choose an inertial frame which coincides with the rest frame of the particle at that event. The acceleration measured in these identical frames at this specific event will coincide with proper acceleration. This is an equivalent definition of proper acceleration. And it is this proper acceleration that is constant in hyperbolic motion.
Geometrically, this instantaneous rest frame of the particle is the Minkowski analogue of the Frenet basis in the Euclicdean plane. This is why proper acceleration is the analogue of curvature and given generally by:
$$
a=\frac{d^2x}{d\tau^2}\frac{dt}{d\tau}-\frac{d^2t}{d\tau^2}\frac{dx}{d\tau}
$$
which you can check explicitly for hyperbolic motion.
Hope this helps.
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Collapse of wavefunction in hydrogen atom under measurement of $L^2$ If a hydrogen atom is in the state $\Psi = \frac{1}{\sqrt{3}}(\psi_{100} + \psi_{211}+\psi_{21-1})$
and the square of the angular momenta was measured to be $2\hbar^2$. After the measurement will the state be
$\Psi' = \frac{1}{\sqrt{2}}[\psi_{211}+\psi_{21-1}]$ or $\frac{1}{\sqrt{2}}[Y^{1}_{1}+Y^{1}_{-1}]$ with $Y_m^l$ a spherical harmonic function?
| After the measurement the state of your Hydrogen atom will be $\frac{1}{\sqrt{2}}[\psi_{211}+\psi_{21-1}]$. With the measurement you are projecting onto the states with angular momentum quantum number $l=1$ as the eigenvalues of $L^2$ are $l(l+1)\hbar$.
The state of your Hydrogen atom remains of course in a "state of the Hydrogen atom". Which contains the Spherical Harmonics, but also information about the radial component. Therefore it is not correct to write the remaining state solely in Spherical Harmonics.
You can make that explicit by writing $\psi_{n,l,m}(r,\theta,\phi)$ and then you see that you would loose the $r$ if you take your proposition with solely Spherical Harmonics.
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Number of free parameters in $SU(5)$ GUT model Lately, I have been studying the potential of scalar fields in this theory. In general, what is the point of this GUT if, there, more free parameters have been added?
The standard Higgs potential in the Standard Model with only 2 free parameters (Higgs mass and self-coupling)is
$$V(φ) = \frac{1}{2}m^2φ^2+\frac{λ}{4!}φ^4.$$
For the $SU(5)$ GUT, we have (ignoring possible odd terms),
$$V(Φ,φ) = \frac{1}{2}m_{Φ}^2\operatorname{tr}(Φ^2)+\frac{a}{4!}\operatorname{tr}(Φ^2)^2+\frac{b}{4!}\operatorname{tr}(Φ^4)\\ +\frac{1}{2}m_{φ}^2φ^2+\frac{λ}{4!}φ^4+αφ^\dagger φ\operatorname{tr}(Φ^2)+βφ^\daggerΦ^2 φ,$$
so, in total, 7 free parameters.
I mean, it is clear with the mass and self-coupling of both scalar fields, but why do we need 3 more terms? Are they important?
| This is simply the most arbitrary renormalizable, gauge invariant potential for SU(5) with the given field content. The minimization conditions of the potential eliminate some of those arbitrary coefficients in terms of the vacuum expectation values and masses of the Higgses in the 5, and 24 rep.
While you do end up with more freedom here, note that in the fermion sector, the Yukawa couplings to the up and down type quarks as well as the leptons (all united in the 5 and 10 reps) are more constrained, and you have a smaller number of parameters there.
| {
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In firing a single photon at the center divider of a double slit, does it ALWAYS go through the slits? If we think of a single photon approaching the slits as a wave function, and we fire the photons at the midpoint of the two slits, one at a time, then I would think the probability function is highest at this midpoint. That would mean that, depending on the separation of the two slits, some of the photons would impact the wall separating the slits. If I fire a photon at a thin piece of material, why would it not at least occasionally 'hit' the material, (with a possible photoelectric effect). How is it that ALL the photons avoid this 'wall' and pass through one slit or the other?
| You are correct. To the left of the "wall with slits" we typically approximate that we are illuminating the wall/slits with a plane wave. If the wall is much bigger than the diameter of the beam then what matters is the fraction $\eta$ of the slit area compared to the total beam area. It is only this fraction $\eta$ of photons that make it through the slits and allow us to perform an interference experiment. If you put a big power meter before and after the slits you will see a power reduction by an amount $\eta$. If you're performing an experiment where you try to send photon or electron waves through the slit if you have a flux of 1 photon or electron per second but $\eta$ is 0.01 then it will take on average 100 s for a single point to be detected on your detection screen.
| {
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Why does high frequency have high energy? The electromagnetic spectrum's wavelengths all travel at the same speed, $c$. Also, the wavelength $\lambda$ and frequency $\nu$ are related by $c = \lambda \cdot \nu$. Since all moving particles here would have the same speed, why would higher frequencies have more energy?
| Planck's constant is a conversion factor which changes the units of energy from a classical basis (as joules, for example) to a quantum basis which has wavelength (and therefore frequency) in it: E = hc/(lambda).
Note that if we write the laws of physics in natural units then energy has units of 1/(seconds) or "per second", which also gives you the tie-in to frequency i.e., "cycles per second*.
| {
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When are my fluid approximations wrong? I did some classical approximations of the Navier Stokes equations, fluid is:
*
*non-viscous
*incompressible
*irrotational
When are these approximations wrong? and particularly is there a "general method" to evaluate in a theoretical way "the error" of an approximation?
For example, for a given fluid with a given velocity flow, what will be the order of the terms that I neglect?
I see some methods using dimensionnal analysis, but it wasn't very clear for me...
| I want to focus on the part of the question that deals with viscous effects: The dimensionless version of the conservation of linear momentum is largely equal to its dimensional version, except the term that deals with the stress tensor: In the dimensionless version, the divergence of the stress tensor is divided by the Reynolds number. Thus, for high Reynolds number flows, the shear stresses in the fluid get negligible. (The surface stresses on the boundary do not.) See, for example, Schlichting, Gersten (2017): Boundary-Layer Theory, Section 4.1 "Similarity Laws", equation 4.4.
| {
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How can both of these equations for pressure be correct? Consider the Gibbs equation:
$$du=Tds-pdv$$
Identifying partial derivatives, one obtains:
$$-p=\left( \frac{\partial u}{\partial v} \right)_T$$
But you can also show that:
$$p=T\left( \frac{\partial s}{\partial v}\right)_T -\left( \frac{\partial u}{\partial v} \right)_T $$
In fact for an ideal gas, the latter partial derivative is $0$ and therefore it is the first term the one that determines its pressure. But how come both of these equations are true, at the same time?
|
Consider the Gibbs equation:
$$du=Tds-pdv$$
Identifying partial derivatives, one obtains:
$$-p=\left( \frac{\partial u}{\partial v} \right)_T$$
No.
$$-p=\left( \frac{\partial u}{\partial v} \right)_s$$
But you can also show that:
$$p=T\left( \frac{\partial s}{\partial v}\right)_T -\left( \frac{\partial u}{\partial v} \right)_T $$
But how come both of these equations are true, at the same time?
Because you are using the wrong expression for $p$. You should use:
$$-p=\left( \frac{\partial u}{\partial v} \right)_s$$
You can then consider:
$$
T=\left( \frac{\partial u}{\partial s} \right)_v = T(s,v)
$$
to see that we can write $s = s(T,v)$. Then you can compute $\left( \frac{\partial u}{\partial v} \right)_T$ by considering the derivative of $u(s(T,v),v)$ with respect to $T$ at constant $v$.
| {
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Avoiding heat death in an accelerated expanding universe? Would there be any way to avoid heat death in an accelerated expanding universe?
https://en.m.wikipedia.org/wiki/Heat_death_of_the_universe
Is the heat death of the universe completely unavoidable in an universe with an accelerated expansion dominated by dark energy like ours?
Or could it be avoided according to current knowledge, observations and experiments in physics?
| We don't know exactly how the universe will face its fate, whether through big rip or heat death, although heat death seems more plausible.
Note that heat death exactly isn't caused by accelerated expansion by dark energy but is a manifestation of thermodynamic's second law. It states that all systems will eventually reach the state of maximum entropy or disorders, so nothing ordered structures like stars, planets maybe even atoms would remain. The universe will become uniform everywhere.
Considering we can't escape our universe, or some entirely new theory points to something else entirely changing, the physics as we know, NO We can't dodge the fate of our universe,
| {
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Why can the coriolis force potential be written as $ E_\text{cor}=m\dot{\theta}\begin{vmatrix} X & Y \\ \dot{X} & \dot{Y} \end{vmatrix}$? I found the following formula for the Coriolis force written here:
$$ E_\text{cor}=m\dot{\theta}\begin{vmatrix}
X & Y \\
\dot{X} & \dot{Y}
\end{vmatrix}=m\dot{\theta}\ (\dot{Y}X-\dot{X}Y)$$
I have the following questions:
*
*Does this matrix have a certain name? Is it related to Jacobian or Hessian?
$\begin{vmatrix}
X & Y \\
\dot{X} & \dot{Y}
\end{vmatrix}$
*Where does it come from?
*Also this seems like a commutator relation: $(\dot{Y}X-\dot{X}Y)$, is that true?
*Can a similar matrix be used for the centrifugal force?
| This can be seen as a consequence of the fact that we can write the triple product $\vec{v} \cdot (\vec{\Omega} \times \vec{r})$ as a determinant, rearrange it, and then do an expansion by subminors:
\begin{align*}
\vec{v} \cdot (\vec{\Omega} \times \vec{r}) &= \begin{vmatrix} v_x & v_y & v_z \\ \Omega_x & \Omega_y & \Omega_z \\ x & y & z \end{vmatrix} \\
&= \begin{vmatrix} \Omega_x & \Omega_y & \Omega_z \\ x & y & z \\ v_x & v_y & v_z \end{vmatrix} \\
&= \Omega_x \begin{vmatrix} y & z \\ v_y & v_z \end{vmatrix} - \Omega_y \begin{vmatrix} x & z \\ v_x & v_z \end{vmatrix} + \Omega_z \begin{vmatrix} x & y \\ v_x & v_y \end{vmatrix}
\end{align*}
In the case where $\Omega_x = \Omega_y = 0$ and $\Omega_z = \dot{\theta}$, this reduces to the formula you found.
As far as whether this can be used for the centrifugal potential, that's of the form $(\vec{\Omega} \times \vec{r}) \cdot (\vec{\Omega} \times \vec{r})$, and so the trick above does not work so elegantly (since it's not a vector triple product like the Coriolis term is.)
| {
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How one can use Wick's theorem for the product $A:\mathrel{B^{n}}:$? I try to use Wick's theorem in the case that some products we deal with are already normal ordered.
My guess is that it could be something like
\begin{equation}
A:\mathrel{B^{n}}:~=~:\mathrel{AB^{n}}:+nA^{\bullet}B^{\bullet}:\mathrel{B^{n-1}}:\tag{1}
\end{equation}
I tried to prove that by induction but I failed, maybe the formula is similar and I am somehow close, or maybe my intuition totally fails. How one could approach such a problem?
Also, how would that Wick's expansion look in the general case: $$A_{1}\cdots A_{n}:\mathrel{B_{1}\cdots B_{m}}:~ ?\tag{2}$$
| *
*There is usually a second implicitly written operator ordering besides the normal order. [This plays a role in e.g. eq. (2).] E.g. in the context of 2D conformal field theory, there is typically an implicitly written radial ordering ${\cal R}$.
*Then eqs. (1) and (2) become examples of a nested Wick's theorem discussed in my Phys.SE answer here.
*Eq. (1) is correct, because the only possible terms are a term with no contraction and $n$ terms with a single $AB$ contraction.
*Eq. (2) becomes a sum of all possible $AA$ and $AB$ (but not $BB$) contractions.
| {
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Question about adjoint representation and Lie algebra My question is: In Polchinski II, section 11.4, page 62, said that
"the adjoint representation is the antisymmetric tensor, which is contained in the product of two vector representations".
May I ask why is this?
| *
*More generally, Ref. 1 is considering an orthogonal group
$$G~=~O(V)~=~\{g~\in{\rm End}(V) \mid \forall v,w\in V: \beta(gv,gw)=\beta(v,w)\}$$
over an $\mathbb{F}$-vector space $V$ with a non-degenerate $\mathbb{F}$-bilinear symmetric form $\beta:V\times V\to\mathbb{F}$.
*The adjoint representation $${\rm Ad}: G\to {\rm End}(so(V)),$$ is defined as
$$ {\rm Ad}(g)m~=~g\circ m\circ g^{-1}. $$
*Here the corresponding Lie algebra
$$\begin{align} so(V)~=~&\{m~\in{\rm End}(V) \mid \forall v,w\in V: \beta(mv,w)+\beta(v,mw)=0\}\cr
~\cong~&V\wedge V \end{align}$$
is isomorphic to the antisymmetric tensor product of $V$,
cf. OP's question.
References:
*
*J. Polchinski, String Theory Vol. 2, 1998; section 11.4 p. 62.
| {
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Does horizontal acceleration affect gravity? If we apply 1G horizontally in some object, will this constant force equal to G affect the time of falling? If the force does not affect gravity, why gravity is prioritized over this force if both are equal?
Edited: For the ones who didn't understand what i mean well, i mean why we can't switch them and say gravity is the horizontal force, and the force we applied is gravity? why the trajectory does not change as shown in this image?:
| We have Newton's second law F = ma which is a vector equation. If we look at the horizontal and vertical components, gravity acts vertically and your 1g force acts horizontally. So the vertical acceleration is determined by gravity. The 1g force has no vertical component so does not affect the vertical motion.
In your terms, gravity is prioritized in determining the time of fall because it's the only force in the vertical direction.
| {
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Interesting relationship between the 2D Harmonic Oscillator and Pauli Spin matrices I have an isotropic 2D Harmonic Oscillator in cartesian coordinates
\begin{equation}
H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{1}{2} m\omega^2 (x^2 + y^2)
\end{equation}
In terms of the usual creation and annihilation operators for the $x$ and $y$ modes, $n_x$ and $n_y$ this can be written as
\begin{equation}
H = \hbar\omega(n_x^\dagger n_x + n_y^\dagger n_y + 1)
\end{equation}
Now we can 'construct', three operators that commute with the Hamiltonian, apart from the (trivial) number operators $N_x$ and $N_y$:
\begin{align}
V_x &= a_x^\dagger a_y + a_y^\dagger a_x\\
V_y &= i(a_y^\dagger a_x - a_x^\dagger a_y)\\
V_z &= a_x^\dagger a_x - a_y^\dagger a_y
\end{align}
Now these operators can been proved to satisfy the commutation relations of angular momentum. In fact,
\begin{equation}
V_i = a^\dagger \sigma_i a
\end{equation}
Where $a = \begin{pmatrix} a_x & a_y \end{pmatrix}$ and $\sigma_i$ are the Pauli matrices. This is really surprising to me as I don't see why and how this must be true. Can someone shed some light on this? I think for $2s+1$ dimensions, we can 'construct' such operators using the matrix representations of spin $s$. However, I haven't seen a proof of this.
| It is well known the operators
$$
\hat C_{ij}=a_i^\dagger a_j\, ,i, j=1,\ldots, n
$$
span the Lie algebra $\mathfrak{u}(n)$. If you choose $\hat C_{ij}, i\ne j$ and use $h_i= \hat C_{ii}-\hat C_{i+1,i+1}$, you get instead the Lie algebra $\mathfrak{su}(n)$. Yours is just the special case with $n=2$.
The $n=3$ case is discussed in details in
Fradkin DM. Three-dimensional isotropic harmonic oscillator and
SU$_3$. American Journal of Physics. 1965 Mar;33(3):207-11,
and indeed the connection with harmonic oscillator for general $n$ goes back to
Jauch JM, Hill EL. On the problem of degeneracy in quantum mechanics. Physical Review. 1940 Apr 1;57(7):641.
Note that general representations (not of the fully symmetric type) of $\mathfrak{su}(n)$ having Young diagrams with $m<n$ rows usually start with $U(m\times n)\supset U(m)\otimes U(n)$ so the bosons $a^\dagger_{\alpha i}$, $\alpha=1,\ldots m$, $i=1,\ldots,n$,
etc and the operators
$$
\hat C_{ij}=\sum_\alpha a^\dagger_{\alpha i}a_{\alpha j}\, .
$$
| {
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What gives rise to mass gap for gluons, even if they are massless? It is known that QED does not have a mass gap.
On the other hand, at the heuristic level, QCD has a mass gap.
But photons and gluons are both "massless". Could anyone explain (at least at the conceptual level) what the fundamental difference between gluons and photons is?
| The difference is gluons carry their own charge and thus are bound in glueballs, and while gluons are massless, glueballs are not. It is glueballs, not gluons, which give a mass gap.
| {
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When a football/rugby ball (prolate spheriod) is dropped vertically, at an oblique angle, why does it bounce at an angle? A prolate spheroid like a rugby ball or American football, dropped vertically with no rotation, will bounce vertically if it is oriented along one of it's axes of symmetry (in other words, if its centre of mass is directly above the point of impact on the ground).
However, if the ball is tilted some amount, the ball bounces with some horizontal component to its motion (as well as some angular velocity, but I'm not concerned with that for now).
What's going on here to make this happen? At the point of impact, the normal force from the floor is of course purely vertical, so where does the horizontal component come from?
Intuitively, (and probably incorrectly) I have a picture in my head of a force being applied in a direction determined by the vector given by the point of impact on the ground and the centre of mass, but I think that's probably the wrong way to think about it.
Is the horizontal force coming from friction, that acts on the ball during the duration of its contact with the ground? I guess during the impact, the ball has a tendancy to want to push the point of contact away from the center of mass, in order for the center of mass to continue travelling downwards, and perhaps it's the friction in response to this that causes the horizontal component to the bounce?
| If the football is at an angle, the pointed end is not below the center of gravity. The point strikes the ground first. The ground pushes up on the point. This creates a torque that starts the ball rotating.
Since the point is on the ground, it may stick or dig in. Or there may be friction. The rotation may direct the point sideways, but forces from the ground push back. This sideways force on the ball give the ball a sideways velocity. When the ball bounces up, it is at an angle.
| {
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What is the state of an entangled photon after its twin is absorbed? Let's two photons are entangled in polarization after a laser beam passes through a Betha Barium Borate crystal. They take different paths and one of them (1) is absorbed in a black sheet. What is the state of the leftover photon (2)? Is it in superposition of polarization h/v or it must flip spontaneously in a certain polarization? What if the black sheet atoms absorb photons only with a certain polarization (say h)? Will the absorbed photon (1) take h polarization in the process of absorption and hence the second twin flip to v?
| When two photons are entangled in polarization, the state of each photon is dependent on the state of the other. In the scenario you describe, if one of the entangled photons (photon 1) is absorbed by a black sheet, the state of the remaining photon (photon 2) will change instantaneously. However, the exact state of photon 2 after the absorption of photon 1 will depend on the details of the situation.
If the black sheet absorbs photons of both horizontal and vertical polarization, then photon 1 will be absorbed in a superposition of horizontal and vertical polarization. In this case, photon 2 will also be in a superposition of horizontal and vertical polarization after photon 1 is absorbed.
If, on the other hand, the black sheet absorbs photons of only one polarization (say, horizontal), then photon 1 will be absorbed in that polarization. In this case, photon 2 will be in the opposite polarization (vertical) after photon 1 is absorbed.
Finnaly, the state of the remaining entangled photon (photon 2) will change instantaneously when the other photon (photon 1) is absorbed by the black sheet. The exact state of photon 2 will depend on the details of the situation, such as the polarization of the photons and the properties of the black sheet.
An example of the state of an entangled photon after its twin is absorbed is the phenomenon of "quantum teleportation." This is a process in which the state of one entangled photon is transferred to the other entangled photon, even when the photons are separated by large distances. In this case, the state of the photon that is absorbed will be "teleported" to the remaining photon, effectively making the two photons "exchange" their states. However, this process is probabilistic and depends on a number of factors, such as the efficiency of the teleportation process and the state of the entangled photons before the absorption occurs.
| {
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Conservative Force: Translational Invariance I have a question about the following.
Why if there are two masses, $m_1$ and $m_2$ respectively, and the only force acting on them is from their mutual interaction which is conservative and central, the following is true?
$U(\vec{r_1},\vec{r_2})=U(\vec{r_1}-\vec{r_2})$
(It says this is from Translational invariance.)
Also, it says since it is central, meaning it is along the line joining them, why is it
$U(\vec{r_1}-\vec{r_2})=U(\lvert\vec{r_1}-\vec{r_2}\rvert)$?
| Imagine the situation you have described. Let the potential be denoted as $U(\vec{r_1},\vec{r_2})$. Note that this picture shouldn't change if I decide to observe what they do in one place or the other. Therefore, the potential should be invariant if I decide to displace the particles in the same direction by the same amount. This means the potential should satisfy the following behaviour:
$U(\vec{r_1} +\vec{a},\vec{r_2} +\vec{a}) = U(\vec{r_1},\vec{r_2})$
for any arbitrary vector $\vec{a}$. This means that the potential should be a function of $\vec{r_1} - \vec{r_{2}}$ as this is invariant under displacements. Physically, this corresponds to the fact that the relevant information for the equations of motion is encoded in the distance between the two particles. If that wasn't true, the motion would depend on the position of my experiment.
When a force is central, this means that there is no "angular dependence". The systems should look the same if I rotate them by equal amount. Therefore, in the potential you shouldn't have any terms depending on the angle between the two particles. This along with our previous result imply that the potential $U$ should be a function of $|\vec{r_1} - \vec{r_{2}}|$ as this has no relative angular dependence between the particles.
Hope this helps.
| {
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Is there a "fundamental problem of thermodynamics"? The "fundamental problem of mechanics" can be boiled down to finding and solving the equation of motion of a system. Similar statements can be said for quantum mechanics for the Schrödinger equation and for electrodynamics and Maxwell's equations, etc. But is there such a thing for thermodynamics? Is there a formulation that allows for this kind of perspective?
| I think the best description of what is the "fundamental problem" in Thermodynamics remains the one by H.B. Callen in his textbook (Thermodynamics and an Introduction to Thermostatistics. John Wiley & Sons):
The single, all-encompassing problem of thermodynamics is the determination of the equilibrium state that eventually results after removing internal constraints in a closed composite system.
I would add that the system should not necessarily be closed. Even isothermal or isobaric conditions are allowed.
Notice that Equilibrium Thermodynamics solves the problem by showing that a few special functions of the state variables, the so-called fundamental equations (either the thermodynamic potentials or Massieu's functions), allow the solution of the fundamental problem under different external conditions via maximum or minimum principles.
Another observation is that such a problem is related to but different from the central problem of Equilibrium Statistical Mechanics, which is the determination of the fundamental equation from the knowledge of the microscopic interactions. However, a clear boundary between the two approaches is that within Equilibrium Thermodynamics, all the microscopic degrees of freedom have already been used and eliminated from a description in terms of the few macroscopic variables necessary to describe a system at equilibrium.
Finally, let's note that an essential part of the fundamental problem of thermodynamics, according to Callen's definition, is the identification of the correct set of state variables for each class of thermodynamic systems. Therefore, it remains a problem to be solved every time one would like to treat a new class of systems with the methods of Equilibrium Thermodynamics.
| {
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What's the smallest signal to noise ratio for which a signal has been extracted? Suppose we have some physical variable $y$ that is changing in some way and we want to detect this change in the presence of noise (e.g. white noise) in that same physical quantity. For example $y$ is an electric field amplitude and there is black body radiation or something like that. If we know of some pattern to be expected in the change of $y$ then this helps to detect it against the noise.
For example $y$ could be an oscillation at a very well-defined frequency and we know the phase, so we can use a lock-in amplifier (also known as phase-sensitive detection). Or, $y$ could be a binary message encoded using an error-correcting code (e.g. transmission from space probes). By such ingenuity one can extract very low-level signals against very noisy backgrounds. What is the best anyone has ever done? What limits have been encountered?
Added remark
I was aware when asking that this is a somewhat ill-defined question because for many types of noise you can do better by averaging if the signal lasts long enough. But typically the signal is not permanent. I am asking about what really happens, especially in the presence of thermal noise, when people want to communicate or measure. For example, what is the noise at LIGO (in the relevant frequency band) before signal processing? Here for example I mean the noise in the photocurrent at the light detectors (converted into equivalent strain noise).
I see there is already a helpful answer from one user so please don't close this question! I think further helpful answers will emerge.
| WSPR, a "ham radio" digital radio transmission protocol with redundancy and error-correction, can be routinely decoded at a level 29dB below the noise floor. This is good enough to permit error-free radio communication between any two points on earth on much less than one watt of radiated power, as long as ionospheric bounce conditions prevail.
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Do we need a quantum gravity theory to model an hydrogen atom on earth? The hydrogen atom is a quantum mechanical system. However, it is also attracted by the gravitational pull of the earth. Therefore, do we need quantum gravity to model its behavior correctly? Conversely, can we study hydrogen atoms on earth to obtain new information about quantum gravity?
| We don't need a quantum theory of gravity to model a hydrogen atom in the gravitational field of the Earth. In general, it's understood how to model quantum fields on a fixed gravitational background (see, e.g., the book by Birrell and Davies) — at least at a mathematical level; experimentally it's very hard to probe those kinds of situations. The hydrogen atom on Earth would be a special case of that framework, where the quantum degrees of freedom (the hydrogen atom) and the gravitational field are also non-relativistic. In that case, the formalism simplifies and becomes the Schrödinger equation with a potential due to the Earth's gravitational field, like
\begin{equation}
i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2 \Psi + mg z \Psi + V\Psi,
\end{equation}
where the second term on the right hand side, $mgz$, is the potential energy due to being on the Earth's surface. Meanwhile, $V$ describes any other sources of potential energy.
We would need a quantum theory of gravity to describe the gravitational field of the hydrogen atom itself. Actually even that isn't quite true, since there are actually approximate methods to deal with the interaction of matter and gravity quantum mechanically, as well, so long as the gravitational interaction is small. However, at large enough energies (near the Planck scale — much, much larger energies than we can probe experimentally), the gravitational interactions of the hydrogen atom become strong. This is really the regime where we need quantum gravity.
| {
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Why the double slit pattern is a blurred image instead of two strips? I am complete dilettante in physics, when i was first introduced to quantum mechanics by a You tuber named Arvin ash, he explained about the double slit experiment with quantum particles, there in his video when the which way path of the particles is detected formed a double slit pattern it was just a two strips in the video he made, but when i gone through a paper regarding Young's double slit experiment with single photons the double slit pattern was just a blurred image with no interference, but why is that instead of a two strip pattern the double slit pattern is just a blurred image with no interference?
| Without interference, the energy intensity at a point on the screen is proportional to $A^2+B^2$, where $A$ and $B$ are the amplitudes of the waves coming through each slit individually. With interference, it's proportional to $A^2+B^2+2AB\cos θ$, where $θ$ is the phase angle between the waves from the two slits.
If $A\approx 0$ or $B\approx 0$ at a point, then $AB\approx 0$, and the intensity will be about the same whether there is interference or not. So while it is possible to set up your light source and slits such that the light from one slit scarcely overlaps with the light from the other on the screen, you are not really doing the double-slit experiment if you do that.
Any introduction to the double-slit experiment that says "with a detector at the slits, you get two stripes, and without the detector you get an interference pattern" is incorrect.
| {
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How to go from Lagrange equations to d'Alembert's principle? All sources I know show how to use d'Alembert's principle and/or Hamilton's principles to derive Lagrange equations. It is also common to use d'Alembert's principle to derive Hamilton's principle (see Lanczos "the variational principles of mechanics", p.112) But what about the opposite direction? If you only have Lagrange equations, how can we derive d'Alembert's principle?
| Lets look at this example
Pendulum 2D space
the kinetic energy is
$$T=\frac m2 (\dot x^2+\dot y^2)$$
the potential energy is:
$$U=m\,g\,y+\lambda\,\underbrace{(x^2+y^2-l^2)}_{\text {holonomic constraint eq.}}$$
thus the equations of motion
$$\ddot x=-\lambda\,\frac{2\,x}{m}$$
$$\ddot y=-g-\lambda\,\frac{2\,y}{m}$$
how to eliminate the generalized constraint force $~\lambda~$ from the equations of motion ?
the EOM's
\begin{align*}
\begin{bmatrix}
\ddot{x} \\
\ddot{y}\\
\end{bmatrix}=-\frac\lambda m\,\underbrace{\begin{bmatrix}
2\,x \\
2\,y \\
\end{bmatrix}}_{\mathbf{C}_c^T}\tag 1
\end{align*}
from the constraint equation
\begin{align*}
&z=x^2+y^2-l^2=0\quad\Rightarrow\quad
\dot z=2\,x\,\dot x+2\,y\dot y=0\\
& \dot z=\mathbf{C}_c\,\begin{bmatrix}
\dot x \\
\dot y \\
\end{bmatrix}=0
\end{align*}
to eliminate the constraint force we multiply from the left equation (1) with
\begin{align*}
&\mathbf{J}^T=\begin{bmatrix}
-\frac yx & 1 \\
\end{bmatrix}\quad, \mathbf{J}^T\,\mathbf{C}_c^T=0
\end{align*}
this is the "d'Alembert principle".
Fazit:
from the holonomic constraint equation you obtain the constraint matrix $~\mathbf{C}_c~$ and from here ein orthogonal
matrix $~\mathbf J~$ .
notice that
\begin{align*}
\begin{bmatrix}
\dot{x} \\
\dot{y}\\
\end{bmatrix}\mapsto\mathbf{J}\,\dot{y}
\end{align*}
thus $~y~$ is the generalized coordinate , and you can transformed the kinetic and potential energy and obtain the EOM's with out the generalized constraint force
How to obtain the Jacobi matrix $~\mathbf J~$
with:
\begin{align*}
&x^2+y^2=l^2\quad\Rightarrow x^2=l^2-y^2\quad ,2\,x\dot{x}=-2\,y\dot{y}\\
&\dot{x}=-\frac{y}{x}\dot{y}\quad,\dot{y}:=\dot{y}\quad\Rightarrow
\mathbf{J}=\begin{bmatrix}
-\frac{y}{x} \\
1 \\
\end{bmatrix}=\begin{bmatrix}
-\frac{y}{\sqrt{l^2-y^2}}\\
1 \\
\end{bmatrix}
\end{align*}
Newton equation of motion
the generalized coordinate if $~y~$
\begin{align*}
&m\,\mathbf J^T\,\mathbf J\,\ddot{y}=\mathbf J^T\begin{bmatrix}
0 \\
-m\,g \\
\end{bmatrix}
-m\,\,\mathbf J^T\,\frac{\partial \mathbf v}{\partial y}\,\dot{y}\quad,
\text{where}~\mathbf v=\mathbf J\,\dot{y}\\
&\Rightarrow
\end{align*}
\begin{align*}
& \ddot{y}=-\left({\frac {y{{\dot{y}}}^{2}}{{l}^{2}-{y}^{2}}}+{\frac { \left( {l}^{2}-{y}
^{2} \right) g}{{l}^{2}}}\right)
\end{align*}
| {
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Why is the degeneracy factor for Bose Einstein distribution set to 1 automatically? In https://scholar.harvard.edu/files/schwartz/files/12-bec.pdf, the article says
"With Bose-Einstein statistics, we determined that using the grand canonical ensemble the expected number of particles in a state i is"
$$\langle{n}_{i}\rangle=\frac {1}{e^{\beta(\varepsilon _{i}-\mu )}-1}$$
And it goes on and derive the relationship between the total number of particles and the number of particles in the ground state.
$$N=\sum_{n_x,n_y,n_z=0}\frac{1}{e^{\beta\varepsilon_{1}(n_x^2+n_y^2+n_z^2)}\left(\frac{1}{\langle{n_0}\rangle}+1\right)-1}$$
However, as I understand, the equation for the expected number of particles in a state i is
$$\langle{n}_{i}\rangle=\frac {g_i}{e^{\beta(\varepsilon _{i}-\mu )}-1}$$
where $g_i$ is the degeneracy of energy level $i$. My question is why can I assume $g_i=1$ in this case since wouldn't that affect the answer?
| The expected number of particles in a given state $i$ is given by $$\langle n_i\rangle = \frac{1}{e^{\beta(\epsilon_i-\mu)}-1}$$
If there are $g(\epsilon)$ states which all have energy $\epsilon$, then the expected number of particles with energy $\epsilon$ is given by
$$\langle n(\epsilon)\rangle = \frac{g(\epsilon)}{e^{\beta(\epsilon-\mu)}-1}$$
In other words, in your second expression $i$ does not label a state but rather an energy level; $g_i$ is then the number of states with that energy.
| {
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What would a standing wave of light look like? I want to know what a standing wave of light would like and what properties it might have that are interesting.
| It arguably doesn't "look" like anything. The standing wave in an optical resonator (e.g., a laser cavity) is trapped in the cavity. It isn't entering your eye. If we're talking about a laser, then the part that escapes the cavity and ultimately enters your eye† arguably is no longer a standing wave.
A diagram that represents a standing wave of light would look the same as a diagram representing any other standing wave.
† Do not stare into beam with remaining eye.
| {
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Is the operator identity $[X,F(P)]=[X,P]F'(P)$ always true? I came up with the operator identity in my QM textbook
$$
[X,F(P)]=[X,P]F'(P)
$$
where $X,P$ are Hermitian operators whose commutator commutes with them: $$[X,[X,P]]=[P,[X,P]]=0.$$ $F(x)$ is some well-behaved function.
In the book, the identity is proved by verifying $$[X,P^n]=[X,P]nP^{n-1}$$ by induction and then expanding $F(x)$ into power series. However, the identity still works under conditions where this is quite impossible. For example,
$$
[X,P^{-2}]=-2[X,P]P^{-3}\\
[X,\sqrt P]=\frac12[X,P]P^{-1/2}
$$
is also true at least in some cases. (the second identity requires $P$ to be positive) As for the position and momentum case, it is beyond doubt because in the $p$-representation $X$ is $i\hbar\dfrac{d}{dp}$ and the identities are obvious. But difficulties occur when the commutator is not a constant.
Question Can we prove the identity is indeed true in such cases, or are they actually not true and there are some more restrictions on the function $F(x)$?
| As it finally emerges from your graduated comments, you certainly ought to have indicated that your Hilbert space goes beyond the image of all functions of X and P! In the language of the field, this center Y is a "constant", to the extent X and P commutations don't affect it. You need not fuss with weird operators: Indeed, the identity holds for the simplest Heisenberg group, trivially. But, in physics, a "constant" is an element of the center of the Lie algebra!
In any case, your operators
$$
X=e^{-i\phi} (-\partial_\theta + i \cot \theta ~ \partial_\phi), \qquad P=\cos\theta ,\\
\leadsto ~~~~ [X,P]\equiv Y =e^{-\phi} \sin\theta , \leadsto [Y,X]=[Y,P]=0,
$$
do satisfy the identity, for F Laurent-expandible, avoiding the singularities inherent in the weird operators you chose (take θ small avoiding 0). Many operator identities of this form parallel similar Sylvester's formula identities for matrices.
| {
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Why does a chimney sometimes create a "draft"? Sometimes, while lighting a fire in a wood stove in a basement, the smoke does not exit through the chimney like it normally does. Rather, a large amount of the smoke seems to get "pushed back" into the room instead of exiting through the chimney.
Why does this occur?
| Airflow around buildings and over walls can compress air at the top of the roof at higher pressure and bring low pressure at the sides of the house and the shielded zones.
If the chimney has very cold conditions at the top, and compressed air on the roof, the chimney can draft backwards, especially if the house has leaks towards the lee side of the wind, where the pressure is low.
https://www.youtube.com/watch?v=WZrg_uxGyaU
| {
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About a battery's positive terminal Does the positive terminal of the battery mean that the concentration of electrons is more there. In some places people say that current moves from positive to negative and they call it a convention what does this really mean and others say it flows from negative to positive it's really confusing.
| The names 'positive' and 'negative' were assigned to charges long before protons and electrons were discovered and even before batteries were invented. The type of charge that glass acquired when rubbed with silk was called 'positive', and the type that amber acquired when rubbed with fur was called 'negative'.
In the early 1800s batteries were invented and found to have opposite charges on their terminals. Hence their designations as 'negative' and 'positive' terminals. When wires were connected across the terminals, heating effects and (in the 1820s) magnetic effects were found and (in conjunction with the running down of the battery) attributed to a flow of charge through the wires. But no-one could know whether it was positive or negative charge or even both that was flowing. [The Hall effect had not been discovered, and few believed that atoms existed, let alone that there was such a thing as atomic structure.] So the decision was made to assume arbitrarily that it was positive charge that flowed through a metal wire from the positive battery terminal to the negative terminal. The 'hand rules' (invented in the later 1800s) that we can use in electromagnetism are based on this so-called conventional current.
In the 1890s, the negatively charged particles that we now call 'electrons' were discovered. [Remember: negatively charged meant having the same kind of charge as amber rubbed with fur!] Within the next few years atomic structure came to be understood as electrons surrounding a nucleus, and electric current in a metal wire, as a flow of electrons. They flow from the negative battery terminal (with an excess of electrons) through the rest of the circuit to the positive terminal (with an electron deficit). Life would have been easier for 20th and 21st century high school students if, back in the 1700s, the decision had been made to call the type of charge that amber acquired when rubbed with fur 'positive'!
| {
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Maximum Range for Projectile Motion This is a question concerning a trick I observed while solving for the angle responsible for maximum range of a projectile.
What I have observed is :
If you draw two lines, one opposite to the acceleration faced by the body(i) , another in the direction we want to find the projectile(ii), the angle for maximum range is always half the angle between the two lines .
In other words , the body should be thrown along the angle bisector of the two lines mentioned above to get the maximum range .
We note that this trick works for all the normal projectile cases because we know the angle should be 45° .
For an example , let's say a body is projected and faces acceleration due to gravity , 'g' in both vertical and horizontal directions. Line (i) will be at 135° from the horizontal opposite to the resultant acceleration √2g and line (ii) will be along 0° from the horizontal. So 67.5° is the angle for maximum range which is also the result derived after differentiating the range w.r.t the angle .
For projections along/down inclined planes as well , the trick seems to work. If x° is the angle of the inclined plane,line (i) is along y-axis and line (ii) ,x° from the horizontal. The angle for maximum range is thus (90°-x)/2 = 45°-x/2 from the inclined plane .
For down the incline , line (i) is as usual 90° from the horizontal and rotating line(i) through 90°+x° gives us line(ii) . The angle for maximum range is thus 45°+x/2 from the inclined plane .
I would like to know if there is some solid mathematical proof behind this trick . Why does it seem to work everytime ? Are there any cases where it doesn't?
I am sorry if I am missing something trivial .
| The proof of this is a straightforward generalization of the one for maximizing horizontal distance given downward acceleration. You mention differentiating with respect to the angle so I'm not sure which part you got stuck on.
Nevertheless, orient your axes WLOG so that the distance you want to maximize is $x$. Let the launch angle be $\theta$ and the acceleration angle be $\phi$. The position of the projectile at time $t$ is then
\begin{align}
x &= v_0 \cos\theta t + \frac{1}{2} a \cos \phi t^2 \\
y &= v_0 \sin\theta t + \frac{1}{2} a \sin \phi t^2.
\end{align}
The projectile lands at the greater solution of $y = 0$ so you can plug in this time to get
\begin{align}
x &= v_0 \cos\theta \left ( \frac{-2v_0 \sin\theta}{a \sin\phi} \right ) + \frac{1}{2} a \cos\phi \left ( \frac{-2v_0 \sin\theta}{a \sin\phi} \right )^2 \\
&= \frac{2v_0^2 \sin\theta \sin(\phi + \theta)}{a \sin^2 \phi}
\end{align}
after some trig identities. A sanity check is that $\sin 2\theta$ appears for $\phi = 3\pi / 2$ (downward acceleration) which has a maximum in the right place. More generally, we maximize $x$ by solving
\begin{align}
\frac{dx}{d\theta} = 0 \Leftrightarrow \frac{2v_0^2 \sin(\phi + 2\theta)}{a \sin^2\phi} = 0.
\end{align}
Within the relevant quadrant, this is solved by $\theta = \frac{2\pi - \phi}{2}$ which reproduces your examples above.
I'm not sure if it reproduces your prose description of the rule because at one point you say "perpendicular to the acceleration" and at another you say "opposite to the resultant acceleration".
| {
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Timescale for unstable stratification to resolve itself The ocean is typically vertically stratified, that is the deeper you go the higher the density should be. This can be because of increased salinity or decreased temperature.
If that is not the case then we have an unstable stratification and this instability is called Rayleigh-Taylor instability.
My question is what is the timescale for this stratification to resolve itself? If say at 7m depth the density is 0.1% higher than at 8m, then how much time it takes to get resolved by convection? Is it in minutes, hours, days?
| In real oceans this is complicated by currents, eddies, wind driven surface flows, thermohaline gradients, …, but in an idealized two-layer system with a sharp boundary and a density difference of about 0.1%, convective mixing over a 1 m distance should take less than a minute.
For an upper and lower layer with densities $\rho_u$, $\rho_l$ and dynamic viscosities $\mu_1$ and $\mu_2$, Rayleigh–Taylor turbulence has characteristic, length, time, and velocity scales:
$$l_c=\left(\frac{\nu^2}{g A}\right)^{1/3},\qquad
t_c=\left(\frac{\nu}{g^2 A^2}\right)^{1/3},\qquad u_c=\frac{l_c}{t_c}=\left(\nu g A\right)^{1/3}$$
where $g$ is the local acceleration due to gravity, $\nu$ is the average kinematic viscosity, and $A$ is the Atwood number:
$$\nu=\frac{\mu_u+\mu_l}{\rho_u+\rho_l},\qquad A=\frac{\rho_u-\rho_l}{\rho_u+\rho_l}$$
For warm seawater, $\mu\approx 0.001$ Pa s and $\rho \approx 1.02\,\textrm{lg/m}^3$ are reasonable values. A $0.1$% density difference gives $A=5\times 10^{-4}$ and
$$l_c\approx 1\,\textrm{mm,}\qquad t_c\approx 0.4\,\textrm{s,}\qquad u_c\approx 2\,\textrm{mm/s}$$
So it should take the order of a second or so for the Rayleigh-Taylor turbulence to start. Once started, the buoyancy acceleration of water from the lower layer rising in the upper layer is
$a_B = 2 A g = 0.0098 \textrm{m/s}^2$.
In the absence of viscous drag and turbulence, the time needed rise a distance $h\approx 1$m will be
$$t_L = \sqrt{\frac{2h}{a_B}} = \sqrt{\frac{h}{A g}} \approx 14\,\textrm{s}$$
and its final vertical velocity would be
$$v_f = \sqrt{2 a_B h } = \sqrt{4 A g h} \approx 0.14 \,\textrm{m/s}$$
It turns out ignoring viscous drag is not a terrible approximation in this case. The viscous de-acceleration (force per unit mass) of turbulent eddies of size $l_t$ rising with velocity $u$ is (Lawrie Eq. 4.31)
$$a_v \sim \frac{\nu u}{l_t^2}$$
The wavelength of the maximally unstable Rayleigh-Taylor mode is $\lambda_{mx} = 4 \pi l_c$, and the viscous de-acceleration for $u\sim v_f$ and $l_t \sim \lambda_{mx}$ is $0.002\,\textrm{m/s}^2$. Not completely negligible, but still significantly less than the buoyancy acceleration $\sim 0.010\,\textrm{m/s}^2$, so our rough estimate for the mixing time should still be about right.
Whenever turbulence is involved, however, we should worry about any simple theoretical analysis, so it is nice that our rough estimates are consistent with this video from Stuart Dalziel's Cambridge DAMTP Lab showing the mixing in real time in a $0.5$ m tall, $0.4$ m wide, $0.2$ m thick tank with $A=7\times 10^{-4}$.
Figure 5.12 (p. 69) of Lawrie shows a nice sequence of timed photos in the same tank.
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Does fractional freezing always happen in mixtures that are prone to it? Let's assume that we have a mixture that is prone to fractional freezing, such as water with table salt, or water with alcohol.
If we place the mixture in a sufficiently cold environment, will we always have fractional freezing, with ice forming from only the water? Or are there further conditions which have to be met? Are there cases where we will get a homogenous ice block of the mixture, as opposed to a separation?
As some background, I remember reading somewhere that it only happens when the conditions are "right", but don't remember what these "right" conditions are. Also, in the kitchen, I have observed that at least for stock, what I get is frozen cubes, not pure water ice floating on top of a cooled layer of concentrated stock.
| Not sure if this answers your question, but if the temperature is below the freezing point for both components of the solution, and the solution cools down fast enough, there will not be fractional freezing.
Also, some mixtures can form an "eutectic system", which has a freezing point that is lower than that of both its components. In such a system there will also be no fractional freezing.
| {
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How do we maintain polytropic processes? In polytropic processes for an ideal gas,
$$PV^{\alpha}=constant$$ where $\alpha \neq 0,1,\gamma$
And $\gamma$ is adiabatic exponent of gas
So, how these processes are maintained?
What things are done to initiate this process?
| You add or remove heat to change the temperature along the polytropic path in such a way that the exponent $\alpha$ remains constant. You have $$d\ln{P}+\alpha d\ln{V}=0$$and $$d\ln{P}+d\ln{V}=d\ln{T}$$So $$dln{T}=(1-\alpha)d\ln{V}$$or$$TV^{\alpha-1}=const$$
| {
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Is momentum conserved in a traversable wormhole? Wikipedia says:
A wormhole (Einstein-Rosen bridge) is a hypothetical structure connecting disparate points in spacetime, and is based on a special solution of the Einstein field equations.
If a traversable wormhole existed, could its ends move relatively to each other in space? What would be the momentum of the energy-matter escaping from one end, in relation to the momentum of the energy-matter entering the other end?
Imagine there is a traversable wormhole, with two ends, A and B, each in a separate galaxy A* and B* where these ends do not move relative to its galaxy but the galaxies A* and B* are moving (through space) away from each other with speed v=1000 m/s. When I enter the end A with speed v=1 m/s, will I emerge from end B moving relatively to galaxy B* with speed v=1000 m/s or with speed v=1 m/s?
Can the same question be asked about white hole/black hole pair?
| General relativity with one test particle (i.e you) can be described by an action principle. The action only depends on the metric and your position, so it's translationally invariant. Hence by Noether's theorem, it has four symmetries, corresponding to translation in space and time. The conserved quantities associated with space translation can be reasonably associated with the total momentum of you and the gravitational field. So yes, for an appropriate definition of momentum, any scenario in GR such as a wormhole will conserve momentum.
| {
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Third principle of thermodynamics and the unattainability of absolute zero Consider a $S-T$ diagram (entropy-temperature) and consider cooling a substance by doing a series of succesive isothermal and reversible adiabatic processes between two volumes $V_{1}$ and $V_{2}$.
Now when cooling the substance from $T_{1}$ to $T_{2}$ in the reversible adiabatic process we can write:
$$S(0, V_{1})+\int_{0}^{T_{1}}\frac{C_{V}}{T}dT = S(0, V_{2})+\int_{0}^{T_{2}}\frac{C_{V}}{T}dT$$
letting $T_{2}=0$ will lead to:
$$\underbrace{\int_{0}^{T_{1}}\frac{C_{V}}{T}dT}_{>0} = \underbrace{S(0, V_{2})-S(0, V_{1})}_{=0}$$
a contradiction showing that the third principle of thermodynamics implies that absolute zero cannot be achived.
Is this reasoning correct?
| Another perspective: The equation is simply not defined for $T_2=0$, since this would result in dividing by 0. You could at most let $T_2$ go against 0 for both sides, but then you have to prove that you can commute the limit with the integral. (In general e.g. by using Lebesgue's convergence theorem).
tl;dr: Correct math is important.
PS: Another example where such intuitive "proofs" can go wrong are infinitesimal rotations, but I can't re-find the paper which I have in my mind.
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Is there any intuition for why bisecting an angle gives the optimal angle to throw a ball? Let's say that I am holding a ball and have zero height, and there is a wall with a height of $h$ that is a distance $d$ away. At what angle should I throw the ball such that getting it over the wall requires minimal speed?
We can actually derive this using calculus. Obviously, the lowest speed that would clear the wall for a given angle would barely clear the wall. If the ball is thrown at an angle $\theta$ from the vertical and just barely clears the wall, then it has fallen by $d \cot\theta - h$ from a straight-line path by the time it clears the wall. The time it takes to fall this much is ${\sqrt\frac{2d \cot\theta - 2h}{g}}.$ In this time, if the ball was thrown at speed $v,$ then it would travel a distance of $d$ horizontally, which is equal to $v \sin\theta{\sqrt\frac{2d \cot\theta - 2h}{g}}.$ Therefore, $d = v \sin\theta{\sqrt\frac{2d \cot\theta - 2h}{g}}.$ Solving for $v$ gives that $v = \frac{d}{\sin\theta{\sqrt\frac{2d \cot\theta - 2h}{g}}}.$ We want to minimize $v,$ so we want to maximize the denominator, which is $\sin\theta{\sqrt\frac{2d \cot\theta - 2h}{g}}$. Therefore, we are attempting to maximize its square, which is $\sin^2\theta{\frac{2d \cot\theta - 2h}{g}}.$ We can factor out positive constants from the expression since we are trying to maximize it to get that we have to maximize $\sin^2\theta(d \cot\theta - h),$ which is $d\sin^2\theta \cot\theta - h \sin^2 \theta = d\sin\theta \cos\theta - h\sin^2 \theta.$ We can write this in terms of $2\theta$ as $\frac{d}{2}\sin(2\theta) - \frac{h}{2}(1 - \cos(2\theta)).$ Again, we can remove the factors of $\frac{1}{2}$ and get $d\sin(2\theta) - h(1 - \cos(2\theta)),$ which we are trying to maximize. Expanding this out gives $d\sin(2\theta) - h + h\cos(2\theta)).$ Again, since we are trying to maximize this, we can remove the $-h$ term, giving us $d\sin(2\theta) + h\cos(2\theta).$ Differentiating and setting this equal to $0$ gives $d\cos(2\theta) - h\sin(2\theta) = 0.$ Dividing through by $\cos(2\theta)$ gives $d - h\tan(2\theta) = 0.$ Therefore, $h\tan(2\theta) = d,$ so $\tan(2\theta) = \frac{d}{h}.$ If $\alpha$ is the angle that points straight at the top of the wall, then $\theta$ is exactly half of $\alpha.$ Is there any intuitive proof for why the optimal angle to throw a ball is exactly half of the angle from the vertical that points straight at the top of the wall?
| starting with
$$x=v\,\cos(\theta)\,t\\
y=v\,\sin(\theta)\,t-g\frac{t^2}{2}$$
solve those equations for $~t~,y~$ you obtain
$$y(x,v,\theta)={\frac {x{v}^{2}\sin \left( 2\,\theta \right) -g{x}^{2}}{{v}^{2}\cos
\left( 2\,\theta \right) +{v}^{2}}}
$$
$$t={\frac {x}{v\cos \left( \theta \right) }}=t_l$$
from here with
$$y(x=d,v,\theta)=h\quad \Rightarrow\quad~\text{initial velocity } \\
v^2=-\frac 12\,{\frac {g{d}^{2}}{ \left( \cos \left( \theta \right) h-d\sin
\left( \theta \right) \right) \cos \left( \theta \right) }}
=v_I^2$$
the velocity "that hit the wall" is
$$v_w=\sqrt{\dot x^2+\dot y^2}\bigg|_{\left(v=v_I~,t=\frac{d}{v_I\,\cos(\theta)}\right)}\\
v_w=\sqrt{-\frac 12\,{\frac {g \left( -4\,\sin \left( \theta \right) dh\cos \left(
\theta \right) +{d}^{2}+4\, \left( \cos \left( \theta \right)
\right) ^{2}{h}^{2} \right) }{\cos \left( \theta \right) \left( \cos
\left( \theta \right) h-d\sin \left( \theta \right) \right) }}
}$$
the velocity $~v_w~$ is minimum at
$$\frac{d}{d\theta}\,v_w=0\quad\Rightarrow\quad
\theta_m=\arctan\left({\frac {h+\sqrt {{d}^{2}+{h}^{2}}}{d}}\right)\quad\Rightarrow~\text{the initial velocity}\\
v_I^2={\frac {g \left( {d}^{2}+{h}^{2}+h\sqrt {{d}^{2}+{h}^{2}} \right) }{
\sqrt {{d}^{2}+{h}^{2}}}}
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/744320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Feynman's QED 36000 revolutions per inch In Feynman's book, QED, on page 27 he says "As long as the photon moves, the stopwatch hand turns (about 36,000 times per inch for red light); when the photon ends up at the photomultiplier, we stop the watch.
There are about 36,000 wavelengths of red light in one inch. So is he saying that the directions of the little arrows represent the phase angle of the EM oscillations at the point of scattering? Or is it a wave solution for the wave function? Or what determines how rapidly the angle of the little arrow da changes with position dx.
| Feynman's stopwatch is the phase of a complex number, $e^{i k x}$, where $k=2\pi/\lambda$ is the wavenumber and $\lambda$ is the wavelength of the wave. One rotation of the complex number occurs when $k x = 2\pi$; $N$ rotations occur when $kx = 2\pi N$.
Therefore, given $\lambda=36000^{-1}\ {\rm inch}$ (or in other words, $36000$ wavelengths per inch), and $x=1\ {\rm inch}$, we can calculate
\begin{equation}
N = \frac{kx}{2\pi} = \frac{x}{\lambda} = \frac{1\ {\rm inch}}{36000^{-1}\ {\rm inch}} = 36000
\end{equation}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/744539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do atoms emit an electromagnetic wave (infrared radiation)? From what I understand, when an object has a certain temperature, its atoms vibrate and this atomic vibration accelerates the electrically charged particles and this generates infrared radiation.
To generate infrared radiation, it is therefore necessary to accelerate electrically charged particles, but since atoms are electrically neutral, how can their acceleration generate infrared radiation?
| It is not the atoms composing an object that emit radiation, called black body radiation, it is changes in the quantized energy levels within a solid or a liquid that will be emitted as photons.
From what I understand, when an object has a certain temperature, its atoms vibrate
Its atoms change energy levels in the lattice of the solid and this generates statistically what is called black body radiation, which at room temperatures is mostly in the infrared. Maybe the answers to this question will help .
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/744698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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