Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
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.
| Note that a (simple) pendulum performs (simple) harmonic motion and oscillates about it's mean position. For circular motion, the force must always be directed towards the center (centripetal). If you draw the force vectors at different positions (one position is shown in the image) as the bob moves note that the force is not always along the radial direction, i.e. towards O.
The image that you have posted likely depicts the extreme position of a pendulum (as $F_{net} = mg \sin\theta$), where the bob is momentarily/instantaneously at rest (another way to think here would be that the system has all potential energy at this very instant and no kinetic energy). At this very instant, the bob's velocity is zero and hence there is no centripetal force. However the tangential force (mg$\sin\theta$) will drive the bob out of that position and accelerate it to some non-zero speed. At this new position, the tension in the string/thread will now perform two functions-
*
*Balance the mg$cos\theta$ / perpendicular component of gravity.
*Provide the centripetal acceleration.
Hence the changing value of tension in the string cancels the component of gravity parallel to the string and provides the centripetal force.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/245223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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When thermal IR gets reflected from an object, does it change its wavelength (frequency) I'm working with thermal infrared (IR) cameras to detect human thermal radiation.
I notice I can easily distinguish non-human objects throughout the camera's field of view, though all are at same room temperature.
I assume this is due to reflection of thermal IR 'bouncing' around the room.
Indeed, we observe our large white board to be an excellent thermal IR 'mirror.'
CONCERN:
Therefore, I'm concerned radiation reflection may affect frequency of thermal IR emanating from a human subject.
QUESTION:
When thermal IR gets reflected from an object, does it change its wavelength (frequency) at all?
| Reflecting from an object doesn't change the frequency.
However the object may not reflect all frequencies equally - that's why things appear colored in white light.
The human is at a different temperature to the room and so will emit at a different wavelength than the other objects at room temperature. It is possible that a mirror (or something sufficently reflective) will give the same signal as the human if it is reflecting IR emitted from the target.
In practice this isn't an issue because the camera is measuring the power emitted for an area of target. When this IR energy spreads out from the human and is then further spread out when reflected from the mirror it is going to be much weaker.
| {
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"url": "https://physics.stackexchange.com/questions/245314",
"timestamp": "2023-03-29T00:00:00",
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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?
| What has just been proven is the existence of gravitational waves, not gravitons. Beside, if graviton exists, they are likely to be a "pseudo particle" like the photon, i.e., mostly a quantized emission of wave packet.
As a wave, by construction the downstream part is late compared to the upstream part, and because of expansion, it will have slightly more length to cross than the uptream part at anytime, which accumulates with distance, resulting into the increase of the wavelenght. All kinds of waves thus "red"shift.
| {
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Swing: why does the body position modify the amplitude? When a person swings, why does the amplitude of oscillations increase if the person changes the body position ?
That is, when descending and approaching the vertical position, if the person extend his legs and changes the body position almost horizontally, the amplitude increases.
I can sense that it has to do with the center of gravity, but i don't understand where the added energy comes from ?
| You are not adding energy, but changing the rotational inertia (of the person relativ to the center point of the rotation/swinging).
In the bottom point of the swinging, where the locational energy is 0, the total energy of the system is calculated as rotational inertia * rotational speed, and that is constant (without adding energy through applying forces). If the person moves body parts = mass further away from (or nearer to) the rotational center, the rotational inertia changes, and correspondigly, the rotational speed changes, and the product is constant.
For a detailed explanation of rotational inertia (or moment of inertia), see https://en.wikipedia.org/wiki/Moment_of_inertia
| {
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Force with different distance dependence Recently, I have learnt about the central force motion. I came across some forces with different distance dependence, for example, k/r^2(common) one, k/r^4, or even the combined one k/r^2 + k'/r^4 (k, k' are force constant).
I remembered when I first learnt the inverse square law, I was convinced that the inverse square law came from the density of the field line on a surface. However, how the forces with 1/r^3, 1/r^4 exists? Can they be derived by classical means?
I have searched for different information but I cannot get a good answer. I hope someone can help me, or relevant reference will do.
Thank you!
| While electric forces are 1/R^2, that's only between two charges. If you look at distributions of mixed charges, the force laws include dipole, quadrupole,
octopole... and while those are a bit more complex than simple central forces (because they depend on orientation), the scaling with distance includes 1/R^N
terms (N being two or more).
Yukawa's potential for the strong atomic force came from an insight, that central forces were mediated by virtual particles (the photon, for electric force), and those particles made the Faraday-style 'lines of force' picture into a special case that depicts 1/R^2 forces due to massless virtual particles. He then calculated the mass of the pion, based on the known properties of nuclear binding, and on the uncertainty principle linking mass to lifetime. If that insight is correct, all fundamental forces in 3-dimensional space are likely to be inverse square law with an (optional) exponential decay factor.
Yukawa's particle, the pion, was found. His insight seems correct, so far.
| {
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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.
| While being admittedly geometric (um..... they're trig functions), this diagram concisely shows the relationship between the units. This has been said in a variety of different ways in the comments and other answers:
The same principal applies to the cross product. The diagram would be similar but look even more "geometric".
| {
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"timestamp": "2023-03-29T00:00:00",
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Questions about Biot-Savart law and Ampere's law A textbook I'm studying with described finding vector magnetic potential $\vec{\text{A}}$ from Biot-Savart law as below.
$\vec{\text{H}_2}=\int_{\text{vol}}\frac{\vec{\text{J}_1}\times\hat{\text{a}_{\text{R}12}}}{4\pi\text{R}_{12}}dv_1$
$=\frac{1}{4\pi}\int_{\text{vol}}\vec{\text{J}_1}\times(-\nabla_2\frac{1}{\text{R}_{12}})dv_1$
$=\frac{1}{4\pi}\int_{\text{vol}}[(\nabla_2\times\frac{\vec{\text{J}_1}}{{\text{R}_{12}}})-\frac{1}{\text{R}_{12}}(\nabla_2\times\vec{\text{J}_1})]dv_1$
$=\frac{1}{4\pi}\int_{\text{vol}}(\nabla_2\times\frac{\vec{\text{J}_1}}{{\text{R}_{12}}})dv_1$
($\because\nabla_2\times\vec{\text{J}_1}=0$)
*
*$\vec{\text{H}_2}$: magnetic field intensity on point 2 ($x_2$, $y_2$, $z_2$)
*$\vec{\text{J}_1}$: current density on point 1 ($x_1$, $y_1$, $z_1$)
*$dv_1$: infinitesimal volume on point 1 (equal to $dx_1dy_1dz_1$)
*$\text{R}_{12}$: distance between point 1 and 2 (equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$)
*$\nabla_2$: del operator which takes derivation on $x_2$, $y_2$, $z_2$
It also described about verifying Ampere's law with $\vec{\text{A}}$ as below.
$\nabla\times\vec{\text{H}}=\frac{1}{\mu_0}\nabla\times\nabla\times\vec{\text{A}}=\frac{1}{\mu_0}[\nabla(\nabla\cdot\vec{\text{A}})-\nabla^2\vec{\text{A}}]$
$\rightarrow\nabla_2\cdot\vec{\text{A}_2}=\frac{\mu_0}{4\pi}\int_{\text{vol}}\nabla_2\cdot\frac{\vec{\text{J}_1}}{{\text{R}_{12}}}dv_1$
$=\frac{\mu_0}{4\pi}\int_{\text{vol}}[\vec{\text{J}_1}\cdot(\nabla_2\frac{1}{\text{R}_{12}})+\frac{1}{\text{R}_{12}}(\nabla_2\cdot\vec{\text{J}_1})]dv_1$
$=\frac{\mu_0}{4\pi}\int_{\text{vol}}\vec{\text{J}_1}\cdot(\nabla_2\frac{1}{\text{R}_{12}})dv_1$ ($\because\nabla_2\cdot\vec{\text{J}_1}=0$)
$=\frac{\mu_0}{4\pi}\int_{\text{vol}}\vec{\text{J}_1}\cdot(-\nabla_1\frac{1}{\text{R}_{12}})dv_1$ ($\because\nabla_1\frac{1}{\text{R}_{12}}=-\nabla_2\frac{1}{\text{R}_{12}}$)
*
*$\nabla_1$: del operator which takes derivation on $x_1$, $y_1$, $z_1$
As you can see, there are some expressions that result zero just because the variables are different to each other ($x_1$, $y_1$, $z_1$ and $x_2$, $y_2$, $z_2$).
This made me wonder, why distinguish point 1's coordinate as ($x_1$, $y_1$, $z_1$) and point 2's as ($x_2$, $y_2$, $z_2$)? Are they not in same coordinate system?
If they are in same coordinate system, why can't point 1's coordinate variables take derivation on point 2's and vice versa?
And if they aren't, how can $\text{R}_{12}$ exist that connects point 1 and point 2 when they are in different coordinate system?
| Those two points are in the same coordinate system, but different and independent points nonetheless. Taking the derivative $\nabla_2$ with respect to point 2 means "how much does the given expression change if I vary point 2". And point 1 does not change if you vary point 2, therefore the derivatives $\nabla_2 x_1, \nabla_2 y_1, \nabla_2 z_1$ are zero.
| {
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Are objects generally neutral or charged? Basically the questions relates to what is taught about electroneutrality and has 2 parts:
*
*Is the overall charge in the universe zero?
*Are objects the world generally neutral or slightly charged, constantly passing a little bit of charge from one to the other on contact?
| The answer involves the current model of the universe. The current model, the Big Bang model, assumes that everything is charge neutral from the beginning of he appearance of the universe almost 14 billion years ago.
The model incorporates the standard model of particle physics and uses it to project the generation of the currently observed universe.
In this model there are conservation laws, some of which are strict, and conservation of charge is one of them. This at atomic level means that there should be as many electrons as protons in the universe, to keep it neutral. Neutrality exists in the everyday level, since it is the statistically most probable state: due to the attraction of positive to negative charges extra energy must be supplied to separate them and keep them separate.
Are objects the world generally neutral or slightly charged, constantly passing a little bit of charge from one to the other on contact?
The answer to this is more subtle. At the microlevel of atoms and molecules the electrons exist quantum mechanically in orbitals about the nucleus. These orbitals have shapes which allow for localized electric fields that may generate repulsion ( from electrons) or attraction ( from the less shielded protons in the nucleus) dependent on the shape.
The five d orbitals in ψ(x, y, z)2 form, with a combination diagram showing how they fit together to fill space around an atomic nucleus.
This allows attraction between atoms with the correct orientation in space, and the same happens with molecular orbitals. Like complicated LEGO bricks they "fit" into complex patterns (latices in solids), in the process releasing some electromagnetic energy when they settle at the best fit.
Thus even though objects are generally neutral, the charge distributions allow for attractive or repulsive behavior at close contact.
| {
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No buoyancy inside liquid
The answer to this question is given as (a) 5m.
Due to the difference in pressure at the curved parts there will be a net upward force of buoyancy.
And how can that be calculated without knowing the shape?
The reasoning is further illustrated in the image below.
Am I missing something?
| The idea is to find the force that the water exerts on the object. If the object were not glued to the bottom, but were instead surrounded by water, the upward force would be given by Archimedes principle. However, since there is no water under the contact area, the Archimedes force of the water is reduced by the contact area times $\rho g h$, where h is the depth to the bottom. So this takes care of the force of the water. In addition, there is a tensile force on the bottom of the object equal to 2000 N. This is enough information to solve the problem.
Added Supplement
The weight of the object is $W=\rho g V$, where $\rho$ is the density of the object. If the object were fully surrounded by water, according to Archimedes principle, the buoyant force of the surrounding water on the object would be $\rho_wgV$, where $\rho_w$ is the density of the water. However, since there is no water under the contact patch, the upward buoyant force exerted by the water on the object would only be $$F=\rho_wgV-\rho_wghA$$where A is the area of the contact patch. If $T$ is the tension in the glue, the net upward force on the object is given by:
$$F_{net}=(\rho_w-\rho)gV-\rho_wghA-T=0\tag{1}$$
The glue fails when T=2000N. This gives one an equation for calculating the depth h at which the glue fails. Try a value of $h = 5\,$m in this equation, and see whether the equation is satisfied.
| {
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"timestamp": "2023-03-29T00:00:00",
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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?
| CASE I : No friction
If the surface is friction less then, the ball wont move neither rotate, it will be motion less
CASE II : Friction
We know that
$$
\tau= I\alpha
$$
$$\tau=\mu mgr $$ and $$ I \alpha K$$
K is radius of gyration
Above statements provide the equations related to ball's rotational motion
Regarding transnational motion $$ F=frictional force
$$
$$ F=ma$$
$$ a= f/m = \mu mg/m=\mu g $$
Thus above equations clarify the transnational motion, and the motion will be backwards :)
So I will go with Option 3
This is my first activity on physics.stackexchange.com so I hope I answered your question well :)
| {
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How does an electron move through a metal or a semiconductor? I understand that atoms in a metal or semiconductor are bonded and that they will have several eigenstates where electrons can reside. When a voltage is applied electrons in eigenstates will move to other eigenstates and generate a current. However, what is this called? What is the mechanism? Are there several mechanisms? I would like to know some textbooks that address this at a quantum mechanical level. Would I need to delve into quantum electrodynamics?
| To expand the previous answer.
You can model conduction properties with the Kubo formalism, but also using the Landauer-Büttiker formalism. It basically states that every conduction problem can be considered as a scattering problem. A key difference between Kubo and Landauer is that the first one is time-dependent problem, while the second is not. If you want to dig for some details, I think the books by Datta are more or less standard. I like pretty much the discussion by Di Ventra: Electrical Transport in Nanoscale Systems, and it covers both formalisms.
| {
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Einstein-Infeld-Hoffmann (EIH) Lagrangian for a Test-Particle as Limit of Schwarzschild-Geodesic Consider a test particle of mass $m$ which is in orbit around a spherical-symmetric body with mass $M$. It therefore has a position as described by the coordinates $r,\phi$, and its motion can be described by the Lagrangian $L$ of the Einstein-Infeld-Hoffmann-Equations:
$$L = \frac{mv^2}{2}+ \frac{GmM}{r}+\frac{mv^4}{8c^2} + \frac{3GmMv^2}{2c^2r}-\frac{kmM\left(m+M\right)}{2c^2r^2},\tag{1}$$
where $v$ is the particles velocity.
But the orbit of the test-particle can also be described by the Schwarzschild-Metric and the corresponding Lagrangian $\mathcal{L}$
$$\mathcal{L} = -\frac{1}{2}\left[-\left(1-\frac{2 G M}{c^2 r}\right) c^2 \dot{t}^2 + \left(1-\frac{2 G M}{c^2 r}\right)^{-1}\dot{r}^2 + r^2 \dot{\varphi^2}\right].\tag{2}$$
Where the dot denotes the derivative with respect to the proper time of the particle $\tau$ along the world line.
I know that the Newtonian Lagrangian for a testparticle can be derived by requiring $\frac{v}{c}\rightarrow 0$.
Since $L$ simply adds some extra terms to the Lagrangian, it should be possible to do something similar here.
But what kind of expansion is needed to arrive at $L$ from $\mathcal{L}$?
| TL;DR: OP's Lagrangians (1) & (2) are not directly related beyond the 0PN approximation.
*
*OP's eq. (2) is (up to normalization) a gauge-fixed version of the square root action
$$\begin{align} L_0~=~~~~&-mc\sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}\cr
~\stackrel{\text{static gauge}}{=}&-mc\sqrt{g_{00}-v^2 +{\cal O}(1PN)}\cr
~=~~~~&-mc^2+\frac{1}{2}mv^2+ \frac{GMm}{r} +{\cal O}(1PN)
\end{align} \tag{2'}$$
for a massive point particle, cf. e.g. this related Phys.SE post. To 0PN order eq. (2') agrees with OP's Lagrangian (1) up to the rest energy $E_0=mc^2$, which can be ignored.
*Let us mention that OP's version of the EIH Lagrangian (1) is under the extra assumption $\frac{m}{M}\ll 1$, so that $\frac{V}{v}\simeq \frac{m}{M}\ll 1$. The EIH Lagrangian is derived in the 1PN approximation.
*The EIH Lagrangian consists not just of the matter action of the point particle(s), but it also contains a term from the metric tensor field. The latter is some gauge-fixed, linearized, truncated remnant of the Einstein-Hilbert action. $\leftarrow$ This in principle answers OP's main question in the negative.
*For a derivation of the EIH Lagrangian, see e.g. Refs. 1 & 2.
References:
*
*M. Maggiore, Gravitational Waves: Volume 1: Theory and Experiments, 2008; eq. (5.54).
*W.D. Goldberger & I.Z. Rothstein, An Effective Field Theory of Gravity for Extended Objects, Phys. Rev. D 73 (2006) 104029, arXiv:hep-th/0409156; eq. (40).
| {
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Forces that exert torque on a rigid body in rotation when angular momentum is not parallel to angular velocity I'm confused about the rotation of a rigid body, when the angular momentum $\vec{L}$ is not parallel to the angular velocity $\vec{\omega}$. Consider a barbell with two equal masses that rotates around a vertical axis $z$ not passing through its center with angular velocity $\vec{\omega}$.
Taking a generic point $P$ on the $z$ axis as pivot point to calculate momenta, the total angular momentum $\vec{L}=\vec{L_1}+\vec{L_2}$ is not parallel to the rotation axis $z$, thus $\vec{L}$ follows a precession motion and, from the theorem of angular momentum, there must be a torque $\vec{\tau}$ on the system, exterted by external forces: $\vec{\tau}=\frac{d \vec{L}}{dt}\neq 0$.
What are the forces exerting this torque?
Weight has non zero torque $\vec{P}$ and it is an external force, but there is also the reaction of the support that must exert an opposite torque $\vec{R}$, since the barbell stays in this position during the rotation. $\vec{P}$ and $\vec{R}$ are opposite but not equal, in particular
$$\vec{P}+\vec{R}=\frac{d\vec{L}}{dt}\neq 0 $$
Is this correct?
| You have to approach problems systematically, and not intuitively. Like I stated in a previous (accepted) answer, resolve everything on the center of mass, and only in the end transfer the quantities to a different point (like P) to get the results you want.
I start with the kinematics. Use $\ell_1$ and $\ell_2$ for the horizontal distances and $h$ for the vertical height above point P.
$$ \begin{aligned}
\vec{r}_{1C} & = \begin{pmatrix}-\ell_1 & h & 0 \end{pmatrix} & \vec{r}_{2C} &= \begin{pmatrix} \ell_2 & h & 0 \end{pmatrix}\\
\vec{\omega} & = \begin{pmatrix} 0&0& \Omega \end{pmatrix} & \vec{\alpha} & = \begin{pmatrix} 0&0& 0 \end{pmatrix}\\
\vec{v}_{1C} & = \begin{pmatrix} 0&0&\Omega \ell_1 \end{pmatrix} & \vec{v}_{2C} & = \begin{pmatrix} 0&0&-\Omega \ell_2 \end{pmatrix} \\
\vec{a}_{1C} & = \begin{pmatrix}\Omega^2 \ell_1&0&0 \end{pmatrix} & \vec{a}_{2C} & = \begin{pmatrix} -\Omega^2 \ell_2&0&0 \end{pmatrix}
\end{aligned}$$
Now find the momentum at the center(s) of mass
$$ \begin{aligned}
\vec{p}_{1C} & = \begin{pmatrix} 0 & 0 & m \ell_1 \Omega \end{pmatrix} & \vec{p}_{2C} &= \begin{pmatrix} 0 & 0 & -m \ell_2 \Omega \end{pmatrix} \\
\vec{L}_{1C} & = \vec{0} & \vec{L}_{2C} & = \vec{0}
\end{aligned} $$
Note point masses do not have angular momentum.
The support for each mass consists of two forces and one torque. These are defined at the support(s) and need to be transferred to the center(s) of mass
$$\begin{aligned}
\vec{F}_1 & = \begin{pmatrix} R_{1x} & R_{1y} & 0 \end{pmatrix} & \vec{F}_2 & = \begin{pmatrix} -R_{2x} & R_{2y} & 0 \end{pmatrix} \\
\vec{M}_1 & = \begin{pmatrix} 0 & 0 & -\tau_1 \end{pmatrix} & \vec{M}_2 & = \begin{pmatrix} 0&0& \tau_2 \end{pmatrix} \\
\vec{M}_{1C} & = \begin{pmatrix} 0& 0 & \ell_1\,R_{1y}-\tau_1 \end{pmatrix} & \vec{M}_{2C} & = \begin{pmatrix} 0 & 0 & \tau_2 - \ell_2\,R_{2 y} \end{pmatrix}
\end{aligned}$$
This is the moment that rotates the momentum vectors.
The equations of motion are
$$\begin{aligned}
\vec{F}_{1C} - m g \hat{j} & = m \vec{a}_{1C} & \vec{F}_{2C} - m g \hat{j} & = m \vec{a}_{2C} \\
\vec{M}_{1C} &= I_{1C} \vec{\alpha} + \vec{\omega} \times I_{1C} \vec{\omega} = \vec{0} & \vec{M}_{2C} &= I_{2C} \vec{\alpha} + \vec{\omega} \times I_{2C} \vec{\omega} = \vec{0}
\end{aligned}$$
Each mass moment of inertia about the center of mass is zero for a point mass. Since $\vec{\alpha}=\vec{0}$ the right hand side of the torque equation is zero. The result is the support forces as
$$\begin{aligned}
R_{1x} &= m \ell_1 \Omega^2 & R_{1y} & = m \ell_2 \Omega^2 \\
R_{1y} & = m g & R_{2y} &= m g \\
\tau_1 &= m g \ell_1 & \tau_2 &= m g \ell_2
\end{aligned} $$
As you can see these forces are non-zero. The horizontal forces keep the masses going in circles, the vertical forces react to the weight, and the torques support the weight also. From all of this you see that the location of P does not matter. The value of $h$ does not appear in any result.
| {
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Reversible process, equivalence of two definitions? There are two common definitions of a reversible process:
A reversible processes is quasistatic with no dissipation.
And
A process where an infinitesimal change in conditions would reverse the direction of the process.
Is it possible to show the equivalence of these two definitions in the general case and if so how? Also are there any specific cases where the equivalence is easily shown?
| The second definition includes involvement of both the system and the surroundings. For this definition, I like to envision that the surroundings has available the following sets of tools:
*
*Tiny weights that can be added or removed from a piston, when the piston is at different elevations
*An array of constant temperature reservoirs in a continuous sequence of slightly different temperature.
With these sets of tools, we can reversibly transition a system from one state to another while at the same time allowing the combination of system and surroundings to be returned to the original state with no significant change in anything else.
Consider the quasistatic (reversible) expansion with no dissipation you were describing. To carry this out, you remove each tiny weight from the piston at a sequence of increasing elevations. The weights correspond to the surroundings. To compress the system back to its original state, you gradually place the weights back onto the piston, each at their existing elevations. This enables both the system and the surroundings to be returned to their original states, without significantly affecting anything else. Of course, in the limit, the weights have to be infinitecimal in size.
The second definition is what we really mean by reversible, since, in the end, nothing has changed. However, the first definition can also be used to produce the same change in the system, even though the surroundings may not have been handled reversibly. Such a reversible process in which the focus is exclusively on the system irresepctive of how the surroundings is handled is referred to by Moran et al in their book Fundamentals of Engineering Thermodynamics as an "internally reversible process."
| {
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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?
| The differential equation to start with is
$$\ddot{\varphi}(t) = -\frac{g}{r}\sin(\varphi(t)) ~ , ~ \dot{\varphi}(0) = -\omega_0 ~ , ~ \varphi(0) = \theta_0 $$
where $\varphi$ is the angle of the pendulum, $\omega_0$ the initial angular velocity and $\theta_0$ the initial angle. The angles are aligned to have 0° on the bottom, 90° at the right, 180° at the top at -90° al the left.
The function for the angular velocity $\omega$ as a function of the angle $\varphi$ is then
$$\omega(\varphi, \theta_0, \omega_0) = \sqrt{\frac{2 g\cdot (\cos (\varphi )-\cos (\theta_0))}{r} +\omega_0^2}$$
and the integral of the inverse is then the time elapsed as a function of the angle:
$$t(\varphi, \theta_0,\omega_0)= \int_{\varphi }^{\theta_0} \frac{ \text{ d}\theta}{\omega(\theta, \theta_0, \omega_0)}$$
To get a stable integral we have to set $\theta_0$ to $\pi$, so we assume that the pendulum in question will have enough energy to make a whole revolution and $\pi$ will be covered on the path. The solution we get in the end will also be valid for smaller and zero initial velocities, as we will see later.
Because we still want to start from an arbitrary position and with an arbitrary velocity we define our true initial velocity as $\omega_i$ and the $\omega_0$ at amplitude $\pi$ as a function of $\omega_i$:
$$\omega_0 = \sqrt{\frac{-2 g\cdot \cos (\theta_i )-2 g+r \cdot \omega_i^2}{r}}$$
Now we expand the equation to get the time as a function of angle $\varphi$, initial angle $\theta_i$ and intial velocity $\omega_i$:
$$t(\varphi, \theta_i,\omega_i)= \int_{\varphi }^{\theta_i} \sqrt[-2]{\frac{2 g\cdot (\cos (\theta )+1)}{r} +\frac{-2 g\cdot \cos (\theta_i )-2 g+r \cdot \omega_i^2}{r}} \text{ d}\theta$$
To get the angle as a function of time instead the time as a function of the angle we first define
$$\kappa = 2 g\cdot \cos (\varphi )+r\cdot \omega_i^2-2 g\cdot \cos (\theta_i) ~ \text{ , } ~ \xi = 2 g+r\cdot \omega_i^2-2 g\cdot \cos (\theta_i)$$
and transform the integral to the elliptic form (which gives us a division by zero at $\omega_i=0$):
$$t(\varphi , \theta_i, \omega_i)=\frac{2 \sqrt{\frac{r\cdot \omega_i^2}{\xi}} \cdot \text{F}(\frac{\theta_i}{2}|\frac{4 g}{\xi} )}{\omega_i}-\frac{2 \sqrt{\frac{\kappa}{\xi}}\cdot \text{F}(\frac{\varphi }{2}|\frac{4 g}{\xi} )}{\sqrt{\frac{\kappa}{r}}}$$
which can now be inverted, and the division by zero at $\omega_i=0$ disappears:
$$\varphi(t, \theta_i,\omega_i)= 2 \text{ Am}(\frac{2 r \cdot \text{F}(\frac{\theta_i}{2}|\frac{4 g}{\xi})-\sqrt{r} \cdot t \cdot \sqrt{\xi}}{2 r}|\frac{4 g}{\xi})$$
Now we have an explicit formula for the angle as a function of time and the arbitrary initial angle and initial velocity. The first time derivative of the angle, the angular velocity as a function of time is then
$$\omega(t, \theta_i,\omega_i) = \frac{\sqrt{\xi}\cdot \text{Dn}(\frac{2 r \cdot \text{F}(\frac{\theta_i}{2}|\frac{4 g}{\xi})-\sqrt{r}\cdot t\cdot \sqrt{\xi}}{2 r}|\frac{4 g}{\xi})}{\sqrt{r}}$$
which also holds for zero and greater than zero initial velocities and all starting angles.
$\text{Am}()$ is the Jacobi Amplitude, $\text{Dn}()$ the Jacobi Elliptic Function and $\text{F}()$ the Elliptic Integral of the 1. kind.
This needs about 1/6th of the computing time a regular differential would take.
If the inital velocity should be negative (counterclockwise) one would need to flip the x-axis, since situation A and B are in principle equivalent:
The result is the same as with the brute force differential, but can be achieved with less CPU time with the same precision goal:
Brute force differential:
Explicit function:
Plot 1
Plot 2
more details
| {
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Microscopic Definition of Heat and Work If I am given a statistical system, then I can define state-variables like Energy, Entropy or other observables, and then I can (at least for equilibrium states) give the infinitesimal change of energy as:
$$ d E = T dS + K dx $$
Here x means any observable and K means the depending force, for example if x is the volume $V$, then K is minus the pressure $-p$. What I read all the time is
$$ d E = \delta Q + \delta W $$
Is there a general microscopic way to define what part of the above formula is $\delta W$ and what part is $\delta Q$ ?
For example, for reversible processes, $\delta Q = T dS$ and $\delta W = Kdx$. But what if I'm looking at an arbitrary process?
| Let's consider an exchange dE of energy. Using the statistical definition of $E=\sum p_i\epsilon _i$ as the average value of the energies of the microscopic states :
\begin{equation}\label{eq:dE}
dE = \sum\epsilon_i dp_i + \sum p_i d\epsilon_i
\end{equation}
We can see that the change in average energy is partly due to a change in the distribution of probability of occurrence of microscopic state $\epsilon_i$ and partly due to a change in the eigen values $\epsilon_i$ of the N-particles microscopic eigen states.
Now taking the statistical definition of entropy as the average lack of information $S=-k_B\sum p_i ln(p_i)$. Using $\beta = 1/k_B T$ and noting that $\sum dp_i=d\sum p_i = 0$, one can write:
\begin{equation}
\begin{array}{ccccc}
TdS &=& -1/\beta (\sum dp_i ln(p_i) &+& \sum p_i\frac{dp_i}{p_i})\\
&=& -1/\beta (\sum dp_i ln(\frac{e^{-\beta\epsilon_i}}{Z}) &+& d\sum p_i)\\
&=& -1/\beta (\sum -\beta\epsilon_i dp_i - lnZ\sum dp_i )& &\\
&=& \sum \epsilon_i dp_i
\end{array}
\end{equation}
So here we can identify the change in entropy at constant temperature (change in the distribution probability over the microscopic states) as the first term in equation for dE. We have decided to call this term heat and note it $\delta Q$ .
| {
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Is Newton second law covariant or invariant? Is Newton second law covariant or invariant between two inertial frames, moving with uniform traslational motion with respect to each other?
If it is invariant then, indipendently from the frame, $\vec{a}=\vec{a'}$ and $\vec{F}=\vec{F'}$ (of course $m=m'$) and this means that $\vec{F}=m\vec{a}$ has exactly the same form in both the frames.
I'm totally ok with $\vec{a}=\vec{a'}$, but how to be sure that $\vec{F}=\vec{F'}$ (without using the fact that $\vec{F'}=m\vec{a'}$)?
Moreover is this true if the two frames are are oriented differently (but the orientation is constant), as showed in the picture?
| To add more details to jimjo's answer, I would like to explain the "at most" in my comment
Vector, at most, can be covariant.
Three-vectors are only covariant under rotations, but if you include boosts then three-vectors transform in a non-covariant way. Therefore, Newton's second Law is non-covariant under the full Lorentz Group. To get a covariant equation you need to add "a zeroth component" to Newton's 2nd Law:
$$
\dot p^\mu=f^\mu
$$
where now both $p$ and $f$ are four vectors - that is, covariant by definition. For more details, see Principle of covariance or this post of mine.
| {
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Why don't we hear sound reflecting from buildings, mirrors, etcetera? We can see buildings, doors, cars etc. as light falls on it gets reflected to us.
but why doesn't the same thing happen with sound? I mean why don't we hear sound reflecting that much?
| It is because of our reaction time ,speed of sound and disturbance in atmosphere. As our reaction time is 1/10th of a second and speed of sound is about 343.2 metres per second. So sound travels 34.32 metres in 1/10th of a second. To distinguish between 2 sounds the distance between them must be 34.32 metres and if we want to notice reflection of sound(echo) then we must count the distance back and forth i.e 34.32metres and half of it would be 17.16 metres. So to distinguish between original sound and its reflection,the distance between person and reflecting surface must be at least 17.16 metres ,but again sound loses its energy on travelling and in an urban area where so many noises are disturbing the medium of travel..it becomes very difficult to notice such phenomenon. Though we notice it everyday in form of reverbs created in day to day places,but they are hardly distinguishable.
| {
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Why must we use mode locked lasers? Quick question:
If I have a laser cavity with a bunch of harmonics under the gain curve:
Why do I not always get a pulsed laser? EG:
I mean since these harmonics exist in the cavity arn't they always in phase because the ends are closed and they create standing waves in phase to each other? Where does the out of phase aspect come from ? Mirror imperfections -> implying the length of the cavity is not constant? I mean I am wondering, why not all lasers are pulsed, since there is always going to be more than one frequency under the gain curve to some degree.
Thanks!
| I think you are asking if the various harmonics in a laser cavity are in phase with each other, and if not then why not.
It's pretty straightforward. Consider two laser wavelengths in a cavity of length L: one is $\frac {L}{N}$, and the other is $\frac {L}{N+1}$. Though the two can have the same phase at the ends of the cavity, they usually don't.
Moreover - and more importantly for Q-switched lasers- the different modes of a laser are typically uncoupled. That means that even though the frequency relationships between the modes are fixed, their phase relationship is not. A Q-switch couples the modes basically by allowing them to be amplified only if they are in phase at a specific moment, once per round trip cycle in the cavity.
| {
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Energy of electrons in wire The Drude Model helped me quite a lot to visualize how current could flow in a circuit. However, there is still a point that I cannot grasp in the explanation given by some people when they talk about energy. When considering a round trip of one electron along a circuit, it is usually said that :
The electron gains energy when going through the battery and loses that energy along the wire as it collides with the lattice of the conductor.
My problem with this reasoning is that it is implying that the electron has more energy when it leaves the battery than anywhere else in the circuit. However, according to the Drude model, when the electrons collide with the lattice, they indeed lose kinetic energy but they regain kinetic energy as soon as they start accelerating thanks to electric fields produced along the wire. This causes electrons to have an average drift speed. In other words, their kinetic energy is constant on average, which is at first glance, in contradiction with the reasoning given above. What am I missing here?
| 1) Battery constantly moves electrons from the positive terminal to the negative terminal creating non-zero positive and negative charges at the respective terminals. These net non-zero charges create electric field inside the conductor (E=V/d).
2) This electric field inside the conductor accelerates electrons which from time to time collide with the lattice loosing their energy.
3) So, all in all, chemical energy of battery is transformed, first into kinetic energy of electrons, and secondly into heat.
| {
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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?
| It is possible as soon as one is sure to be very distant from every body (gravitational source) in the universe. This is because all inertial forces behave as gravitational forces. If one is confined to stay in a closed room and observes the motion of bodies therein, he/she cannot decide whether the observed accelerated motion is due to a gravitational field or to fictitious inertial forces. However if he/she is allowed to check the distribution of masses outside the room he/she can decide which of the two options is the correct one.
| {
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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.
| If the force being applied is small, it can be applied to either block, and the static friction force will keep them accelerating together. If it is large then the maximum static friction force will be exceeded, and the upper block will slide along the lower with constant kinetic friction force f = μN, where N is the weight of the upper block. The block where the external force is applied will accelerate at: a = (F – f)/M, and the other will accelerate at: a = f/m.
| {
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Photoelectric effect on charged plate As far as I know, to observe the photoelectric effect, one has to expose a metal surface to high-energy radiation. But what happens if the surface has a surplus of electrons? What is the energy needed to knock of those? Would a lower frequency be sufficient to produce the same effect?
| When light falls on the photoelectric surface, photoelectric radiations are emitted, when the energy is more than the work function i.e when more electrons are made to fall on the photoelectric surface,it will go beyond the work function and electrons will be emitted normally.
No,there will be not the same effect because, $f=\frac1\lambda$,where wavelength is inversely proportional to the frequency.
| {
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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 ?
| If I'm understanding your question properly, the block in that situation is over-constrained (i.e. statically indeterminate - you cannot calculate the unknown forces from the laws of statics alone) - you cannot assume that F equals the limiting friction - both it and the tension are unknown.
When you apply a sideways force to the block, it starts to move, and as it moves both the tension (due to stretching of the string) and the friction (due to the stretching of bonds between the two surfaces, and other effects) increase. The block will move and eventually reach an equilibrium in which it is static and the sum of those two forces has a magnitude equal to the force you've applied. You need to know the two "stiffnesses" in order to determine the values of each force.
As you increase the applied force, then, at a level determined by the two "stiffnesses", the friction force will reach the limiting friction and will increase no further.
| {
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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 $T$?
| G is actually not independent of everything.
It does depend on factors like density of space, rate of universe expansion etc.
Its just scientists have not found this dependence.
For time there have been many constants which were later were found to be dependent.
For example at time of Kepler
$T^2 =k R^3$
Later,
k was found to depend on Mass(by Newton) of our sun, and new constant G was defined.
Its not far when G would break to a number of factors.
| {
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Tree Level Feynman Diagrams of Electron Positron Interaction Consider the interaction $$e^-+e^- \rightarrow e^-+e^-$$ The following is a tree level Feynman diagram for this: We can also make the paths of the two electrons on the right hand side cross over and this is a distinct diagram.
However lets consider the interaction $$e^-+e^+ \rightarrow e^-+e^+$$ The above diagram would be a tree level diagram in this situation (swap the top electron for a positron). However I am led to believe that in this case making the path of the electron and positron on the right hand side cross would not give a distinct diagram.
Why is this the case and generally when does a 'cross-over' give rise to a distinct diagram? Thanks :)
| 'Crossover diagrams' exist when the final products are indistinguishable. When using Feynman diagrams for computations, you need to add all distinct diagrams that depict the same process, i.e., have the same external lines. Both the crossed and uncrossed diagrams for the electron-electron process have the same final products, since you cannot tell the electrons apart. You can have one electron with momentum $p_1$ and another with $p_2$, or vice versa, and the process would look exactly the same to the experimenter. But the diagrams themselves are distinct, and need to be added.
However, for the electron-positron diagram, the electron and positron are distinguishable objects, and the crossed diagram is a different process altogether. $e^-(p_1)+e^+(p_2)$ is different from $e^-(p_2)+e^+(p_1)$ and the crossed diagram should not be added.
| {
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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?
| Lookup Optical Tweezers.
The limitations of the technique are due to the damage thresholds; see laser ablation.
This idea has been used extensively in science fiction, especially when implemented as solar sails. It's even practical for some applications, as noted in the article. But laser propulsion from the ground suffers losses due to the atmosphere.
The photonic laser thruster is the hot new technology.
| {
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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.
| your mass is independent of gravity (ignoring relativity). in space (or on the moon) your scale would not show what it shows on Earth, so it can't be measuring your mass. a scale measures weight, but sometimes converts to mass via
$$ Weight = 9.8 N/kg * Mass$$
(near surface of Earth)
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Why do positron generators use solenoid magnets to focus positrons instead of FODO structures? When positrons are generated from a particle beam hitting a dense target, why do we use solenoid electromagnets to focus the resulting positrons? As far as I can see, a FODO (focus-open-defocus-open, focus means quadrupoles focus on x plane, but defocus on y plane) array of quadrupole magnets can focus them, too.
| Solenoid magnets can focus in both the X and Y planes, unlike quadrupoles. They are also simpler devices that don't require the precise alignment that a FODO array would require. The limitation on solenoid magnets is that they are only effective at focusing low-energy beams, like the positrons from your spallation source or primary electrons after being emitted from a cathode.
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Relation between the electromagnetic wave and quantum wavefunction I have been thinking about this for a while. I think I misunderstood something about the basics of quantum waves.
Let's look at light diffracted in conditions similar to the double slit experiment. The intensity pattern we can observe on the screen is due to the wavelike propagation of photons. We can then construct a wavefunction, whose square will give us the probability of a photon being measured at a certain point in space. The maxima of this probablity distribution will resemble the intensity maxima on our screen. The intensity of light is the square of the amplitude of electric wave associated with it.
So it seems to me that these two things tell us the same thing. The quantum wavefunction will give us the probability of a photon being at a point in space, which will be proportional to the intensity of our light (more probability means more photons over time, which means more energy). We would get the same results from calculating the electromagnetic wave at each point, right? So what is different in these two things? Does the electromagnetic wave resemble the quantum wavefunction? Is it okay to think of the EM wave as a result of the probability distribution calculated from QM? (the stronger the EM wave, the more photons can be measured, which means bigger probability... etc)
I have been reading forums and I can't find satisfying answers to this question.
| Photons don't have an associated wave function. You either use the (fully classical) Maxwell's equations for the electromagnetic field (without photons) or Quantum Electrodynamics (which doesn't work with wave functions at all).
For more details, see What equation describes the wavefunction of a single photon?.
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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?
| As far as I know, the only clue at the time that the speed of light would be invariant were Maxwell's Equations where "something" shows up as a constant. However, speed of light being invariant in all inertial reference frames is very counter-intuitive. One might rather expect physics to be slightly different in different frames, which is what the MM experiment was looking for.
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Spin Orbit Coupling Hamiltonians I am really struggling with something fundamental.
I keep coming across two versions of the hamiltonian for spin orbit coupling:
$H_{soc}=\frac{\mu_B}{2c^2}(v \times E) \cdot \sigma $
$\mu_B =$ bohr magnetron
$v =$ velocity
$E = $ Electric Field
$\sigma = $ pauli matrices
and
$H_{soc} = \alpha L\cdot S$
$\alpha =$ constant
$L = $ Orbital Angular Momentum
$S = $ Spin angular momentum
Are these equivalent? If not what situations are they referring to.
| The first formula is generally valid while the second applies to spherical symmetry.
| {
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Do any elements form stable doubly-charged negative ions? It is perfectly possible for an atom - particularly on the electronegative end of the periodic table to form negatively-charged ions by attracting an electron, and these species can be stable, requiring (a small but positive amount of) energy to detach the extra electron.
Is the same thing possible with doubly-charged anions? I know hydrogen can form $\mathrm H^{2-}$, but that one has a finite lifetime. Are there elements that can acquire a second extra electron and keep it until they're perturbed?
| Assuming I understand you correctly the quantity you refer to is the second electron affinity i.e. the energy absorbed in the gas phase reaction:
$$ X^- + e \rightarrow X^{2-} $$
(it's the energy absorbed because a negative second electron affinity means energy is released)
If so, there are no elements for which the second electron affinity is negative. The nearest is sulphur with a second electron affinity of $+532$ kJ/mol followed by oxygen at $+844$ kJ/mol. So there no elemental $X^{-2}$ anions that are stable in the gas phase.
| {
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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?
| I think that I understand what you are asking and I think that it is a good question.
If the centre of mass of the disc and the point mass moves a distance $x$ then the work done by the force is $Fx$ in both cases and yet the disc has gained some extra kinetic energy because it is rotating.
If the centre of mass of the disc moves a distance $x$ the point of application of the force has moved a distance $x+R\theta$ where $\theta$ is the angle through which the disc has rotated.
So the work done by the force is $F(x+R\theta) = Fx + FR\theta$.
For the disc the $Fx$ term is the work done increasing the translational kinetic energy $\frac 1 2 mv^2$ of the disc (and the point mass) and the $FR\theta$ term is the work done increasing the rotational kinetic energy $\frac 1 2 I \omega^2$ of the disc.
So the answer is that although the centres of mass of the disc and the point mass have the same acceleration the applied force moves further and thus does more work when applied to the disc than when applied to the point mass.
| {
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Israel-Wilson-Perjés Solutions I'm searching for a reference that gives explicitly the field strength (or at least the gauge fields) of the Israel-Wilson-Perjés Solution, using complex harmonic functions for the metric.
In "Gravity and Strings" by Ortin, one can find the required formula at pag. 280, but I'm not able to verify that it is solution to the equation of motion, therefore I suspect that there is some misprint in the formulas for the gauge fields.
EDIT: Indeed there was a misprint in the first edition of the book, that is corrected in the second edition.
| There was a misprint in the first edition of the book, now corrected in the second edition.
| {
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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?
| Another way to look at this is to ask yourself "how is the water at the bottom to know about the ship above it?" It can't; a given water volume "communicates" only with its immediate neighbors. Now it's possible that the water volume at the bottom experiences different pressures from different sides. This will just push it away along the pressure gradient, until all over the bottom all water volumes experience the same pressure again. That may happen while the ship is passing over a spot, because of the waves created by the ship, but is only minor and temporary. Any water displaced by the ship just flows away until the water pressure is the same again at equal depths all over the length of the canal.
Since the water pressure of the bottommost layer is what the bridge must carry, and doesn't change, what the bridge must carry doesn't change either. Q.e.d. ;-).
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What is the QED model about half wave plate? Surely classical electrodynamics (CED) has a good and well-known answer about the change of light polarization in a HWP (or Quarter wave plate). I tried to find how does this look like from point of view of QED but couldn't find anything by googling. And nevertheless as QED is a superior to CED theory it must somehow explain it. For QED an interaction of the photon changes its helicity abruptly from +1 to -1 (left to right polarization or vice versa). How this fact can explain the length of the plate?
| A wave plate is a passive component, and can be modelled as a unitary operator on the quantum state. The state is a superposition of left and right circular polarized photons, and the operator gradually alters the relative phases. The total distance then determines the final polarization state.
| {
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Non coherence of Fermions and Bosons through $U(1)$ I "know" the textbook answer why we cannot write,
$$
|\psi\rangle = a|j=\tfrac{1}{2}\rangle + b|j=1\rangle
$$
as "each term in the quantum superposition transforms differently under $U(1)$",
$$
U(2\pi)\ |\psi\rangle = -a|j=\tfrac{1}{2}\rangle + b|j=1\rangle
$$
and "hence the phase difference is unobservable"...
However, I do not understand what is meant by the italicised parts.
Why is it that we cannot observe the phase difference?
The question has arisen from studying notes on a course in Symmetries of Quantum Mechanics focusing on the group theoretical basis of QM.
| This is the prototypical example of a superselection rule.
The operator $U(2\pi)$ commutes with all observables (because it represents a full rotation, and is hence physically a "do nothing" operator), and yet is not a multiple of the identity (because it is -1 on the fermionic and 1 on the bosonic parts of the Hilbert space). Therefore, the representation of the algebra of observables is reducible by Schur's lemma, and the invariant subspaces are precisely the bosonic and fermionic subspaces.
Because the bosonic and fermionic subspaces (let's call them $H_b,H_f$) are invariant, all cross matrix elements for all observables vanish: $\langle b\vert A \vert f\rangle = 0$ for all $\lvert b\rangle\in H_b,\lvert f \rangle \in H_f$ and all observables $A$.
From this we can directly see that relative phases between bosonic and fermionic states are unobservable: For any $\lvert \psi \rangle = \lvert b\rangle + \mathrm{e}^{\mathrm{i}\phi}\lvert f\rangle$, $\langle \psi \vert A \vert \psi\rangle$ is independent of $\phi$, so there is no possible way to determine such relative phases experimentally - they are "unobservable".
| {
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Are there negative energy states in QED? I was reading Weinberg I, when I came upon the following statement$^1$ (slightly edited by me):
\begin{align}
(\not p+m)u=ie\not A\\
(\not p-m)v=ie\not A \tag{1}
\end{align}
The minus sign on the r.h.s. of the equation of $v$ shows that the $v$ are the famous "negative-energy" solutions of the Dirac equation. [...] Of course, for moderate external fields there are no negative-energy states in the theory.
What does Weinberg mean by this? For large enough $A$, is there some ket with $(H-E_0)|\varphi\rangle=-E_\varphi|\varphi\rangle$in the theory? (here, $E_0$ is the vacuum energy). Is this a flaw of the theory? Why moderate external field?
My thoughts
*
*I believe this has nothing to do with negative norm states of QED, because these are gauge-dependent while $E_\varphi$ is not.
*I believe this has nothing to do with bound states, as external fields have nothing to do with these, so the requirement for moderate external field wouldn't make sense.
*Lastly, I think it might be related to $H$ being unbounded below. The QED interaction is $A\bar \psi\psi$, which is cubic in the fields. Therefore, "$H\to-\infty$ as $A\to-\infty$". But $H_\mathrm{QED}$ being unbounded would be something awful: the theory wouldn't have a ground state, so I don't think this is the answer either.
$^1$: The Quantum Theory of Fields, Volume 1: Foundations, page 567.
| It does involve bound states. For an electron in the Coulomb field $A(r)=-\frac{Ze}{r}$ that arises from a nucleus (of mass $m$), the lowest energy is $E_0=m\sqrt{1-(2Z\alpha)^2}$ where $\alpha$ is the fine structure constant. This is positive for $Z\le\frac{1}{2\alpha}$, which is what Weinberg means by "moderate" fields.
The assumption so far was that the nuclues is pointlike. When $Z$ is equal to or greater than this critical value $\frac{1}{2\alpha}$, we must instead view the nucleus as a finite-sized object. When this is done, you'll see that the energy levels become negative, up until the "continuum limit" value of $-m$.
| {
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Can I re-wind gravity? Suppose I perform an arbitrary simulation where I integrate the motions of a collection of particles which interact only gravitationally. Suppose I use a time reversible integrator (to be specific, let's say leapfrog, which is also symplectic, in case that's important). The fact that the integrator is termed 'time reversible' is highly suggestive that I should be able to run my simulation 'in reverse' simply by choosing time steps that are the negatives of the time steps used to run the simulation 'forwards'. But is this actually true? Does it matter how I calculate forces (accelerations)? For instance, does it matter if I'm using direct summation of $GM/r^2$, or a tree algorithm such as a Barnes-Hut tree? One last simplification, let's suppose I have a computer capable of arbitrary floating point precision so that we can ignore roundoff error.
| TL;DR: Yes. Although in reality you don't have unlimited floating point precision, and this will almost always break time-reversibility.
I should point out that not all integrators are time-reversible. For example, predictor-corrector schemes, and most schemes that deal with constraints. The Verlet method, however, is time-reversible, even for large time-steps. It is fairly trivial (but cumbersome) to show this by applying a forward integration step and then a backward one and getting back to where you started.
Related: What does the time-reversibility of Verlet (or other) integration mean?
In reality, however, limited floating-point precision will introduce small errors that will grow exponentially (due to the Lyapunov instability) and break the time-reversibility.
Regarding force field optimisations, they will only break time-reversibility if the computed forces depend on the history of the trajectory or are in any way stochastic. This is not the case for a naive implementation of Barnes-Hut.
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Where does the energy of a photon trying to escape a black hole go? I've heard "light cannot escape a black hole" explained several ways. One is that if a photon inside the event horizon tries to escape a black hole it loses energy to gravity. As it loses energy its wavelength gets longer and longer until its energy is zero.
Where does that energy go and how is it transported?
| A photon that escapes from black hole's neighborhood does work on the black hole. The photon causes the black hole not to be in the photon's gravity well after the photon has escaped. In other words the photon increases the potential energy of the black hole.
The following paragraph may not be science, I just want to say something sane, as opposed to "photon that tries to escape loses energy":
A photon that tries to escape, tries to do all the things that a photon that escapes does: it tries to escape, it tries to redshift, it tries to do work on the black hole, it does not do those things, it tries to.
What does a upwards directed photon exactly at the event horizon do? Well, it stays at the event horizon, so no change of potential energy of photon happens, and no redshift of photon happens.
| {
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D2O distribution in water I just learned that about 1/1000 of hydrogen is deuterium. Does that mean 1/1000 of water is D2O, which is heavier and sinks to the bottom in a glass of water? And consequently is there a layer of D2O at the bottom of the ocean?
thanks!
| It's rarer than that, more like 1 in 6000. Because each molecule of water has two atoms of hydrogen, then about every 2 in 6000 (1 in 3000) has a single atom of deuterium (DHO). And would be closer to 1 in every 6000$^2$ for a molecule of D2O.
The linked question Deuterium density in seawater gives sources that show deuterium is well-mixed in the ocean. The difference in weight between D2O and H2O is similar to the difference in weight of N2 and O2. Currents, turbulence, and diffusion are sufficient to keep those molecules well distributed in our atmosphere, so I would not expect separation of D2O to occur spontaneously in water.
| {
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According to the doppler effect , if my source is faster than sound , will I hear the sound backwards? In a hypothetical situation,say I have a sound box with a very large volume and let's say it is placed pretty far away from me. If it travels towards me at a speed higher than that of sound in the same medium, air for instance, then will I hear the song playing backwards?
according to the equation f' = [v+vo/v+vs]f
where all symbols have their usual meaning , I get a negative f` for vs>v . What does a negative frequency indicate?
I understand a similar question has been asked however the answers didn't satisfy me.
I still believe that before the first pulse reaches the observer , the second pulse is created and so on until the observer hears the 'n'th pulse and then the pulses preceding it.
| Lets take a simpler case: instead of a song, the fighter pilot emits two pulses separated by some time. After the first pulse is emitted, he will indeed outrun it - he is after all going faster than the sound speed. He will then emit a second pulse in front of the first (and outrun that as well).
As a distant observer, you will indeed then hear the second pulse first, and in that sense there is an inversion of the time sequence of signals.
Of course we are ignoring practical considerations, but surely this can be observed: a toy boat moving faster than the wave speed in water, with an agitator hanging off to the side some distance so as not to interfere greatly with the wake that it creates.
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Light bulbs flashing on when disconnected For the 2nd time I've had a lightbulb flash on in my hand after removing from the power socket. The first time the overhead light in my hallway was out for a couple of days before I got around to changing it. This was a cfl bulb and after I unscrewed it and was making my way down the ladder it flashed back on in my hand for about 3 seconds and went back out. I decided to put it back in the fixture and it went back to working. It's now approximately 4 months later and again the bulb went out. I unscrewed from the socket, had the tube end in my hand coming down the ladder when it again flashed on briefly and went back out. I tried putting it back in the fixture and again it has gone back to working. I don't know what to make of this as I've been unable to find any explanation as to how this can happen. What could make a disconnected bulb light up in my hand without any power source? Any thoughts?
| Flurescent light bulbs are known for the way they light up when subjected to static electricity. Here is a video of an instructor using a Van de Graaff generator to light up an old fashioned fluorescent light. CFLs are just old fashioned fluorescent bulbs bent into a different shape with a DC power supply tacked on.
| {
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Intuition on Gibbs measures I am (roughly) aware of the way Gibbs measures are used to solve physical systems (e.g. the Ising model). We can basically boil it down to pinpointing a Hamiltonian.
My question is, consider a system with Ising-reminiscent, cascading dynamics (like opinion formation). In that case, how would we define the energy of an arrangement? What would the temperature be?
In short, can we generalise the concepts that show up in Gibbs measures, and if so how?
I sincerely apologise if my question is non-sensical. Please let me know if this is the case.
| Many researchers are thinking about this question. I'm afraid that there is no straightforward answer. That is the main reason for non-equilibrium statistical physics to be so hard. We can't even pinpoint a good measure. Think of it the other way around: If the dynamics of your system are Hamiltonian and if we know this Hamiltonian, then we know how to describe it's thermodynamic properties: use the Gibbs measure. The rest is often really hard because we don't really know what to do.
There are situation when it is not even possible to define a Hamiltonian in the first place. It may also happen that the Hamiltonian is known but the system is coupled to a non-termal bath (or two baths at different temperatures). Then Gibbs goes out of the window as well. Non-equilibrium physics is just much more diverse than equilibrium. There are no tools that can be applied in general.
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How does acceleration feel compared to gravitational pull? I was debating a variation of this Phys.SE question with a friend. The original question is:
"If you had your eyes closed, could you distinguish between standing still on earth and being in a spaceship which is constantly accelerated at 1g?"
Yes. When you are "standing still" on earth, you actually have velocity due to the fact that the earth is spinning. If gravity suddenly disappeared, you would fly off into space in a straight line, according to your velocity. Since you remain on the surface, there must be a gravitational force constantly pulling you down, constantly changing the direction of your velocity (velocity has both magnitude and direction). Acceleration is defined as the change of velocity, so as you are "standing still", you are actually being constantly accelerated. Now the question has been reduced to "does being accelerated in space feel the same as being accelerated on earth?"
What if the earth and all other objects in space were standing still? You could still feel the earth's gravitational pull. Would it feel different than being accelerated in space?
I imagine being accelerated feels different than not being accelerated. I also imagine you can feel a gravitational pull of a planet even if it is not spinning. This is about all I could muster.
| Gravity and uniform acceleration ar not the same. and with sensitive equipment the man in a box can tell the difference.
imaging he has two plum bobs, on strings. if he measures the difference between the top of the strings and the bottoms, he will notice in uniform acceleration the strings are parallel, But in Gravity the strings point to the center of mass. so the bottoms are slightly closer together.
Also in rotational acceleration the strings point away from the center of rotation.
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What are the accelerations of blocks? I've talked with 2 teachers about this situation:
one teacher said he was completely sure that B have twice the acceleration of A, the other said he was completely sure they have same acceleration. Can you have a better look on it? What do think? Consider it has no friction.
| The accelerations are the same. They can't be different, since otherwise $B$ would move faster than $A$ and the rope wouldn't be tight anymore.
The reason your teacher thought that they are different is a common mistake: under the assumption $F_{AB} = -F_{BA}$ and $F=ma$ (force on A should be force from B and vice versa) you might think that same $F$ and different $m$ should result in different $a$, e.g. $a_{A}=\frac{F_{AB}}{m_{A}}=\frac{m_{A}-m_{B}}{m_{A}}g^{*}$ and $a_{B}=\frac{F_{BA}}{m_{B}}=-\frac{m_{A}-m_{B}}{m_{B}}g^{*}$ (where $g^{*} = g * sin(25°)$ in your example).
The reason that's wrong is the following: the resulting force, that is what's left if you substract the gravitational forces on both masses, $F=({m_{A}-m_{B}}) g^{*}$, has to accelerate BOTH masses. Image how the blocks would move without gravity, they are just two masses bound together behind each other that move in the same direction (= "the same direction as the rope"). Gravitational forces on $B$ are already "taken care of" by some part of the gravitional force on $A$ via the pulley. So if you (or the earth) pull on $A$ you have to move both $A$ and $B$.
So you get $a = \frac{F}{m_{A}+m_{B}} = \frac{{m_{A}-m_{B}}}{m_{A}+m_{B}}g^{*}$ (for both blocks).
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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?
| Yes, but where is the problem? :)
If you have a resistor in your series circuit, then the current is reduced - in the whole circuit, of course, not just in the resistor. The current is the amount of charges per time. There are just less, in a certain time, but they still have to pass through all the other elements of the circuit too.
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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?
| The heat is continually being released to the atmosphere, and the layer is continually getting thicker. The heat has to be conducted from the water-ice interface to the ice-atmosphere interface through the layer of ice. And, as the ice gets thicker, the rate of heat being conducted slows down. And the rate of ice formation slows down. So the amount of time taken for the heat to be conducted and for a new incremental layer of ice to be formed is not the same for each incremental layer. Those are $x^2$'s in the equation, not x's.
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Utilising Black Holes as a potential energy source I'm aware of the Penrose process and the basic physics behind that. Also, I know that the Blandford-Zjanek process (That is potentially responsible for the relativistic jets). Aside from these two, and Hawking Radiation, what other methods or theories are there for extracting energy from a black hole, or the phenomena associated with one? Of course, all these ideas neglect the engineering challenges - assume an infinite budget and an advanced civilization. Thanks!
| Great question. Black holes are some of the brightest objects in the universe. While we think they require the Blandford-Znajek (BZ) mechanisms to produce things like Relativistic Jets, the bulk of the light (emission) they produce is just the efficient thermalization of gravitational energy when material falls into (`accretes' onto) them.
The simplest way to think about this, is how much energy must be (generally and approximately) be released for material to accrete onto a black hole. The binding (gravitational potential) energy is:
$$\varepsilon \approx \frac{1}{2}\frac{GMm}{R}$$
The radius down to which matter can accrete is roughly the Schwarzschild Radius,
$$R_s = \frac{2GM}{c^2}$$
So the energy of material at the Schwarzschild radius is roughly,
$$\varepsilon \approx \frac{1}{2} \frac{GM}{R_s} \approx \frac{1}{4} mc^2$$
This means that something like 1/4 of the entire mass-energy of accretion material is available to produce emission. Generally, from more precise modeling, the 'efficiency' (fraction of energy available) is more like $\sim 10\%$. Still, if you compare this to something like nuclear fission which only converts less than $1\%$ of the mass to energy, then black holes are outrageously efficient!
So even without complicated BZ/Penrose like processes, just normal accreting black holes are extremely effective at emitting energy. So we'd just need to put super-efficient solar panels (etc) around a BH and we're set.
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Gravitational wave equations worked out Is there a website where gravitational wave equations are worked out numerically?
I would like to experiment with mass configurations but can't find examples.
| The topic is a bit too broad. If you want to look at gravitational wave waveforms coming from e.g. binaries then Post-Newtonian approach is usually sufficient and there are many codes for that out there.
If you want gravitational wave emission worked out in any system it might become a bit troublesome.
I would suggest looking into lalsuite developed by the LIGO collaboration, and especially the python interface to it. It is not too easy to use but you can plot binary waveforms with it.
It does require you to have a bit of patience as well as programming skills.
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Diminishing solar temperature and its effects on earth This is a hypothetical question; considering both the earth and the sun as black bodies. If the temperature of the sun decreased N times, what would be the effect on the radiation intensity received on earth?
(i am inclined to say it would decrease N^4 times using the Law of Stefan-Boltzman)
| Yes, the amount of radiation (i.e. power) would decrease $N^4$ times, since the whole geometry of the system does not change, only the power emitted by the sun.
But the temperature of the earth would decrease only $N$ times, since the ratio of the temperatures remains equal - again because of the unchanged geometry. The earth radiates a power proportional to its $T^4$ times $4\pi$ - the whole angle. The sun only hits the earth in a certain solid angle, depending on the distance and radius of the earth.
So the ratio of this angle and $4\pi$ has to be the ratio of the $T^4$, thus also the ratio of the $T$ does not change.
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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?
| The term "gradient" implies a change in some quantity versus a change in second quantity, usually over a distance. It's very much like a slope. For example, the gradient of a roof line on a house is given as rise/run like 15 cm/m or 5 inches/foot. The gradient of potential is the electric field magnitude, with SI units of volts/meter.
A magnetic field is measured in tesla or sub-units thereof, so the gradient of the field would be tesla per meter. It tells you how much the field changes when you change position in the field.
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Is it possible to low pass filter the amplitude of a sound wave? Is it physically possible to block or attenuate noise above a certain amplitude, but leave other lower amplitude noises unhindered?
| Remember that noise is unwanted frequency signal superimposed into the original frequency. This means to reduce noise, you need to block the unnecessary frequency components. The effect of noise increases while increasing the amplitude because amplitude is a measure of loudness. So the amplitude (loudness) of the noise also increase as you increase the amplitude of the sound wave. That doesn't mean you can make the noise disappear by doing something on the amplitude. By reducing the amplitude, you also reduce the amplitude of noise and the noise will stay "silent". In order to reduce noise you must use a filter and the purpose of filter is to select a particular range of frequency and block the remaining frequencies. You cannot "block" the noise by selecting a particular amplitude. But the effect of noise can be hidden up to a certain limit by decreasing the loudness or amplitude. If the noise is too hgih and it's amplitude is comparable to the audio amplitude, it's better to switch off the player.
| {
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Why does imaginary periodic time of the Rindler space of black hole give Hawking temperature? The metric of a Schwarzschild black hole in Rindler coordinates is
$$ds^2= -\frac{\rho^2}{(4MG)^2} dt^2 +d\rho^2 + d\mathbf x_\perp$$
where $\rho$ gives us the distance to the horizon.
If we switch to imaginary time $\tau=it$, the metric becomes
$$ds^2= \frac{\rho^2}{(4MG)^2} d\tau^2 +d\rho^2 + d\mathbf x_\perp.$$
It is said that to prevent a conical singularity, time must be periodic with period $8\pi GM$.
If we identify the period with $\beta=1/(k_B T)$, this leads to
$$T=\frac{1}{8\pi G M k_B}$$
which is Hawking temperature.
Is there any intuition of why this works? There are no fields present in the calculation, so it's not like the standard proofs. Is it just a coincidence?
| It is not a coincidence. It has to work like that. The deficit angle has to be zero.
It's most convenient to see it in the Feynman's path integral approach to quantum mechanics. One works in the Euclideanized spacetime to calculate the temperature $T=1/\beta$ partition sum. Let us consider the full finite-size black hole; the Rindler geometry is a local description of the near-horizon geometry of a very large black hole.
There are contributions to the path integral without any black hole, and those with a black hole. The latter must combine to the smooth "cigar" geometry without any deficit (or excess) angle because Einstein's equations have to hold everywhere, even in the Wick-rotated setup, and there's no source at the horizon. The correct periodicity (absence of deficit angle) imposes the relationship between the temperature and the size of the black hole.
These ideas were clarified by Hawking and Gibbons in 1977, building on insights by Perry and Gibbons, see e.g.
http://motls.blogspot.com/2012/01/gibbons-hawking-and-euclidean-path.html?m=1
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Question about the apparent loophole in principle of least action: boundary condition vs initial condition In Lagrangian formalism, given two points $(x_1,t_1)$ and $(x_2,t_2)$, we ask the question which paths $x(t)$ make the action $S=\displaystyle \int_{t_1}^{t_2}L\ \mathrm dt$ stationary and satisfy the boundary condition $x(t_1)=x_1,\ x(t_2)=x_2$. This question is equivalent to solving the Euler-Lagrange equation
$$\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial q}$$
with boundary condtion $x(t_1)=x_1,\ x(t_2)=x_2$.
My question is why we are authorized to use the Euler-Lagrange equation to solve the initial condition problem $x(t_1)=x_1,\ \dot{x}(t_1)=v_1$.
It seems that these are two different problems. One problem is to find a path satisfying the boundary condition $x(t_1)=x_1,\ x(t_2)=x_2$ and make the action stationary.
The other problem is to find a path with initial condition $x(t_1)=x_1,\ \dot{x}(t_1)=v_1$ and I even don't know how to put other requirements such that its equation of motion is the Euler-Lagrange equation.
How can you prove these two problems are equivalent if you can make the second problem clear? Or maybe it is an axiom that we require the initial condition problem is solved by Euler-Lagrange equation. I'm confused about the logic of Lagrangian formalism.
| The issue is that the underlying classical physics is determined by equations of motion (EOMs) (i.e. Newton's 2nd law), which are common for initial value problems (IVPs) and boundary value problems (BVPs). For BVPs , the EOMs can often alternatively be formulated as Euler-Lagrange (EL) equations of a stationary action principle. The latter approach does never work for IVPs, because one does not have the correct boundary conditions to deduce EL equations via integration by parts. See also this related Phys.SE post and links therein.
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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?
| A probable explanation for this effect is simply that the bottom of the pot might be a bit bulged out, as to form only one point of contact around which the pot then can rotate relatively freely (with little friction).
As you stir the water inside the pot, the moving water molecules exert a frictional force on the walls of the pot, dragging it in the same direction as the water is moving.
For the same reason, the speed of a fluid inside a pipe is distributed in a quadratic manner, where the maximum speed is in the center of the pipe and the minimum speed at the boundary to the pipe itself. Frictional forces slow down the molecules closest to the stationary pipe.
| {
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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!!!
| I agree that friction in the drive mechanism reduces thrust, rather than opposing the motion of the car. However, this is not the case for wheels which are not in the drivetrain - ie where there is front/rear wheel drive instead of 4-wheel drive. Friction in non-drivetrain wheel mechanisms are then sources of resistance to motion. If the car has rear-wheel drive, then the static friction on the rear wheels is forwards and the friction on the front wheels is backwards.
In both types of wheel there is also "rolling resistance" which is the net horizontal component of mostly vertical forces caused by deformation of the tyre.
https://en.wikipedia.org/wiki/Rolling_resistance
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What ds>dQ/T mean? I read the derivation on page 216 over here.
First, it considers an irreversible process between states 1 and 2, followed by a reversible process between states 2 and 1.
From my interpretation, equation 8.31 means that the entropy change of a reversible process is greater than that of an irreversible one between the two states. From equation 8.32, did they generalise and change the RHS to a reversible integral?
(We know that $\delta S = \frac{\delta Q}{T}$ in the case of equality, hence the inequality must be the case of an irreversible process. Is that how to interpret it?)
Then I read in my class notes that the inequality can be removed by adding an entropy generation term:
$$\delta S = \frac{\delta Q}{T} + c$$ where c is the entropy generation term.
For an irreversible process, $c$ is positive. Where does entropy generated go, the system or surroundings?
| I think you have a typo in your question. A reversible process will have a smaller entropy change than an irreversible process. Your interpretation that the equality refers to a reversible process, while the inequality refers to in irreversible process is correct. Looking at the specific equations in that notes document, the integral in 8.31 applies to an irreversible process, while the integral in equation 8.32 is more general, and could apply to either a reversible or irreversible process (reversible only when both sides are equal).
Sometimes it's a bit ambiguous where the distinction between system and surroundings is. A car engine for example is taking in air, and releasing exhaust. The surrounding air could be considered as part of the system. In realistic (irreversible) systems that undergoe a thermodynamic cycle, the extra entropy manifests as heat released to the environment. If this weren't the case, and the system was isolated from the surroundings, then entropy would increase in the system to a maximum, and the system would halt.
| {
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Intensity fluctuations at the output of a single mode fiber coupled to a He Ne laser I have coupled a Thorlabs HNL050L-EC - HeNe, 632.8 nm, 5 mW, Polarized Laser to a 2 meter long single mode fiber patch chord using a Thorlabs F230-FC-B aspheric lens. While I am certainly able to obtain a pure single mode Gaussian at the output, the total output intensity seems to be fluctuating over time scales of about a second. In some sense, the mode appears to be "breathing". The aspheric lens has been mounted on a stable mount, and the fiber is at the correct wavelength. I have also verified that the fluctuations are over and above the intrinsic fluctuations from the laser itself. Has anyone had this issue before? If so, what is the cause and what could be the best way to work around it to get a stable single mode Gaussian output?
P.S Please drop a comment if you require any further details to diagnose this issue.
| My bet is that your fiber is very short (something like one meter or so) and that the fluctuations you see on the output mode are due to cladding modes, i.e. a part of the injected light propagating into the cladding of the fiber instead of the core. The resulting fluctuations are due to external perturbations of the fiber (thermal fluctuations or you touching the fiber and stuff).
Usually these modes are attenuated over long distances but I've already had this problem when using short fibers.
Try to shake your fiber or heat it up with your hands to see if it's doing something on the "breathing" that you see. If so then it could be those cladding modes. They should disappear when using a longer fiber though (for instance 5 meters).
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Finding relation between angular acceleration and liner acceleration directed in one direction
In the above given configuration what could be the relation between the angular acceleration of the rod $\alpha$ and the acceleration $a$ of the $2m$ block.
My attempt : What I thought is that the string connected to the rod will move with the same acceleration as that of the block (using constraint relation or simple intuition). The only thing I can think of after that was to take the component of acceleration perpendicular to the rod. As we have in usual cases $\alpha L=a$ so we can have here $\alpha L=acos37^{\circ}$ but that is not getting me to the answer provided. Where am I wrong. Do the component of acceleration along the rod has any role here ?
| The rod will only rotate upwards due to the force that is normal to it's length, acting from the tether on the rod,
$$F_{t,\perp} = F_t\cos\theta$$
Here $F_t = (2m)g$ is the force from the tether, equal to the weight of the block, and $\theta$ is the angle. The rod also feels a downward force, gravity, that has a normal component
$$F_{G,\perp} = mg\cos\theta$$
The net force is then
$$ F_\perp = F_{t,\perp} - F_{G,\perp} =2mg\cos\theta - mg\cos\theta = mg\cos\theta$$
We then have the link $a_\perp = F_\perp/m = L\alpha$, where $a_\perp$ is the acceleration perpendicular to the length of the rod, $L$ is the length and $\alpha$ is the angular acceleration. We then have
$$\alpha = \frac{a_\perp}{L} = \frac{g}{L}\cos\theta$$
In the mean time, the acceleration of the tether (and then also the block) is the vertical component of $a_\perp$: $a_y = a_\perp\cos\theta$, giving
$$a_{block} = g cos^2\theta$$
Which in turn gives the final relation:
$$ \alpha = \frac{a_{block}}{L\cos\theta}$$
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If we threw a baseball from the ISS, could we deorbit the ball? Clearly this is a hypothetical question.
Say we bring a star baseball player into NASA, prep them appropriately for a mission in space, and fly them up to the International Space Station. They go on a spacewalk with a baseball, and at the apoapsis (highest, slowest point in the orbit) throw it retrograde as hard as they can. Could they decelerate the baseball enough that its periapsis (lowest, fastest point in the orbit) dips into Earth's atmosphere enough to de-orbit the ball over time?
(Let's assume that the ball must de-orbit within about 10 years or less. 10,000 years is too long. Also, let's neglect any loss in mobility that a space suit might cause.)
| You don't need to throw the ball!
At the altitude of the ISS the atmosphere is thick enough that it loses 50-100m of altitude every day due to the drag. At that rate over your ten year timescale the ISS would lose 180 to 360km. When you take into account the increased drag at lower altitudes ten years is enough to bring the ISS crashing to a fiery end.
So just put the ball in your pocket and wait.
Actually this is just as well because the orbital velocity of the ISS is a bit over 17000 miles per hour and the record speed for throwing a baseball (not in a spacesuit!) is only a shade over 100 mph.
| {
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Momentum conserving delta-function in the transfer matrix of quantum-field-theoretic scattering theory The $S$-matrix vanishes unless the initial and final states have the same total $4$-momentum, so it is helpful to factor an overall momentum-conserving $\delta$-function:
$$\mathcal{T}=(2\pi)^{4}\delta^{4}(\sum p)\mathcal{M}.$$
Here, $\delta^{4}(\sum p)$ is shorthand for $\delta^{4}(\sum p^{\mu}_{i} - \sum p^{\mu}_{f})$, where $p^{\mu}_{i}$ are the initial particles' momenta and $p^{\mu}_{f}$ are the final particles' momenta. In this way, we can focus on computing the non-trivial part of the $S$-matrix, $\mathcal{M}$. Thus we have
$$\langle f|S-\mathbb{1}|i\rangle = i(2\pi)^{4}\delta^{4}(\sum p)\langle f|\mathcal{M}|i\rangle.$$
My question:
Since the only way to implement the $4$-momentum conservation is by integrating over the delta-function, does this mean that $\langle f|S-\mathbb{1}|i\rangle$ is integrated over to find the probability? What is the integration variable in this case?
| Yes, there is an integral, which comes from the LSZ reduction formula,
$$
\langle f|i\rangle\sim \int \mathrm dx\ \mathrm e^{ikx}\square_x G(x)
$$
where $x=(x_1,x_2\cdots,x_n)$, $k=(k_1,k_2,\cdots,k_n)$ and $G$ is the $n$-point function. If you go to momentum space you'll get that integrand depends on $x$ only through exponentials, and therefore there is a global delta function.
For more details, see here or chapter 10 in Srednicki's book.
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Will the two capacitors be charged in this circuit? Is such a circuit possible to exist with the two capacitors charging?
Because if we considered the outer loop using Kirchhoff's law we get:
ℰ = q1/C1 + q2/C2
But since the two capacitors are initially uncharged, the two terms of the equation increase, that is, q1 and q2 increase, while they equal ℰ which is a constant.
How's that possible? and will the capacitors be charged? also how could we calculate the time constant for it?
| No, that circuit cannot exist in that regime. You are neglecting the internal resistance of the wires between the voltage source and the capacitors, and if the capacitors are discharged (in which case the voltage over them is zero) that's no longer a good approximation. You therefore need to insert a small resistance on either side of the voltage source, which will then govern the $RC$ charging time of the capacitors.
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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?
| Yes. With the qualification that chaos describes the behaviour of an ideal (continuous?) mathematical model, whereas leaves and the air through which they fall are real, I think it is a chaotic system. Leaves falling from the same place but with a small change in orientation can land in very different places, and intermediate orientations do not necessarily land in between. As the difference in orientation is changed by smaller and smaller amounts (ad infinitum), the path of connected landing points becomes more convoluted.
I think that increasing the height of the tree will also increase the radius of landing sites on the ground, without limit.
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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?
| Wouldn't a wave smaller then the smallest be "in the same theoretical universe of discourse" mathematically negative? Looks like a good explanation for a time traveling Sci-Fi xD
Also, wave may be able to reach a higher speed then it's correlation to length. For example if traveling on a referent that is also traveling. A wave inside another one would be 2 time faster if we can alter it's spatial referent, for which quantum intrication could be the key.
Don't forget that all those equations have or are directly correlated to a constant. Those are human made, at least human labelled.
Thinking about it in separate questions was a good lead, what would happen over gamma ray speed/frequency, to an electromagnetic signal, imagining we have a transmitter that can achieve that?
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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).
| Your vector sum drawing is incorrect. Considering the simple case of a perfectly circular orbit, there is a tangential velocity vector and a perpendicular (inward) acceleration vector. Strictly speaking, you can't add them because they are different quantities (acceleration vs velocity).
At some point in the orbit, the object has an initial velocity V in some direction we'll call X, and an inward acceleration in the perpendicular direction we'll call Y. One quarter orbit later, the acceleration vector has accelerated the object from zero to V in the Y direction, but it has also rotated (still pointing inward), decelerating the object from V to zero in the X direction. At this point, there is no more Y component to the acceleration.
If you break down the acceleration and velocity vectors into cartesian components, you find that each describes a sinusoid, peaking at some point, crossing zero one quarter cycle later, reaching an inverse peak one quarter cycle after that, crossing zero again after another quarter cycle, and returning to the initial peak at one full cycle. If you add up the area under the curves, you get zero - the negative exactly equals the positive, so there is no net change, so no net increase in velocity.
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Resistance from $I$-$V$ graph
I thought that the answer was obviously A as the gradient remains constant...however, the answer is B...who so?
| Remembering that resistance = $\frac V I$ work the resistance at $I=1$ and $I=2$.
For the resistance to be constant the current-voltage characteristic must be a straight line and go through the origin.
There is another parameter which is useful in some instances and that is called the incremental resistance $\frac {\Delta V} {\Delta I}$ which is related to the gradient of the graph. In this case the incremental resistance is constant at 1 ohm.
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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?
| The primary difference is that the electrons in metallic hydrogen are nearly completely degenerate.
Degenerate electrons cannot be dissipatively scattered and lead to the "metallic" characteristics of extremely high electrical and thermal conductivity.
To first order, to produce metallic hydrogen you need to make sure that the electron kinetic energy at the Fermi energy of the system is much greater than $kT$. For a pure hydrogen gas and non-relativistic degeneracy, this leads to
$$ \left(\frac{3}{8\pi}\right)^{2/3} \frac{h^2 n_e^{2/3}}{2m_e} \gg kT$$
This means that either very high electron densities, or low temperatures, are required.
"Metallic hydrogen" can exist as a fluid or a solid. In liquid metallic hydrogen, both the protons and electrons are "free". To produce a solid requires that the ratio of the coulomb energy of the protons is much larger than $kT$ so that they "freeze" into a crystalline lattice. This in turn is a requirement that needs even higher densities or even lower temperatures. Thus under these conditions the electrons are indeed completely degenerate,
| {
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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?
| Your original equation is incorrect. For a process occurring in a closed system, the equation should read$$\Delta U=Q+W$$where U is the internal energy of the system, Q is the heat added to the system, and W is the work done by the surroundings on the system, $W=-\int{PdV}$. If the process takes place at a constant pressure, then $W=-P\Delta V$, and the change in internal energy becomes:$$\Delta U=Q-P\Delta V$$. But, from the definition of enthalpy, we have $\Delta H=\Delta U+\Delta (PV)$. So, finally, $$\Delta H=Q$$
So, for a process carried out at constant pressure, if the heat added to the system is positive (endothermic), $\Delta H$ is positive and if the heat added to the system is negative (exothermic, heat removed from system), $\Delta H$ is negative.
| {
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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.
| At a certain depth a liquid exerts the same pressure on all sides.
So if you want to calculate the force exerted by water on sides w and l then you have to put in the value of depth in the formula.
Whatever the answer will come will be the pressure exerted by water on the sides w and l for that particular depth.
As for the ratio F/A is concerned, it states the amount of force F applied on the area A.
For eg. if the pressure is 5 pascals (SI unit of force) then it means that 5 Newtons of force is being applied on 1 m^2 of area.
Summary : the pressure on sides w and l vary with the depth so at different depths the pressure will be different. And you have to calculate them separately.
Hope it helps !
| {
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CFT conformal weight vs. scaling dimension I was wondering if anybody could clarify what the difference between the conformal scaling dimension $\Delta$ and the conformal weight $h$ is?
Is it correctly understood that $\Delta$ is related to the transformation properties of a field and $h$ is an eigenvalue of the Virasoro operator $L_0$? Or is it that $\Delta$ is for general dimensions, while $h$ is used in 2 dimensions?
I seem to have confused myself while reading Francesco et. al.'s Conformal Field
Theory.
| Two dimensional CFTs separate into a left-moving sector and a right-moving sectors. The Virasoro generators $L_n$ act on the left-moving sector and ${\tilde L}_n$ act on the right-moving ones. Operators (or states due to the state-operator map) are labelled independently by representations of the left- and right-moving Virasoro. In particular, $h$ and ${\tilde h}$ are the eigenvalues of $L_0$ and ${\tilde L}_0$. These are called the conformal weights. $\Delta$ is the eigenvalue of the dilatation operator which is $D = L_0+ {\tilde L}_0$. In other words, $\Delta = h + {\tilde h}$.
In higher dimensions, there is no separation into left- or right- moving sectors. There is no analogue of $L_n$ and ${\tilde L}_n$. However, the dilatation operator $D$ still exists and representations are labelled by its eigenvalues, $\Delta$.
| {
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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?
| The result is problematic that the compressibility factor is greater than 1.
The well-known Van der Waals equation is, $$(P+\frac {n^2a}{V^2})(V-nb)=nRT$$
And ideal gas EOS is, $$P_{ideal}V=nRT$$
where n is molar number and R is universal gas constant. You can use particle number if you prefer that way.
From VDW equation, we get $$ P+\frac {n^2a}{V^2}=\frac{nRT}{V(1-\frac {nb}V)}=\frac {P_{ideal}}{1-\frac {nb}{V}}$$
Using Taylor expansion and taking the first order,$$ P+\frac {n^2a}{V^2} \approx P_{ideal}(1+\frac {nb}{V})$$
Therefore, we get $$ P \approx P_{ideal}(1+\frac {nb}{V})-\frac {n^2a}{V^2}$$
Because, $$\frac nV =\frac {P_{ideal}}{RT}$$
We get, $$ P \approx P_{ideal}(1+\frac {n}{V}(b-\frac {a}{RT}))$$
or $$ P \approx P_{ideal}(1+\frac {P_{ideal}}{RT}(b-\frac {a}{RT}))$$
For argon at 15 degrees C and 1.013bar, this gives compressibility 0.99899. Though it is higher than reference 0.99925, it is less than 1.
| {
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What do we mean by speed of light dependent on direction? I have a statement in textbook saying:
When the speed of light is independent of direction, the secondary waves are spherical.
When is it dependent on direction and how will the secondary waves defer?
| There are some materials, called Birefringent in which the index of refraction depends on the direction in which light is moving (often also dependence on the particular polarization of the light). This is due to the crystallin structure of the material which allows faster propagation in one direction or another.
Another example would be a magnetized plasma where the magnetization means that the charged particles will respond different in-vs.-against the direction of the magnetic field, and thus propagating light rays will be affected differently.
These are some really nice slides talking about both anisotropic plasmas, and crystals.
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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 one dimensional Maxwell distribution for the $i-$component of the velocity vector is
$$f_{1D }(v_i) = \left(\frac{m}{2 \pi k T}\right)^{1/2} \exp\left(-\frac{m v_i^2}{2kT} \right)$$
Let's drop the $i$ and $1D$ subscripts for simplicity. You are looking for the average of the absolute value of $v$, $|v|$. To find $\langle |v| \rangle $, we have to perform the integration
$$\int_{-\infty}^{\infty} |v| \ f(v) \ d v$$
Now, since $f(v)$ depends only on the square of $v$ and we are in one dimension, $f(|v|) d|v| = f(v) d v$ and we have
$$\left(\frac{m}{2 \pi k T}\right)^{1/2} \int_{-\infty}^{\infty} |v| \exp\left(-\frac{m |v|^2}{2kT} \right) \ d |v|$$
Since the absolute value is present, this is equal to (let's change notation for clarity, $|v|=u$)
$$ 2 \left(\frac{m}{2 \pi k T}\right)^{1/2} \int_{0}^{\infty} u \exp\left(-\frac{m u^2}{2kT} \right) \ d u$$
The integral can be solved easily integrating by parts, and yields $\frac{kT}{m}$, so that:
$$ \left(\frac{m}{2 \pi k T}\right)^{1/2} \left(\frac{2 kT}{m}\right)$$
Simplifying, we obtain:
$$\langle |v| \rangle = \left( \frac{2kT}{\pi m} \right)^{1/2}$$
Q.E.D. :-)
P.S. Notice that the average of $v_i$ (with no absolute value) would be $0$ for symmetry. For clarity:
$$\langle v \rangle = \int_{-\infty}^{\infty} v \ f(v) \ d v = 0$$
$$\sqrt{\langle v \rangle^2} = v_{rms} = \sqrt{\int_{-\infty}^{\infty} v^2 \ f(v) \ d v }$$
$$\langle {|v|} \rangle = \int_{-\infty}^{\infty} |v| \ f(v) \ d v $$
| {
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Model/formula for bouncing ball I'm programming an animation of a bouncing ball, and I want it to be as realistic as possible.
I fully understand the physics while the ball is rising and falling: It accelerates downward at 9.8 meters/second/second.
But once it hits the ground, I'm lost. I know that it experiences some compression which translates it's velocity upward again, at which point gravity is again the only force acting on it. But I don't know how to model the compression and deflection of the ball hitting the ground.
(Also, I'm only dealing with strictly vertical linear velocity and acceleration. The ball isn't traveling horizontally or rotating at all. And I'm assuming negligible loss to air resistance.)
I know there are constants to deal with: how much the ball resists compression and how much energy is lost during the bounce, but I don't know how they are related.
Can anybody give me a one- or two-paragraph explanation and/or (a) formula(e) and/or point me to a resource where I can read more? (yes, I've already spent an hour or so googling it)
I'm happy to continue reading about it until I understand, I just don't know where to turn at this point.
Thanks in advance!
| Visual Solutions (Now Altair) makes VisSim Software. Here is a demo block diagram that they have used to simulate a bouncing ball:
The $1/s$ blocks are integrator blocks from the VisSim library.
The plot of the bouncing ball with these parameters is shown below.
If you are interested in running this, the demo comes with the install and you can download a free 60 day trial here
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How to imagine WiFi signal propagation? When thinking about how the WiFi signal propagates through a household, can I use the following thought experiment?
Assume absolute darkness. Place a strong lightbulb where the WiFi access point is. The illumination that reaches various places in the house is approximately proportional to the strength of the WiFi signal in that place.
How precise is this mental image? I know that the radio waves can penetrate some objects / walls that the light cannot. Is this at least somewhat representative?
| It's hard to access how 'accurate' an analogy is (i.e. how is this being quantified?). But, I think, there is a simple - better analogy:
WiFi is more like sound in a house. The transmitter is a speaker. If its a good, loud speaker, you will still easily be able to hear it in the next room - through a wall. A few walls inbetween and it gets very faint. Depending on the materials in the walls, opening and/or closing doors may or may not make much of a difference. And there may be odd corners or directions where the sound gets louder, or gets extra soft.
| {
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Metric that is Minkowski plus sum of null vectors In GR exercises I've often seen metrics of the form $g_{ab} = \eta_{ab} + k_ak_b$ where $k_a$ is null with respect to $g$ (or equivalently $\eta$). I'm happy doing calculations with such metrics, but I wondered what the physical interpretation of such a metric is?
If $k$ is in some sense small then this is a linearised limit, but I don't think this always has to be the case.
| The reason to consider such metrics was not a particular interpretation, but that for this ansatz one could hope to find some exact solutions. It is known as the Kerr-Schild metric, and its role in the process of finding exact solutions has been described by Kerr in Wiltshire, Visser, Scott (eds.), The Kerr Spacetime.
One can, of course, try to develop some sort of physical interpretations of special solutions, A.J.S. Hamilton, J.P. Lisle, The river model of black holes, Am.J.Phys.76:519-532 (2008), arxiv:0411060v2 would be an example for the slightly different ansatz
$ds^2 = -dt^2 + (dr+\beta dt)^2 + r^2(d\theta^2 + \sin^2(\theta)d\varphi^2)$
. But I doubt it makes sense, a physical interpretation should be applicable to more than a few exceptional solutions, see my article http://ilja-schmelzer.de/papers/river.pdf for details.
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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:
| The equations are entirely equivalent, as can be proven using Gauss' and Stokes' theorems.
The integral forms are most useful when dealing with macroscopic problems with high degrees of symmetry (e.g. spherical or axial symmetry; or, following on from comments below, a line/surface integrals where the field is either parallel or perpendicular to the line/surface element).
The differential forms are strictly local - they deal with charge and current densities and fields at a point in space and time. The differential forms are far easier to manipulate when dealing with electromagnetic waves; they make it far easier to show that Maxwell's equations can be written in a covariant form, compatible with special relativity; and far easier to put into a computer to do numerical electromagnetism calculations.
I would think that these three points generalise to any system of differential vs integral forms in physics.
| {
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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?
| If the event horizons overlap you get one big horizon. EM forces can not counteract gravity if the curvature is too large since the force required to counteract gravity becomes infinite at the horizon. You can see this in the equation
$$F=\frac{G\cdot M\cdot m}{r^2\cdot\sqrt{1-r_s/r}} $$
which becomes infinite at the horizon $r_s$. Since from the outside perspective everthing that fell into the black hole got stuck on the horizon because of time dilation, the matter that makes up the black hole is layered outside the horizon, so if both horizons would touch, the two black holes would have to merge.
Also see http://arxiv.org/pdf/gr-qc/0411060v2.pdf:
"For a Schwarzschild (non-rotating, uncharged) black hole, the river falls radially inward at the Newtonian escape velocity, hitting the speed of light at the horizon. Inside horizons, the river of space moves faster than light, carrying everything with it."
In other words, you can't hold two black holes stationary if their horizons overlap.
| {
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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?
| It probably depends what interpretation of quantum physics you subscribe to. That sounds approximately right for the Copenhagen interpretation, in which you aren't allowed to analyze where the wave function comes from.
For those who appreciate more what de Broglie, Einstein, Bell and others have put into quantum physics, there's always the interpretation that a real, physical system underlies the probabilistic description, a system with a localized component (particle) interacting with a non-localized one (waves). In this case the wave-like properties really do come from the wave part and the particle-like properties from the particle part. See bouncing droplets, for example: https://www.youtube.com/watch?v=nmC0ygr08tE, http://math.mit.edu/~bush/wordpress/wp-content/uploads/2015/08/Bush-PHYSICS-TODAY2015.pdf
| {
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How to deal with zero uncertainties? Suppose you measure quantity $x$ with an uncertainty ${\rm d}x$. Quantity $f$ is related to $x$ by $f=x^2$ . By error propagation the uncertainty on $f$ would be ${\rm d}f=2x{\rm d}x$. If a certain point $x$ equals zero then the uncertainty on $f$ would be zero, even if $x$ carries an uncertainty. Is there a special procedure in these cases?
| Use the second derivative (or third, or whatever). The reason we use that formula is that
$$
df \approx \frac{df}{dx} dx
$$
is the first order Taylor approximation to df. If the first order term vanishes, you should include higher terms:
$$
df \approx \frac{df}{dx} dx+\frac{1}{2}\frac{d^2f}{dx^2} dx^2+...
$$
In your case, with $f=x^2$, and $x=0$, we'd have
$$
df \approx dx^2
$$
| {
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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?
| The standard reply is: get your video camera out, take a movie of dropping a (empty) coffee cup on the kitchen floor, then play the movie backwards. There is no physical way of telling which way time is running from your movie or the equations behind the event (Newtons Laws).
Rather than thinking of an absolute arrow of time, so the coffee cup will break apart and NEVER reassemble itself later, think in terms of probabilities. It's almost certain the cup will break up and in doing so, it will give you an indication of the direction of time, but there is a tiny, tiny probability that it may reassemble itself and give you an indication that time is running backwards. But because the coffee cup is composed of zillions of atoms, all of which would need to be restored to their former positions, this is very unlikely.
| {
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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?
| Momentum and position are conjugate variables in classical mechanucs, which means they satisfy the Poisson bracket relationship. When quantum mechanics was invented the Poison bracket relation was replaced by the operator commutation relationship which results in the relation under consideration.
| {
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How is the Rayleigh criterion connected to the Abbe limit? I am interrested whether one can derive a formula for the point resolution (like Abbe did) of an optical system from the Rayleigh criterion (without the use of small angle approximation i.e. $\rm{sin}(\alpha)=\rm{tan}(\alpha)$ which is not really suitable e.g. for microscopy).
And if so whether formula is directly comparable to the Abbe limit for point (or rather line) resolution.
The Rayleigh criterion is given as:
$$\theta_{min}=1.22\frac{\lambda}{D}$$
where $\theta_{min}$ is the smallest resolvable angle, $\lambda$ is the wavelength of the used lightsource and $D$ is the diameter of the used aperture (or of the used lens).
And the Abbe limit is given as:
$$d=\frac{\lambda}{2\,n\,\rm{sin}(\alpha)}=\frac{\lambda}{2\,\rm{NA}}$$
where $d$ is the smallest resolvable distance, $n$ is the refractive index of the medium between the object and the optical system, $\alpha$ is the biggest scattering angle (incident on the optical system) and $\rm{NA}$ is the numerical aperture.
| As @Felix and @Caterina are still not satisfied I will add my 2 cents, for which I hope they are correct.
As far as I am aware Rayleigh developed his criterion on diffraction of light on slits, whereas Abbe was working on microscopy. Therefore, you have a refraction index in one and not in the other. However, you could bring the Rayleigh version close to the Abbe's. Rayleigh stated the following:
\begin{equation}
\theta_{min} \approx 1.22 \frac{\lambda}{d},
\end{equation}
where $\theta_{min}$ represents the minimum angular radius of an Airy disk as seen from the centre of the circular aperture, $\lambda$ the wavelength of light and $d$ the diameter of circular aperture. Now this is angular separation and we have to bring it to spacial separation via $x_{min} = R \sin\theta_{min}$ and we get
\begin{equation}
x_{min} \approx 1.22 \frac{\lambda R}{d},
\end{equation}
where $R$ is distance between the slit and imaging screen. Now we can convert the $R/d$ into the $\sin \alpha$ via $\sin \alpha = \frac{d/2}{R}$, and we get
\begin{equation}
x_{min} \approx 1.22 \frac{\lambda}{2\sin\alpha} \approx 0.61\frac{\lambda}{\sin\alpha},
\end{equation}
At this point we can introduce that there is a different refractive index on one side of the slit and $\sin\alpha$ goes to $n\sin\alpha$.
\begin{equation}
x_{min} \approx 0.61\frac{\lambda}{n \sin\alpha} \approx 0.61\frac{\lambda}{NA}.
\end{equation}
And this is very close to Abbe's limit of
\begin{equation}
x_{min} = 0.5 \frac{\lambda}{NA}.
\end{equation}
So all in all it's just how you define the minimum distance at which you can still separate two sources.
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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.
| All that you have written is correct.
Let's go further to the next point 4:
Once he comes to the cylinder's wall and stands on it, he gets the same angular speed as the cylinder, and he will also get a centrifugal force and rotational gravity as in the films.
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Can we use radiation pressure to push/levitate a human? I want to know, at least in theory, are there any safe em wavelength that could affect human body as net force and do no harm?
To put it bluntly I want to know is it possible for levitate a human on Earth with radiation pressure?
I just think maybe some range of infrared will giving momentum as push force more than heat. And if it has a large quantity it could be enough to push large object.
| The force of radiation pressure is $F=P/c$ for absorbed radiation or $F=2P/c$ for reflected radiation, with $P$ the power and $c$ the speed of light. If you want to generate 750 N of force, you need (a) a radiation source of 100 GW and (b) a mirror that reflects so well that nor the mirror, nor you will be vaporized in an instant.
Update: CuriousOne remarked correctly that you don't need a power source of 100 GW. Rather, use two mirrors to make a laser cavity with 100 GW of irradiance on the mirrors, but the actual power that you need to pump in is only a fraction 1e-3 or so for good visible or infrared laser mirrors, i.e. a sustained 100 MW. That's still a rather substiantial laser; I don't think they exist.
An alternative approach would be to use microwave mirrors, which can be designed to have losses on the order of 1e-6, or even 1e-9 in the case of superconducting mirrors. Microwave sources that can deliver high power exist (everyone has one in their kitchen). Taking an optimistic 1e-9 loss at the mirror, you would only need to feed it with 100 W of microwave power. A problem here is that low-loss microwave cavities are typically bottle-like enclosures, not separate mirrors.
For any low-loss (high-Q) cavity, the length and orientation of the cavity walls or mirror must be kept stable within very tight tolerances, which is at odds with the idea of levitating a human.
| {
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Coupling a spinor field to a preexisting scalar field? So I'm not a physicist, but I'm thinking about a mathematical problem where I think physical insight might be useful.
We're working on a Riemannian manifold $(M,g)$ (positive definite metric) with a distinguished smooth function $f$. The metric and the smooth function are related by a tensor equation $$Rc(g)+\nabla^2f=\displaystyle\frac{1}{2}g$$ (The second summand is the Hessian wrt the Levi-Civita connection.)
I want to study spinor fields $\psi$ which solve some kind of Dirac equation but somehow involve this function $f$. (Is it appropriate to say I want to "couple" $\psi$ and $f$?) I thought perhaps physicists had thought about such things and may have insights.
So, my question: Are there natural equations (from a physics POV) to write down for $\psi$ that involve $f$?
I'm aware of the Yukawa interaction (thanks Google), which is a Lagrangian you can write down for undetermined scalar and matter fields, but in this case the scalar field is fixed ahead of time, so I don't know how that figures in.
Any thoughts at all are appreciated.
| Your first equation looks a bit like GR with a dilaton. IIRC, the analogous equations in supergravity will naturally have spinors coupling to the dilaton.
| {
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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?
| No, the hot object itself experiences no relativistic effects. The particles in the object however may well experience time dilation or length contraction. In fact in very hot gases and plasma, ideal gas laws may not always apply due to relativistic effects.
These effects are especially noticeable in gases once the kinetic energy or fermi energy (for fermionic gases) exceeds $mc^2$ by a significant amount.
Relativistic Gas Problems
| {
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Conserved quantity in a spacetime with Killing vector I am trying to prove that that the expression $Q=-\frac{1}{\kappa}\int_{S_\infty} \nabla^i \xi^k \mathrm{d}\sigma_{ik}$ is a conserved quantity for a spacetime with Killing vector $\xi^i$ where $S_\infty$ is the 'boundary at infinity' of the spacelike hypersurface $\Sigma$.
Using the Gauss Theorem, I have written $Q$ as follows:
$$\begin{align*}Q&=\frac{1}{\kappa}\int_{\Sigma} \nabla_c \nabla^c \xi^k \mathrm{d}\sigma_{k}\\&=\frac{1}{\kappa}\int_{\Sigma} R^{kb} \xi_b \mathrm{d}\sigma_{k}\end{align*}$$
Is there any way of seeing why this last expression is constant?
Thanks!
| I will just sketch this out. The Killing equation $\xi_{a;b} + \xi_{b;a} = 0$ tells you that $\nabla^i\xi^k$ is antisymmetric in indices. I am going to write this as the commutator between $U$ the vector along which we differentiate $\xi$ with $\nabla_U$ so in compact notation we have
$$
Q = -\frac{1}{\kappa}\int[U, \xi]d\sigma
$$
I now use Stokes' rule to get this on the volume, where now the commutator is $[V, [U, \xi]]$ and this equals $R(U, V)\xi + \nabla_{[U,V]\xi}$. This gives
$$
Q = -\frac{1}{\kappa}\int \left(R(U, V)\xi + \nabla_{[U,V]}\xi\right) dVol
$$
This leads to the equation you have. However, in general the $\sigma$ is an area two-form and that $Vol$ is a volume three-form. This means the indices in tensor notation on $[V, [U, \xi]]$ have permutation symmetry and the integral is then over the permutation of this commutator. Then by Jacobi theorem this is zero and all that is left is a constant of integration.
| {
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Difference between sunrise and sunset? Other than knowing which direction is east and which direction is west, or observing for a sufficient timespan (to determine the direction of motion), is there any way of telling whether what one is seeing is a sunset or a sunrise? A priori it seems not but I was wondering if there are some subtler effects beyond Rayleigh.
Note. I just came across an article that mentions that the green flash can occur only at sunset, and provides additional references:
Broer, Henk W.
Near-horizon celestial phenomena, a study in geometric optics. Acta Appl. Math. 137 (2015), 17–39.
| It's not something you can directly see, but various amateur radio propagation modes turn on around dawn and turn off around dusk. Read up on the ionosphere and solar ionization.
I imagine this could lead to spectral voids in setting sunlight, but I expect the effect to be too small to see (maybe not too small to measure).
| {
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Components of Velocity in polar co-ordinates Consider a point moving along a curve in a plane. The position of a point P on a coordinate system can be specified by a single vector $\vec{r}$=$r\hat{r}$. A rough sketch describing the situation is shown below:
In order to find the velocity of the particle we take the first derivate of $\vec{r}$ with respect to time. Thus: $\vec{v}=$$d\vec{r}\over{dt}$, which implies
$$\vec v=\frac{d}{dt}(r\hat r)=\hat r\frac{dr}{dt}+r\frac{d\hat r}{dt}.$$
Now the author of my textbook has written that the expression
$$\frac{d\hat r}{dt}=\frac{d\hat r}{d\theta}\frac{d\theta}{dt}.$$
Furthermore, he writes that $d\hat{r}\over{d\theta}$$=$$\hat{\theta}$
This step has confused me and from the diagram I am not able to convince myself why this equality holds. Any insights or arguments will be helpful in clarifying my query.
| Consider the picture below.
In Cartesian coordinates
$$\hat r=\cos\theta\hat i+\sin\theta\hat j,$$
and
$$\hat \theta=-\sin\theta\hat i+\cos\theta\hat j.$$
Therefore
$$\frac{d\hat r}{d\theta}=\hat\theta$$
| {
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Why does Bernoulli's equation only apply to flow along a streamline that is in viscid, incompressible, steady, irrotational? I am learning about hydrofoil on this website.
In a later video I watched, I learned that in the process of deriving Bernoulli's equation, $$constant=P/d+gh+1/2v^2$$ has to multiplied through by density. To keep the left side constant, fluid density has to be constant and thus is incompressible.
But what about other qualities like in-viscid, and irrotational? What do they mean? And why are they necessary?
| Bernoulli's equation is really like an energy conservation equation: if you multiply both sides by the mass flow $\dot{m}$ (also assumed constant) you get:
$$\frac12 \dot{m}v^2+\dot{m}gh+\dot{m}\frac{p}{d}=C$$
The terms are all energy per unit of time. The first one, $\frac12 \dot{m}v^2$, represents translational kinetic energy (per unit of time) of the fluid. But there's no term included for rotational kinetic energy (after all, fluid running through conduits rarely rotate!) So using Bernoulli we assume only translational motion of the fluid.
The equation applies only to inviscid fluids because fluids with significant viscosity experience viscous energy losses, which are not conserved: the energy lost due to viscous friction would have to be supplied, for example by extra pressure, to prevent deceleration ($\dot{m}$ decreasing).
| {
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Positron losing energy but is not destroyed In August 1932, Anderson photographed a positron originating from cosmic rays as it entered a bubble chamber, passed through a 6mm lead plate (in the process loosing energy, as apparent from the changed radius of its path through the chamber).
My question is, as it passed through the 6mm lead plate it did do some interaction with the lead: it lost energy. Why was the positron not annihilated? In its path it had to pass an estimated 1.6 billion electrons.
What is the proposed mechanism for it to interact with the lead, but avoid annihilation?
| Likely the dominant process for the positron in the lead plate is simply Bhabha scattering. This is no surprise. We must consider the lab energy of the positron.
At center-of-momentum energies less than twice the muon rest mass, the electron and positron annihilation is limited primarily to radiation of multiple photons. At energies above the eV scale, which is definitely the case for this positron going through 6 mm of lead, the probability of this process occurring is very small. Typically, the positron will need to scatter until it loses enough energy to annihilate.
For a relevant read, see Prantzos, N., C. Boehm, A. M. Bykov, R. Diehl, K. Ferrière, N. Guessoum, P. Jean, et al. “The 511 keV Emission from Positron Annihilation in the Galaxy.” Reviews of Modern Physics 83, no. 3 (September 29, 2011): 1001–56. doi:10.1103/RevModPhys.83.1001.
| {
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Does "IF isentropic THEN reversible" only holds for adiabatic processes? I know that: IF adiabatic and reversible THEN isentropic
First question: does the implication IF isentropic THEN reversible hold for adiabatic processes?
Second Question: if yes to the above, are there other processes other than adiabatic for which it is true?
| A process does not have to be adiabatic or reversible to be isentropic. If the difference in entropy between State A and State B is zero, since entropy is a state function, any oddball path (either reversible or irreversible) that can take you from State A to State B will be isentropic.
| {
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Meaning of the phase space in statistical physics I have a silly question about the phase space. I am confused with the meaning of points in phase space. Does the each point in phase space represent concrete particle of the system, or does it represent the whole state of the system? Our teacher told us, that we use the phase space to describe the development of each particle. It is not right, isn't it?
| A point in phase space represents the state of the whole system.
For example, if you have a system of $N$ particle with coordinates $\vec r_1, \dots, \vec r_N$ and momenta $\vec p_1, \dots, \vec p_N$, its general state will be a point in a $6N$ dimensional phase space:
$$\vec X = (\vec r_1, \dots, \vec r_N, \vec p_1, \dots, \vec p_N)$$
| {
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Why is oil a better lubricant than water? How come mineral oil is a better lubricant than water, even though water has a lower viscosity?
When two surfaces slide over each other with a gap filled with a fluid, the different layers of the fluid are dragged at different speeds. The very top layer touching the top metal surface will have the same speed as the surface itself, while the bottommost layer is stationary. The speed in the layers between is distributed linearly and there exist friction forces between those layers that slow the movement. Those frictional forces should be reduces however, if a fluid with a lower viscosity is chosen.
How come this is not so?
Does it have to do with water's polarity, so that it sticks to surfaces in a different way than oil?
|
Why Oil is Slippery
Explaining why oil is slippery requires a look at its chemical
properties. First, oil is non-polar, which means it does not have a
positive or negative charge. Some molecules, like water, have a
“charge distribution,” which means the molecule acts almost like a
battery, part of it has a positive charge and part of it has a
negative charge. The result, because positive is attracted to negative
and vice versa, is that water and other “polar” molecules stick to
each other. Oil doesn’t have this problem, so one oil molecule can
slide past another more easily than one water molecule can slide past
another.
Adding to the slipperiness of oil is its tendency to form distinct
layers through forces called Van der Waals forces, or more
specifically London Dispersion forces (a type of Van der Waals force).
These forces, which are the weakest known in science, can help old
things together, which would increase friction. However, oils have the
unique property of forming forces only within layers because the
molecules are essentially planar. Planar just means that molecules are
flat as the diagram below emphasizes and only take up space in two
dimensions rather than three. Without projections to attach to, forces
can only be distributed within the plane and so there are no forces to
bond one layer to the next. Thus, two layers of oil don’t bond to one
another to any great degree. ...
*
*http://www.petroleum.co.uk/oil-as-a-lubricant
I hope this answers your question
| {
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Physics Chain Problem
A chain of length d lies on a table, $d/4$ of it hangs loose on the side of it. Friction coefficient is 0.2. It is released and begins to slide. What is the final velocity with which it falls of the table?
My attempt:
$Wf = \bigtriangleup Ug = - \int_0^{0.75d} \! \mu_k g \frac{m}{d}(\frac{d}{4}+x) \, \mathrm{d}x = -0.4687\mu_k mgd$
Which is the change and loss in energy during the slide. My plan was to equate the initial gravitational potential energy to the final kinetic energy plus the energy lost from friction (calculated above), however there is no way to define the friction for the rope since it has two parts, with apparently distinct potential energies. Is there another way to do it?
| Yes, the work-energy method can be used. The particular difficulties you noticed can be dealt with by considering the chain in 2 parts : that part hanging down (which loses PE but experiences no friction) and that part remaining on the table (which does not lose PE but experinces friction).
Suppose the mass of the chain is $M$. In the initial position, the CM of the overhaning section is $\frac{d}{8}$ below the table. The section remaining on the table has $0$ PE. So the initial PE of the chain relative to the table is $-\frac14Mg(\frac{d}{8}) = -\frac{1}{32} Mgd$. When the last link of the chain slips off the table then the CM of the whole chain is $\frac{d}{2}$ below the table so the PE is $-\frac12 Mgd$. The decrease in PE is
$(\frac12-\frac{1}{32})Mgd = \frac{15}{32}Mgd.$
When length y of the chain remains on the table then the friction force on it is $\frac15Mg\frac{y}{d}$ where $\frac15$ is the coefficient of friction. The work done against friction is
$\int \frac15 Mg\frac{y}{d}dy = \frac15 Mg[\frac{1}{2d}y^2] = \frac{9}{160}Mgd$
where the limits of integration are from $y=\frac34d$ to $0$.
The KE of the chain when the last link slips off the table is therefore
$\frac12MV^2 = (\frac{15}{32}-\frac{9}{160})Mgd$
$V^2=\frac{33}{40}gd$
where V is the velocity at this point.
Alternatively you can apply Newton's 2nd Law :
$F = M\frac{dv}{dt} = M\frac{dv}{dy}\frac{dy}{dt} = -Mv\frac{dv}{dy}$.
When length y remains on the table the force accelerating the chain is $Mg\frac{d-y}{d}$, while the friction force retarding motion is $\frac15Mg\frac{y}{d}$.
The equation of motion is
$-Mv\frac{dv}{dy} = Mg(1-\frac{y}{d}) - \frac15Mg\frac{y}{d} = (1-\frac65\frac{y}{d})Mg$
$2v\frac{dv}{dy} = (2\frac65\frac{y}{d}-2)g$
$v^2 = (\frac65\frac{y^2}{d}-2y)g$.
The limits of integration are $v=0$ to $V$ and $y=\frac34d$ to $0$. So
$V^2 = (\frac65(0-\frac{9}{16}d)-2(0-\frac34d))g = \frac{33}{40}gd$.
| {
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Why do we add intensities for coherent sound sources? When there are N coherent sound sources playing the same note at equal loudness, their sound waves add up to make a sine wave of the same frequency but most likely different amplitude (can be anything from zero to N times the amplitude of one wave), depending on their relative phase shifts.
Can you prove that on average, the square of the amplitude of the resultant wave is N times the square of the amplitude of one wave? This would explain why we add intensities of coherent sources (intensity is proportional to the square of pressure) to get the average intensity of the resultant sound.
This page explains what happens for 2 waves, the new amplitude is
...and the average value of this is 2I_0.
I need to generalise this for N identical randomly shifted sine waves.
Thanks!
| Yes. Let's consider the phase shifts to be random. Then we can consider each amplitude to be an independent random variable $A_i(t)$. Each one has a variance of
$$\text{var}(A_i) = \langle A_i^2 \rangle - \langle A_i \rangle^2 = \langle A_i^2 \rangle \propto I_i$$
because the average amplitude should be zero, and intensity is proportional to amplitude squared. The variances of independent random variables add, so
$$I = \text{var}(A) = \sum \text{var}(A_i) = \sum I_i$$
as desired.
You might not like the 'random variable' phrasing, but it's not essential. At heart, the idea is that the $A_iA_j$ cross terms in $(\sum A_i)^2$ must average to zero. I call that property 'independence', you can call it something else.
| {
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Can an electron make quadrupole gravitational waves? A gravitational wave is a quadrupole wave. Now when an electron is accelerated it usually emits a photon. But can an electron also emits an gravitational wave? If so how does it 'make' an quadrupole wave?
| The answer by Physics Guy says:
Since the electron moves in spacetime and has mass, it produces gravitational waves. that can be derived from General Relativity.
This is wrong. Motion and mass are not sufficient criteria.
The answer by Árpád Szendrei says:
Now in your case, as an electron alone (as an isolated system) does not satisfy this, it would not emit gravitational waves. To create gravitational waves that are detectable at our current level of capabilities, you need a system of two objects, like two black holes, merging.
This is also wrong. If the electron is accelerated, then it is not an isolated system.
In general, the criterion for an object to emit gravitational waves is that its mass quadrupole moment has to vary with time. In most cases, an accelerated electron would satisfy this criterion, so the answer to the question is yes.
| {
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