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Help! An 8 year old asked me how to build a nuclear power plant I would really like to give an explanation similar to this one.
Here's my current recipe:
(i) Mine uranium, for example take a rock from here (picture of uranium mine in Kazakhstan).
(ii) Put the rock in water. Then the water gets hot.
(iii) [Efficient way to explain that now we are done with the question]
This seems wrong, or the uranium mine would explode whenever there is a rainfall. Does one need to modify the rock first? Do I need some neutron source other than the rock itself to get the reaction started?
As soon as I have a concrete and correct description of how one actually does I think I can fill in with details about chain reactions et.c. if the child would still be interested to know more.
|
This seems wrong, or the uranium mine would explode whenever there is a rainfall.
A natural nuclear "reactor" probably existed at Oklo, Gabon
The natural nuclear reactor formed when a uranium-rich mineral deposit became inundated with groundwater that acted as a neutron moderator, and a nuclear chain reaction took place. The heat generated from the nuclear fission caused the groundwater to boil away, which slowed or stopped the reaction.
Does one need to modify the rock first?
No, you just need enough of the right kind of rock in close enough proximity.
Nowadays, on our planet, most of the right kind of rock (containing lots of U235) has turned into the wrongish kind of rock (mostly U238 and U234) by the natural process of nuclear decay.
So you need to separate out the right kind of stuff (nuclear fuel) from your rock. This is done by a complicated process (gas centrifuge).
Do I need some neutron source other than the rock itself to get the reaction started?
The rock produces neutrons. You usually need a moderator to slow your neutrons down, water will do.
| {
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"url": "https://physics.stackexchange.com/questions/244158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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Beam Splitter: looking for a "not-typical" second quantization but full-quantum description In all the books of Quantum Optics I read, the theory of beam-splitter (BS) is presented in more or less the same way, e.g. introduction of the transmission-reflection matrix, case study of the single photon and vacuum at the BS inputs, or example with the coherent states, or Hong-Ou-Mandel effect.
However, I'm looking for a quantum description of the BS in terms of the interaction between the photons and the atoms, e.g. with the Fermi Golden rule.
Please, can someone point me to some paper and/or book where I could find such description?
| @TrulyIgnorant
I personally think this is a great question, and surprisingly one that has not yet been fully answered to my knowledge! There are two works that come very close to putting together a first-principles derivation that is fully quantum-mechanical, i.e. by quantizing the modes of a beam-splitter, showing their connection to a Hamiltonian and taking a limit that reproduces the unitary transformation you're referring to.
The papers are:
"Quasi mode theory of the beam splitter-a quantum scattering theory approach" by B. J. Dalton, Stephen M. Barnett & P. L. Knight
combined with
"A quantum scattering theory approach to quantum-optical measurements" B. J. Dalton, Stephen M. Barnett & P. L. Knight
In my advisor's group we're currently working on a tutorial that will go through this quantization and derivation to make it hopefully more clear - I'll try to post back here after we're done. A related question is how precisely to quantize a cavity coupled to the world - Sergio Dutra worked on this problem and distilled it very nicely in his book "Cavity Quantum Electrodynamics".
| {
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"timestamp": "2023-03-29T00:00:00",
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Fourier transform of Hamiltonian for scalar field In the Srednicki notes (http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf) page 36 he goes from
$$H = \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x) $$ to
$$H = \int d^{3}p\frac{1}{2m}P^{2}\tilde{a}^{\dagger}(p)\tilde{a}(p) $$
Where $$\tilde{a}(p) = \int \frac{d^{3}x}{(2\pi)^{\frac{3}{2}}}e^{-ipx}a(x)$$
I tried doing this by saying
$$H = \int d^{3}x \int \frac{d^{3}p}{(2\pi)^{3}}e^{-ipx} \tilde{a}^{\dagger}(p) \left(\frac{P^{2}}{2m}\right)e^{ipx}\tilde{a}(p) $$
But then I'm unsure how to proceed with commutators. Does $P^{2}$ commute with $e^{ipx}$? What about with $\tilde{a}(p)$?
Any help would be greatly appreciated.
| Starting from
$$
H = \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x)
$$
and
$${a}(x) = \int \frac{d^{3}p}{(2\pi)^{\frac{3}{2}}}e^{ipx}\tilde{a}(p)$$
(the second of which follows by inverting the expression above which defines the momentum space $a$ in terms of the position space $a$.
Plugging in for both operators, we have
\begin{align*}
H &= \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x)\\
&=\int d^{3}x
\int \frac{d^{3}p'}{(2\pi)^{\frac{3}{2}}}e^{ip'x}\tilde{a}^{\dagger}(p')
\left( \frac{- \nabla^{2}}{2m}\right)
\int \frac{d^{3}p}{(2\pi)^{\frac{3}{2}}}e^{ipx}\tilde{a}(p)\\
&=
\int d^{3}p'\int d^{3}p\tilde{a}^{\dagger}(p')\tilde{a}(p)
\int \frac{d^{3}x}{(2\pi)^3}
e^{ip'x}\left( \frac{- \nabla^{2}}{2m}\right)e^{ipx}
\end{align*}
after some rearranging. Then, taking the derivatives, this becomes
\begin{align*}
H
&=
\int d^{3}p'\int d^{3}p\tilde{a}^{\dagger}(p')\tilde{a}(p)
\int \frac{d^{3}x}{(2\pi)^3}
e^{ip'x}\left( \frac{p^2}{2m}\right)e^{ipx}
\end{align*}
which we can rearrange as
\begin{align*}
H
&=
\int d^{3}p'\int d^{3}p\frac{p^2}{2m}\tilde{a}^{\dagger}(p')\tilde{a}(p)
\int \frac{d^{3}x}{(2\pi)^3}e^{ip'x}e^{ipx}.
\end{align*}
Recognizing the last integral as a representation of a delta function and then evaluating the integral over the primed momentum coordinates gives us our result:
\begin{align*}
H
&=
\int d^{3}p'\int d^{3}p\frac{p^2}{2m}\tilde{a}^{\dagger}(p')\tilde{a}(p)
\delta^{(3)}(p-p')\\
&=
\int d^{3}p\frac{p^2}{2m}\tilde{a}^{\dagger}(p)\tilde{a}(p).
\end{align*}
| {
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How can you calculate the thrust-to-weight ratio of a gun? Is there some formula I can apply to the weight of a gun, along with the force of thrust it creates, and calculate if it will be able to lift itself off the ground or not? Without actually handling the gun, of course.
| Given the pieces mentioned in your question: the weight of the gun and force it creates, the answer is a trivial yes. If the force is greater than the weight, it could lift.
The problem is in calculating that force. A rocket is designed to produce force continuously for a period, and given the fuel and some other parameters makes it possible to calculate the average thrust over that period. The gun produces force only for a short period of time. As an example, it might generate a force in excess of its weight for only a couple hundredths of a second because the forces diminish as the gas in the firing chamber expands.
| {
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Units don't match in the torsional spring energy! According to Wikipedia's description of torsion springs and according to my understanding of physics the energy of a torsional spring can be written as
$$U=\frac{1}{2}k \varphi^2$$
where $k$ is a constant with units of $\rm N\,m/rad$.
I am freaking here because if the energy of a torsional spring is really $k \varphi^2$ than the units are
$\rm (N\,m/rad) \cdot rad^2=Joule\cdot rad$. ??
What on earth am I missing here?
| An angle is just the ratio of the length of a circular arc to its radius, so the radian has units of length/length, which means it's a dimensionless quantity.
| {
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Does continuity equation hold if the flow is accelerated? I'm Studying the streamline flow, specifically the continuity equation Bernoulli's Principle.
Consider the following system where a liquid flows through the pipe of a uniform area of cross section A, from high pressure P2 to low pressure P1, both of which are maintained.
<--------length = r----------->
_______________________________
P2 P1
_______________________________
flow ->
Based on the assumptions that
*
*Difference in pressure is what causes the liquid to flow $P_2 > P_1$
*The continuity equation $A_1v_1 = A_2v_2$ holds
*The Bernoulli's Principle holds
Then my analysis is as follows
The Bernoulli's equation gives
$$ P_2 + \frac{1}{2}\rho v_2^2 = P_1 + \frac{1}{2} \rho v_1^2$$
Therefore, the speed of liquid element will be more at $P_1$ and less at $P_2$ since $P_2 > P_1$
But then
$$A_1v_1 = A_2v_2$$
and since the Area of cross section is uniform, the velocities must be same, which is contradicting the result from Bernoulli's equation!
There is something wrong with my analysis or assumptions, but I can't figure out what it is
| In pipe flow two terms are added to Bernoulli's equation. Major loss and minor loss terms. Major loss is due to wall friction and minor losses are due to shape change in the pipe. (Valves, elbows etc.)
In the system that you draw the pressure difference overcomes the wall friction. The wall friction increases with velocity of the fluid, therefore, increasing pressure difference increases the fluid velocity.
| {
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The force felt by a pilot when a package is dropped? Here is a physics question that I wish to prove mathematically.
A 100,000 kg aircraft drops a 1000 kg package of supplies over an arctic research station. What approximate force is felt by the 100 kg pilot at the instant of the release?
Is there a reactive force on the plane when the package is dropped?
| He'll feel the release of potential energy, which is equivalent to
$$1000kg*height*gravitational~acceleration$$
The pilot would feel the same effect as if he was dropping an object from his hand weighing 1 kilogram.
| {
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Why does moonlight have a lower color temperature? Moonlight has a color temperature of 4100K, while sunlight has a higher color temperature of more than 5000K.
But objects illuminated by moonlight don't look yellower to the eye. They look bluer. This holds for indoor scenes (like my hall) and for outdoor. I find it counter-intuitive that moonlight has a lower color temperature. I thought the sun is the yellowest natural source of light we have.
Is that because of the poor color sensitivity of the eye in dim light? In other words, moonlight is actually yellower, but our eyes can't see the intense yellow color?
If one were to use a giant lens to concentrate moonlight to reach the brightness of sunlight, will objects illuminated by this light appear yellower to the eye than the same objects under sunlight? Has anyone done such an experiment? I looked, but couldn't find any.
Alternatively, if I take a long-exposure photo of a landscape illuminated by the full moon, and another one illuminated by sunlight, and equalise the white balance and the exposure, will the moonlit photo look yellower?
| I thought the sun is the yellowest natural source of light we have.
Whatever made you think that? It's pretty much by definition the whitest source of light we have since sunlight is what we judge all colors in.
Now direct sunlight is pretty close to a black body emission spectrum. The moon, not so much. It's reflected light, so assigning an actual color temperature is not going to reflect reality all that much since a color temperature implies a whole spectral distribution, and the moon has the spectrum of the sun modified by its wavelength-dependent average reflection coefficient. Its own temperature is such that its own black body radiation is not going to significantly contribute to the visible part of the spectrum.
The reddish color of a lunar eclipse is attributed to scattered light from the earth's atmosphere rather than "afterglow" from the immediately preceding full moon.
So I doubt that a "color temperature" will make much sense for the moon.
| {
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"url": "https://physics.stackexchange.com/questions/244922",
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Current operator in continuum model of graphene For the graphene hamiltonian with NNN hopping, the wavefunctions are of the form:
$(\psi_A ,\psi_B)^T$. The current from A(i) to B(j) site in the lattice model is given by:
\begin{equation}
J_{ij}=\mathrm{i}t(c^{\dagger}_ic_j-c^{\dagger}_jc_i)
\end{equation}
where $t$ is the hopping parameter.
1) How can this operator be generalised to the continuum model? Is it same as the general way in which Dirac current is defined?
2) What does the following convey:(Does it in some sense capture the A to B current?)
$\hat{O}\propto\mathrm{i}\langle\psi_B^{\dagger}\psi_A-\psi_A^{\dagger}\psi_B\rangle$
| The general way to find the current opperator is to gauge the U(1) symmetry and take the derivative of the Hamiltonian with respect to the gauge field $a_{ij}$ or $a_\mu$ and then turn off the gauge field:
$$\text{lattice: }J_{ij}=\left.\frac{\partial H}{\partial a_{ij}}\right|_{a_{ij}\to 0},\text{ continuum: }J^\mu=\left.\frac{\partial H}{\partial a_\mu}\right|_{a_\mu\to 0}.$$
On the lattice, after gauging the U(1) symmetry, $H=-\sum_{ij} t (e^{\mathrm{i} a_{ij}} c_{i}^\dagger c_{j}+h.c.)$, so the derivative with respect to $a_{ij}$ gives $J_{ij}= -t(\mathrm{i} c_{i}^\dagger c_{j}+h.c.)$. In the field theory, $H=\int \mathrm{d}^d x\; c^\dagger(\mathrm{i}\partial_\mu+a_\mu)\gamma^\mu c$, so the derivative with respect to $a_\mu$ gives $J^\mu=c^\dagger \gamma^\mu c$. You only need to figure out how to relate the lattice Hamiltonian to the continuum Hamiltonian. A simple method is to Fourier transform to the momentum space, expand around the gapless momentum points, and transform back to the real space.In general, all operators like $\mathrm{i}(\psi_i^\dagger \psi_j-\psi_j^\dagger\psi_i)$ can be interprete as a current.
| {
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What Keeps a Pendulum Moving In a Circular Path?
From the figure, we know that $F_{net} = mg\sin\theta$. Now, this force $\vec{F_{net}}$ is in the direction of the velocity $\vec{v}$ of the bob, both are tangent to the path. Therefore, the net acceleration $\vec{a_{net}}$ has no component perpendicular to the path, that is along the length $l$. I read that if acceleration is in the direction of velocity, then a body must be moving in a straight line, but such is not the case. Why? Also the bob is moving in a circular path and it should be experiencing centripetal force. What might be providing that force? The tension in the string is cancelled by the component of gravity parallel to the string.
|
What might be providing that force? The tension in the string is cancelled by the component of gravity parallel to the string.
I think a correction to the force diagram is required.
One should show the tension in the string and along the string a force called centripetal force is necessary to keep the bob in a circular path of radius equal to length of the pendulum.
No doubt the tangential net force is driving the pendulum.
So, T-mg Cos (theta) = centripetal force(provided for keeping the body on its
circular path ).
The force of tension varies and its maximum at the lowest point ;
Suppose one releases the bob from horizontal position i.e. theta= 90 degrees then the tension at the lowest point comes to about 3mg .
| {
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What direction are the decay particles of muons emitted in? If I were to run an experiment where I could measure the neutrino as well as the positron emitted when a muon decays, what direction should i be looking in?
| The decay of a (anti-)muon to a positron is
$$ \mu^+ \longrightarrow e^+ + \nu_e + \bar{\nu}_\mu .$$
As there are three light decay products their directions (in the rest frame of the anti-muon) are only weakly correlated.
The usual experimental procedure for dealing with this is to work with a beam of highly relativistic (anti-)muons so that all the decay products go in roughly the direction of the (anti-)muon beam. This also has the advantage of increasing the neutrino interaction cross-section up to a level where it is merely tiny rather than well-nigh-infinitesimal.
All that said, even a very small and basic muon-neutrino measurement requires a considerable accelerator, a custom focusing system called a "horn", and a highly specialized detector system. It's not really suitable for a one-person project, nor for a hobbyist budget.
| {
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What do we exactly mean by a "topological object" in physics? I have been working on topological defects like monopoles, etc. for some time. One think that I have not been able to understand is the physical meaning of the phrase "topological object". I have tried to find answers in many books on topological defects, gauge field theories, etc. but most of these books start with some Lagrangian and start talking about kinks and stuff like that. I have not been able to get a clear picture of what a topological object is physically? Is it just a mathematical construct or has some deep physical meaning (I am sure it has one) ? What is the difference between a topological object and a non-topological one. I understand that, in topology, we study properties under continuous deformations, stretching, twisting, etc.....so there is a context here but I do not understand its significance in physics. I need a very clear physical picture of this...I wouldn't mind some math though.
| The broad meaning is: an object which depends on global properties of the system, rather than depending on the metric or other local properties. Therefore these objects are described via topological concepts like homotopy group, fundamental group and so on.
| {
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If gravitons are 'real' and analogous to photons are they also being 'stretched' by the universe's expansion? Since photon wavelengths are stretched by our expanding universe, appearing to us as a redshift, would graviton wavelengths similarly be stretched? For that matter, do gravitons even have a wavelength like photons?
| Suppose, there are gravitons, and space is expanding.
Gravitons stretching with stretching space, would loose strength in same way as lower frequency (red shifted) light has lesser energy. That would change laws of gravity constantly. At the least, it would keep changing gravitational constant. That does not seem to be the case. There are no colors in gravity.
Gravity appears to be apart from all other stuff. It is property of space that becomes evident in presence of mass/energy. The space has this property everywhere but it becomes intense (curved inward) due to presence of mass/energy. It seems to be a property, and not any particle. Most fundamental property (gravity) of most fundamental entity (empty space).
We keep going smaller and smaller, the particles must end at some level and we will have to deal with emptiness in any case. It is just a matter of - at what level.
| {
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Could you use polarization filters to make a privacy screen? I remember seeing that brusspup video where the polarization filter on the monitor was removed and put it in his glasses, causing only the wearer to see the screen.
(, and)
I was thinking, would another polarization filter set on top of a phone screen to re-twist the polarized light and another in glasses achieve a display that others see only white, while the wearer can see the actual contents?
If not this idea, is there any possible way?
| Putting a polarization filter of any kind (linear or circular) at any orientation onto your iPhone screen will never be able to turn its display from normal-looking to all-white. Look at it this way: If you have a display on your iPhone screen that has black areas (say, black text on a white screen), then that means that no light photons at all are coming out of those black areas. No sort of polarizer at any orientation is going to be able to turn those black areas to white since there is nothing for them to work with - there is no light energy at all emerging from those areas so putting a polarization filter over those areas has no effect. Black remains black.
It you want your iPhone screen to look all-white to others, then the only way is to take out one of the polarization filters of your iPhone's LCD display just like the people in your linked video removed a polarization filter from their LCD monitor.
You might also want to check out this other question where the effect of removing the polarization filter from an LCD screen is discussed: What happens if you remove the polarization filter from a computer monitor?
| {
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Shape of water on top of a thin sheet of stretched plastic Consider a thin sheet of plastic (a square sheet for simplicity) that is stretched taught in a plane parallel to the ground. If a volume of water is then placed on top of the thin plastic sheet, then the water will further stretch the plastic and create a slight depression (assuming the plastic doesn't break). My question is: What shape will this volume of water be (or what shape is the bottom curved surface of the water) after it is allowed to settle? You can check out this video to see an example of what I mean https://www.youtube.com/watch?v=eeSyHgO5fmQ . The guy in the video says that it is almost a perfect paraboloid, but I don't see why it should be. This seems like a problem that can be solved using the calculus of variations, but I am stuck as to what the constraint/s should be.
| Let's first recall the wave equation for a membrane
$\partial_{tt} u = c^2 \Delta u$ where $u(x,y)$ is the vertical displacement. See, e.g., https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane.
One can recognize the left-hand side $\partial_{tt} u$ as the acceleration of a membrane element and $\Delta u$ is the vertical force provided by a stretched membrane. A derivation of this is given, e.g., in
http://www.math.iit.edu/~fass/Notes461_Ch7.pdf.
For a static problem we need only the force term which is the Laplacian.
The water lying on the membrane provides a vertical force on a membrane surface element proportional to the water column height above it which is the vertical displacement of the membrane $u(x,y)$ counted from the water surface level, $F = \rho_{water} g u(x,y) dx dy$. So equating the vertical forces on each surface element of the membrane one arrives at the eigenvalue equation
$\Delta u = \lambda u$. A solution with azimuthal symmetry $u(r)$, non-singular at $r=0$, is the Bessel function $J_0$. Of course all this is formally valid only for small displacements as this is all just linear theory.
| {
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How to interpret the units of the dot or cross product of two vectors? Suppose I have two vectors $a=\left(1,2,3\right)$ and $b=\left(4,5,6\right)$, both in meters.
If I take their dot product with the algebraic definition, I get this:
$$a \cdot b = 1\mathrm m \cdot 4\mathrm m + 2\mathrm m \cdot 5\mathrm m + 3\mathrm m \cdot 6\mathrm m = 4\mathrm m^2 + 10\mathrm m^2 + 18\mathrm m^2 = 32 \mathrm m^2$$
Dimensional analysis tells me that this is in meters squared, if I understand correctly.
Doing the cross product, however, I get this:
$$a \times b = \left[
\begin{array}{c}
2\mathrm m \cdot 6\mathrm m - 3\mathrm m \cdot 5\mathrm m\\
3\mathrm m \cdot 4\mathrm m - 1\mathrm m \cdot 6\mathrm m\\
1\mathrm m \cdot 5\mathrm m - 2\mathrm m \cdot 4\mathrm m\\
\end{array}
\right]
= \left[
\begin{array}{c}
-3 \mathrm m^2\\
6 \mathrm m^2\\
-3 \mathrm m^2\\
\end{array}
\right]
$$
This doesn't make sense to me either.
I don't know if I'm thinking about this in the right way, so my question is this: when dot or cross-multiplying two vectors, how do I interpret the units of the result? This question is not about geometric interpretations.
| The dot product of two unit vectors can safely be considered a dimensionless quantity, from a dimensional analysis perspective — a unit vector is what you get when you divide a vector by its magnitude, and the dot product is linear in terms of the magnitudes of both vectors, so all of the units cancel out — and for the reason that you can take its arccosine to obtain the angle between the two vectors, and the cosine of an angle is dimensionless.
| {
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Physical meaning of $Tr(\rho ^2)$ If $\rho$ is the density matrix of a system then $Tr(\rho ^2) \leq 1$. If the equality holds the system is in a pure state and it is in a mixed state otherwise. But, what is the physical meaning of $Tr(\rho^2)$ ?
$Tr(\rho) = 1$ for all valid density matrices. This stems from the normalization constraint that total probability must be one. Is there any such interpretation for $Tr(\rho ^2)$ ?
| You can interpret $1-Tr[\rho^2]$ as a kind of entropy. If you think of $\rho$ as a classical mixture of quantum states, then $Tr[\rho^2] = \sum_i p_i^2$, where $p_i$ is probability of being found in eigenstate $i$. The more dispersed the states, the smaller the quantity. So for a completely mixed state you have $1/n$, whereas for a pure state you have 1.
| {
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Airplane on a treadmill - Variant Thought Experiment This thought experiment is in a way related to the (in)famous airplane on a treadmill problem.
If you take a ball and place it on a treadmill, will the ball:
*
*Move backwards relative to the ground at the same speed as the treadmill (as if placing any other non-circular object on the treadmill)?
*Roll in place without moving relative to the ground (the speed of the treadmill is converted directly into rolling motion of the ball)?
*Exhibit some other behavior such as rolling while also moving backwards?
For this problem assume that there is no slippage between the treadmill and the ball (sufficient friction to make full contact at all times), and assume that the ball has mass.
I know the answer is not #1. I am not sure if the answer is #2 or #3. If the answer is #3, what factors affect the movement of the ball? Is it the mass of the ball, the speed or acceleration of the treadmill, and/or other factors?
| The treadmill will apply both a non-zero torque and a net force to the ball. Thus the ball will both move backwards and roll. To know precisely how much of each the ball does we need to know more about the initial setting:
Does the treadmill start from rest and then accelerate? How does it accelerate?
Then using the Principle of Least Action I believe we can model exactly the dynamics of this scenario.
| {
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What causes polarization current density in plasma? Plasma current density is defined as $\textbf{j}=-\mathrm{e}n_e\textbf{u}_e$ where $\mathrm{e}$ is electron charge, $n_e$ is electron concentration and $\textbf{u}_e$ is electron velocity. What exactly causes polarization current density and why we do not consider other types of current densities (free-charge current density, magnetization current density) in plasma?
| A polarization current can be driven by something that causes a time-varying electric field, like a gradient drift which forces positive and negative charges to drift in opposite directions. This can be similar to a displacement current, but it is generally driven by bulk particle motion.
Side Note:
The current density is not just due to the electron drift. Currents are also not a Lorentz invariant. The current density in the macroscopic sense is defined by:
$$
\mathbf{j} = \sum_{s} \ n_{s} \ q_{s} \ \mathbf{v}_{s}
$$
where $n_{s}$ is the number density, $q_{s}$ is the electric charge, and $\mathbf{v}_{s}$ is the bulk flow velocity of particle species $s$. So as you can see, the current density is not just derived from the electron flow, though electrons often do carry currents due to their higher mobility than ions.
| {
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Quantum versus classical computation of the density of states If I consider for instance N non interacting particles in a box, I can compute the energy spectrum quantum mechanically, and thus the number of (quantum) microstates corresponding to a total energy between $E_0$ and $E_0 + \delta E$. In the limit of large quantum numbers, the result is well known to coincide with the available volume of the phase space of the corresponding classical system of N newtonian free particles in a box, namely
$$
\Omega(E_0,V,N; \delta E)_{\textbf{quantum}} \to \frac{1}{h^N} \int_{E_0<E<E_0 +\delta E} d^{3N}x d^{3N}p
$$
in the limit of large quantum numbers.
My question is the following. Is there any proof, besides this specific example of the quantum gas in a box, that the quantum expression is always going to approach the classical one in phase space, for any given physical system (and thus for some generalized coordinates), provided some classical limit is used?
This does not seem a trivial statement to me, and I can't find the proof in textbooks.
| For particles in a 1-D potential $V(x)$, we can connect the volume of phase space to the number of quantum states via the WKB approximation. Under the usual WKB assumptions, it can be shown (see, e.g., Liboff or Griffiths) that to have a well-defined wavefunction we must have
$$
\int_{x_1}^{x_2} p(x,E) \, dx = \left( n + \frac{1}{2} \right) \frac{h}{2},
$$
where $n$ is an integer, $p(x,E) = \sqrt{ 2m(E - V(x))}$, and $x_1$ and $x_2$ are the classical turning points for energy $E$ (i.e., $V(x_1) = V(x_2) = E$.) The curves $\pm p(x,E)$ are of course the curves that the particle would take in phase space classically; and so the area enclosed by the (closed) classical trajectory in phase space must be $(n+ \frac{1}{2}) h$ for some integer $n$.
Consider now all the states between $n$ and $n + \Delta n$. These states span an energy range between $E$ and $E + \Delta E$. The area between these curves is the allowed volume of phase space with energies between $E$ and $E + \Delta E$; and this is obviously the difference between the corresponding areas enclosed by the energy-$E$ curve and the energy-$E + \Delta E$ curve
$$
\int_{[E, E + \Delta E]} dp \, dx = \left( n + \Delta n + \frac{1}{2} \right) h - \left( n + \frac{1}{2} \right) h = (\Delta n) h.
$$
This is the shaded area in the diagram above. But by the WKB quantization conditions, there are simply $\Delta n = \Omega(E; \Delta E)$ states in this energy range. Thus,
$$
\Omega(E; \Delta E) = \frac{1}{h} \int_{[E, E + \Delta E]} dp \, dx
$$
as expected.
It might be possible to generalize this to systems in higher dimensions, but I am not familiar enough with the higher-dimensional versions of these quantization rules to know for sure.
| {
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Angle of a pendulum as an explicit function of time What would be the solution for the angle $\varphi(t)$ and angular speed $\omega(t)$ on pendulum without the small angle approximation - not as differential, but as an explicit function of time?
The pendulum starts at angle $\theta_i$ with the arbitraty initial angular velocity $\omega_i$. What is the speed after time $t$ in the form of
$$\varphi(t) = f(t, \theta_0, \omega_0)$$
when the angle is arbitrarily high, or even at a looping pendulum?
| You can predict that as the initial angle of deflection increases so does the period: Look at the special case for a bob held by a stiff wire, when the angle is exactly $180^0$ then the bob will remain in the upright position for ever (ignoring any small perturbations). Continuity suggests that as the initial angle of deflection increases so does the period.
| {
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Double slit with opposite circular polarizers Let's say I'll send linearly polarized light onto double slit but in front of one slit I'll have quarter wave plate and before the other I'll have 3/4th wave plate (half+quarter? minus quarter?) oriented in such way that the light will get polarized clock-wise before going into first slit, and counter clock-wise before going into the other.
Will I still get interference pattern (not just diffraction patter from each separate slit)?
| Circular polarisation can be though of as a horizontally polarised wave together with a vertically polarised wave which is $\frac \pi 2$ out of phase with the horizontally polarised wave.
With the two slit arrangement that you have described let the horizontal components of the circularly polarised light from each slit be in phase with one another and these components would produce an interference pattern with a maximum at the centre.
The vertical components of the circularly polarised light from each slit would be $\frac \pi 2 + \frac \pi 2 = \pi$ out of phase with one another and so would produce a minimum at the centre.
So the two components from each slit can be thought of as producing two interference patterns one of which is shifted by half a fringe relative to the other.
So no fringes would be seen.
| {
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Wigner-d matrices for higher (than 1/2) spins I’ve been reading
¨Halzen, F., and A. D. Martin. Quarks and Leptons. New York: Wiley Text Books, January 1984. ISBN: 9780471887416¨,
and I’d like some clarification of a concept, please: I’m looking at Problem 2.6, and the question asks us to show that the rotation matrices are represented by certain values, depending on the j-value used, for the equation: $d^{j}_{m’m}(\theta)=\langle{jm’}\lvert{e^{-i{\theta}{J_2}}}\lvert{jm}\rangle$. $J_2$ is represented as a rotation generator, $\theta$ is obviously the angle of rotation, $j$ represents the Eigenstate, and $m$-values are the different available states for each $j$-value. The question said that if $j={1 \over 2}$, we’d have:
$$j = {1 \over 2}
\begin{cases}
d_{++} = d_{--} = \cos({1 \over 2}\theta) \\
d_{+-} = -d_{-+} = \sin({1 \over 2}\theta)
\end{cases}$$
, where $\pm$ designates $m = \pm {1 \over 2}$- values.
For $j=1$, I’m supposed to be able to find that:
$$j = 1
\begin{cases}
d_{01} = -d_{10} = -d_{0-1} = d_{-10} = \sqrt{1 \over 2} sin(\theta) \\
d_{11} = d_{-1-1} = {1 \over 2} (1 + \cos\theta) \\
d_{-11} = d_{1-1} = {1 \over 2} (1 - \cos\theta) \\
d_{00} = \cos\theta
\end{cases}$$
Now, I’ve finished the first part, where we need to find the values for $j= {1 \over 2}$; I basically just used Euler’s Rule ($e^{ix} = \cos (x) + i\sin (x)$) to break down the exponential, and then considered odd and even-integer solutions.
I have several questions about the second part of the question:
*
*How do I, in general, manipulate a higher-spin system, using a given operator? I feel like this has something to do with SU(3) or expanding SU(2) to SU(3), or perhaps I’m mistaken?
*I looked at the solution to the first part in the book, and the authors used the Pauli Spin Matrices in order to solve it. Aren’t I doing the same thing with my own solution (as I described above), just without the explicit use of the Pauli Spin Matrices?
| *
*Your feeling looks very misguided. Whatever you do, stay away from SU(3) for rotations. The rotation group and its Lie algebra are always linked to SO(3) ~ SU(2), to avoid formal forays into double covers and half angles. Read up on the spin matrices for any representation of the very same group (any spin).
*There are, in fact, simple systematic generalizations for the simple Euler-like exponential of the Pauli matrices, for spin 1 and for all representations of SU(2), but this is distinctly egregious overkill, for your purpose. Wigner's little d rotation matrices in the spherical representation solves the problem simply and in full generality. The text you are referring to assumes the reader has taken a good QM course where all this is covered quite nicely.
*In any case, since it is safely late to do homework for one, recall the expression for $J_2$ for spin one,
$$
J_2 = \frac{1}{\sqrt{2}}
\begin{bmatrix}
0 &-i &0\\
i &0 &-i\\
0 &i &0
\end{bmatrix} \equiv iK,
$$
which you may easily exponentiate
$$
e^{-i\theta J_2}=e^{\theta K}= I + (\sin\theta) K + (1-\cos\theta) K^2 ~,
$$
since you can confirm that $K^3=-K$, so $K^2$ behaves like i in combinatoric terms in the series for the exponential, when multiplied by K.
Now, since
$$
K^2= \begin{bmatrix}
-1/2 &0 &0\\
0 &-1 &0\\
0 &0 &-1/2
\end{bmatrix} ,
$$
the net rotation in the exponential is just
$$
e^{-i\theta J_2}= \begin{bmatrix}
\frac{1+\cos\theta}{2} &-\sin\theta /\sqrt{2} & \frac{1-\cos\theta}{2}\\
\sin\theta /\sqrt{2} &\cos \theta &-\sin\theta /\sqrt{2}\\
\frac{1-\cos\theta}{2} &\sin\theta /\sqrt{2} &\frac{1+\cos\theta}{2}
\end{bmatrix} .
$$
You are done. From this, you simply read off the spherical basis matrix elements you have for the destination Wigner d-matrix stated.
| {
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Ohm's law deviation In terms of superconductivities and diodes (I do not know anything else except these), Ohm's law deviate from a linear relation. I search many titles or tags for this but I did not understand properly how it becomes. I wonder somethings related with this.
*
*What is the basic of this deviation?
*How it is deviate?
*Why there is linear relation for metal conductors, if is it true?
| Ohm's law is a misnomer. It is not actually a true law, in the sense of Coulomb's of Ampère's; rather it is a 'rule of thumb' that applies pretty well in most circumstances. You will certainly not get a nobel price for finding an exception! A more general form of Ohm's law is
$$\mathbf{J} = \sigma \mathbf{E},$$where $\mathbf{J}$ is the current density, $\sigma$ the conductivity and $\mathbf{E}$ the electric field.
Now in this form, you can easily find a classical derivation, due to Drude, (which is totally inaccurate but does make some good qualitative predictions
| {
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What is $\langle \phi | H | \psi \rangle$ in QM? I know that $\langle \phi | \psi \rangle$ is the probability of going from the $\psi$-state to the $\phi$-state, and that $\langle \phi | H | \phi \rangle$ is the expectation value of the energy for the $\phi$-state.
But how should I interpret $\langle \phi | H | \psi \rangle$?
| This is a scalar value that is a projection of the state $H|\psi \rangle$ on the state $|\phi \rangle$. The state $H|\psi \rangle$ results from the action of the operator $H$ on the state $|\psi \rangle$. If the state $|\psi \rangle$ is an eigenstate of the operator $H$, the expression can be rewritten as $E \langle\phi|\psi \rangle$. If the state $|\phi \rangle$ is also an eigenstate of the operator $H$, we have $E \delta_{\phi,\psi}$, meaning we get zero if the states are orthogonal and an expectation value of the energy if they are conjugate.
If both states are not eigenstates, we can elaborate on it using the resolution of identity in terms of the Hamiltonian eigenstates: $1=\sum_i |i \rangle \langle i|$. By multiplying the identity from both sides of the Hamiltonian, the resulted expression reads: $$\sum_{ij} \langle\phi|i \rangle \langle i|H |j \rangle \langle j|\psi \rangle=\sum_{ij} E_j \delta_{i,j} \langle\phi|i \rangle \langle j|\psi \rangle=\sum_{j} E_j \langle\phi|j \rangle \langle j|\psi \rangle$$
Thus we have a sum of products of the two states projections on all eigenstates of the Hamiltonian multiplied by a corresponding energy.
| {
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Why don't stars re-emit the photons they absorb, thus restoring a continuous emission spectrum? If you shine white light through a gas, electrons can absorb sufficiently energetic photons to reach higher excited states. This produces gaps in the spectrum and it's how Helium was discovered. So goes the story.
But an excited electron now exists above a gap in a lower shell. It should fall back down, re-emitting a photon of the same characteristic frequency.
So why doesn't this plug the gaps in the spectrum?
| When the electron falls to its ground state, it will emit the photon in a random direction.This means that the photon might not travel in the direction as the rest of the white light.
In the absorption spectrum of the sun, there will be some photons of the characteristic frequency of helium, but significantly less than the rest of the spectrum, meaning a gap is created.
| {
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Is it always possible for an observer to realize to be in a non-inertial frame? Galilean relativity principle states that two frames moving with uniform linear motion cannot be distinguished. But is it always possible to realize to be in a non-inertial frame?
In a rotating frame it is surely possible for the observer to realize that because of Coriolis force, which cannot be explained, even supposing the presence of a source of force somewhere.
But in a frame moving linearly, for istance, with acceleration $A$? Coriolis term is not present, so what is the way for the observer in the frame to realize that he is in a non-inertial frame?
| In a non-inertial frame, observers will see fictitious forces with no reaction pair. For example, in a frame accelerating linearly forward, there appears to be a force acting backwards, and one cannot find the reaction (or source) of this force.
| {
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Understanding role of friction in block on block problems Consider two blocks stacked on top of one another. There is friction between the blocks but there is no friction between the lower block and the table. So the only frictional force that tends to retard the lower block is due to the friction between the blocks right? And the force that accelerates the body on top is only the frictional force?
Also, I fail to understand how there are different conditions for relative motion to occur between the blocks when the force is applied on the lower/upper block.
| As long as the two blocks move together as one whole block, it does not matter you apply accelerating force to the upper or lower block.
But, if the accelerating force is greater than a threshold, the two bodies move in relative to each other, in this case, the threshold value and the resulting movement depend on which block you apply the force.
| {
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Why does limiting friction have to act when a block tied to a wall is pulled? A block lying on a rough surface, is connected to a wall by a mass less, inextensible string and an unknown amount of force is applied to the block opposite to the side of the wall. Now, if it is given that there is some tension in the string and the block is stationary, does it necessarily imply that limiting static friction is acting? Apparently it does. My question is, if:
$$F = f_{limit} + T \rightarrow F = (f_{limit} - c) + (T+c) \rightarrow F = f_n + T_n$$
So why is it not possible that the tension in the string increases to value greater than that when limiting friction acts and the frictional force acting is lesser than the limiting friction ?
| Because the tension force of a string is a reaction force (like the friction force). This means there is no tension force until the string isn’t stretched. The string will be non-stretched until the force F is lesser than the limiting static friction force. There is a same state for friction force. The friction force is lesser than the limiting static friction force until the force F is lesser than the limiting static friction force (for example, when F is zero, the friction force doesn’t exist.)
| {
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Why do constants have dimensions? I am just a beginner in dimensional analysis, and I see that $G$, the universal gravitational constant, is independent of everything. Speed, for example, depends on distance and time, but $G$ does not depend upon anything.
Then why is $G$'s dimensions not $M^0 L^0 T ^0$, as it is not dependent on $M$, $L$ or $T$?
| Speed doesn’t depend on distance and time. We measure it by measuring distance and time. We should understand the difference between physics and mathematics. We use mathematics because we want to live better in nature. Speed is a abstract concept that we created it. We never can discover how the nature works or what are the rules nature based on?
About G, we defined it and so it will have dimensions based on our contracts. It doesn’t depend on its dimensions, but it is measured by them.
| {
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relationship between torque and potential energy for electromagnetism It is well known that the energy of a magnetic dipole in a magnetic field is taken as
$$U = - \bf{m}.\bf{B}$$
The dipole also experiences a torque
$$\bf{\tau = m \times B}$$
In classical mechanics, the torque is given as
$$\bf{\tau} = \bf{r \times F}$$
The force is derivable from a potential energy $V$, i.e. $\bf{F = -\nabla}$$V$; you can write the torque as $\bf{\tau = -r \times \nabla}$ $V$.
The dynamics of an object carrying an electric current is governed by
the Lorentz force, a velocity-dependent force that cannot be derived
from a potential energy function
Source: Daniel R Stump, Am J Phys 66, No. 12 December 1998 pp 1042-1043
Keeping the above quote in mind, is it permissible to use this relationship in the electromagnetic case?
| A conservative force is one for which the curl is zero, i.e: force $F$ is conservative if and only if it satisfies the following condition,
$$\nabla \times F=0. $$
This is true for electrostatic forces. However, if at a point in space the magnetic field is changing with time, then the electric force will, in general, be non-conservative. The changing magnetic field induces a curl in the electric field.
The Lorentz force is given by,
$$\vec{F} = q(\vec{E} + \vec{v}×\vec{B})$$
$$\nabla \times F = q(\nabla \times E-B (\nabla. v) + (B.\nabla)v -(v.\nabla)B)$$
As you can see, in general, this will not be equal to zero. Therefore, we cannot write the Lorentz Force as the gradient of a scalar potential energy function, unless in the electrostatic case. But it can be derived from a velocity dependent potential function:
$$U= q\phi - q\vec{v}.\vec{A} $$
where
$\phi$ a scalar function o position
$v$ is the velocity of charge on which the Lorentz force law is applied and
$\vec{A}$ is the vector potential
| {
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Can lasers lift objects? I have been fascinated by a very intriguing question - Can lasers push objects up?
I have done the below math to find out
Lets say we have a $1000~\text{mW}$ laser and we would like to lift an object of weight $100~\text{g}$.
By definition: $1~\text{W} = 1 \frac{~\text{J}}{~\text{s}}$
That means the laser is emitting $1~\text{J}$ of energy per second.
On the other hand energy required to lift an object off the ground is given by $m \cdot g \cdot h$.
Putting in the number and lets say we want to solve for
$0.1~\text{kg} \cdot 9.8 \frac{~\text{m}}{~\text{s}^{2}} \cdot h = 1~\text{J}$
So, $h \approx 1~\text{m}$.
You see, if we had a $1000~\text{mW}$ laser we could lift an object of $100~\text{g}$ weight up to 1 meter in one second.
I can't see anything wrong with the above math. If this is correct, can anyone tell me then why on Earth we use heavy rockets to send objects into space?
| Your approach is incorrect. You cannot do this calculation by considering that the energy absorbed by the object is converted into a change in gravitational potential energy. For one thing the object would just get hot and radiate away most of the energy and for another this is a dynamical problem, you have to be able to accelerate the object upwards.
What is important is the product of the power per unit area of the laser and the area over which it is incident. More precisely, to "levitate" an object by shining a laser onto its underside requires that the force exerted upwards by the laser is equal to the force $mg$ acting downwards. A general expression one could use is
$$\frac{1+r}{c}\int \vec{S} \cdot d\vec{A} \geq mg,$$
where $\vec{S}$ is the time-averaged Poynting vector of the laser, with a magnitude equal to the power per unit area in the beam, and the component of this normal to the surface is integrated over the surface area of the object to be levitated. The term $r$ is the reflectivity of the surface. $r=0$ for a black surface, but the upward force would be doubled for a perfectly reflective surface with $r=1$.
Hence assuming I had a completely black cube of surface area $A$ oriented so that a surface was perpendicular to a laser beam with Poynting vector $S$:
$$ \frac{SA}{c} \geq m g$$
$$ m \leq \frac{SA}{cg}$$
and if $SA = 1$ W, then $m \leq 3.4 \times 10^{-10}$ kg is the mass which it could accelerate upwards against gravity.
| {
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What is the moment of inertia really? Is moment of inertia or second moment of inertia, simply the resistance of a body to rotate it over an axis? What is radius of gyration? What if the axis is via the center of mass or somewhere different? can you give me please an overview of these issues with SIMPLE words, and without nonsense, like maths who nobody will ever remember. I need the SENSE how the brain comprehends these stuff in simple terms.
|
simply the resistance of a body to rotate it over an axis?
Gosh, I dislike the word resistance in this context since resistance is, in general, dissipative and, in particular, resistance to rotation would imply that an isolated object that is rotating would eventually stop.
Think of moment of inertia (rotational inertia) about an axis as a measure of an object's opposition to change in rotation (about that axis) not as a resistance to rotation itself.
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Different definitions of the parity transformation for the Dirac spinors There are two definitions of the parity transformation acting on the Dirac spinors: $\Psi_P = \eta \gamma^0 \Psi$ with $\eta = i$ ($P^2=-1$ as in Srednicki) and $\eta=1$ ($P^2=+1$ as in Peskin & Schroeder).
Both definitions result in the same parities of the sesquilinear forms such as $\bar\Psi \Phi$.
However, the bilinear form $\overline{\Psi_C} \Phi$ (à la pseudoscalar diquark) is scalar under $P^2=-1$ inversion and pseudoscalar under the one with $P^2=+1$, since $\Psi_{CP} = -\eta^* \gamma^0 \Psi_C$; $C$ is charge conjugation.
Does it mean that the two definitions are physically inequivalent (and $P^2=-1$ is incorrect)?
Or...
Do I miss something that makes $\overline{\Psi_C} \Phi$ a pseudoscalar even for $\eta = i$?
| For non-hermitian products of Dirac field operators the parity is not well defined and depends on the phase $\eta=\pm1,\,\pm i$ of the parity transformation $\eta \gamma^0$. For example, $I_P = -\eta^2 I$, where $I = \overline{\Psi_C}\Phi$.
In the $S$-matrix elements, however, all phases go away eventually, because creation and annihilation operators come in pairs there and $|\eta|^2=1$. Thus, for many practical calculations, it appears reasonable to omit the phase $\eta$. Then, we can consider $I$ as a pseudoscalar even for $\eta=\pm i$.
| {
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Quick question on weight/mass (In the US, just to clarify)So, from a physics perspective weight and mass are different, but when people are talking about weight in everyday(non-physics) situations ("how much do you weigh" etc.), are they actually talking about mass and it's just common to refer to it as 'weight'?
Expanding on that, when you step on a scale, i've read it displays your mass(after conversion from weight as it's displaying the "results"). Seeing as pounds is a measurement of weight, why will it use pounds as a unit of mass?
Maybe i'm getting things completely confused, any help is appreciated.
| When you are looking at the car speedometer you are actually measuring AN ANGLE between the zero and dial hand. Nevertheless, you read SPEED units.
This is because scale is CALIBRATED in units of speed.
Similarly, when you are looking at mercury thermometer, you are actually measuring the DISTANCE. Nevertheless, you read TEMPERATURE units.
This is because device is CALIBRATED in units of temperature.
This is the common case in metrology: you are transforming physical quantities one to another and finally get easy observable and CALIBRATE it.
| {
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Why doesn't a swing make a full revolution on a swingset in a park? I'm familiar with the concept of swinging of a swing in a park, but I'm confused why it doesn't it swing at a complete circle around the center bar?
| You mean go all the way around? It could if you had enough force to overcome gravity and like a tether ball swing all the way although most humans do not have the strength to apply the force needed to push another or them selves to a full revolution around the bar of a swing with out a jerk, but if the chain was replaced with a solid bar to prevent jerking in mid stride it could be done.
| {
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Do gravitational waves affect light? Gravity "bends" light, predicted with theory of relativity and subsequently observed: how does gravity and gravitational waves achieve this effect, and shouldn't this effect be present wherever there's gravity, for example shouldn't there be a "shimmering effect" when observing distant stars/galaxies as their emitted light is "bend" this way and that (like heatwave-shimmer)?
| Gravity "bends" light, predicted with theory of relativity and subsequently observed: how does gravity and gravitational waves achieve this effect
I conceive the question as one referring to General Relativity (GR) which is an entirely classical theory, and not to a (thus far missing) theory of Quantum Gravity.
The point of view of General Relativity is that free falling reference frames are true inertial frames (ie where Newton's first law holds). Thus, viewed from within a free falling frame, light goes on a straight line: a person inside a free falling elevator, will see a straight light ray. Then, another person, standing on firm ground on the outside the free falling elevator, will thus see a bend line.
From the above, it is clear that everything that has a gravitational field will bend light rays. Masses clearly come with a gravitational field. And because of the equivalence of mass and energy, $E=m c^2$, everything that has energy, has a gravitational field, too.
However, the energy density of a gravitational wave is tiny. So the bending effect will be tiny.
Nevertheless, the "shimmering" you refer to can be observed if the gravitational waves are humungous: this is just what LIGO did with gravitational waves stemming from the merger of black holes.
| {
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Deviation of free falling objects (Coriolis effect) using conservation of angular momentum I read this pdf on non inertial frame, in particular I have a question on the deviation of free falling object due to Coriolis effect.
Consider a ball let go from a tower at height $h$. The displacement due to Coriolis effect, calculated with formulas in Earth system, is $(4.19)$, after it there is explanation of the effect that uses the conservation of the angular momentum of the ball in a inertial frame.
$$x =\frac{2\sqrt{2}ωh^{3/2}}{3g^{1/2}} \tag{4.19}$$ Just before being
dropped, the particle is at radius $(R+h)$ and co-rotating, so it has
speed $(R+h)ω$ and angular momentum per unit mass $(R+h)^2ω$. As it
falls, its angular momentum is conserved (the only force is central),
so its final speed v in the (Eastward) direction of rotation satisfies
$Rv = (R+h)^2ω$, and $v= (R+h)^2ω/R$. Since this is larger than the
speed $Rω$ of the foot of the tower, the particle gets ahead of the
tower. The horizontal velocity relative to the tower is approximately
$2hω$ (ignoring the $h^2$ term), so the average relative speed over
the fall is about $hω$. We now see that the displacement $(4.19)$ can
be expressed in the form (time of flight) times (average relative
velocity) as might be expected.
But $$v_{average} t_{flight}=h \omega \sqrt{\frac{2h}{g}}$$
Which differs by $\frac{2}{3}$ from $(4.19)$. Is that due to the approximation made?
I also don't understand completely why the average relative velocity $v_{average}$ is taken to be half the relative velocity found. Isn't this valid only for constant accelerated linear motions?
| The error is just to consider an average speed $h\omega$.
When the particle is at height $z$, its horizontal (relative to the Earth) speed is $v=2z\omega$, ignoring higher order terms in $z$.
During the time interval $dt$ the particle falls $dz$ with vertical speed $u(z)$. Hence
$$dt=\frac{dz}{u(z)}=\frac{dz}{\sqrt{2gz}},$$
where $u(z)=\sqrt{2gz}$ can be obtained from Torricelli's formula.
The horizontal distance traveled during this $dt$ is
$$dx=vdt=2z\omega\frac{dz}{\sqrt{2gz}}.$$
Integrating from $0$ to $h$ we obtain the total horizontal displacement
$$x=\sqrt{\frac 2g}\omega\int_0^h\sqrt zdz=\frac{2\omega}{3}\sqrt{\frac{2h^3}{g}}.$$
| {
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Were the Michelson-Morley results a surprise? How unexpected were the Michelson-Morley experiment results?
Did physicists have theoretical reasons to predict that the speed of light would result to be invariant?
| It was a completely unexpected result at the time. The principle of the MM experiment hinged on the hypothesis that Maxwell's equations of electromagnetism were valid only in a special frame of reference called the aether frame.The speed of light was equal to its standard value only in this frame and its speed in any other inertial frame had to be given by the Galilean velocity transformation.It was thought that the aether was an all pervasive medium through which the Earth moved. So the claim was that, by the Galilean velocity transformation, the speed of light measured on the Earth should vary with the direction in which light traveled. This was what the Michelson-Morley experiment failed to establish. The negative result led to the postulate of the special theory of relativity that the speed of light is the same in all inertial frames.
| {
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How does the viscosity of a non Newtonian fluid (ooblek) affect its resistance to electricity? I know the conductivity of water is based on whatever is dissolved into the solution, hence pure water does not conduct electricity.
However, these ions in solution must also be free to move around.
If I keep a non Newtonian fluid in motion it will become hardened. Would this stop the ions from moving enough to impede it from conducting electricity?
| The thickening of oobleck and similar materials is due to a phenomenon called dilatancy. This happens because shearing the suspension forces the water to flow at very high shear rates through the restricted gaps between the solid particles, and that requires a very high shear stress.
However the water itself is not thickened in any way, and its electrical conductivity is unaffected. There will be no effect of shear rate on the conductivity for a suspension like oobleck.
| {
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Rolling without slipping in absence of friction force I'm confused about a rolling without slipping situation. Suppose to have a disk of radius $R$ on a floor, and exert a horizontal force at a certain distance $r$ from the center of mass.
I would like to see in which situation I obtain that the disk rolls without slipping without static friction force on the ground, that means
$$F=ma_{cm}$$
$$F r=I_{cm} \frac{a_{cm}}{R}$$
Which gives $$r=\frac{I}{mR}$$
So actually for any force I exert horizontaly there is a point of application of the force at a distance $r$ from the center, which does not depend on the force, for which I get the rolling with no slipping with no need of friction force?
If this is correct I'm confused about the physical meaning. In this case I get $a_{cm}=\frac{F}{m}$, which is strange because the motion of the disk is not only linear but also rotational.
Compare the situation with a point that moves because of $F$: its acceleration is $\frac{F}{m}$ but the point does not rotate. Here the disk moves at $\frac{F}{m}$ and rotates too!
How can the same force $F$ produce "more motion" (forgive this expression), such as a rotation, in one case?
| There's one thing in you question that doesn't quite add up and that is when you say exert a horizontal force at a certain distance r from the center of mass. If you are pushing a disk anywhere you push it (on the curved surface at least) is going to be a distance r from the centre of mass because the whole surface is a distance r from the centre of mass. I interpret the question to be asking at what height above the floor, h, do you want to push it so that it rolls without static friction. In that case your second equation becomes $$F\left ( h-R \right )=I_{cm}\frac{a_{cm}}{R}$$ since (h-R) is the distance perpendicular to F to the centre. Now, substituting the expression for $F=ma_{cm}$ from the first equation, and $I=\frac{1}{2}mR^{2}$ for a disk about the centre you get
$$ma_{cm}\left ( h-R \right )=\frac{1}{2}mR^{2}\frac{a_{cm}}{R}$$
which is independent of F and reduces to $$h=\frac{3}{2}R$$
Having said all that Duncan Harris's answer is probably the one you were looking for.
| {
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Atmospheric Pressure inside a closed room
Even though they’re too tiny to see, all the molecules of air in the atmosphere above your head weigh something. And the combined weight of these molecules causes a pressure pressing down on your body of 10,000 kg per square metre. This means that the mass of the air above the 0.1 square metre cross section of your body is 1,000 kg, or a tonne.
I would agree with the argument that the atmospheric pressure is a result of the weight of the air above me were I standing in an open area. I do not understand how, by this model of atmospheric pressure, the reason of atmospheric pressure can be explained in a closed room say.
Sourcehttp://www.physics.org/facts/air-really.asp
| From Pascal's law, we know that pressure is isotropic, which means that at a given location in a fluid, it acts equally in all directions. So, at a given location, the horizontal force per unit area acting on a small vertical surface is the same as the vertical force per unit area acting on a small horizontal surface.
Usually, a room is not hermetically sealed, so it is not totally separated from the atmosphere. Any connection between the room and the atmosphere allows the pressure to equalize (by air flowing in or out). As we said above, pressure acts horizontally also, so air can come through a vertical crack just as easily as through a horizontal crack. In a house, there are typically vents in the attic which allow communication with the atmosphere.
If the room were totally hermetically sealed from the atmosphere, then you could impose any air pressure you wanted inside the room. It would not have to match the outside atmospheric pressure. But, the forces on the walls could get pretty large between inside and outside as a result of the pressure difference, and you would have to be pretty careful so that the room didn't implode or explode. When tornadoes occur, the atmospheric pressure outside drops substantially, and people are recommended to open the windows (to allow the pressures to equalize) in order to avoid the windows blowing out (or even worse).
| {
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How much load does an aqueduct support? Recently, I found out about /r/InfrastructurePorn, and I found a particularly interesting photo of the Gouwe Aqueduct in Gouda, NE:
It seems like the bridge that is supporting the boat wouldn't be able to do it. Is the weight of the actual boat being supported by the aqueduct?
| Yes -- because if the boat displaces xx tons (however many tons it is) of water, then if the ship weren't there then the water would be (where the boat is now) -- and it's IMO clear that the weight of that water is being supported by the aqueduct.
I guess the boat looks relatively heavy, but it's mostly hollow.
Another way of looking at it is that the aqueduct needs to carry enough water (width and depth) to float a boat of that size.
If the boat is (eyeballing it) 60m long with 3 decks, that displaces (googles for motor yachts of comparable size) about 1000 tons order of magnitude.
I think that is a lot, compared to e.g. a big truck, which weighs only 30 tons -- but an ordinary road bridge obviously carries several trucks, and that bridge is short and stubby -- so its ability to carry the weight is not implausible.
| {
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Producing gravitational waves in labs Can gravitational waves be created on very small region of vacuum with quadruple movement of atom or subatomic particles?
| Yes, even tiny objects produce gravitational waves as they move. It's just that their gravitational waves will be way too tiny to measure.
Just consider that the recent gravitational wave detection was caused by 2 black holes weighing 36 and 29 times the mass of our sun. Even those enormous black holes only caused a tiny change a thousand times smaller than the width of a proton. Surely the movement of an atom will cause a far smaller gravitational wave. Too small to detect.
| {
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Meaning of a certain value at Fourier Transform Define the Fourier Transform of a certain signal in the time domain FT[$x(t)$]=$X(j\omega)$
$X(j\omega)$ = $\int$ $x(t)$ $e$^($j\omega$$t$)$ $dt
I'd like to ask what is the meaning of the value obtained from $X(j\omega)$ with certain frequency $\omega$
for example if we have a voltage signal of 1 Volt and found that
$$X(j100) = 100$$
- the unit is weird for me which will be voltage*sec- what does that 100 mean here?
Also: Why there is a factor of $\frac{1}{T}$ difference between the units of Fourier series and Fourier transform ?
I've asked at Signals processing/Math stack exchange but no answer
I've read this answer but it says:
The multiplication by $T$ in the limit is to account for the differences in definition between the Fourier series and Fourier transform: the series representation typically has a factor of $\frac{1}{T}$, while the transform does not. I don't know that there is a lot of insight to be gained via this analysis, but it shows that the series and transform representations are intimately related.
which didn't satisfy me.
|
I'd like to ask what is the meaning of the value obtained from X(jω)
with certain frequency ω
Consider for a moment, the synthesis equation where we 'construct' $x(t)$ out of a weighted 'sum' (integral) of the orthonormal basis functions of time: $e^{j\omega t}$
$$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}\omega\:X(\omega)\:e^{j\omega t}$$
Here, it is clear that $X(\omega)$ is the amount or weight of $e^{j\omega t}$ that goes into 'constructing' $x(t)$ from these basis functions. We haven't specified what $X(\omega)$ is but we assert that, for a large class of $x(t)$, there is an appropriate $X(\omega)$ such that the above holds.
As you may have already concluded, there is a way to find $X(\omega)$ given an $x(t)$ which is
$$X(\omega) = \int_{-\infty}^{\infty}\mathrm{d}t\:x(t)\:e^{-j\omega t}$$
which (left as a fun exercise for the reader) can be verified by substituting the later expression into the former expression.
Since, in the first expression, we're integrating with respect to $\omega$ which has units of $\mathrm{s}^{-1}$, it must be that $X(\omega)$ has the units of $x(t)$ multiplied by $\mathrm{s}$.
Why there is a factor of $1/T$ difference between the units of Fourier
series and Fourier transform ?
If we were to make a discrete approximation of the first expression, it would be something like
$$\tilde{x}(t) = \frac{1}{2\pi}\sum_{n = -\infty}^\infty \frac{2\pi}{T} \:X(n\frac{2\pi}{T})e^{jn\frac{2\pi}{T}t} = \sum_{n = -\infty}^\infty \:\frac{X(n\frac{2\pi}{T})}{T}e^{jn\frac{2\pi}{T}t} = \sum_{n = -\infty}^\infty \:a_n e^{jn\frac{2\pi}{T}t}$$
which is periodic with period $T$ (thus the tilde over). Using the usual method of integrating the product of both sides with $e^{-jm\frac{2\pi}{T}t}$ over a period $T$, we arrive at
$$\int_0^T\mathrm{d}t\:\tilde{x}(t)\:e^{-jm\frac{2\pi}{T}t} = Ta_m$$
since the integral on the right hand side vanishes for $n \ne m$. And so,
$$ a_n= \frac{X(n\frac{2\pi}{T})}{T} = \frac{1}{T}\int_0^T\mathrm{d}t\:\tilde{x}(t)\:e^{-jn\frac{2\pi}{T}t}$$
So, the $\frac{1}{T}$ comes from the fact that the $a_n$ are the 'sampled' Fourier transform divided by the period $T$ as shown above.
| {
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How is a mass, suspended vertically by two springs in parallel, kept stable? Consider a mass suspended vertically from above by two springs in parallel with different spring constants. Wouldn't the tension be different in each spring? How is this system kept in equilibrium?
| I've simulated the case
The two springs had the same initial length, and the block in the picture is in equilibrium. See how it is deviated towards the spring with bigger stiffness to decrease the resultant moment, and stop the rotation of the block.
| {
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Free Expansion Of and Ideal Gas We know that in free expansion of an ideal gas, no heat enters or leaves the system.
We also know that
$P_\text{initial}V_\text{initial}=P_\text{final}V_\text{final}$
is valid.
If heat exchange is zero, then we can call this process to be adiabatic.
Then why the following is not valid?
$P_\text{initial}{V_\text{initial}}^γ=P_\text{final}{V_\text{final}}^γ$
Also, if I am wrong above, are isothermal free expansion and adiabatic free expansion different?
|
Are isothermal free expansion and adiabatic free expansion different?
No. They are the same.
Your mistake is in thinking that $PV^\gamma = \text{constant}$ applies to a free expansion. That expression is for a reversible (i.e., isentropic) adiabatic process. A gas that has undergone a free expansion has more entropy after the expansion is complete than it did before the expansion started. Free expansion is not isentropic, and therefore $PV^\gamma = \text{constant}$ does not apply to free expansion.
| {
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What is the minimal G-force curve in 2-dimensional space? Given two parallel roads, which need to be connected, what shape of curve would produce the minimum overall horizontal G-force(s) on travelers?
Is it a $sin$ or $cos$ wave?
Is it a basic cubic function?
Is it something else?
I'm working on an engineering project, not actually involving roads, but the road analogy is easier to understand than my actual project (it involves more complex topics, like aerodynamics, which would just confuse the problem needlessly).
| The answer is two arcs. One arc with a constant gee loading in one direction and then flipping to the opposite direction. This is called the bang-bang method, and it is no very smooth, but the gee forces never exceed the specified maximum.
Given a path $y(x)$ the instantaneous radius of curvature at each x is
$$ \rho = \frac{ \left(1+ \left(\frac{{\rm d}y}{{\rm d}x}\right)^2 \right)^\frac{3}{2} }{ \frac{{\rm d}^2 y}{{\rm d}x^2} } $$
The lateral acceleration is $a_L = \frac{v^2}{\rho}$ so we are comparing paths using the parameter $\gamma = \frac{L}{\rho}$
Here are some possible curves (use $L$ for the transition length, and $W$ for the step width)
$$ \begin{align} y(x) &= \tfrac{W}{2} \sin \left(\frac{\pi x}{L} \right) & \text{harmonic}\\
y(x) &= \begin{cases} -\frac{L^2-W^2}{4 W} + \sqrt{ \left( \frac{(L^2+W^2)^2}{16 W^2}-\left( \frac{L}{2}+x \right)^2 \right)} & x<0 \\ \frac{L^2-W^2}{4 W} - \sqrt{ \left( \frac{(L^2+W^2)^2}{16 W^2}-\left( \frac{L}{2}-x \right)^2 \right)} & x>0 \end{cases} & \text{arcs} \\
y(x) &=- \tfrac{W}{2} \frac{{\rm erf}\left(\frac{2 \pi x}{L} \right)}{{\rm erf}(\pi)} & \text{smooth} \end{align} $$
Above ${\rm erf}(x)$ is the error-function
$$\begin{align} \frac{L}{\rho(x)} & = \frac{4\pi^2 L^2 W \sin\left( \frac{\pi x}{L} \right)}{ \left( \pi^2 W^2 \cos^2 \left(\frac{\pi x}{L}\right)+4 L^2\right)^\frac{3}{2}} & \text{harmonic} \\
\frac{L}{\rho} & = \pm \frac{4 L W}{L^2+W^2} & \text{arcs} \\
\frac{L}{\rho(x)} & = \frac{ 16 L W x \pi^\frac{5}{2} {\rm e}^\frac{8 \pi^2 x^2}{L^2} {\rm erf}(\pi)^2}{\left( L^2 {\rm e}^\frac{8 \pi^2 x^2}{L^2} {\rm erf}(\pi)^2+4 \pi W^2\right)^\frac{3}{2}} & \text{smooth} \end{align} $$
The peak for the harmonic is $\frac{L}{\rho} = \frac{\pi^2 W}{2 L}$ at $x=\frac{L}{2}$ which is always a higher value than the arcs solution.
The peak for the smooth is not easy to find analytically, but for some test cases I looked at it was much higher than the arcs solution.
| {
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Hamiltonian or free energy corresponding to 2+1D Kuramoto-Sivashinsky model I am trying to understand if the deterministic 2+1D Kuramoto-Sivashinsky equation
$$
\partial_t h =
-\nu \nabla^2 h - K \nabla^4 h + \frac{\lambda}{2} (\nabla h)^2,
$$
where $\nu$, $K$, $\lambda$ are constants and $h=h(x,y,t)$ is a time-dependent real field in two spatial dimensions,
can be seen in the light of a statistical field theory. It seems to me that this equation corresponds to a non-conservative system (similar to the Korteweg–de Vries equation, see wiki). Thus, my question is the following:
Is there a Hamiltonian or a free energy that has been written down and analysed for the deterministic Kuramoto-Sivashinsky? If not, has another similar equation been analyzed in terms of statistical field theory (phases, free energy, phase transitions, critical behavior etc.)?
| Comments to the question (v2):
*
*On one hand, the Kuramoto-Sivashinsky (KS) equation is a dissipative differential equation (DE). Each term has an even number of spatial derivatives. It's a non-linear version of the heat equation. Dissipative systems rarely have variational action formulations nor Hamiltonian formulations.
*On the other hand, in the Korteweg de Vries (KdV) equation, each term has an odd number of derivatives. The KdV equation is not a dissipative DE. It has both a Lagrangian and a Hamiltonian formulation. The energy is a conserved quantity.
| {
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May a point rotate about itself? Suppose we have two equivalent rigid cylinders. Cylinder 1 is moving (translating) with constant velocity of v. Cylinder 2 is rotating without slipping and its center’s velocity is constant and equal to v. So, the motion equations of both centers of cylinders are same (x=vt). If we consider centers of cylinders, their kinematics are same. My questions are: 1. What is the difference between these two points (centers of cylinders)? 2. Can we define rotation for a point about an axis that crosses that point?
| "What is the difference between these two points (centers of cylinders)? "
1) there is no difference in the motion of the points (lines that extend down the cylinder, actually). They each translate in space with the center of the cylinder as you might expect.
2) "Can we define rotation for a point about an axis that crosses that point?"
I assume your asking if a point can rotate about its 'own axis'. Since points are mathematical objects of zero dimension, with no internal structure, it is mathematically non-sensical to talk about them rotating 'about their own axis'; there is no internal axis for them to rotate about. Actually, points and the real number line are a little pathological/non physical when you consider them carefully... they are 0 and 1 dimensional objects, but everything you've interacted with is 3D, even if small or thin. Take a look at the wikipedia page for a point for more information.
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Eddington-Finkelstein coordinates, how to tell which is ingoing and which is outgoing? The Eddington-Finkelstein coordinates in case of Schwarzschild metric are defined as
\begin{align}
u&=t-r^*\\
v&=t+r^*
\end{align}
where $$r^*=r+2GM\ln\left|\frac{r}{2GM}-1\right|$$
The question is that how to understand which one is ingoing and which one is outgoing. Why $v$ is ingoing and $u$ is outgoing?
| $du,dv$ are light-like, i.e. they could, in principle, be viewed as some affine parameters of some light-rays. However, we will focus (in the spirit of the usual coordinate-nature analysis) on what is the nature of either $u,v$ constant. I.e., we want to know what is the nature of $u,v$ constant hypersurfaces and derive the nomenclature from this.
We will take it as fact that u,v can be viewed as parametrizing some light congruence.
Now the question is what is the nature of congruence parametrized by $u,v$. Let us consider some finite $t=t_0$ and $r_*=r_0$. If we are investigating $u,v$ constant hypersurfaces, then $t>t_0$ means surely $r_*>r_0$ for $u=t-r_*$ constant. I.e., the $u=const.$ surface is farther out at a later time. Hence, the lightcone is interpreted as outgoing with respect to the centre $r_*=0$ for $u=$const. and we call the respective Finkelstein coordinate the outgoing coordinate.
The opposite is true for $v=t+r_*$. $t>t_0$ means surely $r_*<r_0$ for $v=$ const., or that for $v$ constant we have a lightcone ingoing with respect to the centre $r_*=0$.
| {
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How do we know WHEN to get the result from quantum computers? So I always hear that you can't disturb quantum computers because this will ruin the particles superstate.
Well, how do we know WHEN to retrieve the result from the calculation?
How can we determine when the calculation is finished, without observing it?
| As explained on page 7 of Deutsch's first paper on quantum computation, you can set aside one of the qubits of the computer as a flag for completion of the computation. The qubit starts with the value 0 and the computer doesn't interact with it unless the computation terminates, at which point it sets the bit to 1.
| {
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Why current in series circuit is the same? I have read in the internet that the charges do not have any other path to go and they must go through the same in a series circuit,hence the current is same.
It was quite convincing but what confused me was:
"A resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element. Resistors act to reduce current flow..."(according to the Wikipedia).
This means that the resistors slow down the rate of flow of charges. By definition, electric current is the rate of flow of charges. Then must not the current be reduced in a resistor even when the amount of charge is same?
| What this means is that the resistor reduces the current compared to a circuit that didn't have the resistor in it.
Say you have circuit with a bulb and a battery in which 0.5 A of current flows. If you then introduce a resistor in series with the bulb the current everywhere in the circuit will be less than 0.5 A. The current entering and leaving the reistor will always be equal.
| {
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Time taken for a layer of ice to form The book I have gives the following derivation:
Let the temperature of the atmosphere be $-\theta$ and the temperature of the water be $0$.
Consider unit cross sectional are of ice, if layer of thickness $dx$ forms in time $dt$ with $x$ thickness of ice above it,
heat released due to its formation is $dx\rho L$ where $L$ is latent heat. If this quantity of heat is conducted upwards in time $dt$,
$$dx\rho L=K\frac{\theta}{x}dt$$
Therefore, the time taken $$t=\frac{\rho L}{2K\theta}(x_{2}^2-x_{1}^2)$$
What I don't understand is why the same amount of time should be taken for the heat to be conducted and for a new layer of ice to be formed. In other words, why is it that the next layer of ice forms only after the heat is released into the atmosphere?
| You assume that the temperature difference between the air at $-\theta^\circ C$ and the water directly under the ice $0^\circ C$ is constant.
So looking at the thermal conduction equation $\dot Q = K A \frac {\theta}{x}$ if you increase the thickness of the ice $x$ by a factor of two you reduce the rate of heat flow $\dot Q$ by a factor of two.
This is because $K, A$ and $\theta$ are constant.
So it will take twice as long to freeze a thickness of water $\Delta x$ when the thickness of ice is $2x$ than to freeze the same thickness of water when the thickness of ice is $x$.
Your analysis does not not include the additional but smaller factor of having to reduce the temperature of the water near the ice to $0^\circ C$.
If water behaved as most liquids it would not start to freeze until the temperature of all the water was $0^\circ C$ the heat being transported through the water by convection to achieve such cooling.
Since water is anomalous in that it has a maximum density at $+4 ^\circ C$, the water under the layer of ice has to be cooled to $0^\circ C$ by conduction of heat through the water and then the ice.
| {
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Twin paradox in special relativity: length contraction Can the concept of twin paradox be applied to length contraction as well? meaning that the twin which is in spaceship will have its meter rod "actually" contracted while he will see his brother's meter rod contracted which is in fact will be an "apparent" effect.
| This might or might not be responsive to the question you intended to ask:
Suppose you've got a meter stick. Over a period of time, I apply identical forces to the front and back ends of that meter stick, causing them to accelerate in the same direction. Therefore the entire stick, being a rigid body, accelerates in that direction. After a while, the forces stop, so the stick is now moving at a fixed velocity.
In your frame (the frame the stick was in to begin with), the length of the stick can't change, because we applied identical forces to the front and back, so the distance between the front and the back can't change. On the other hand, the Lorentz contraction tells you that the (now-moving) stick is shorter in your frame than it is in its own frame. This means the moving stick, in its own frame, is now more than a meter long. It has stretched.
(In its own frame, the moving stick says that it has stretched because the front started accelerating before the back did.)
But there is a limit to how far you can stretch a stick. So if you get the velocity high enough (no matter how gently you accelerate it to that velocity) the stick must snap. And of course all observers must agree that it has snapped. I think we can count that as a real effect, which no observer can dismiss as merely "apparent".
I'm still a little unclear on whether we can give a completely non-relativistic explanation of why the meter stick snaps. I once asked a closely related question here.
| {
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non constant acceleration problem
The acceleration of an arrow from a bow falls from $6000m/s^2$ to zero when it leaves the bow after travelling a distance $x=0.75m$. Assuming that this acceleration can be expressed by the linear equation $a=6000[1-(4x/3)] m/s^2$ determine the speed of the arrow when it leaves the bow.
How long does it take for the arrow to leave the bow?
to answer the first part of the question
$v dv = a dx$
$\int_U^V{v}$ dv = $\int_0^x{6000(1-4x/3)}$ dx
$\frac{(V^2 - U^2)}{2}$ = $2000(3x-2x^2)$
$V^2 = 4000(3x-2x^2) + U^2$
$V = \sqrt{4000(3x-2x^2) +U^2}$
substitute $x= 0.75$ and $U=0$ giving,
$$v=67.1 m/s$$
For the second part of the question I have got this far
$v=\frac{dx}{dt}$
$dt = \frac{1}{v} dx$
$\int_0^T{dt} =\int_0^x \frac{1}{v} dx$
$T = \int_0^x \frac{1}{\sqrt{4000(3x-2x^2) +U^2}} dx$
this integral is nasty, giving a complex solution but works out when I solve it giving the correct answer of $$t=0.0176s$$
Can anyone see a more efficient method for solving this problem? Is there a substitution I could use to make the integral easier? or am I missing an aspect of this question.
| This is the setup described in the equation:
The acceleration is defined in terms os the displacement of the bow $x$ by:
$$ a = 6000 \left(1 - \tfrac{4}{3}x\right) \tag{1} $$
So initially $x=0$ and when we substitute this into equation (1) we get $a = 6000 \text{ms}^{-2}$. When the arrow leaves the bow so $x=\tfrac{3}{4}$ and we get $a=0$. So far so good.
But suppose we choose a different definition for the variable $x$ as shown below:
So now $x$ starts at $\tfrac{3}{4}$m and when the arrow leaves the bow $x=0$. If we define $x$ this way then the equation for the acceleration becomes:
$$ a = -8000x \tag{2} $$
let's just check this: at the start $x = \tfrac{3}{4}$m and putting this into equation (2) gives $a = 6000 \text{ms}^{-2}$. When the arrow leaves the bow $x=0$ and equation (2) gives $a=0$.
So equation (2) gives us the acceleration with our redefined meaning for $x$. But equation (2) is just the equation of motion for a simple harmonic oscillator:
$$ \frac{d^2x}{dt^2} = -kx $$
So the motion of the arrow is going to be given by an equation:
$$ x = \tfrac{3}{4}\cos\left(2\pi \frac{t}{\tau}\right) $$
where you can calculate the period $\tau$ by solving equation (2).
| {
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Derivation of TKNN's main result from Kubo formula I have a question about a small but meaningful (to me at least) step in the original TKNN paper (http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.49.405). I understand the construction of the magnetic translation operators and the use of the Kubo formula to get to the following formula (Eq. 5):
\begin{equation}
\sigma_\mathrm{H} = \frac{ie^2}{2\pi h}\sum \oint dk_j \int d^2r \left(u^*\frac{\partial u}{\partial k_j} - \frac{\partial u^*}{\partial k_j}u\right).
\end{equation}
However, the following few lines read: "For nonoverlapping subbands $\psi$ is a single-valued analytic function everywhere in the unit cell, which can only change by an $r$-independent phase factor $\theta$ when $k_1$ is changed by $2\pi/aq$ or $k_2$ by $2\pi/b$. The integrand reduces to $\partial \theta/\partial k_j$."
If you can say $u_k(r) = \left| u(r)\right| e^{i\theta(k)}$ then $u^*\frac{\partial u}{\partial k_j} - \frac{\partial u^*}{\partial k_j}u = 2i\frac{\partial \theta}{\partial k_j}\left|u\right|^2$, and the rest of the results follow.
But why must there exists an $r$-independent phase factor when translating to the boundary of the magnetic Brillouin zone? The values of $k_1$ and $k_2$ are only defined modulo $2\pi/aq$ and $2\pi/b$ to begin with. In the nonmagnetic situation where you have the standard Bloch relations, you find $\psi_{k+Q} = \psi_k$. Clearly the difference in the case of the magnetic unit cell is related to the gauge potential, but I'm having trouble seeing the connection. Specifically, I'm trying to understand how the Berry's phase arrises in this derivation. Could anyone flesh out the origin of $\theta$ as it relates to this derivation in particular?
| The OP's main concern is about:
But why must there exists an r-independent phase factor when
translating to the boundary of the magnetic Brillouin zone? The values
of $k_1$ and $k_2$ are only defined modulo $2π/a$ and $2π/b$ to begin with.
Answer: for "$k_1$ and $k_2$ are only defined modulo $2π/a$ and $2π/b$", the wave function of $\psi(r;k_1,k_2)$ and $\psi(r;k_1+2\pi/a)$ and $\psi(r;k_1, k_2+2\pi/b)$ must describe the same physical states. Therefore, they must differ at most a global phase factor, that is, a phase factor at most depend only on $k$, not $r$.
Otherwise, suppose $\psi(r;k_1+2\pi/a, k_2)=e^{i\theta(r)}\psi(r; k_1,k_2)$, although the modulo $|\psi|^2$ is unchanged, however some physical quantity, such as the expected momentum, which is defined as $\left<\psi|-i\hbar\nabla|\psi\right>$ will change because the r-dependent phase factor.
| {
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Thermal de Broglie wavelength - definition The thermal de Broglie wavelenght is often defined by the formula
$$\lambda=\frac{h}{\sqrt{2\pi mkT}}$$
but equally frequently is it defined as de Broglie wavelength for a free ideal gas of massive particles in equilibrium, but in this case we obtain
$$\lambda=\frac{h}{\sqrt{3 mkT}}$$
Even though the second version has some theoretical predictions, in the literature the first one is used.
I also find papers where authors used approximation $\pi\approx 3$, and claim that $E_K=3kT$ instead of $\frac{3}{2}kT$.
My question is : What is the reason ? Why we used $2\pi$ instead of $3$ ? Is it motivated historically or we can define this value within some other theory using its formalism ?
| I see two ways in which the thermal de Broglie wavelength is defined. In both cases we can get it from the probability distribution and partition function of an ideal gas. We will consider a 3D ideal gas with non-relativistic dispersion.
First Way
Consider the partion function of an ideal gas:
\begin{align}
Z &= \int_{p_x=-\infty}^{+\infty}\int_{p_y=-\infty}^{+\infty}\int_{p_z=-\infty}^{+\infty} e^{-\frac{1}{2mkT}(p_x^2+p_y^2+p_z^2)}dp_xdp_ydp_z\\
&= \left(2\pi mkT \right)^{\frac{3}{2}}
\end{align}
Note that this has dimensions of momentum cubed. Noticing this, we can define the characteristic thermal momentum
\begin{align}
p_T = \sqrt{2\pi mkT}
\end{align}
We can consider the de Broglie wavelength of a particle with this characteristic momentum to get the first definition of the thermal de Broglie wavelength:
$$\lambda_T = \frac{h}{p_T} = \frac{h}{\sqrt{2\pi mkT}}$$
Second Way
Consider the average energy of an ideal 3D gas. This can be found from the equipartion theorem to be
\begin{align}
\langle E \rangle = \frac{3}{2}kT
\end{align}
A relation for the de Broglie wavelength is
$$
\lambda = \frac{h}{\sqrt{2mE}}
$$
Considering the de Broglie wavelength of a particle with energy equal to the average thermal energy of a 3D ideal gas we get the second definition of the thermal de Broglie wavelength:
$$
\lambda_T = \frac{h}{\sqrt{3mkT}}
$$
A Note on a Possible Third Way
A third way which would make sense would be to calculate the average de Broglie wavelength of all of the particles in an ideal gas:
$$
\langle \lambda \rangle = \iiint \frac{h}{p} e^{-p^2/2mkT} dp^3 = \frac{2h}{\sqrt{2\pi m kT}}
$$
We see that this is within a factor of 2 of the first definition.
Summary
The three approaches differ by factors of order unity so they all refer to similar length scales. In the end the thermal de Broglie wavelength is largely a notational convenience so we don't need to carry around factors of $\frac{h^2}{mkT}$ all over the place so we shouldn't worry too much about the pre-factor. But it is nice to know where the different conventions come from. Though it is largely a notational convenience it does clearly have physical significance since it is related to $\langle \lambda \rangle$.
I have never really seen the third way presented. I have seen the first way presented by far the most often. I think this is because the partition function appears all over the place so very commonly the specific factor $2\pi mkT$ arises so it is nice to give this quantity a name. The second approach may be presented more often in introductory approaches to statistical mechanics. This convention is most problematic because it very clearly relies on the specific problem of a 3 dimensional ideal gas.
| {
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gradient strength units in MRI This may not be the appropriate forum for this but seemed to be the closest.
I am trying to understand some concepts around MRI physics and it is common to use external magnetic fields created using gradient coils to manipulate the main magnetic field strength at different locations.
Now, the books talk in terms of gradient amplitude and the units they typically use is mT/m (microtesla/metre). I am not sure why there is this per meter as it is just the gradient amplitude should it not just be microtesla or teslas? Why is it defined per unit distance?
| A 'gradient' measures how quickly something changes with respect to something else. In this case, it's how much the magnetic field strength changes per unit length.
| {
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What is the best way to calculate impact time with collisions? I've been teaching myself physics and I've been wondering about the impact time in collision calculations. The scenario I've been using to learn is an object with a mass of 4000 kilograms colliding with a human being, while travelling at 17m/s. The object has a surface area the size and shape of a human elbow (which I very roughly guesstimated to be around 20cm2.
When calculating the force of this impact I need the momentum and the duration of the impact. The momentum is easy enough to calculate, but how is the duration of impact worked out? I know that it isn't referring to how long the objects are in contact, as this would mean that swords would harmlessly rub against a person if they were swung. I assume then, that the time is referring to how long it takes one object to impart the force of it's momentum into the other object.
How am I supposed to do this? The obvious way is to measure it, but given I'm an art student I can't exactly go around driving cars into people to measure how long it takes them to react to the impact. So far I've just been using .1 seconds, but I feel like this is far too slow.
| In general you need to establish some sort of stiffness, or more importantly an natural frequency for the system of two bodies. You can hear impacts and distinguish between slow thuds with fast pings.
For example if a short duration force has a harmonic shape (with frequency $f = \frac{\omega}{2 \pi}$) and peak force $F_{\rm max}$ then the total impulse is
$$ J = \int_{-\frac{\pi}{2 \omega}}^{\frac{\pi}{2 \omega}} F_{\rm max} \cos(\omega t) = \frac{2 F_{\rm max}}{\omega}$$
This means that the peak force is
$$F_{\rm max} = \frac{J\,\omega}{2} = \pi J \,f$$
where $J$ is the total change in momentum (impulse) and $f$ is the natural frequency of the impact (in Hertz). Typically the impulse is expressed in terms of the reduced mass of the two bodies $\mu = \frac{m_1 m_2}{m_1+m_2}$ and the impact speed $v_{\rm imp}$ and the coefficient of restitution $\epsilon$ : $$ J = (1+\epsilon)\, \mu\, v_{\rm imp} $$
Combined you have
$$ \boxed{ F_{\rm max} = (1+\epsilon) \mu\, v_{\rm imp}\; \pi \,f }$$
| {
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Decompression sickness In decompression sickness why does nitrogen gas come out as bubbles? Is it because under high pressure the nitrogen gas was converted into liquid nitrogen and during the process of coming to the surface (depressurisation)the liquid nitrogen evolved out as gas bubbles in the arteries and veins?
I have tried to compare the bubble formation to the concept that gas cylinders contain liquid LPG that comes out as gas.
A follow up question: If my concept is right i.e. the gases are being liquefied at high pressure (under water) how does our body deal with the liquid oxygens and carbondioxides? As far as I know there is no mechanism to transport liquid oxygen and carbondioxide in blood.
| Nitrogen is not liquefied in these conditions at all.
What happens is that when pressure increases, nitrogen's solubility in blood increases (this is a general property of gases: their solubility in liquids always increases with pressure)
When a diver decompresses properly (i.e. quite slowly) the nitrogen is released from the blood slowly (due to the now reduced solubility of the nitrogen) and without consequences to the decompressing diver.
But when the diver decompresses too fast or suddenly what happens is analogous to what happens when you suddenly open a carbonated drinks can: the sudden drop in pressure causes the carbonated drink to release its $CO_2$ suddenly and bubbles then form.
In the case of the suddenly decompressing diver, small nitrogen bubbles formed in the blood stream cause the symptoms of caisson disease.
| {
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Why does a pot start rotating when coffee is stirred inside? I usually make Turkish coffee as my morning coffee. I have a small somewhat rounded pot with handle on one side. I noticed that when I pour water in and start stirring, pot has a tendency to start rotating in the same direction as I'm stirring.
Why is that?
| You have made a model of a viscous fluid coupling which was used in a number of four wheel drive vehicles to transfer torque. The system relies in the fact that adjacent planes of moving liquid experience a viscous force between them.
| {
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Is static friction the only force causing a car to move (without sliding)? A car is moving without sliding means that the friction between wheels and the ground is static friction. This is the force causing an object to move forward, therefore, its direction is the same as the moving direction of the car.
My question is: For the horizontal forces acting on any moving(without sliding) car, $\ F_{fs}$ always oriented forward, what's the backward force to balance $\ F_{fs}$ so as to keep the car moving uniformly? Is that the Force produced by engines?
Just to clarify, I am referring to auto cars (with engines).
Any help or thoughts are appreciated!!!
| A free body diagram will show a car in motion has air drag force, gravity force and friction force on it. The net force keeps the car accelerated, decelerated, or moving at constant speed.
Friction force is due to relative motion between wheel and ground. Engine output spins the wheels (torque from power train system balances the torque produced from the frictional force).
Air drag force is proportional to car velocity to the power 2. It works opposite to the car moving direction. Zero air drag occurs when car is in stall. On horizontal surface, this is the only force on the car external opposite to the frictional force.
If the car is on slope, it can either move itself (down slope) or it need extra force from the frictional force.
There are many areas that consumes engine output such as frictions inside power delivery system and wheel sliding friction.
| {
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Why does the classical Doppler formula make a distinction between movement of the source and movement of the receiver? I've tried rewriting the Doppler formula to include only the relative velocity between the source and the receiver of sound waves. However, when I compare the results with the results of the formula that includes "velocity of the receiver" and "velocity of the source", they are not the same. Is that because the medium (the air) constitutes an absoulte reference frame when it comes to sound waves or is there a different explanation?
| That's correct. Imagine a sound emitter and receiver traveling at twice the speed of sound in the same direction with the emitter behind the receiver. Even though the relative velocity is zero, the sound will never reach the receiver because sound doesn't travel fast enough. At this point, the frequency calculated from Doppler shift formulas would have either diverged or gone negative, indicating a problem with reception.
Light in a vacuum, on the other hand, has a different formula precisely because there is no absolute reference frame. Here, only relative velocities matter.
| {
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Relativistic Mass and Potential Energy In the chapter 16–4 of the Feynman Lectures, Feynman employs a thought experiment to explain why "if two particles come together and produce potential or any other form of energy; if the pieces are slowed down by climbing hills, doing work against internal forces, or whatever; then it is still true that the mass is the total energy that has been put in."
It is clear in the case of kinetic energy and heat. After all it is expressed directly in the formula for mass:
\begin{equation}
\label{Eq:I:16:10}
m_u=\frac{m_0}{\sqrt{1-u^2/c^2}}.
\end{equation}
If two bodies come together in an inelastic collision, then we can see how the particles that make up the new body, increase its mass due to their increased kinetic energy (heat).
But suppose the two masses are two electrons and as they approach each other they slow down to a halt due to repulsion. If we imagine them at the moment in time in which their velocity is 0, their combined mass is still more than twice the rest mass of an electron. But where is it?
Is the Special Relativity formula for mass incomplete? is the extra mass hiding in $m_0$? Does it hide somewhere else?
| Energy is stored in electric and magnetic fields. There's a little bit stored at each point in space. It's proportional to the square of the electric field at that point. The total energy stored that way is just the integral (i.e. sum) of that the energy at each point.
As you move charges around, you change the field values. If you push two like charges together (i.e. electrons), the field between them goes up; that corresponds to an increase of energy in the field.
Summary: The energy you put into pushing them together was pushing against their repulsion, and went into increased energy of their electric field.
| {
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Spinning top fixed point I have seen many explanations about the movement of a spinning top. The explanations were in a varied level, from basic newtonian mechanics to Lagrangian formalism. But I do not understand why some people consider different fixed points. In same cases it is the point of contact with the surface and others consider some point in the "middle" of the spinning top. My question is whether this ambiguity is a misinterpretation (of those authors), a free choice to describe the movement or a difference caused from different spinning tops?
| When a top rotates, it rotates about its centre of mass. The centre of the mass is a point on the axis of rotation. Since the axis is also stationary as is the centre of mass, therefore all the points in the axis are eligible to be considered fixed about which the top is rotating.
Besides,I would prefer to use the term axis instead of a fixed point.
| {
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Luminosity of an accretion disk? With reference to Black holes in particular, how can you approximate the luminosity of an accretion disk? It is possible to quantify the temperature at a given point, but as the disk is not a black body, and this temperature is at a specific point, I am unsure how to equate this to luminosity - surely you could not do so using the Stefan-Boltzmann constant?
| The simplest answer is that you integrate the emission from annuli radiating at various temperatures. Explicitly, the luminosity per wavelength in this approximation is
$$L_\lambda = 2 \int_{r_{\rm in}}^{r_{\rm out}} 2 \pi r [\pi B_\lambda(r)] dr$$
where the overall factor of 2 is for the two sides of the disk, $2 \pi r dr$ is the area of each annulus, $B_{\lambda}$ is the Planck function, which depends on the temperature at the given radius, and $\pi B_\lambda$ is the flux that arises from integrating the thermal emission over solid angle. According to the model by Shakura and Sunyaev 1973, where this is all explained in much greater detail, the temperature roughly goes as $r^{-3/4}$. To get the bolometric luminosity, integrate $L_\lambda$ over wavelength.
This works well for radiatively efficient, thin disks.
The situation gets trickier for thicker disks that can't cool as efficiently.
| {
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Is a falling leaf an example of a chaotic system? Let´s assume is a wind still day in autumn. When a little change is made in the initial motion of a leaf at the time it falls off a tree, the resulting path of motion of the leaf is very different from the path that would develop if these changes wouldn´t have been made.
All the leafs though reach the ground within a maximum radius (wich is a function of the height of the tree) caused by chance effects. Can we, because we know that the leaves land within a certain area, still say that a falling leaf is a chaotic system? Or do we have to consider an infinite high tree, and consider the combined system of the air and the falling leave?
| It depends on your exact definition of chaos:
*
*We certainly have a strong sensitivity to initial conditions (butterfly effect), which is the one property of chaos everybody seems to agree upon.
*We do not have topological mixing.
*The falling to the ground is only a short-lived transient compared to the non-chaotic lying on the ground. So, at most, we have a chaotic transient.
To get something that is chaotic by all definitions (I am aware of), we would have to extend the system a little bit, for example:
*
*look at the entire tree;
*allow the tree to re-grow leaves;
*somehow deal with the leaves that reached the ground and letting their fate influence the system. For example, we could have a literal feedback process by having the rotting leaves act as a fertiliser that makes leaves on the respective side of the tree grow a tiny bit faster.
| {
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What is beyond gamma rays and radio waves in the electromagnetic spectrum? The electromagnetic spectrum is commonly referred to as consisting of;
radio-waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays - of increasing frequency from left to right.
But is it possible to get radiation of higher wavelength than radio waves, or lower wavelength than gamma rays - does it even exist? Or could they be produced?
Most interestingly, from the Planck–Einstein relation, $E = hf$, how high of an energy could you get for a very very high frequency radiation?
| Beyond radio waves are mega-giga-super-long waves as the wavelength approaches infinity; the longer they are, the more they dissolve into nothingness because once half the wavelength $\frac{λ}{2}$ gets more or less bigger than the radius of the entire universe, they cannot really interact with anything. In other words, they practically cease to exist and their energy approaches 0.
Beyond gamma rays are mega-giga-ultra-death rays that are more and more ionizing. Wavelength could be made arbitrarily short as long as it is larger than 0, but good luck gathering amounts of energy to generate such rays. Main difference is that "usual" gamma rays just knock off electron clouds from atoms, break bonds and cause radiolysis of matter, while mega-giga-ultra-death rays instantly turn any state of matter into hot plasma.
| {
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Why doesn't orbital body keep going faster and faster? If we consider the change in velocity during an infinitesimal interval of an orbit:
where body B is orbiting body A, we can see that the magnitude of the resultant vector (the green arrow) is greater than the magnitude of the original tangential velocity. Why doesn't this magnitude keep increasing indefinitely?
As I understand elliptical orbits, they speed up and slow down, but according to the diagram, I would expect them to keep speeding up monotonically.
(The answers to the duplicate question do not answer my question).
| Since you still seem puzzled I'll try a different tactic here:
You're showing the tangential velocity (A) and the radial acceleration (B) and adding them to get the green arrow. What you're missing is that this occurs in a gravity field. As the initial path climbs away from the object it's orbiting it loses velocity. This shows up as a third arrow pointing opposite of A and is of exactly the right amount to keep things balanced and your object peacefully in orbit rather than flying to infinity.
| {
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What is the difference between metallic hydrogen and plasma of hydrogen? As far as I know, metallic hydrogen and plasma seems share some common properties, eg:conduct electricity, nuclei share electrons, high temperatures...
So my question is, is metallic hydrogen in fact plasma state of hydrogen? If not, what is the difference between metallic hydrogen and plasma of hydrogen?
| In metallic hydrogen, the protons share an approximately fixed location relative to each other - energetically, a lattice is more favorable than an randomized state. Because of this, the substance is not a plasma - in a plasma, the positive and negative charges both flow freely.
Metallic hydrogen only exists at very low temperatures, and very high pressures. According to the phase diagram here, pressures above 2 Mbar, temperatures below 100 K (note the dotted line - this is a very fuzzy region of the phase diagram).
| {
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Why does the sign of Delta H indicate whether the reaction is exothermic or endothermic? If $\Delta H = Q + W$ (assuming constant conditions), then there are two terms involved in calculating $\Delta H$, only one of which measures heat gained/released. It is possible then for $\Delta H$ to be negative (if $W$ were very negative), even with $Q$ being positive? (And vice versa too.)
So why do we say that the sign of $\Delta H$ indicates whether a reaction is exo- or endo- thermic?
| The symbol $\Delta$ (here) simply means difference, e.g.:
$$\Delta H=H_2-H_1$$
So, enthalpy of the end state minus enthalpy of the initial state.
In the case of an exothermic reaction the system has lost enthalpy, so:
$$H_2<H_1$$
Thus, for an exothermic reaction:
$$\boxed{\Delta H=H_2-H_1<0}$$
It is possible then for $ΔH$ to be negative (if $W$ were very negative), even with $Q$ being positive? (And vice versa too.)
Yes.
| {
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Difference between inviscid and viscous flow In my lecture notes, I have a load of examples and I want to sort out which egs are viscous flow and which are inviscid flow. It is not always said if the flow is viscous or inviscid. Please can someone tell me what other things determines the difference rather than just the words.
I do know that a gas bubble inviscid flow since there is no viscosity but gas isn't even a fluid which confuses me.
| Quite simply, a viscous flow is a flow where viscosity is important, while an inviscid flow is a flow where viscosity is not important.
Gases and liquids alike are considered fluids and any fluid has a viscosity. So a gas bubble surely has a viscosity, albeit relatively low compared to some liquids; liquids are generally more viscous by a factor of 1000. Especially if the gas bubble is moving in a more viscous liquid, generally we must consider the viscosity of the liquid, but may neglect the viscosity of the bubble. The result of this is that no velocity gradients are present inside the bubble.
Determining if a flow is (in)viscid in my opinion is best characterized through the Reynolds number, $\mathrm{Re}$. If $\mathrm{Re}\ll1$, the flow may be considered viscous, i.e. Stokes flow. If $\mathrm{Re}\gg1$, the viscous forces may be negligble compared to inertial forces, much like in turbulence. Note that for $\mathrm{Re}\gg1$, if there are any boundaries in the flow, near any of those boundaries a viscous boundary layer may be formed, which is considered a viscous flow. So in reality, inviscid flow doesn't exist but is a useful model for certain applications.
| {
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Pressure on the sides of a container? Sorry if this is an incredibly basic question for these categories. Basically, I don't understand these types of problems. I'm sure it's something really simple I'm missing.
Let's say there's an open swimming pool with width $w$, length $l$, depth $d$, and density $\rho$ (equal to water's density).
So basically, I'm quite sure the formula I should be using for this is $P = \rho gd$. This turns into $\frac{F}{A}=\rho gd$
How would I find the force of the water exerted onto the sides ($w$ and $l$)? In a problem like this, what would the area $A$ represent? If I wanted to find the pressure on a $w$ side, would I use $w\times h$? I try this, but it doesn't work. I get $F = \rho gwh^2$. But this answer is double the actual answer. It seems like I'd need to integrate (that's what I tried first), but it didn't work either.
There has to be some really basic concept I don't understand.
| Pressure is indeed F/A however force varies upon the depth of the water therefore to find the force you need to
"divide each area into small bands of areas, each having its own pressure (rho)gh, then add up those little forces you have got to gain the total force on the wall"
Which this is actually the concept of integration
Therefore you will get (integrate)(rho)gh x wdh from 0 to h
So from this you would get 0.5rho g w h^2
If you aren't familiar with integration, look up distributed loadings, it gives the same answers with less time :D
| {
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De-merits of the application of Laplace's Equation to find electrostatics potentials QUESTION: What are the de-merits of the application of Laplace's Equation to find electrostatics potentials?
Our professor told us that the answer was as follows:
It can be used only when the charge distribution is at the boundaries
and not on the region between the boundaries.
But I don't think that it is correct. I think he has just reversed the answer. According to my opinion, the correct answer should be that:
It can only be used to calculate the electric potential at any point
inside the region between the boundary, as specified by the boundary
conditions.
Which one is correct?
Please help.
| Well, if there were charge between the boundaries, you would be solving Poisson's equation rather than Laplace's equation. However, boundary conditions for the potential function are also crucial, because there could always be distant point charges that modify the field in the region of interest, without changing the distribution of charge on the boundary.
| {
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Where are the photons coming from? Particles and Antiparticles can annihilate, and they are completely destroyed in the process, which creates photons.
From wikipedia:
An unstable atomic nucleus with an excess of neutrons may undergo β− decay
n → p + e− + -νe
neutron decays into proton, electron, and electron antineutrino.
Unstable atomic nuclei with an excess of protons may undergo β+ decay
p → n + e+ + νe
proton decays into neutron, positron, and electron neutrino.
So, combining these two reactions gives us
n → n + e- + e+ + -ve + ve
neutron decays into neutron, electron, positron, electron antineutrino, and electron neutrino.
When the electron and the positron or the electron antineutrino and the electron neutrino collide and are annihilated, photons are produced. So, taking it a step further
n -> n + ?γ
neutron decays into neutron and ? photons.
So my question is, sorry for the long lead-up, where are these photons are coming from? Also sorry if I broke physics :P
(Note: I know that in β− decay, the neutron actually emits a virtual W- particle which decays into an electron and an electron antineutrino, but thought that wouldn't be relevant. It's here if it is, though!)
(Another note: Sorry for the bad symbolization, SE doesn't seem to be accepting all of the characters)
| I had similar stupid doubt.
It's coming from binding energy.
The equation you give:
Unstable atomic nuclei with an excess of protons may undergo β+ decay
$$p → n + e^{+} + \nu_{e}$$
proton decays into neutron, positron, and electron neutrino.
How do you think a proton can be converted to neutron which has greater mass?
+ you get positron and neutrinos.
The energy comes from binding energy of nucleons in an multi-proton atom.
When you sum up to
n→ n + $\gamma$
You are actually showing energy released from binding energy.
| {
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estimating deviations from ideal gas behaviour How can one estimate the pressure at which argon atoms show deviations from ideal gas behaviour due to the finite size of the atoms?
I have tried Taylor expanding the hard sphere gas equation:
$$P'(V-b)=NkT $$
to get $P'=P(1+b/V)$ to first order, where $P$ is the ideal gas pressure. However, I don't know if this is the right approach or just what to do next really. Could someone point me in the right direction please?
| For a bit more empirical approach, try the compressibility factor. With this factor, the ideal gas equation becomes PV = znRT, where z is described in very great detail by https://en.wikipedia.org/wiki/Compressibility_factor.
For places on the "z plot" that differ substantially from a value of z=1, you will find that argon starts behaving more and more non-ideally. This typically occurs at a high value of reduced pressure, with a low value of reduced temperature, where reduced pressure is P/Pc and reduced temperature is T/Tc (subscript c represents critical pressure and temperature respectively).
| {
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Why do liquids exert pressure on the sides of a container? What makes a liquid push against the walls of a container if the liquid is completely static?
I was thinking a comparable situation would be a bin full of baseballs. Unless the balls were perfectly stacked they would be rolling off one another and the walls of the bin would stop them. Is it correct to assume the same is happening in liquid on a larger scale, or is something else going on?
It seems like if that was the case the pressure on the walls would be much less than on the bottom.
| The reason is because liquids don't have preferred directions.
It's true that if you squeeze a regular solid by pushing down on it, it'll push back up on your hand but it won't exert any force to the sides (though it might bulge out a bit). If you model a solid as a cubic lattice of masses connected by springs, this makes sense, because only the vertical springs get compressed. A solid has enough order to 'remember' which way you pushed on it.
A liquid doesn't have this long-range order: pushing down on a block of water just makes the water shear to the side. On a microscopic level, you can't push the atoms together only in one direction because directional correlation decays fast; all you can do is push them together in general. Then the liquid responds by pushing out in all directions too.
Your example with a bin of baseballs is in between the two cases, but I think it's closer to a solid. Most of the balls are locked in place by the weight of the balls above, making a lattice. If you pop a small hole in the side of the bin, the balls won't flow out, they're jammed.
| {
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Why does the interference pattern change with given relation when the source slit changes? Why should the dependence relation be like $${\frac{s}{S}}<{\frac{\lambda}{d}}$$ for the interference condition to be seen?
Where $s$ is the width of the source slit and $S$ is the distance between the source slit and the double slit. $\lambda$ is the wavelength of light and $d$ is the distance between double slits.
Why is it that the interference condition is not seen when the relation is equal or just greater than it?
| Answer
Interference pattern is formed by diffracted-on-double-slits beams. When your condition is not met beams of different orders of diffraction superpose and interference pattern becomes blurred and disappears.
Explanation
Angular spacing of the fringes:
$$(tan)\theta_{fringes} = \frac{\lambda}{d}$$
Angular spacing between two light beams (let's call them beam A and beam B) from borders of a source slit entering double slits:
$$(tan)\theta_{beams} = \frac{s}{S}$$
While $\theta_{beams}<\theta_{fringes}$ you can distinguish zero- and first-order-diffraction beams and interference pattern is visible.
When a relation between two angles tends to $\theta_{beams}=\theta_{fringes}$ first-order-diffraction of beam A superposes on beam B. And first-order-diffraction of beam B superposes on beam A. They become indistinguable and so does interference pattern.
When $\theta_{beams}>\theta_{fringes}$ different diffraction orders superposes and no interference patterns is observed.
| {
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Why does the 'Jacobian of at least one combination of $n$ functions shall be different from zero'? I've started reading The Variational Principles of Mechanics by Cornelius Lanczos; here is the concerned excerpt from p. 11:
The generalized coordinates $q_1,q_2,\ldots, q_n$ may or may not have a geometrical significance. It is necessary however that the functions $$x_1= f_1(q_1,q_2,\ldots, q_n),\\ .......................\\ .......................\\ z_N= f_{3N}(q_1,q_2,\ldots, q_n).$$ shall be finite, continuous and differentiable, and that the Jacobian of at least one combination of $n$ functions shall be different from zero. These conditions may be violated at certain singular points, which have to excluded from consideration. ...
While I could get that the functions must be 'finite, continuous and differentiable' but couldn't get the condition that the 'Jacobian of at least one combination of $n$ functions shall be different from zero'.
Can anyone tell me what is the necessity of this condition? What does this actually mean? Or why do the functions need to follow this?
| The conditions about
*
*(i) differentiability of the functions and
*(ii) the maximal rank of the corresponding rectangular Jacobian matrix
are regularization conditions imposed to simplify the mathematical analysis of the physical problem, in particular to legitimate the possible future use of the inverse function theorem. In the affirmative case, the functions are called independent. See also this related Phys.SE post.
Physical systems that do not meet these regularization conditions are more difficult to analyse.
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Maxwell velocity distribution, in 1D or otherwise I learned from my textbook that Maxwell's velocity distribution gives: $$v_{rms} =\sqrt{\frac{3kT}{m}}$$ $$v_{avg} = \sqrt{\frac{8kT}{\pi m}}$$
Presumably this is for a three dimensions. This confuses me because for one-dimensional distributions the values are $$v_{x,rms} =\sqrt{\frac{kT}{m}}$$ $$v_{x,avg} = \sqrt{\frac{2kT}{\pi m}}$$
The fact that the 1D rms is $\frac{1}{\sqrt{3}}$ times the 3D rms is intuitive (and the site explains why.) I don't see why the 1D average, though, is $\frac{1}{2}$ the 3D average. Is the site correct? And what is the reason?
| The definition of $v_{avg}$ for 3 dimension is $$v_{avg}=\int_0^\infty |v|f(v)dv$$ However, it needs to be clear that $$v^2=v_x^2+v_y^2+v_z^2$$
For 3D, the distribution can be derived from 1D distribution but a little different. $$f_{3D}(v) = \left(\frac{m}{2 \pi k T}\right)^{3/2} \exp\left(-\frac{m v^2}{2kT} \right)$$
After similar integration as above, you can get the same result as in your post.
The 3D average is not straightforward but you have to go through definition, which in general consistent with what we do in daily life.
| {
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Speed of Sound in matter So basically when it comes to the speed of sound, it is said that speed of sound in media is based on two main factors - 1)elasticity and 2)density from the formula
V= $\sqrt{E/\rho }$
where E is the modulus of elasticity and $\rho$ is the density.Now this represents that Speed of Sound is inverse to the density. Then how come is the speed of sound more in solids than in gases (which is a denser medium). I think it is because the elasticity might override the density part, but I am not able to understand the inverse relation cause it seems that closer the atoms are to each other (ie denser) there must be quicker flow of energy from one point to other. I agree that this might hinder the intensity, but how come does it affect the speed of sound?
| Adiabatic bulk modulus of air $=1.4\times 10^5$ Pa and Young's modulus of steel $=1.8 \times 10^{11}$ Pa.
Density of air $= 1.2 $ kg m$^{-3}$ and of steel $8050$ kg m$^{-3}$.
The interaction between the atoms within steel is via the bonds whereas the interaction in air are by molecules colliding with one another and limited by the speed at which the molecules move. So the speed of sound in steel is greater than that of air even though it is much denser.
What you have is two parameters one which is related to the restoring force (elasticity) and the other related to the mass of the material (density).
The molecules are vibrating and a greater restoring force means that they return back to their mean position quicker (potentially increasing the speed of sound) but if the mass is larger then this means that their return back to their mean position will be slower (potentially decreasing the speed of sound).
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Rotational movement at the equator What is the centripetal acceleration if you are moving with $100\frac{km}{h}$ on the equator if you are are moving a) east b) west?
Shouldn't the acceleration be the same given with $a_{cp}=\frac{v^2}{r}=\omega^2r$. Has the angular speed of earth $\omega_e=\frac{2\pi}{r}$ any influence on the cetripetal acceleration while moving east or west?
Calculate the Coriolison acceleration ($\vec{a_c}=\frac{\vec{F_c}}{m}$value and direction) if you move with $100\frac{km}{h}$ a) east b) west c) north d) south.
Using $\vec{F_c}=-2m(\vec{\omega}\times\vec{v})$ we get that $\vec{a_c}=-2(\vec{\omega}\times\vec{v})$. It's not that hard to calculate the value for a) and b) because $\vec{\omega}$ and $\vec{v}$ are perpendicular. In a) we have that the direction is perpendicular away from the surface of the earth and in b) perpendicular to the center of the earth. But what happens in c) and d)? The angle changes so the value changes also over time... The direction stays the same so thats not a problem to calculate. Can it even be calculated or?
| For the first part of your question, you have to realize that your net velocity (the one that you plug into the expression for centripetal force) is the vector sum of the surface velocity and your velocity relative to the surface. If you were running West as fast as the earth turns East, you would "stay in place" and the sun would appear to stop moving in the sky. At that point there would be no centripetal force relative to the earth (although there would still be some relative to the sun...)
As for Coriolis acceleration; you have the expression. When you travel due North or South on the equator, the cross product is zero: there is no Coriolis force. Move away from the equator, and a force appears that is proportional to the sine of the latitude. The direction of the force changes as you cross the equator - if you look at global wind patterns, you will see that the two are related.
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What are the differences between the differential and integral forms of (e.g. Maxwell's) equations? I would like to understand what has to be differential and integral form of the same function, for example the famous equations of James Clerk Maxwell:
How to know where to apply each way?
Excuse the ignorance, but always confused me the head.
Edit
Remembering that, I know that the concept of integration and derivation, is something like:
| I would like to point out also, the differences between the Maxwell-Faraday equation, and Faraday's law itself ( emf = -d/dt(magnetic flux)). Faraday's law, combined with the Maxwell-Faraday equation, can show that the total emf due to ANY change in flux is equal to Emf =int (e+Vxb).dl
which corresponds to transformer emf, and also motional emf (due to magnetic force)
The Maxwell-Faraday equation is only due to an induced electric, but Faraday's law is due to transformer + motional emf.
Sorry if this isn't relevant, but I thought it was worth mentioning since this also includes the Lorentz force which is just as important as Maxwell's equations. Actual derivation here:
https://en.wikipedia.org/wiki/Faraday's_law_of_induction
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What is the nature of equilibrium between water and air in a sealed container that is heated above 100C? In the case of a single substance (water), looking at the phase diagram is enough to conclude what happens upon heating:
(source: WolframAlpha)
But what if I have water and air (or some other gas) in the same sealed container? Would the air saturate at some point and prevent further water from turning into gas?
(Wolfram have a "Heating Water and Air in a Sealed Container" demonstration)
EDIT:
According to Wikipedia:
Superheated water is stable because of overpressure that raises the
boiling point, or by heating it in a sealed vessel with a headspace,
where the liquid water is in equilibrium with vapour at the saturated
vapor pressure.
But is there a quantification of this "equilibrium"?
| Let's say that you have a vessel containing water and air and you start heating it. The temperature of the water and air inside will start to rise and so will the pressure, because the air would like to expand (but volume is fixed and water is almost incompressible). Since the boiling point of a substance depends on both pressure and temperature (for example water can boil at ambient temperature in a vacuum), this will prevent water from boiling until its vapor pressure overcomes the air pressure. This is how a pressure cooker works.
When vapor pressure overcomes the air pressure, if nucleation can occur (and this is almost always the case in everyday ife) the water will start to boil, part of it will be converted into vapor and the pressure will rise again. Now the external pressure will be equal to the vapor pressure and water won't be able to boil anymore.
At a certain point, if the container doesn't break because of the high pressure, you will reach water's critical point at $374$ °C - $218$ atm. Above this point, water will cease to exist as two separate phases and a single gaseous phase will be present.
| {
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Two Black Holes held stationary by EM forces If two black holes with large enough mass (so that the tidal forces are minimal and the intersection is large) that are held apart by like charges (saddle point stability). Imagine the black holes in a vacuum placed such that there event horizons are overlapping. Is this overlapping space transversable? Why not if the answer is no?
| Traversable - Overlapping (actually intersecting) region would not be Traversable even if the gravity at some parts of the region may be zero. For exampple, between earth and moon, gravity will be zero at some point. That does not mean something in that region can go out of earth/moon system. As soon as an observer leaves that region, it either falls towards moon, or towards earth, or towards that region.
Same thing would apply in case of two black holes but with extreme forces/speeds.
If you mean "intersecting" when you say "overlapping", then - overlapping EH does not necessarily mean a single black hole. EH is nothing physical, it is just a region of space around singularity.
Suppose, both the black holes can, each hold necessary amount of charge, and suppose the charge does not abandon its normal property when fallen into a singularity, then it should be possible for EM forces to counter gravity, even if their EH are intersecting, but the singularities have to be outside the EH of one another.
Intersection of event horizons would not cause faster than c condition for the singularities. Faster than c would be caused only when something physically enters event horizon. Two event horizons are not physical, singularities are physical which are still outside each other's event horizons.
| {
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Modern interpretation of wave-particle duality As far as I understand, in the early days of quantum theory there was quite a lot of debate over how to interpret what it meant for a quantum mechanical object to exhibit both wave-like and particle-like properties.
Is it correct to say that the modern interpretation (as in the one arrived at at the end of the original construction of quantum mechanics) is that there is an intrinsic uncertainty in the measurement of properties of a quantum mechanical object, such as its position, momentum, etc. As such one can at best describe the state that it is in by a wave function that encodes all the statistical information about the possible values of its observables. Hence there is no wave-particle duality - the wave-like properties arise due to the intrinsic uncertainty arising in the measurements of a particles physical observables.
Is something like this a correct understanding at all?
| I think what you wrote is fine, and interpretation matters a lot to answer this question. However, I think the notion of "wave particle duality" is still meaningful. There are two ways a wave function can evolve, via time evolution given by the Schrodiner equation, or randomly via external measurement. When not being measured, a particle obeys the wave equation. When it's position is measured, it is found all in one place, as a pointlike Dirac delta function. The word "wave" in "wave particle duality" refers to the deterministic wave, while the word "particle" refers to the random measured particle.
| {
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Why are the electric force and magnetic force classified as electromagnetism? I confuse the four kinds of fundamental interactions, so I think the electric force and magnetic force should not be classified as a big class called electromagnetism.
Here is my evidence:
*
*The Gauss law of electric force is related to the surface integration but the Ampere's law corresponds with path integration.
*The electric field can be caused by a single static charge while the magnetic force is caused by a moving charge or two moving infinitesimal current.
*The electric field line is never closed, but the magnetic field line (except those to infinity) is a closed curve.
| The classical electromagnetic effect is perfectly consistent with the lone
electrostatic effect but with special relativity taken into consideration.
The simplest hypothetical experiment would be two identical parallel
infinite lines of charge (with charge per unit length of $ \lambda \ $
and some non-zero mass per unit length of $\rho \ $ separated
by some distance $ R \ $. If the lineal mass density is small enough
that gravitational forces can be neglected in comparison to the electrostatic
forces, the static non-relativistic repulsive (outward) acceleration (at the instance
of time that the lines of charge are separated by distance $ R \ $)
for each infinite parallel line of charge would be:
$$ a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} $$
If the lines of charge are moving together past the observer at some
velocity, $ v \ $, the non-relativistic electrostatic force would appear to be
unchanged and that would be the acceleration an observer traveling along
with the lines of charge would observe.
Now, if special relativity is considered, the in-motion observer's clock
would be ticking at a relative rate (ticks per unit time or 1/time) of $ \sqrt{1 - v^2/c^2} $
from the point-of-view of the stationary observer because of time dilation. Since
acceleration is proportional to (1/time)2, the at-rest observer would observe
an acceleration scaled by the square of that rate, or by $ {1 - v^2/c^2} \ $,
compared to what the moving observer sees. Then the observed outward
acceleration of the two infinite lines as viewed by the stationary observer would be:
$$ a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} $$
or
$$ a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho} $$
The first term in the numerator, $ F_e \ $, is the electrostatic force (per unit length) outward and is
reduced by the second term, $ F_m \ $, which with a little manipulation, can be shown
to be the classical magnetic force between two lines of charge (or conductors).
The electric current, $ i_0 \ $, in each conductor is
$$ i_0 = v \lambda \ $$
and $ \frac{1}{\epsilon_0 c^2} $ is the magnetic permeability
$$ \mu_0 = \frac{1}{\epsilon_0 c^2} $$
because $ c^2 = \frac{1}{ \mu_0 \epsilon_0 } $
so you get for the 2nd force term:
$$ F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R} $$
which is precisely what the classical E&M textbooks say is the magnetic force (per unit length)
between two parallel conductors, separated by $ R \ $, with identical current $ i_0 \ $.
| {
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Question on doing the integral for Fermi golden rule Today in the lecture, my professor did something which confused me
As an example, we consider the photoelectric effect, in which an
electron bound in a Coulomb potential is ionized after interacting
with an external electromagnetic field. [...] The absorption rate is given
by
$$ \Gamma_{i\rightarrow
f}=\frac{4a_{0}^{3}e^{2}}{m^{2}\pi\hbar^{4}c^{2}}\frac{(\mathbf{A}_{0}\cdot\mathbf{p}_{f})^{2}}{(1+(p_{f}a_{0}/\hbar^{2})^{4}}\delta(p_{f}^{2}/2m-E_{i}-\hbar\omega)
$$
This gives us the rate for a precise final momentum $p_{f}$ .
Typically, what we want to know is the rate of electrons detected in a
solid angle $\mathrm{d}\Omega$
$$
\frac{\mathrm{d}\Gamma}{\mathrm{d}\Omega}=\int_{0}^{\infty}p_{f}^{2}\,\mathrm{d}p_{f}\,\Gamma_{i\rightarrow
p_{f}} $$
Doing the $p_{f}$ integral, we have $$
\frac{\mathrm{d}\Gamma}{\mathrm{d}\Omega}=\frac{4a_{0}^{3}e^{2}p_{f}}{m\pi\hbar^{4}c^{2}}\frac{(\mathbf{A}_{0}\cdot\mathbf{p}_{f})^{2}}{(1+(p_{f}a_{0}/\hbar^{2})^{4}}
$$
where $p_{f}=\sqrt{2m(E_{i}+\hbar\omega)}$.
but I don't get how the integral just multiplies a term $m \, p_f$ to the first expression. I thought that the $\delta$ function replace $p_f$ with $\sqrt{2m(E_{i}+\hbar\omega)}$.
| The delta function $\delta(x)$ has unit area, but the function $\delta(2x)$ is "half as wide" and thus has half as much area; thus you can pick up extra factors from 'how fast' you cross the peak of the delta function. The general identity is
$$\delta(f(x)) = \sum \frac{\delta(x-x_i)}{\big| df/dx|_{x=x_i} \big|}$$
where the $x_i$ are the roots of $f$. In this case the factor you pick up is
$$p_f^2 \left(\frac{d}{dp_f}(p_f^2/2m - E_i - \hbar \omega) \right)^{-1} = p_f^2 (m/p_f) = m p_f.$$
| {
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What does it mean that the laws of physics are time reversible?
The Universe, as far as we can tell, only operates according to laws
of physics. And just about all of the laws of physics that we know are
completely time-reversible, meaning that the things they cause look
exactly the same whether time runs forward or backward.
*
*http://www.sciencealert.com/physicists-just-found-a-link-between-dark-energy-and-the-arrow-of-time
I am not sure I understand this, how can we reverse a law in time? So for example how does a time-reversible law and a time-irreversible law look like? What is the difference between them?
| Only the mathematical formulas are time reversible. That does not mean that the law itself is reversible. You can understand it this way - Every law/formula has an implicit condition that says - "time flows only forward".
Other way to understand can be that when the coffee cup falls, universe (gravity in this case) makes the cup fall. There is no force/law that makes the cup rise. It becomes even more complex when you consider the process of the cup breaking, and coffee spilling. i.e. there are natural forces/laws that cause the cup to break but there are no natural forces/laws that cause the broken cup to re-construct. Even though, the mathematical formulas may allow the backward flow of events, there are no such forces that would cause it.
| {
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What does a voltmeter actually measure? For time varying fields (even quasistatic ones) the electric field is given by
$${\bf E} = - \nabla \Phi - \frac{\partial {\bf A}}{\partial t}$$
So what does a voltmeter measure? Does it measure a difference in $\Phi$ between two points $a$ and $b$, or does it measure $\int_a^b {\bf E} \cdot d{\bf l}$ or does it measure something else?
| A voltmeter does not measure $\Phi$ (the electrostatic potential), nor the difference $\Phi_A - \Phi_B$ where $A$ and $B$ are arbitrary points in a material. It also does not measure $-\int \vec E \cdot d\vec l$.
A voltmeter measures the difference in the electrochemical potential between point $A$ and point $B$, divided by the elementary charge. This corresponds to $\Delta V = -\int (\vec E +\nabla \mu/e) \cdot d\vec l$ where $\mu$ is the chemical potential.
In some cases, it may correspond to $-\int \vec E \cdot d\vec l$, or it may measure exclusively $\mu/e$. The general case is a mix of both.
An example given by Apertet et al. is the one of a p-n junction at equilibrium. In that case even though there is a built-in electrostatic potential $V_\text{in}$, a voltmeter would read a nil voltage because $eV_\text{in}=\nabla \mu$ and so $\tilde{\mu}=0$ where $\tilde{\mu}$ is the electrochemical potential.
References: Solid State Physics by Ashcroft and Mermin (1976) page 257. Apertet et al., Riess and this university website.
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Where does $\hat{P}\psi(x) = -i\hbar \partial_x \psi(x)$ come from? It's a very basic question, where does the relation $$\hat{P}\psi(x) = -i\hbar \partial_x \psi(x)$$ for any square integrable $\psi(x)$ come into existence? Some texts I found states that the above relation comes as a consequence of momentum being defined as generator of translation. But what is the basis of this definition? If momentum were defined to be generator of other form of symmetry, then it wouldn't have had the form as it does now.
In some other text, it's the other way around. Namely the action of momentum on a wavefunction is defined to be $$\hat{P}\psi(x) = -i\hbar \partial_x \psi(x)$$ and thence it leads to momentum being the generator of translation.
Which one is the correct one? How was such action of momentum on wavefunction historically developed?
| That $\hat{P} = - i \hbar \partial_x$ generates translations comes from a straight-forward computation: if $\psi$ is continuously differentiable, and $\Psi$ as well as its derivative are square integrable, then you can prove that
\begin{align}
i \frac{\mathrm{d}}{\mathrm{d} y} \bigl ( \psi(x - y) \bigr ) \big \vert_{y = 0} = - i \partial_x \psi(x)
\end{align}
holds, and you write $\mathrm{e}^{- i y \cdot \hat{P}} \psi(x) = \psi(x - y)$.
The best physical motivation in my opinion why $\hat{P}$ should be called “momentum” (operator) is via a semiclassical limit using standard techniques. You can use Wigner-Weyl calculus to show that if the potentials vary slowly compared to the wavelength of your wave function, then
\begin{align}
\hat{P}(t) = \widehat{p(t)} + \mathrm{error}
\end{align}
holds true, i. e. the Heisenberg observable $\hat{P}(t)$ associated to momentum is approximately equal to the quantization of the classically evolved momentum $p(t)$. You can make similar arguments for position, angular momentum and other observables. Making this precise is quite difficult.
A simplified, but in my opinion excellent explanation can be found in Ehrenfest's 1927 paper. Unfortunately, most quantum mechanics text books I have seen do a very bad job explaining this point (perhaps because they can't read Ehrenfest's paper, it is written in German), though.
| {
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Accelerations of Cylinders I got a box hanging from a rope that passes through an ideal pulley that attaches to the superior part of a Cilynder and makes it rotate. An exercise asks me two things:
a) determine the magnitude of the acceleration of the block
b) what is the magnitude of the acceleration of the center of mass of the Cilynder?
My question is: shouldnt the acceleration of center of mass of the Cilynder be equal to the block? Since angular acceleration= acceleration/radius
| If you see on the cylinder as a wheel then note that its center moves twice slower than top point attached to the block. Same is for acceleration.
In each moment the cylinder rotates around the point of its touch to the table, so radius from touch-point to the center is twice less than radius to the cylinder top point.
| {
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Why would a spinning space station create a centrifugal force on an astronaut rather than simply spinning around him/her? We often see films with spinning space station that create artificial gravity by having the astronauts pulled outwards by centrifugal force.
I'd like to know if this would really happen, and if so, why is the following scenario not true:
*
*Take an astronaut in open space. He doesn't move.
*Put a big open spinning cylinder around him - surely he still doesn't move.
*Close the cylinder. I still see no reason for him to be pulled outwards.
| In your scenario, your 3 statements are correct, and if nothing changes, your astronaut will not move from its spot as the wall of the cylinder moves past him. However, if somehow the astronaut "attaches momentarily" to the cylinder wall (the floor), then he will acquire the tangential velocity of the spot he attaches to, and this tangential velocity is what keeps him "attached" to the cylinder wall (the floor).
| {
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Non-abelian current commutators There many articles, in which non-abelian current commutators are computed. The general result is that quantum corrections lead to additional term in commutator
$$[J^a_\mu (x), J^b_\nu (y)] \delta (x^0 - y^0) = [J^a_\mu (x), J^b_\nu (y)]_{classical} \ \delta (x^0 - y^0) + A_{\mu \nu} (x, y),$$
where $A_{\mu \nu}$ is anomalous term and is proportional to $\partial_k \delta (x-y)$.
This result is obtained from calculation of diagrams. But is there any easy way, to show that quantum corrections lead to additional term proportional to $\partial_k \delta (x-y)$ with undefined coefficient (functional of gauge fields)?
| Restrictions can be imposed on the anomalous terms from general considerations even without fully solving the Feynman diagrams. In fact, the restrictions on the Schwinger term in the commutator given in the question are explicitly described in detail by Roman Jackiw in his review article: Field theoretic investigations in current algebra (section 2.2.).
His arguments are summarized here:
Since the non-Abelian charges are the generators of symmetry, they must satify the following (equal time) commutation relations with the currents:
$$[Q^a, J_i^b(0)] = f_{ab}^c J_i^c(0)$$
The local versions compatible with this relation must have the forms:
$$[J_0^a(x), J_i^b(y)] = f_{ab}^c J_i^c(x)\delta(x-y)+S_{ij}^{ab}(x)\partial^j\delta(x-y)+...$$
Where the dots indicate the possibility of higher order delta function derivatives. In addition, the function $S_{ij}^{ab}$, must satisfy conservation relations:
$$\partial^jS_{ij}^{ab} = 0$$
to satisfy the transformation equation after integration over $y$.
Now, as Jackiw mentions in equation (2.18), the commutator of the charge density with the $00$ component of the energy momentum tensor can be evaluated using the canonical commutation relations:
$$[\Theta^{00}(x), J_0^a(y)] = \partial_{\mu}J_{\mu}^a\delta(x-y)+J_i(x)\partial^i\delta(x-y)$$
Finally commuting the $00$ component of the energy momentum tensor with the density current commutator results that the only possible Schwinger term should be the one with a single delta function derivative.
| {
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Does Heat Cause Time Dilation? Since heat is defined as the movement of molecules, and because of relativity time slows for faster moving objects, would a hot object be in a slower time frame then a cooler object, because the hot objects molecules are moving faster?
| Just to add some influences of heat on a stationary clock on the surface of a star.
Heat is energy, and energy increases the energy-momentum tensor, so effectively increases the mass. With everything else remaining equal, this would increase time dilation.
On the other hand, heated bodies increase their volume. The same mass over a greater volume leads to less time dilation.
| {
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Why would two inertial frames be distinguishable, if the coordinate of an event perpendicular to the motion would be different? This question arises from the classical example: imagine a laboratory frame, and a space shuttle frame, the space shuttle moving in the laboratory frame with a constant velocity, let's say in the $x$ direction. Then the $y$ and $z$ coordinate of an event should be the same in both frames, otherwise we would have an experimental method to distinguish between the two frames.
What I would like to know is:
*
*What do physicists mean by inertial frames not being distinguishable? @jim and @ACuriousMind's remarks make this clear, that the equivalence of inertial frames is no more and no less than physical laws taking the same form in all inertial frames. Also @ACuriousMind pointed out the "distinguishability" is not a good choice of words, with which I agree, because this is what partly causes my confusion, since there is of course most of the time some way to somehow tell two inertial frames apart. (A very trivial example would be that one looks like a space shuttle, and the other looks like a house.) However, my second question remains.
*Why the difference in the $y$ coordinate (and, in fact usually many other quantities in the direction of the relative motion, like force, momentum, etc.) would allow an experimental method to differentiate between inertial frames, while measuring the difference between the $x$ coordinates is not considered to be such a method.
| What is meant is that physical laws are the same between (inertial) reference frames so that if you observe two bodies undergoing an elastic collision then you will experimentally determine that the momentum before the collision is the same as the momentum after the collision. An observer in another frame will also note that in his reference frame, the momentum before the collision will be the same after. However, if a person in the first frame communicated the values of the momenta of the particles, they would be different to his values.
| {
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Effect of a nuclear weapon on the Moon In some science fiction novel, there is the scenario of using nuclear weapons to deflect an Earth-approaching asteroid.
But the question is, is a nuclear weapon really as useful on the moon or an asteroid as on Earth?
The point is that, on the Earth, there is the atmosphere. The energy released by the bomb is absorbed by the atmosphere completely and induces huge blast.
But on the Moon, there is no air to trap the energy. Most energy released will simply escape into the universe.
| Indeed, a nuclear bomb works a bit differently in a vacuum than in the atmosphere. If you want to generate momentum, an atmosphere or some material with a low boiling point (e.g. ice) is probably better than bare rocks in vacuum, because all the heat will be converted directly into gas with a high momentum, without "wasting" energy on heating and evaporating a solid.
There is a video on Youtube on this topic: The Use of Nuclear Explosives To Disrupt or Divert Asteroids, by David Dearborn. Some calculation examples for an astroid with 1 km diameter start around 45 minutes into the video.
Some notable facts:
*
*You generally need a really big bomb (megatons) and for a big asteroid, you may only get a few cm/s of change in velocity, which is enough to prevent an Earth collision if you do it years in advance.
*Bombs that produce a lot of energy as neutrons may work better than the ones that release the same energy as x-rays, because neutrons penetrate deeper into the asteroid surface.
*What will happen depends on the structure of the asteroid. Under some circumstances, you may get some mass ejection from the opposite side of the asteroid, which will counteract the momentum generated at the impact side. For other scenarios, the asteroid may break up in small pieces, some of which are still likely to hit the Earth.
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What is more basic thing to explain phenomenons : Energy or Force? You know many physical happening are explained by considering forces or we can explain these phenomenon using energy (like energy conservation, etc.).
so i was wondering that explaining the physical phenomenon using energy method is more fundamental or explaining through considering forces?
Explaining things differently helps us to view thing differently. So either one of the methods(energy and forces) can gives us more broader vision, i.e. explanation of one phenomenon (through one of the method) can also be applied for the explanation of some other phenomenon(with some minor changes as required) and this is what i mean by fundamental.
thanks.
| As @CuriousOne quoted, the concept of energy and force are both fundamental. The choice depends on how you need to analyse the system. As a simple example, consider a collision between two identical particles. One is at rest and the other has a definite momentum. Obviously, the one at rest is set to motion. There the momentum is transferred from the colliding body to the body at rest. How momentum is transferred? Through a force. How a force helps in transferring the momentum? By transferring the kinetic energy.
All this means is that force and energy are correlated. How to analyse a system is purely the choice of a person. Whatever way you analyse, both tells you the same state of the system.
However energy is the much more basic way that can be used since it just fit in all theories so nicely. No worries. Especially, when we deal with quantum physics, there is only pure energy. Physics is fundamentally a study of energy transactions.
There are four fundamental forces in nature. They have there own fields. Field is a region of influence where it is well defined. It's the same where it's energy can be seen. The four forces are termed interactions. Interaction is possible only if they could transfer energy in some way. Once we realize that they interact, we say that they could exert force on each other. So, energy is more fundamental to use.
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Why does Bernoulli's equation only apply to flow along a streamline that is in viscid, incompressible, steady, irrotational? I am learning about hydrofoil on this website.
In a later video I watched, I learned that in the process of deriving Bernoulli's equation, $$constant=P/d+gh+1/2v^2$$ has to multiplied through by density. To keep the left side constant, fluid density has to be constant and thus is incompressible.
But what about other qualities like in-viscid, and irrotational? What do they mean? And why are they necessary?
| I think a derivation of the Bernoulli equation will help clarify things.
We begin with the Navier-Stokes equations
$$\frac{\partial \vec{u}}{\partial t}+\vec{u}\cdot \vec{\nabla}\vec{u} =-\frac{1}{\rho} \vec{\nabla} p +\nu \nabla^2 \vec{u},$$
where $\rho$ is the density, $p$ the pressure, and $\nu$ the kinematic viscosity. The advective term can be rewritten as
$$\vec{u}\cdot \vec{\nabla}\vec{u}=\vec{\nabla}(\frac{1}{2}\vec{u}\cdot \vec{u})-\vec{u}\times \vec{\omega}$$
where $\vec{\omega}=\vec{\nabla}\times \vec{u}$ is the vorticity.
We can now examine this equation under a variety of different assumptions.
For instance, let's assume the density is constant. Furthermore, we'll take the flow to be irrotational.
By definition this means
$$\vec{\nabla}\times\vec{u}=0\implies \vec{u}=\vec{\nabla}\phi$$
for $\phi$ a scalar function. Finally take the flow the to be inviscid (i.e. $\nu =0$).
With these assumptions the Navier-Stokes equations can be rewritten as
$$\vec{\nabla}\left(\frac{\partial \phi}{\partial t} + \frac{1}{2}\vec{u}\cdot\vec{u}+\frac{1}{\rho}p\right)=0.$$
This implies
$$\frac{\partial \phi}{\partial t} + \frac{1}{2}\vec{u}\cdot\vec{u}+\frac{1}{\rho}p=B(t)$$
where $B$ is just a function of time. This is usually absorbed into $\phi$, but there are examples when one needs to pay attention to the value of the Bernoulli head (eg Whitham 1962).
There are other simplifications (eg $u$ time independent, but possibly rotational) that one can also use to get a Bernoulli equation.
Let me know if you have any questions,
Nick
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/258876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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