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Mass loss in Red Giants via dusty-winds and chromosphere activity I'm reading some literature on mass loss in the RGB/AGB branches and so far I'm getting a lot of information regarding mass loss via dusty-winds/pulsations but almost no explanation of mass loss by 'chromosphere activity'. I asked my mentor about it and he couldn't come up with anything so I'm turning to whomever might be able to throw a few references or explanations my way.
| I guess what is meant is mass loss powered by non-radiative heating occurring above the photosphere. i.e The outer atmosphere is heated by magnetic Alfven waves and causes a wind to stream away from the star. The same magnetic field is responsible for heating a chromosphere - a hotter layer above the photosphere.
Here is a paper by Schroder and Cuntz (2005), though not really at an introductory level.
http://iopscience.iop.org/1538-4357/630/1/L73/pdf/19311.web.pdf
An update: Recent theoretical modelling claims to have reproduced some of the mass loss and chromospheric properties of the K5 giant Aldeberan, through the dissipation of Alfven waves in the chromosphere.
Airapetian et al. (2014) - http://arxiv.org/pdf/1409.3833v1.pdf
| {
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What is the RC time constant in a superconductor? In conventional conductors, the RC time constant is the time required to charge or discharge a capacitor through a resistor by ≈ 63.2 percent of the difference between the initial value and final value:
$$\tau = R \cdot C $$
However, in a superconductor, the resistance is exactly zero.
$$\tau = 0 \cdot C = 0$$
Which would require infinite current because the capacitor is charged in 0 time. So the RC time constant equation above must not be valid for superconductors. How should the RC time constant be defined in a superconductor such that it does not require infinite current?
That is, what is the expression for the time constant of charging a superconducting capacitor by ≈ 63.2 percent?
Edit: I re-framed the question because we were getting bogged down about signal propagation time. Now instead of using special relativity to point out the flaw in the equation, I note that it would require infinite current, which would not be physically realizable.
| The RC constant is not fundamentally related to the speed of signals. It is derived from an equation which already assumes the speed of signals in your circuit is effectively infinite. Therefore there is no relativistic constraint on RC time. To arrive at a formula which takes into account the propagation time you would need to account for the impedance of the wires at least.
Superconductors display a variety of complicated behaviors when put into larger circuits. To give you a more concrete formula would require a more concrete circuit. The simplest possible behavior is that of a perfect inductor.
Also to correct errors in the comments, the speed of signals in a conductor has almost no relation to the "speed of electrons".
| {
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A water pipe from sea level to beyond the atmosphere If a pipe extended from just above the ocean floor to outside the atmosphere, would water be sucked up it by the vacuum beyond the atmosphere? If a hole was made in the pipe, above sea level, how would that affect the flow of water? Would it stop it completely?
| Ask yourself this: why doesn't the vacuum of space just suck away our atmosphere? The reason is because of the earth's gravity, which pulls on the gas envelope around the planet to keep it in place. The phenomenon we call 'air pressure' is also the result of this.
The tube will only fill until its contents are being pulled down by gravity with the same force as the rest of the atmosphere. If the straw starts out with air in it, then the air in the straw will be pulled exactly as hard as any other column of air that size, and so the water will not move at all. If there is no air in it, only about 10 meters of it will fill (like Hennes said), only until the amount of extra water in the tube weighs the same amount as an equivalent area of the atmosphere does.
| {
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Are Lagrangians and Hamiltonians used by Engineers? Analytical Mechanics (Lagrangian and Hamiltonian) are useful in Physics (e.g. in Quantum Mechanics) but are they also used in application, by engineers? For example, are they used in designing bridges or buildings?
| I'm not an engineer myself, but as far as I know, Lagrangians and Hamiltonians have their use in complicated but calculable situations. In order to solve the e.o.m. for a bridge or so, most engineers rely on specific programs for exactly such a purpose that calculate the statics using Newtonian mechanics and a finite element approach.
| {
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Does general relativity fail in conditions with very large gravitational forces? It is said in this wikipedia article (in the 7th paragraph) that where there exists huge masses and very large gravitational forces (like around binary pulsars), general relativistic effects can be observed better:
By cosmic standards, gravity throughout the solar system is weak. Since the differences between the predictions of Einstein's and Newton's theories are most pronounced when gravity is strong, physicists have long been interested in testing various relativistic effects in a setting with comparatively strong gravitational fields. This has become possible thanks to precision observations of binary pulsars.
But here in whystringtheory.com (in the last paragraph), it is said that
For small spacetime volumes or large gravitational forces Einstein has little to offer
I know that in singularities, GR fails and this is a motivation for quantum gravity theories. But the second quote above says in small spacetime volumes or large gravitational forces.
Is there any problem with general relativity in conditions with very large gravitational forces (in big enough volumes of spacetime)?
| The whystringtheory page is written in a popularized style that makes it impossible to tell what they really have in mind. Their statement doesn't make sense if interpreted according to the standard technical definitions of the terms. GR doesn't describe gravity as a force. In the system of units normally used in GR, with G=c=1, force and power are unitless, so there is also no natural motivation for defining something like a Planck force by analogy with the Planck length, etc. Possibly their "force" really means curvature, in which case this could be interpreted as a correct statement that GR breaks down when the Riemann tensor corresponds to a radius of curvature comparable to the Planck length.
| {
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How is energy extracted from fusion? I understand that combining deuterium and tritium will form helium and a neutron. There are three methods to do this (1) tokamak (2) lasers and (3) cold fusion. I would like to know after helium is formed. How is that energy extracted from tokamak and stored?
| As others mentioned it depends on what the product is of the reaction. It's hard enough to get deuterium-tritium reactions to happen efficiently, but if we get the science down it usually gets more efficient as we scale up. If we can get p + B11 -> 3 He4 to work it yields 3 highly energetic He4 ions (alpha particles), which have a +2 eV charge and here is where I am fuzzy:
They have high energy, but are also highly affected by charge, so if the fuel was electrostatically confined, a bit of the potential energy will be given up as it climbs out of the well. Assuming it does escape with most of it's energy still, they easily interact, so by using a funnel that decelerates those alpha particles and induces current, it can directly capture this energy as electrical current, without steam turbines. That said isolating them from the fuel could be tricky.
I mean you can still do the steam turbines but p 11B is aneutronic, so the neutrons are less than 1% of the energy produced.
| {
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Change of orbit with radial impulse This probably is dumb question, but it has been giving me issues for days.
I was thinking about a point of mass $m$ in a circular orbit around a planet. The question is: giving a radial impulse, what is the change in the shape of the orbit? Will it change in another circular orbit? Or maybe will undergo a spiral motion?
|
Will it change in another circular orbit?
The radial velocity of a circular orbit is zero. Since the radial impulse imparts a radial velocity to the point mass, the new orbit cannot be a circular one.
Or maybe will undergo a spiral motion ?
No, assuming two ideal point masses in the Newtonian context and no drag, the new orbit must be a conic section as Deer Hunter points out in a comment.
| {
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Absorption & emission spectra The process of obtaining an absorption spectra involves passing a complete spectrum of light from the material under consideration. The material absorbs the specific wavelength and allows the rest to pass through. But the reason of this absorption is given as the specific quantum gaps between the various electronic energy levels in the material and when an electron jumps to a higher level only a specific wavelength photon is absorbed. But the higher energy state is considered to be unstable and thus the electron will fall back immediately and would again give the wavelength preventing any spectra to be formed. The same doubt is in emission spectra when used to describe various flame colours but the same doubts apply there too.
| You do not state your question clearly
But the higher energy state is considered to be unstable and thus the electron will fall back immediately and would again give the wavelength preventing any spectra to be formed. The same doubt is in emission spectra when used to describe various flame colours but the same doubts apply there too.
I suppose you are asking "why we can see absorption and emission spectra after all, if de-excitation happens almost immediately".
Consider what happens with light absorption : A beam of light falls on the sample and one of its photons with the appropriate energy excites the electron to a higher state. The electron almost instantaneously falls back into the ground state BUT it has now got the four pi spherical angles to be emitted in, instead of in the direction of the beam. If one puts the film/detector well away from the beam direction one will see the emission spectrum. The film in the direction of the beam will be depleted at the frequencies of the absorption and show up as dark lines.
| {
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Calculating the torque at a point when a motor is stopped? So, I'm trying to solve for the torque $\tau_A$ of a motor. I have attached a strong stick to the motor, like so:
I apply a force $F$ on the stick which stops the motor. The distance from the outside edge of the cylinder to the end of the stick is $L$. The torque for the motor is $\tau_A=F(L+r)$, $r$ is the radius.
My friends believe that the torque at point $B$ is $\tau_B=-FL$, but I believe even though the motor is not moving (due to the force), it still applies a torque at point $B$. It would be less than $\tau_A$ since it doesn't push around the point uniformly, but it should be $\tau_B=-FL+\tau_Ac$ ($c$ is a constant). Using their method, they got that $\tau_A=0$, which I believe happened because in calculating the torque at point $B$, they make $\tau_A=0$.
Who is right? How do I calculate how much $\tau_A$ is applied about point $B$ (assuming I'm correct)?
| The first part of your analysis is right. Your hand exerts a torque
$$\tau_{hand} = F(L+r)$$
about the center of the disk. The disk is stationary, so it must have zero net torque on it, so
$$\tau_A = -F(L+r)$$
As long as the system you want to analyze is the disk and rod together, that's the end of the story.
Evidently, your friend wants to analyze the system of just the disk, with force from the rod on the disk considered as an external force. You can do that, but the story is a little more complicated if you do. Although the net force from the stick on the disk is $F$, the force takes the form of a pressure distributed over the area over which the rod and disk are in contact. This pressure creates a force acting in different directions at different points over that contact area. The net torque from this force on the disk about the disk's center of mass is still $-F(L+r)$, which comes from integrating the moment of the pressure over the area of contact. You still have the same equation for $\tau_A$. Anyone whose ultimate conclusion was $\tau_A = 0$ was incorrect.
I find the part about $\tau_B$ confusing because it is not clear what body the torque is acting on and what forces are being considered as making up $\tau_B$.
| {
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How can you weigh your own head in an accurate way? I read some methods but they're not accurate. They use the Archimedes principle and they assume uniform body density which of course is far from true. Others are silly like this one:
Take a knife then remove your head.
Place it on some scale
Take the reading
re-attach your head.
I'm looking for some ingenious way of doing this accurately without having to lie on your back and put your head on a scale which isn't a good idea.
| If you can find a way to calculate the density of your head by looking at the amount of the head that is - for example - bone, brain or space, and take the density of each of them, then you could get an average density for the whole head. Then simply measure the volume of the head by possibly submerging it and work out the mass.
Of course this isn't accurate in the slightest, but at least it's an answer.
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Vectors with more than 3 components
*
*I have some confusion over Vectors, Its components and dimensions. Does the number of vector components mean that a vector is in that many dimensions? For e.g. $A$ vector with 4 components has 4 dimensions?
*Also, how can a Vector have a fourth dimension? How can we graphically represent vectors with more than 3 dimensions? Its hard for me to visualize such a vector, Can anyone point me to some resource where they explain this graphically and in detail?
| The most basic way to imagine 4 dimensions (or more) is as follows:
Picture four straight lines that are all perpendicular to one another. Imagining 3 mutually perpendicular lines is not a challenge, it's the fourth one that's a challenge. The reason for our three dimensional vision is that our retinas are only a 2 dimensional plane, and depth perception adds the 3rd dimension.
One way that I go about imagining a 4 dimensional vector is visualizing a 3 dimensional vector, and attributing some other characteristic as its fourth dimension (usually its color). This also somewhat touches on another part of your question, which is how can a vector have 4 dimensions? A vector doesn't always have to represent a location in space. It can also simply be a set of attributes that form an ordered pair. This is especially common in graphics programming, where color is represented as a vector (each component representing one of the four ARGB channels).
In general relativity (sometimes referred to as Einsteinian coordinates as opposed to Euclidean coordinates), time is a fourth dimension. So the point in time that an object exists and its location in space form a vector in spacetime.
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How does autorefractor work? Autorefractors are being used by eye opticians for eye diagnosis. I searched internet for articles and wiki page as well but I wasn't satisfied. I am interested to know how do they actually work. How the machine is able to focus sharp on retina automatically. How do they figure out spherical/cylindrical aberration for human eye. I even asked optician as well but he tricked out of this position saying it does all happen automatically. LOL!
A detailed answer would be much appreciated. Please don't hesitate to write any technical details or mathematical derivations. I would love to know how these machines actually work.
Thanks in advance!
| the autorefractor projects an image into the eye. the light rays pass through the lens and strike the retina. a small amount of the light bounces off the retina, passes through the lens a second time and exits the eye. Imperfections in the shape of the eye's lens distort and defocus the "return" image. the autorefractor senses the distortions and misfocus and tweaks the projected image with lenses of its own until the "return" image is in focus and distortion-free. the software in the autorefractor then deconvolves the tweaks and back-calculates the corrective lens prescription that the lens in the eye needs to properly focus the image.
| {
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Are quantum effects significant in lens design? Over on Photography, a question was asked as to why (camera) lenses are always cylindrical. Paraphrasing slightly, one of the answers and follow-up comments asserted that quantum effects are significant and that you need to understand QED in order to understand lens design. My intuition tells me that this probably isn't the case, and that modern lens design can be handled via classical optical theory. Is my intuition correct, or are quantum effects really significant in modern lenses?
| Your intuition is correct, you don't need quantum electrodynamics to explain/model/engineer camera lenses. When considering the propagation of light, the results of geometric optics can be interpreted in terms of path integrals, as Feynman does in his QED: The Strange Theory of Light and Matter, but this is not necessary for lens design. Geometric optics itself suffices in most cases, but there are some design tasks which require an incorporation of the wave nature of light (or the application of empirical rules). (Thanks @Matt J for this info)
| {
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What is physics behind of explosion under Atmospheric pressure? An explosion is a rapid increase in volume and release of energy in an extreme manner.
A blast wave in fluid dynamics is the pressure and flow resulting from the deposition of a large amount of energy in a small very localised volume.
The equation for a Friedlander waveform describes the pressure of the blast wave as a function of time:
$$P(t)=P_oe^{-\frac {t}{t^*}}(1-\frac {t}{t^*})$$
where P$_o$ is the peak pressure and t$^*$ is the time at which the pressure first crosses the horizontal axis (before the negative phase).
My Question:
Why does the pressure temporarily drop below ambient after the wave passes?
| If I understand correctly what you're asking, you are referring to the vacuum created from an explosion. Think about it like this: An explosion is rapidly expanding gas from a central point (the explosive material). As that gas expands, it pushes all of the material (the gas of the atmosphere) away from that central point. The front of the gas (shockwave) is what pushes the atmosphere. This leaves a vacuum behind the shockwave. I believe that is the negative pressure to which you are referring. This is best illustrated in the old cold-war era nuclear test footage that shows the bending of the palm trees. When the explosion goes off, the shock front pushes out in a generally 360 degree pattern in 3-dimensional space. This will cause the palm tree to bend away from the explosion. Then, shortly after, you will see the trees bend towards the explosion. This is due to the atmosphere rushing back in to fill the vacuum created by the shock front. These different phases of the explosion are called the positive phase - when the pressure moves away from the blast - and the other is cleverly named, the negative phase - when the surrounding fluid (the atmosphere in this case) rushes back in to fill the vacuum created by the positive pressure phase.
I hope that was the drop in pressure to which you were referring.
| {
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Is there an EMF in a conductor moving at constant speed across the uniform magnetic field If a conductor - a long rod - moves at constant speed across the "lines" of a uniform magnetic field, is there an EMF within this conductor? Or, if a conducting rod rotates at uniform rate, pivoted in the middle or at one of its ends in a uniform magnetic field perpendicular to the plane of rotation, is there an EMF generated within the conductor?
| Faraday's law relates to the amount of energy a charge would gain by going around a loop. This energy per unit charge is called an "EMF". Because there is no change in the flux through any surface in this situation, there is no loop that gives a charge a different energy.
This doesn't mean that there is no path that gives the particle a different energy. As Alfred Centauri's answer illustrates, a conductor moving through a magnetic field is subject to the Hall effect, which will create a voltage between the two ends. This is not a violation of faraday's law, because the closed-loop voltage remains zero.
| {
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1D Ising Model with different boundary conditions The Hamiltonian for one-dimensional Ising model is given by,
\begin{equation}
\mathcal{H} = -J\sum_{<ij>} S_iS_j; \quad i,j=1,2,...,N+1
\end{equation}
where $<ij>$ denotes that there is nearest neighbor approximation. The partition function is given by,
\begin{equation}
\mathcal{Z}=\sum_{\{S_i\}} e^{-\beta \mathcal{H}(S_i)}
\end{equation}
with $\beta = \frac{1}{k_BT}$, where $k_B$ is the Boltzmann constant and $T$ is the temperature.
Now my questions are:
1. How to compute the partition function when $S_{N+1}=+1$ while other spins ($S_i$ for $i=1,2,...,N$) may take value $+1$ or $-1$?
2. How to compute the partition function when $S_{N+1}=-1$ while other spins ($S_i$ for $i=1,2,...,N$) may take value $+1$ or $-1$?
| 1.
\begin{equation}
\mathcal{Z}(N+1,+)= \sum_{S_1}...\sum_{S_N} e^{K(S_1S_2+S_2S_3+...+S_{N-1}S_N}e^{KS_N}
\end{equation}
where $K=\beta J$.
We define new variables,
\begin{equation}
\eta_i =S_i S_{i+1}; \quad i=1,2,...,N-1
\end{equation}
The $\eta_i$s take value:
\begin{equation}
\eta_i= \left\{
\begin{array}{l l}
+1 & \quad \text{if} \quad S_i=S_{i+1} \\
-1 & \quad \text{if} \quad S_i \neq S_{i+1}
\end{array} \right.
\end{equation}
Then the partition function becomes,
\begin{array}
\mathcal{Z}(N+1,+) &= \sum_{S_N}\sum_{\{\eta_i\}} e^{K\sum_{i=1}^{N-1}\eta_i}e^{KS_N} \\
&= \sum_{S_N} \left\{\prod_{i=1}^{N-1}\sum_{\{\eta_i\}} e^{K\eta_i}\right\}e^{KS_N} \\
&= \sum_{S_N} \left(2coshK\right)^{N-1} e^{KS_N} \\
&= \left(2coshK\right)^{N-1} \left(e^K+e^{-K} \right) \\
&= \left(2coshK\right)^{N}
\end{array}
2. In a similar way we can show that,
\begin{equation}
\mathcal{Z}(N+1,+) = \mathcal{Z}(N+1,-) = \left(2coshK\right)^{N}.
\end{equation}
| {
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What causes a black-body radiation curve to be continuous? The ideal black-body radiation curve (unlike the quantized emission seen from atomic spectra), is continuous over all frequencies. Many objects approximate ideal blackbodies and have radiation curves very similar in shape and continuity to that of an ideal black-body (often minus some emission and absorption lines from the atoms in an object, such as radiation curves seen from stars). I am wondering what exactly gives rise to a basically continuous black-body radiation curve in real objects? Since atomic energy states are quantized, it seems real life black-body curves would have some degree of measurable quantization to them (or perhaps the degree of quantization is so small the radiation curves look continuous).
| If the absorptivity of a medium really was discrete, then there would be no way it could emit blackbody radiation. The defining characteristic of a blackbody is that it absorbs light of all frequencies that are incident upon it (and that it is in thermal equilibrium). There is a close relationship (a direct proportionality) between the Einstein absorption and emission coefficients for atomic, ionic and molecular processes which ensures this.
So whilst you can imagine hypothetical materials with discrete absorption spectra caused by "delta function" spectral lines, you cannot also hypothesise that these would emit blackbody radiation - they would not.
In practice the absorption coefficients of real materials are not delta functions at fixed frequencies. Electronic transitions have finite widths - there is natural broadening, doppler broadening, pressure broadening. Real materials also have continuous absorption coefficients caused by photoionisation, free-free absorption, inelastic scattering etc. These effects cause the absorption coefficient to be non-zero at practically all frequencies. In those circumstances, to get a continuum blackbody we simply need to arrange to have enough material present that it is optically thick (that is, it has an optical depth much larger than unity) at all relevant frequencies. If that is so, and the material is in thermal equilibrium (energy levels populated according to Boltzmann factors etc.) then it will emit what is close-to-blackbody radiation.
| {
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What is the probability density function over time for a 1-D random walk on a line with boundaries? If a single particle sits on an infinite line and undergoes a 1-D random walk, the probability density of its spatio-temporal evolution is captured by a 1-D gaussian distribution.
\begin{align}
P(x,t)&=\frac{1}{\sqrt{4 \pi D t}}e^{-\frac{(x-x_0)^2}{4Dt}}
\end{align}
However, suppose there are impassable boundaries on the line; on one side, or on both sides. Are there any boundary conditions for which there exists a closed form probability density function for how this particle will behave over time? Any references to such solutions would be extraordinarily helpful.
EDIT. Attempting to generalize Emilio's result below for an arbitrary initial particle position $-L/2 < x_0 < L/2$.
I had to work it out by example. I found the following "images" were required to account for reflections of an off-center particle at position $x_0$: for the first and second reflections on both sides the new gaussians had to be centered on ($-2L+x_0$, $-L-x_0$, $x_0$, $L-x_0$, $2L+x_0$). From the pattern I think the full solution can be expressed, for all integers $n$, as:
\begin{align}
P(x,t)&=\frac{1}{\sqrt{4 \pi D t}}\sum_{n=-\infty}^\infty e^{-\frac{(x-nL-(-1)^nx_0)^2}{4Dt}}
\end{align}
where the old $x_0$ is now defined as $nL+(-1)^{n}x_0$
| This can probably be solved by the method of images, depending on your precise formulation of the problem. The main idea would be to place image particles at the initial time at positions given by treating your impassable boundaries as mirrors; this makes the probability flow at the boundary zero.
To give a more precise formulation, suppose your problem is
$$
\frac{\partial P}{\partial t}=D\frac{\partial^2P}{\partial x^2}\text{ under }\frac{\partial P}{\partial x}(-L/2,t)=0=\frac{\partial P}{\partial x}(L/2,t)\text{ and }P(x,0)=\delta(x),
$$
where I've initially placed the particle in the middle of the barriers for simplicity but this can be altered. The solution is then given, by linearity, by your expression, added up for $x_0=nL$ for all integers $n$:
$$
P(x,t)=\frac{1}{\sqrt{4 \pi D t}}\sum_{n=-\infty}^\infty e^{-\frac{(x-nL)^2}{4Dt}}.
$$
This can be solved exactly in terms of Jacobi theta functions, which makes the calculations and graphing a lot faster, but does not necessarily (at a first go) make this easier to work with:
$$
P(x,t)=\frac{1}{L}\vartheta _3\left(\frac{\pi x}{L},e^{-\frac{4 D \pi ^2 t}{L^2}}\right).
$$
(For asymetrically placed initial particles, you would have two series of gaussians separated by $2L$, so therefore two theta functions.)
I'm not sure this is very useful by itself, but the method of images is very powerful.
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The stability of D-Brane In "String Theory and M-Theory: a modern introduction" by K.Becker, M. Becker and J.H.Schwarz, they say that BPS D-brane is stable as it preserves half of the Supersymmetry. I really want to understand more about this statement and see detail calculations. What is the mechanism of D-brane stability? Is there any derivation for the instability of space-filling D-brane (so that open string tachyon will be eliminated from the theory)?
Thank you.
| BPS objects are stable because they're the lightest objects with given values of certain conserved charges. So there exists no potential final state that would be lighter and that the BPS state could decay into, by conserving the energy. The excess energy may be invested to the kinetic energy of the final energy but a deficit energy means that the decay is prohibited.
As an analogy, note that the electron has to be stable because there exists no lighter $Q=-e$ object than the electron (and positron).
BPS objects are either those preserving some (enough) supersymmetry; or objects saturating the BPS bound $M=Q$, schematically speaking (for branes, it's the tension equal to the charge density; coefficients should be inserted everywhere). These conditions are equivalent because
$$ \{Q,Q\} = M-Z $$
schematically, for some conserved supercharge $Q$. So the expectation value of $\{Q,Q\}$ in the BPS state is zero – because $Q$ annihilates the state – but it's also equal to $M-Z$ which means that the mass is equal to the charge. For non-BPS states, we have the strict $M\gt Z$. Here, $Z$ is the conserved central charge.
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Understanding the different kinds of mass in gravity On this site, the Phys.SE question Is there a fundamental reason why gravitational mass is the same as inertial mass? has been asked. See also this Phys.SE question. The 'answer' provided on this forum has been that the curvature of spacetime explains both. The answer is still cryptic for me as I am more a concrete thinker.
Newton said $F=ma$. I can use this formula to measure inertial mass. Experimentally I can measure the motion of an object while applying a constant force to it. Newton also said $F=\frac{GMm}{r^2}$. In this case, what simple experiment will allow the measurement of gravitational mass?
| Mass universe is constituted of micro / macro mass natural bodies/systems. It is known that the mass (micro / macro) natural bodies is concentrated mainly in neutrons, protons and electrons as entities with a certain stability, fig. 3.
Fig. 3. (Electro)convergence of the electron/neutron /proton, )[4]
1- Neutron matrix (local substantial body/ local wave induced environmental, mi); 2 – Neutrino flux, JdV,( cilyndrical cloth neutrinos Mi/H1, penetrated the transverse plane of matrix neutron); 3- Rotation moment of neutron matrix, respectively vortices, mi (with speed v, ω ); Pv’, Pv - electroconvergence vortex magnetic moments (coupling) of underground neutrino, Mi, respectively, neutron/ mi;
Neutrino flux, JdV, (2) unbalanced spatial neutron matrix, (1) and maintains a transport phenomenon (on the environment) and displacement (in counter flow) of entities (pre) mass, mi, (3). For reasons of logical presentation, we associate pulsating neutrino mass rotating blade / spine, mi, (3) respectively neutron mass mN =ρ dV, an electric charge, q = Dm, where D is the constant magnetic universal. To describe the dynamics of (micro) local mass systems, N, with mass, mN =ρ dV , is associated with the force of gravity, dVg , and neutrino flux, JdV, ( mass emerge entities induced environmental) magnetic force, J BdV, so that it follows the equation for mass magnetic interaction between natural bodies:
F =pdV+JB (2)
In the event that all substances/neutrino flux, Mi/JdV, of the moving with the same speed v, the resultant current of matter, J = Dv , so that equation (2) can be written as:
F = (g+Dv+B) (3)
Formula (3) represents the equation of gravity for the mass unit of the substance
D crivoi
| {
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What caused scientists to study Black Body Radiation? After spending hours understanding what exactly Black Body radiation and Ultraviolet catastrophe is, I cannot help myself asking what was the reason that scientists such as Wilhelm Wien and Max Planck studied Black Body Radiation in the first place? What intrigued them to study a hypothetical situation? What were they looking for exactly that made them study this phenomenon?
| In the late 19th century physics seemed more or less complete, in the sense that it explained everything that could be measured. However when applied to a black body the accepted physics of the day predicted that the black body would emit an infinite amount of energy, and this was obviously in conflict with experiment. There's nothing hypothetical about this - theory predicted one thing and experiment measured something different.
This is an exciting situation for any ambitious physicist because it means that something must be wrong with the existing theory. Indeed, resolving the problem required the creation of quantum mechanics, and immortality for Max Planck (Wien is less of a household name, but he still has a law named after him ;-).
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What is canonical momentum? What does the canonical momentum $\textbf{p}=m\textbf{v}+e\textbf{A}$ mean? Is it just momentum accounting for electromagnetic effects?
| The whole problem starts when you try to do electromagnetism with the Lagrangian because you can't write the magnetic field in terms of a potential. However we CAN write it in terms of a vectorpotential $\vec{A}$:
$\vec{B} = \nabla\times\vec{A}$.
It seems that this is usefull and can be used to derive the appropriate Lagrangian and Hamiltonian which are given and checked here.
It seems (from the calculations given in the link) that to include the magnetic field, we need to replace our momentum with:
$\vec{p}-q\vec{A}$.
By replacing the momentum by this term, you are able to do Lagrangain and Hamiltonian mechanics (which work with potentials) for magnetic fields (which can't be written in terms of a potential).
For electric fields you can still include them by using the electronic potential.
| {
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Lorentz homogeneous group and observables For generators of the Lorentz group we have the following algebra:
$$
[\hat {R}_{i}, \hat {R}_{j} ] = -\varepsilon_{ijk}\hat {R}_{k}, \quad [\hat {R}_{i}, \hat {L}_{j} ] = -\varepsilon_{ijk}\hat {L}_{k}, \quad [\hat {L}_{i}, \hat {L}_{j} ] = \varepsilon_{ijk}\hat {R}_{k}.
$$
For the splitting of algebra, we can introduce operators
$$
\hat {J}_{k} = \hat {R}_{k} + i\hat {L}_{k}, \quad \hat {K}_{k} = \hat {R}_{k} - i\hat {L}_{k}.
$$
So
$$
[\hat {J}_{i}, \hat {J}_{j} ] = -\varepsilon_{ijk}\hat {J}_{k}, \quad [\hat {K}_{i}, \hat {K}_{j} ] = -\varepsilon_{ijk}\hat {K}_{k}, \quad [\hat {J}_{i}, \hat {K}_{j}] = 0.
$$
So, each irreducible representation of Lie algebra is characterized by $(j_{1}, j_{2})$, where $j_{1}$ is max eigenvalue of $\hat {J}_{3}$ and $j_{2}$ is max eigenvalue of $\hat {K}_{3}$.
Then I can classify objects that transform through the matrices of the irreducible representations,
$$
\Psi_{\mu \nu}' = S^{j_{2}}_{\mu \alpha }S^{j_{2}}_{\nu \beta}\Psi_{\alpha \beta},
$$
where $S^{j_{i}}_{\gamma \delta}: (2j_{i} + 1)\times (2j_{i} + 1)$.
For $(0, 0)$ I have scalar field, for $\left(\frac{1}{2}, 0\right); \left(0; \frac{1}{2}\right)$ I have spinor, for $(1, 0); (0, 1)$ I have 3-vectors $\mathbf a, \mathbf b -> \mathbf a + i\mathbf b$ creating antisymmetrical tensor etc.
Also, for scalar $j_{1} + j_{2} = 0$, for spinor - $\frac{1}{2}$, for tensor - $1$. So, the question: is sum $j_{1} + j_{2}$ experimentally observed? Is it connected with a spin?
| Yes, the comination $j_1 + j_2$ determines the spin of the particle. Note however, that this is an addition of angular mementum which may be complicated.
Furthermore, you can count the degrees of freedom:
In $(j_1, j_2)$, each contribute $2j_1 + 1$ states and we construct a tensor product, so $(j_1, j_2)$ gives $(2j_1 + 1) * (2j_2 + 2)$ degrees of freedom. For the vector we have $(1/2, 1/2) \mapsto 2 * 2 = 4$ degrees of freedom.
If the representation is reducible, i.e. of the form $(j_1, j_2) + (k_1, k_2)$, then you simply add the d.o.f. you get from each pair. The dirac spinor has $(1/2, 0) + (0, 1/2) \mapsto (2) + (2) = 4$ degrees of freedom, the field-strength tensor has $(1, 0) + (0, 1) \mapsto 3 + 3 = 6$ d.o.f.
As a sidenote: the representations $(1, 0)$ do not corresond to vectorlike degrees of freedom, but rather to antisymmetric self-dual tensors. $(0, 1)$ is the antisymmetric anti-self-dual tensor. The vector (and the only way to get a vector out of this) is $(1/2, 1/2)$!
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Would connecting p-type and n-type semiconductors work as a diode? If we placed p-type and n-type semiconductors close enough to be touching (see fig. 1), would this arrangement work as a diode? Please explain.
Fig. 1 - Connecting p-type and n-type semiconductors
| A pn junction is one piece of a semiconductor that receives n-type doping in one section and p-type doping in an adjacent section. If you simply stick two p-type and n-type semiconductors to each other by hand, it will not behave as a diode.
The main reason that a pn junction can behave as a one-directional device is it's built-in potential. Upon formation of the pn junction (In a processas I said above), sharp gradients of carrier densities across the junction result in a high current of electrons and holes and these carriers leave ionized atoms as they cross and a depletion region is formed. An electric field will result from these charges that causes a built-in potential across the junction. You can find more explanation about how a diode works in this question.
If you put a p-type semiconductor in contact with a n-type the above processes can't happen.
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What is the sign of the work done on the system and by the system? What is the sign of the work done on the system and by the system?
My chemistry book says when work is done on the system, it is positive. When work is done by the system, it is negative.
My physics book says the opposite. It says that when work is done on the system, it is negative. When work is done by the system, it is positive.
Why do they differ?
| It is just a matter of convention. It should be consistent throughout.
Case 1: Work done on the system is positive.
Here the first law is written as
$$ \mathrm{d}U = \mathrm{d}Q + \mathrm{d}W \,.\tag{1}$$
If your frame of reference is "system", then the work done on the system ($W$) is positive and the heat that is added to the system is also positive, which means the change in internal energy is also positive by first law of thermodynamics, which means that there is an increase in temperature. This appeals to common sense. Here positive change in internal energy corresponds to increase in temperature
Case 2: Work done by the system is positive
Here the first law is written as
$$ \mathrm{d}U = \mathrm{d}Q - \mathrm{d}W \,. \tag{2}$$
If work is applied to the system, $\mathrm{d}W$ term becomes negative making two negatives positive, which is identical to equation (1) and heat added to the system is still positive here. Rest of the arguments follow as above.
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Steering forces on a bicycle I always notice this weird thing and try to overcome it but cant.
As shown in the image when I ride the bike by just one hand and pull the handle back say from the right side so as commonly the handle should rotate towards right and the bike should turn to right. But that doesn't happen. No matter what I do the handle turns to left and the bike always go to left. And when I try to push it forward so that it turns to left then the handle turns to right and so the bike also turns to right.
As from how much I know about laws of motion I don't know why it happens. Please explain as it is too much weird and interesting for me.
| Pulling the right handlebar rotates the front wheel clockwise. This causes the bottom of the bike to move to the right. If you haven't leaned to the right as part of this maneuver then you are now out of balance and leaning to the left (your body hasn't moved, but since the bottom of the bike moved right you are now leaning left). If you don't correct (by turning left or leaning right) then you'll fall over to the left due to balance, not anything to do with the forward/inward/outward forces on the bike.
| {
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Would a small puddle of water evaporate faster if you spread a dry towel over it? Let's say you spill 10ml of water on the kitchen counter. It forms a small puddle that would evaporate after a while (assuming room temperature and sane humidity).
Would spreading a large, dry towel over the puddle cause the water to evaporate quicker? What I mean is a towel that's much larger than the diameter of the puddle. Say 4x as large.
On the one hand, the water by itself comes into direct contact with the air which causes it to evaporate. On the other hand, assuming the towel absorbs the water almost entirely, it increases the wet surface area significantly, but also "locks" away some of the water in the fabric where it has less contact with air.
| I think it would evaporate quicker for anything we normally call a "towel". The evaporation of the plain puddle is limited by the surface area of the water. A towel provides many capillary pathways for the water to diffuse thru the fabric, eventually presenting a much larger surface are for evaporation. Other fabrics could be hydrophobic and decrease overall water/air surface area, but the point of a towel is to do the opposite.
Assuming the towel is clean and not causing additional impurities to dissolve into the water, the vapor pressure of the water should still be the same. The towel is then just a mechanical support for lots of capillary channels. I don't think the vapor pressure is lower just because the water is held by capillary action between structures like cloth fibers.
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Reflection, transmission, absorption...how to calculate them? I was wondering whether there is an equation that enables me to calculate the reflection, transmission, absorption and polarization, when the electric field everywhere is given?
Consider this: You have solved the full Mie scattering process, so incident field, the field in the sphere and the scattered field are known. How can one calculate those quantities then?
| For example, you can calculate absorption through integration of the Poynting vector over the surface of the sphere. I am not sure there are any standard general definitions of transmission and reflection for diffraction on a sphere.
| {
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How to prove that we are living in a 3+1D world? Is there any scientific experiment that can lead us to conclude we live in 3 spatial dimensions without the premise of the conception of limited dimensions?
Thank you all who helped in the improvement of this question (which was not clear at first).
EDIT:
I know that this can be a little philosophical, but it is also a scientific question.
Let's consider the scenario where the mankind was not ever able to see.
Let's also consider that this limitation could be surpassed thus not limiting us to reach a scientific and technological knowledge "similar" to what we have today.
Would this civilization of blind people reach the conclusion that they are living in a 3D spatial world?
Is the sense of touch enough to reach that conclusion? Is there any scientific experiment that can lead us to that conclusion without the premise of the conception of limited dimensions?
Would it be easier, harder, or just different to reach a conclusion predicted by the M-Theory? (please do not focus only on this last question)
| First , human brain would not be able to benefit from the great amount of information that electromagnetic force provide about the natural world . Human beings will then resort to mechanical means to interact with the environments .They would develop more sophisticated mechanisms that can sense chemical and mechanical stimuli more effectively . I think that they will be required to be more smart too, otherwise , they wouldn't be able survive
Humans would be able to understand that they live in a 3D world because they will understand that there are three perpendicular direction that things can move in . They will differentiate simply between two universes (or places ) that are different in spatial dimensions even if they could't benefit from light .
Anyway , what does string theory and quantum mechanics have to do with this ?
| {
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Opening the fridge door to cool a room I'm well aware that the default answer to this textbook default question is "it doesn't work", but still, I believe it does.
To cool the insides of the fridge, the compressor must do work, and since the efficiency isn't 100% you are constantly warming the whole room to cool it's insides, the winning move here is simply turning the fridge off. However, let's suppose the fridge must stay on, wouldn't it be better to open the door?
In other words: Isn't opened fridge turned on better than closed fridge turned on for the whole room temperature?
| NO. What allows a fridge maintain cool inside is the fact that the walls of the CLOSE fridge prevents heat from outside (the room heat) flows to the inside of the fridge WHILE the rate the fridge's compressor is drawing heat out (from the back or lower grills of the fridge). But if you open the door, the same amount of heat that flows INTO the fridge will be the same as the heat flowing out by the compressor.
If you open the door is like you taking the compressor of your air conditioning unit at home and moving it from its outside location to the center of the family room. All the heat taken from the AC unit bill be blown back into the inside of the house.
There are other details about entropy and maximum heat removal not explain here. But , for the sake of simplicity of your question statement, I offered to provide a simple none elaborated answer.
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About an upside down cup of water against atmosphere pressure There is an experiment we learned from high school that demonstrated how atmosphere pressure worked.
Fill a cup of water and put a cardboard on top of it, then turn it upside-down, the water will not fall out. The explanation said this was because the atmosphere pressure was greater than the water pressure, which holds the water up.
I believed this explanation once, until I found some points that confused me:
1. Is the water pressure in the cup really smaller than the atmosphere pressure?
This is what we have been taught through our life. However, consider an object in the water under sea level, it experiences the pressure from water plus atmosphere pressure. So the water under sea level must be greater than the atmosphere pressure. Even it is contained in a cup, the pressure wouldn't change. Is this True?
I read the webpage which gave the explanation excluding the reason of pressure. If the cup is fully filled, the compressibility of water is much greater than that of the air, also the surface tension of the water keeps the air out of the cup. So the water is held in the cup. This explains the problem. But I still want to ask if the water pressure smaller or greater than the atmosphere pressure in this situation.
2. When the cup is half filled with water, why it still holds?
I saw most of articles or opinions against this. They all agree that the water would not fall only if the cup is entirely filled. But I did the experiment myself, the water stayed still in the cup even it is not fully filled. Actually even with little amount of water, as long as it covers the open of the cup and the cardboard on it, the water stays in the cup.
Even the compressibility of water is much smaller, the air inside the cup provide enough compressibility, how come it still holds?
| Atmospheric pressure is caused by air gravity and air molecules movement, air pressure in the half filled upside-down cup is lower than the air pressure outside due to less gravity, so the air pressure can still hold the water in half filled cup.
| {
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Why does the guy moving on spaceship look younger in twin paradox? If there is no particular absolute choice of frame of reference, the guy who sits on Earth is also moving away from the guy on spaceship perspective and hence time on Earth should also dilate when viewed from the guy on spaceship perspective. But why does the guy moving on spaceship look younger in twin paradox? What am I misunderstanding terribly?
ADDED:: Is time dilation symmetric? If one frame of reference are moving with constant velocity w.r.t other, we have to transformation relations
$$\Delta t = \frac{\Delta t'}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Does each one of them see clock on other tick slowly than their own?
| I assume the inertial twin is A ,while the one in the spacecraft is B
The relative velocity between A and B is v.
The distance B has to travel is from x0 to x1 as seen by A observer= x1-x0.
The same distance is k*(x1-x0) as seen by B observer,where k =(1-(v^2/c^2))^0.5 because of length contraction.
So from the point of view of the observer A, B has to travel a distance equals to 2*(x1-x0).
The time measured in A's frame = 2*(x1-x0)/v.
While from the point of view of the observer B,
The point x1 is travelling towards him at speed of v.
the initial distance between B and x1 = k*(x1-x0).
x1 reaches B and reverses course to move away from him and x0 reaches B again.
The total distance measured in B's frame = 2*k*(x1-x0).
The time measured in the B's frame of reference for this journey = 2*k*(x1-x0)/v.
The whole thing is down to the initial assumption that the points x0,x1 are inertial to the A observer.
This is embedded in the problem.
In fact we can design simpler thought experiments without acceleration ,deceleration and yet the same result will still hold.
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What do you call the period after sunrise when the sky is bright? At sunrise, the sky isn't actually up in the sky yet. Twilight occurs before sunrise, then at sunrise the leading part of the sun crosses the horizon. But, the sky isn't bright yet. It takes some time for the sky to be blue again.
Then, at the closing of the day, the sky darkens before twilight, then sunset.
Basically, there is a narrower time when the sky is bright and blue, rather than being the time between sunrise and evening twilight. Is there a name for these times, or at least a name for its boundaries?
Edit:
Here's a picture to better point out what I mean:
The shaded part after sunrise I label 1 is the part after sunrise, when the sky is still dark but the sun is out in the sky (here's an image).
The shaded part just before sunset I label 2 is the part when the sky is already darkening, but it's not yet sunset as the sun hasn't set below the horizon (an image again).
The large portion of the day I label 3 would be the part when the sky is blue and bright, the term for which I'm asking for, if it exists (here's an image).
| The time after the Sunrise is known as Morn.
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Fukushima vs Thorium This is not a question about traditional nuclear power plants vs the thorium based. But about the Fukushima plant itself and the very negative environmental effects from its meltdown of reactors 1 and 3 vs a Thorium based design under similar natural stressors.
Fukushima leaking radioactive water for ‘2 years, 300 tons flowing into Pacific daily’
This is the basic outline of what happened to cause the melt down.
Immediately after the earthquake, the remaining reactors 1–3 shut down
automatically and emergency generators came online to power
electronics and coolant systems. However, the tsunami following the
earthquake quickly flooded the low-lying rooms in which the emergency
generators were housed. The flooded generators failed, cutting power
to the critical pumps that must continuously circulate coolant water
through a Generation II reactor for several days to keep it from
melting down after shut down. After the pumps stopped, the reactors
overheated due to the normal high radioactive decay heat produced in
the first few days after nuclear reactor shutdown (smaller amounts of
this heat normally continue to be released for years, but are not
enough to cause fuel melting).
http://en.wikipedia.org/wiki/Fukushima_Daiichi_nuclear_disaster
I am just wondering if this was a Thorium reactor, would all of this had taken place in the first place? Because thorium needs to be constantly primed and not cooled.
I am not an engineer nor a paid scientist so please excuse my ignorance. I am just very curious.
Thanks.
| This is not so much a question of nuclear fuel (uranium fuel cycle versus thorium fuel cycle) as the question of reactor design such as choice of liquid vs solid nuclear fuel, choice of coolant / moderator containment vessel design.
For instance molten salt reactors have freeze plug safety feature allowing evacuation of nuclear fuel in passively safe storage in case of overheating. But one can have molten salt reactors on thorium fuel cycle such as LFTR, on the other hand it is also possible with molten uranium chloride based reactors. Also eliminating water as coolant can be done with uranium and thorium fuels.
One could also conceive thorium fuel cycle reactor designs equally prone to the type of accidents as Fukushima.
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How to block neutrons What is a good way to block neutrons and what is the mechanism that allows this? It's my understanding that polyethylene is somewhat effective. Why?
| To expand on dmckee's answer, water is an excellent way to both slow and block neutrons. If you are doing any home experiments with any type of radiation, remember that the human body is mostly water and radiation will happily dump energy into YOU. Be very very careful and protect yourself. Better yet, seek the help of a trained professional or just don't do it.
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Is the symmetrisation postulate unnecessary according to Landau Lifshitz? The symmetrisation postulate is known for stating that, in nature, particles have either completely symmetric or completely antisymmetric wave functions. According to these postulate, these states are thought to be sufficient do describe all possible systems of identical particles.
However, in Landau Lifshitz Quantum Mechanics, in the first page of Chapter IX - Identity of Particles, he comes to the same conclusion without needing to state any ad-hoc postulate.
It goes like this: Let $\psi(\xi_1,\xi_2)$ be the wave function of the system, $\xi_1$ and $\xi_2$ denoting the three coordinates and spin projection for each particle. As a result of interchanging the two particles, the wave function can change only by an unimportant phase factor:
$$
\psi(\xi_1,\xi_2)=e^{i\alpha}\psi(\xi_2,\xi_1)
$$
By repeating the interchange, we return to the original state, while the function $\psi$ is multiplied by $e^{2i\alpha}$. Hence it follows that $e^{2i\alpha}=1$ or $e^{i\alpha}=\pm1$. Thus
$$
\psi(\xi_1,\xi_2)=\pm\psi(\xi_2,\xi_1)
$$
Thus there are only two possibilities: the wave function is either symmetrical or antisymmetrical.
It goes on by explaining how to generalize this concept to systems with any number of identical particles, etc.
Im summary, no symmetrisation postulate was ever stated in this rationale. Is "shifting by an unimportant phase factor" a too strong requirement for ensuring identity of particles?
| Check out the section about Chapter 17 Identical particles in Ballentines, he not only points out why looking at the Permutation operators of two particles in a multi particle setting is misleading but also discusses some errors in previous claims.
| {
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Why does the thickness of a wire affect resistance? For small thicknesses of wire, it's pretty obvious why resistance affects thickness. (The electronics squeeze to get through). But after a certain thickness shouldn't the thickness become irrelevant?
For example if your trying to pour a bucket of water through a straw, the thickness of the straw is obviously gonna be a bottle neck- the bigger the straw, the easier it is for water to get through.
But if you try to pour a bucket of water through a tunnel - the size of the tunnel doesn't really matter, because the tunnel is already big.
So after a certain thickness shouldn't the thickness stop mattering?
| All (non-superconductor) metals have electrical resistance, no matter how thick a wire made of them is, as the conducting electrons are always "bumping into" the atoms in the lattice of the metal, slowing them down. Therefore there is always a resistance that can be reduced by increasing the cross sectional area of the wire (by allowing more free electrons per unit of area).
By contrast once a pipe carrying water becomes of a greater diameter than the body of water flowing through it, there is no resistance that can be reduced by increasing the diameter of the pipe, and thus the analogy between water and electricity fails at that point.
(imagining the water pipe to be filled with saturated sponge makes the analogy closer to the free electrons slowly migrating in a metal)
| {
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Electric potential due to circular disk Relevant diagram is available here.
The circular disk of radius $a$ lies in the $xy$ plane and carries surface charge density of
$\sigma (s, \phi) = s^{2}cos\phi $,
where $(s,\phi)$ are in cylindrical co-ordinates.
The problem is to find potential at a point which is slightly displaced from the $z$ axis at position $ r = z \hat z + \delta s \hat s = z\hat z + \delta x \hat x + \delta x \hat y$, and
$ r' = s cos\phi \hat x + s sin\phi \hat y $
Since potential is given by
$V(r) = \frac{1}{4 \pi \epsilon_{0}} \int \frac{1}{\bf {r} - \bf {r'}} dq$
Here, $dq = \sigma da = \sigma dl_{s} dl_{\phi} = \sigma sdsd\phi = s^3 cos\phi dsd\phi$
and
${r-r'} = z\hat z + \delta x\hat x + \delta x \hat y - (s cos\phi \hat x + s sin\phi \hat y) $
Therefore,
$|r-r'| = \sqrt{z^2 + s^2 - 2\delta x s(cos\phi + sin\phi)} $.
It can be assumed that $(\delta x)^2 = 0$ since $\delta x$ is infinitesimally small.
This means that the final integral for potential is given by
$V(r) = \frac{1}{4 \pi \epsilon_0} \int_S \frac{s^3 cos\phi}{\sqrt{z^2 + s^2 - 2\delta x s(cos\phi + sin\phi)}} dsd\phi$.
Any suggestions on how to proceed with evaluating this integral would be very much appreciated.
Thanks.
| The problem is that you're ignoring the angular dependence of your probe point $\mathbf r$, and that is messing with your integral. If your probe point has cylindrical coordinates $(s,\phi,z)$ and your integration variables are $(s',\phi')$, then the distance between the two is
$$\frac 1 {|\mathbf r-\mathbf r'|}=\frac{1}{\sqrt{s^2-2ss'\cos(\phi-\phi')+s'^2+(z-z')^2}}$$
by the cosine rule (draw it!). If you put this into your integral it will no longer vanish.
(A few pointers on the new integral: the new dependence on $\phi'$ is a bit more complicated. The standard practice is to change variables to $\varphi=\phi'-\phi$. This will leave a simpler denominator, and a factor of $\cos(\varphi+\phi)=\cos(\varphi)\cos(\phi)-\sin(\varphi) \sin(\phi)$ on the numerator. One of the two terms will vanish and the other will yield to a change of variables to $u=\cos(\varphi)$.)
| {
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How do we know photons have spin 1? Electrons have spin 1/2, and as they are charged, they also have an associated magnetic moment, which can be measured by an electron beam splitting up in an inhomogeneous magnetic field or through the interaction of the electrons's magnetic moment with an external magnetic field in spectroscopic measurements.
On the other hand, a photon is neutral - how can one measure its spin if there's no magnetic moment? How do we know it has spin 1?
| I would say this is an empirical fact. In atomic physics you
don't observe optical transitions (e.g. induced by a laser)
without angular momentum transfer. The change in angular
momentum is always $\pm$ 1, that's what the photon can transmit.
See http://en.wikipedia.org/wiki/Selection_rule
The atomic states with different angular momenta are identified
via their different energy levels due to the Zeeman effect in
a magnetic field.
| {
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Why does electricity need wires to flow? If you drop a really heavy ball the ball's gravitational potential energy will turn into kinetic energy.
If you place the same ball in the pool, the ball will still fall. A lot of kinetic energy will turn into thermal energy because of friction, but the gravitational potential energy will still be converted.
Similarly, why doesn't electricity flow without a good conductor? Why won't Electrons flow from the negative terminal to the positive terminal without a wire attaching them?
Electricity flows like a wave and metals have free electrons in the electron cloud that allows the wave to propagate, or spread. But when these free electrons aren't available to propagate the wave, why don't the electrons just "move" like the ball? Why don't the electrons just "move" through the air to the positive terminal?
A slow drift speed means that the electrons most likely will take a long time to propagate the wave of electricity, but they should still get there.
| Electrons do flow without a wire. This is exactly what is happening in a cathode ray tube. So why don't electrons flow from one conductor to another through vacuum or air if there is a potential difference? There is a minimum energy of a few eV required for an electron to exit a metal known as the work function. In a cathode ray tube this energy is minimized by selecting a material with low work function and by heating the anode.
| {
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Why do electrons not bump into impurities in a superconductor? Just a simple question. Why is it, that when a material becomes superconducting, and by that gets zero resistivity, the electrons don't hit impurities in the material?
For the material to have zero resistivity, that means that the electrons can just flow without any disturbance at all?
Is it because of the Cooper pair creation? In that case, why exactly?
| When the electrons pair up this opens an energy gap between the energy of the Cooper pairs and the energy of the lowest quasiparticle excitation. There is a nice article discussing this effect here (NB it's a PDF).
The gap means that you cannot scatter a Cooper pair by an arbitrarily small energy. If the energy is less than the gap energy it won't scatter because there are no available energy states for it to scatter into. That's why the electrons in a superconductor don't scatter off impurities, defects, etc. If you apply enough energy, e.g. a very high voltage, the collisions are energetic enough to scatter the Cooper pairs and the superconductivity breaks down.
| {
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Why was the conversion factor of the metric unit bar chosen the way it was? The unit bar for pressure is clearly a metric unit, but its order of magnitude is a bit strange. In the centimeter–gram–second system of units we have:
1 bar = 1 000 000 baryes = 1 000 000 dyn/cm²
so the bar is not "coherent" with this system (the factor is not one). Also in the meter–kilogram–second (and SI) system we get:
1 bar = 100 000 pascals = 100 000 N/m²
while in the meter–tonne–second system:
1 bar = 100 pièzes = 100 sn/m²
So my question is simply, where does the conversion factor for bar come from, since it seems to not fit into usual systems? According to the Wikipedia article bar this unit was created already in 1909 by British meteorologist Shaw, but not much detail is provided.
Maybe the factor was simply chosen as the power of ten making the unit closest to the atmospherical pressure at sea level (which is 1.01325 bar by convention, and close to 1.01 bar on average)?
| The $\text{pascal}$ seems of a much later date than the $\text{bar}$. In fact, it seems that, at some time, the $\text{bar}$ was adjusted a bit away from the average air pressure on earth (its originally intended definition), to get it "in line" with the SI units, and therefore also with the new or later $1~\text{Pa}=1~{\text{kg}\over\text{m}\cdot\text{s}^2}$.
Meteorologists worldwide have for a long time measured atmospheric pressure in bars, which was originally equivalent to the average air pressure on Earth [...]. After the introduction of SI units, many preferred to preserve the customary pressure figures. Consequently, the bar was redefined as 100,000 pascals, which is only slightly lower than standard air pressure on Earth. [My emphasis.]
Pascal (unit)
As noted in the comments below, this answer (and, perhaps, the Wikipedia quote) might or might not stand up to scrutiny. Further digging in history seems necessary, but I dug a bit and can't find anything really substantiating this reading. However, I also didn't find anything conclusively and explicitly pointing towards a different reading (i.e., that the magnitude of the $\text{bar}$ is the same now as it was when the $\text{bar}$ was adopted initially).
| {
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Why is there a factor of $4\pi$ in certain force equations? I mean to ask why there is $4\pi$ present in force equations governing electricity? Though all objects in universe are not spherical and circular, the constant of proportionality in both equations contain $4\pi$. Why?
| For example, if you mean $k_e=\frac1{4\pi\epsilon_0}$, it comes from a natural "Gauss's law" understanding of Coulomb's law, where the electric field is distributed over the surface of a sphere of area $4\pi r^2$.
$$F= \frac1{\epsilon_0}\frac1{4\pi r^2}q_1q_2 $$
This is also the explanation for the $1/r^2$ term (and in other classical fields), and there are also attempts to extend this sort of reasoning to gravitational fields.
| {
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What are correlated magnetic moments? My book has the following sentence and I don't understand what correlation or lack of correlation means:
At high temperature the magnetic moments of adjacent
atoms are uncorrelated (to maximize the entropy) so the crystal has no net magnetic moment.
The book is touching on second-order phase transitions and it's describing how magnetic transitions are an example of such 2nd order phase transitions.
| It means that the directions of the magnetic moment of neighboring atoms is statistically independent. At high temperature the thermal energy is greater than the magnetic energy; the resulting thermal fluctuations cause any material to become paramagnetic. As you lower the temperature these thermal fluctuations are reduced. Ferromagnetics and anti-ferromagnetics have long range order below the temperature at which thermal fluctuations are suppressed. This is a big topic of course, but I hope this little bit helps. The following book is excellent.
“Magnetism in condensed matter” by Stephen Blundell
Oxford masters series; Oxford University Press, 2001; ISBN: 0198505914
| {
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How far can you scatter light using a prism? If I were to scatter light how far do you think it would disperse? What prism most effectively scatters light?
| A prism refracts light, it doesn't scatter it. I'm confused about what you're asking.
If you're asking about the dispersion that occurs when (for example) white light passes through a prism, that depends on the material of the prism, and it's a fairly involved topic. But the Wikipedia article points out some of the physics behind it.
If you're asking about literally how far the light refracted by a prism travels, that again depends greatly on the material surrounding the prism, but you could have a look at the attenuation coefficient (also extinction coefficient).
| {
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Walking along a moving train, we make 18 steps. Opposite: 11. How long is the train?
A man walks in the same direction as a slowly moving train ($v_{man} > v_{train}$). He counts the train to be 18 steps long. Then he turns around and counts the train to be 11 steps long.
(Suppose both man and train are moving at a constant speed; every step is the same length.)
How long is the train?
For some reason I always end up with two equations and four unknowns... I'd really appreciate a solution!
| Key is to notice that your steps provide you with a unit length as well as a unit time. So, let's measure distance in $steps$ and time in $ticks$, with your speed being $1 \ step/tick$.
The length of the train is $x$ steps, and its speed is $v \ steps/tick$ ($v<1$).
It follows that
$$x \ + \ 18 \ v \ = \ 18 $$
$$x \ - \ 11 \ v \ = \ 11 $$
Adding 11 times the first equation to 18 times the second yields $29 x = 396$. The train is $396/29 \ steps$ long.
You also need to check if indeed $v < 1 \ step/tick$. Leave that to you to demonstrate.
| {
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Does alternating current (AC) require a complete circuit? This popular question about "whether an AC circuit with one end grounded to Earth and the other end grounded to Mars would work (ignoring resistance/inductance of the wire)" was recently asked on the Electronics SE.
(Picture edited from the one in the above link)
Though I respect the AC/DC experts there, I think (with the exception of the top answer) they are all wrong.
My issue is that they all assume that AC requires a complete circuit in order to function. However, my understanding is that a complete circuit is necessary for DC, but not AC. My intuitive understanding is that AC is similar to two gas-filled rooms with a pump between them - the pump couldn't indefinitely pump gas from one room to another without a complete circuit (DC), but it could pump the gas back and forth indefinitely (AC). In the latter case, not having a complete circuit just offers more resistance to the pump (with smaller rooms causing a larger resistance).
Is my understanding correct - can AC circuits really function without a complete loop?
More importantly, what are the equations that govern this?
If larger isolated conductors really offer less AC-resistance than smaller AC conductors, how is this resistance computed/quantified? Would its "cause" be considered inductance, or something else?
| For me, AC is a closed circut all together within the same unique wire
As the + and - are chasing eachother in the same line. As for DC, the - chase the + around the other end of the loop
| {
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Study of Black-body Radiation Why did scientists study black body radiations from something as complicated as a hollow container rather than the radiation from something simple like a thin solid cylinder?
| Since a black body is an exact absorber and during practice it's demanding to make things that is really exactly a good absorber so to make black body we choose a material that is as good an absorber as we can obtain it into a hollow Sphere with a small hole in it. The black body is then the hole not the sphere. It is nearly perfect absorber because any radiation falling on it is multiply reflected within the sphere because the material of the sphere is a good absorber and absorbs most of the radiation at each reflection.
| {
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When is temperature not a measure of the average kinetic energy of the particles in a substance? I had always thought that temperature of a substance was a measure of the average kinetic energy of the particles in that substance:
$E_k = (3/2) k_bT $
where $E_k$ is the average kinetic energy of a molecule, $k_b$ Boltzmann's constant, and $T$ the temperature. (I'm not sure of the 3/2 coefficient.) Then I heard from several folks that this is a simplistic notion, not strictly true, but they didn't explained what they thought was flawed with this idea. I'd like to know what (if anything) is objectionable about this idea? Is it that the system must be macroscopically at rest? Is it that it ignores the quantum mechanically required motion of particles that persists at low temperatures? When is it not valid?
|
temperature of a substance was a measure of the average kinetic energy of the particles in that substance
This is only true for a monatomic gas. Even without quantum mechanics, a classical diatomic gas has three more degress of freedom than a monatomic gas—two rotational and one vibrational—for a total of 6. The equipartition theorem tells us that the kinetic energy will be divided equally among these. Thus the kinetic energy of the gas particles in this case represents only 1/2 of the energy represented by the temperature.
| {
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Strings and their masses How do strings present in particles give mass to them? Is it only by vibrating? I have been trying to find the answer but could not find it anywhere, can this question be answered?
| I presume that you are asking about the mass spectrum of string theories.
The mass spectrum of a Classical string theory, or the mass of a string is (due to Special Relativity) given, by:
$$m=\sqrt{-p^\mu p_\mu}=\sqrt N $$
In natural units $c_0=\ell_s=\hbar=1$. Where $N$ is an operator, called the "Number Operator". In Classical string theories, this is continuous. When we quantise the theory, we realise that the new mass spectrum is actually given by:
$$m=\sqrt{N-a} $$
Where $a$ is called the normal ordering constant. Now, $N$ is going to take discrete values, multiples of $\frac12$.
In Bosonic String Theory, $a=1$. In superstring theories, $a$ depends on the sector you' are talking about; it is $0$ in the Neveu-Schwarz sector, and $\frac12$ in the Ramond sector.
Of course, In GSO Projected theories (i.e. the tachyon is removed (yes, even in the RNS (Ramond-Neveu-Schwarz) Superstring, there are tachyons if you don't GSO Project; although this problem is absent in the GS (Green-Schwarz) Superstring)) , a GSO Projection gets rid of certain states and so on, but let's keep things simple right now.
Now, I've only been talking about open strings. What about the closed strings, which are more important, because the open strings are present only in the Type I Superstring theory (and Bosonic, of course (and probably also Type 0A and 0B (not sure))), whereas the closed strings are there in all string theories?
The transition happens to be relatively simple.
You replace $N$ with / $N+\tilde N$ and $a$ with $a+\tilde a$.
EDIT
I also see that in your post, you say "strings in particles". Actually, the partickles themselves are strings. And they get their mass as per the vibrational modes $\alpha,\tilde\alpha,d,\tilde{d}$of the string with the Number operator $N$ given by
$$ N = \sum\limits_{n = 1}^\infty {{{\hat \alpha }_{ - n}}\cdot{{\hat \alpha }_n}} + \sum\limits_{r/2 = 1}^\infty {{{\hat d}_{ - r}}\cdot{{\hat d}_r}} $$.
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Can Deuterium undergo a nuclear fusion without the presence of Tritium? I've been reading about fusion fuels for a while now, and I understand that in Lithium-Deuterium fuel, the neutrons from the fission reaction bombard the Lithium to produce Tritium and the D-T reaction occurs and we get the energy.
So what about Deuterium-Deuterium fusion ? Is it possible without the presence of Tritium at all ?
If so, then does it release more or less energy than D-T fusion ?
| All three isotopes of Hydrogen can undergo fusion under the right conditions. The main reason to use D or T is that they fuse more easily than H. For example, H-H fusion is primarily what drives our sun, but in the lab D-D or D-T reactions are much easier to initiate.
The D-T reaction gives off 17.6 MeV of energy, D-D actually has 3 different reactions it can undergo (4, 3.3, 23.9 MeV), T-T gives 11.33 Mev, D-H gives off 5.5 MeV and H-H fuses into D giving off 1.44 MeV.
| {
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Where does gravity get its energy from? I would like to know where gravity gets its energy to attract physical bodies?
I know that the law of conservation states that total energy of an isolated system cannot change. So gravity has to be getting its energy from somewhere, or else things like hydropower plants wouldn't be able to turn the power of the falling water into a spinning rotor.
Just to be clear, Lets create an example:
Lets say we have two objects with equal mass close to each other. So gravity does its job and it pulls each other closer, this gets turned into kinetic energy. This is where I'm lost. According to the law of conservation energy can't be created or destroyed and the kinetic energy comes from the gravitational pull so where does the gravitational pull gets its energy.
If that energy isn't being recycled from some where else then that means you have just created energy, therefore breaking the law of conservation.
| The universe is composed of over $90$% dark matter and dark energy. Dark energy comprises about $70$% of this. So far, dark matter and energy remain undetectable and until there are the mechanism that produces gravity, will probably remain a puzzle. If you want to know more in detail you will need to research quantum mechanics and string theory.
| {
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How does electron know when to change into a wave? It is known that electron behaves as a wave also.
How does electron know that it has to change into a wave?
Are there any factors that influence the behavior of electron changing into wave?
| The electron is always a wave.
The electron is wave, as experiments of diffraction and interference showed.
Waves come in an infinity of "shapes". Some kinds of shapes have some properties, and others have other properties. Examples of properties are position and momentum. The two shapes of the electron's wave having these properties are
*
*When the wave is concentrated at a point. In this case, it has a definite position. It doesn't have a definite momentum.
*When the wave is a plane wave, having a definite frequency (more complex shapes can be obtained by superposing various frequencies). In this case, there is no definite position, since the plane wave extends in all space.
The problem is, when does the electron how to be plane wave, or to be concentrated at a point. The answer is strange. If you measure the frequency (or momentum), you will find that the electron is a plane wave. If you measure the position, you will find that the wave is concentrated at a point.
Yes, you understood well. The electron, and any other particle for that matter, has precisely the kind of shape for which the property you want to measure is defined. Measure another property, which is not compatible with the property previously measured, and you will find it has another shape. Now, this may look strange, but this is how it happens.
Wait, there is more, when more particles of the same kind are present. Then, saying it is a wave is not enough. But this is another story.
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Frequency of small oscillation of particle under gravity constrained to move in curve $y=ax^4$ How to find the frequency of small oscillation of a particle under gravity that moves along curve $y = a x^4$ where $y$ is vertical height and $(a>0)$ is constant?
I tried comparing $V(x) = \frac 1 2 V''(0) x^2 + \mathscr O(x^3) = \frac 1 2 kx^2$ (assuming $V(0)$ is ground state and $V'(0)$(slope) remains is horizontal at extrememum, but unfortunately $V''(0) = 0$. I am pretty much clueless. Thanks for your help.
ADDED:: Letting $m=1$ the Lagrangian is $L = \frac 1 2 (\dot x^2 + \dot y^2) - gax^4 = \frac 1 2 (\dot x^2 + (4 ax^3 \dot x)^2)-gax^4 = \frac 1 2 \dot x^2(1 + 16a^2x^6)-gax^4$
The above Lagrangian gives the equation of motion as
$$\ddot x(1+16a^2x^6)+\dot x^2 96 a^2x^5 + 4 gax^3 = 0$$
Since we are considering the system a small oscillation whose potential is of order 4, $\mathscr{O}(x^{k>4})$ can be ignored which reduces into $\ddot x = -4agx^3$.
To solve this,
$$\frac 1 2 \frac{d}{dt}(\dot x^2) = -\frac{d}{dt}(agx^4)$$
which gives $\dot x = \sqrt{k - 2agx^4}$, Assuming the system begins from $t=0$ at $x=x_0$ with $\dot x = 0$, $k = 2agx_0$, which turns the integral into
$$\int_0^{T/4} dt= \int_0^{x_0} \frac 1 {\sqrt{2agx_0^4-2agx^4}} dx=\frac{1}{\sqrt{2ag}x_0}\int_0^1\frac{1}{\sqrt{1-y^4}}dy$$
Which gives $$T = 2 \sqrt 2 \sqrt{\frac{\pi}{ag}}\cdot \frac{\Gamma(5/4)}{\Gamma(3/4)}\cdot \frac 1 {x_0}$$
which is a dubious result. Please someone verify it. Any other methods are welcome.
| Is it fine if I solve it using Newton's laws? You can then maybe convert it to a Lagrangian. I will assume the path on which it oscillates is almost flat for small oscillations
At a displacement of $x$($\lim_{x\to0}$), the gravitational force on it will be
$$F_{mg}=-mg\sin\theta$$
where $$\tan\theta=\frac{dy}{dx}=4ax^3$$
As $x\to0$ , $\tan\theta\to0$ so we can approximate $\tan\theta\approx\sin\theta$
Thus we have
$$\ddot x=-4agx^3$$
You get a differential equation which is solvable(probably a lot easily using lagrangian equations!)
| {
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Why does boiling water in the microwave make a cup of tea go weird? When I boil water in the kettle, it makes a nice cup of tea. Sometimes I need to use a microwave because a kettle isn't available. I boil the water in the mug and it looks pretty normal, but when I drop in the teabag the water froths up and looks foamy. I don't see what the chemical difference is here, so I assume it must be some physical difference. I have noticed this with multiple types of tea and multiple microwaves, the results being consistent so it's not just a weird microwave or something like that.
What is the reaction here and how/why does it occur?
Here is a photo of the 'fizzy' looking tea just after dunking in the teabag.
| http://en.wikipedia.org/wiki/Superheating
I think you are superheating the water and you provide nucleation sites (by means of the tea bag) so it starts to boil.
| {
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Why can't we destroy energy? From a wikipedia article:
In physics, the law of conservation of energy states that the total energy of an isolated system cannot change—it is said to be conserved over time. Energy can be neither created nor destroyed, but can change form; for instance, chemical energy can be converted to kinetic energy.
What is the reason behind this Law: is there a proof that we can't destroy energy? I mean if we couldn't destroy energy in our universe maybe we can do it in other universes the point is why is it a law and not a theory
| It is much more appropriate to call it a law than a theory. A theory is an explanation, whereas a law is based on repeated observation and/or experimentation. If you are concerned that it isn't or can't be proven then you could call it a hypothesis, assumption, or postulate of conservation of energy. Of those three, postulate would probably be the best answer, as conservation of energy is accepted. However, the conservation of energy has been demonstrated so many times in so many ways, and in so many fields, it probably has more experimental support than many of our other scientific laws.
| {
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View from a helicopter rotor: why is the horizon distorted? This video ("rotor panorama") was captured by a camera attached to the rotor head of a radio-controlled helicopter, with the frame rate set to the rotor's frequency. During a long segment of the video, the horizon looks distorted:
What causes this distortion?
| The frame rate and the rotor frequency do not exactly match.the picture seems to be clear some times when the rotor reaches its set rpm. But after that the vibration starts with a different noise.the wobble increases and the distortion too.as Emilio pisanty said there is a vertical displacement due to vibration.
| {
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Why does electric field intensity $E$ can be uniquely determined by its divergence and curl? My question is, the number of following equations
$$\nabla\cdot E=\frac{\rho}{\varepsilon}$$
$$\nabla\times E=-\frac{\partial B}{\partial t}$$
is 4 while the number of unknown variables $E=(E_1,E_2,E_3)$ is 3. Intuitively, the equation is overdetermined and the solution may not exist unless four equations are correlated. Is my intuition right?
| The divergence and curl do not uniquely determine a vector field. For example, if all the derivatives are zero, any constant field is a solution. In order to determine the field, we must include boundary conditions.
As far as your question, you are not looking at equations for the three components of $E$. You are looking at equations for the partial derivatives of those components, of which there are nine ($\frac{\partial E_x}{\partial x}, \frac{\partial E_x}{\partial y}$), etc.
| {
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Can the same car with 100HP and 500HP produce the same acceleration on first gear? My friend and I are debating about the acceleration of cars in first gear. He said that the same car (weight, tires, etc) produces the same acceleration with both 100HP and 500HP. I imagine that the moment when the tires catch the surface of the road both start with the same speed at moment 0, but the one with 500HP will have a higher acceleration in the next moment because it manages to spin the tires faster than the other one.
If someone can explain me a little about this can be great.
| I am his friend :). This argument came up when we were discussing the G forces felt by our bodies in a smaller powered car (let's 90 hp) and a larger power one (400 hp). I said that the G forces are dependent on the amount of grip you can catch with your tires and assuming both cars are exactly the same in (body mass, tires, aerodynamics, front/rear wheel drive train) then you can achieve a fixed maximum G force for both cars (if the lower powered car can achieve skidding in first gear). The maximum G force is felt at the peak of the friction function graph (right before the skidding occurs - when there is a slight drop in the gripping power of the tires). This is the same principle followed in ABS breaks design sistem.
Therefore is my opinion that you will not feel large (peak) G force in the larger power car then in the smaller one because the smaller one can achieve that peak (for a shorter period of time thou) in first gear right before the wheels start skidding. The difference between the two cars will be noticed more on higher gear ratio acceleration. So let's say the "medium" of the G forces will be higher for the larger power car because it can sustain larger G forces for a longer period of time. But a "peak" G force can be felt by both cars.
| {
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Why do meteors explode? A report on the Chelyabinsk meteor event earlier this year states
Russian meteor blast injures at least 1,000 people, authorities say
My question is
*
*Why do meteors explode?
*Do all meteors explode?
| I suspect that there are multiple phenomenologies involved, but for a gravitationally bound pile of aggregate, there is not a lot of cohesion. Thus it is possible that entry dynamic pressure might be high enough to cause the "pancake" behavior described above.
But I suspect that there is another way for more cohesive objects to "explode". It is almost certain that the center of pressure of a random rock is not aligned with and "ahead of" the center of mass along the direction of flight, so torques will arise that (even over short periods) will tend to rotate the object. If the torques are high enough, and the inertia small enough, the accumulating rotational velocity will overcome the strength of materials and the object will come apart. This seems unlikely for very large objects which won't spin up fast enough, and very small objects which are relatively strong. But in the middle there is a class of weak object of appropriate size where the breakup probably can be attributed to rotational velocity and centrifugal force.
| {
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Double Slit Experiment: How do scientists ensure that there's only one photon? Many documentaries regarding the double slit experiment state that they only send a single photon through the slit. How is that achieved and can it really be ensured that it is a single photon?
| In the double slit experiment, if you decrease the amplitude of the output light gradually, you will see a transition from continuous bright and dark fringe on the screen to a single dots at a time. If you can measure the dots very accurately, you always see there is one and only one dots there. It is the proof of the existence of the smallest unit of each measurement which is called single photon: You either get a single bright dot, or not.
So, probably you may ask why it is not a single photon composite of two "sub-photon", each of them passing through the slit separately and then interference with "itself" at the screen so that we only get one dot. However, the same thing occurs for three slits, four slits, etc... but the final results is still a single dot. It means that the photon must be able to split into infinitely many "sub-photon". If you get to this point, then congratulation, you basically discover the path-integral formalism of quantum mechanics.
| {
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CMBR temperature over time? How has CMBR temperature dropped as function of time? A graph would be nice, but I'd be happy with times (age of universe) when it cooled enough to not be visible to human eye, became room temperature equivalent, or reached some interesting temperatures regarding matter in the universe.
If there is a nice formula giving the temperature as function of time, that would be great too.
| The exact formula in the Standard ΛCDM-model is
$$
T(t) = T_0\big(1+z(t)\big) = \frac{T_0}{a(t)},
$$
where $a(t)$ is the cosmological scale factor, which can be calculated by numerically inverting the formula
$$
t(a) = \frac{1}{H_0}\int_0^a\frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a' + \Omega_{K,0}\,a'^2 + \Omega_{\Lambda,0}\,a'^4}},
$$
with $H_0$ the present-day Hubble constant, $\Omega_{R,0}, \Omega_{M,0}, \Omega_{\Lambda,0}$ the relative present-day radiation, matter and dark energy density, and $\Omega_{K,0}=1-\Omega_{R,0}- \Omega_{M,0}- \Omega_{\Lambda,0}$. See this post for more details.
| {
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How are the Weyl & Riemann curvature tensors related to the stress energy tensor in GR? Einstein's vacuum equations, that is without matter, allows the possibility of curvature without matter. For instance, we may consider gravitational waves.
The question is: Is there some link between the Riemann curvature tensor, and/or the Weyl tensor, and some gravitational "physical" quantities (as stress-energy tensor or total energy)?
Of course, at first glance, there is no covariant gravitational stress-energy tensor, so it seems there is no relation, but maybe things are more subtle?
|
Is there some link between the Riemann curvature tensor [...] and some gravitational "physical" quantities*
Maybe you could clarify what you want that would qualify as "physical." Curvature is observable, and IMO is physical. Projects like LIGO are designed to detect gravitational waves. Gravity Probe B was a project that accomplished its purpose of essentially verifying GR's predictions of spacetime curvature in the neighborhood of a gravitating, spinning body. In the simplest terms, curvature can be measured by transporting a gyroscope around a closed path. This is essentially what GPB did.
Of course, at first glance, there is no covariant gravitational stress-energy tensor
But that's only a prohibition on defining a local measure of gravitational-wave energy. For example, in an asymptotically flat spacetime, the ADM energy includes energy being radiated away to null infinity by gravitational waves. If LIGO-like projects succeed, they will measure the energy of gravitational waves.
| {
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Pulley problem - unable to understand the concept Given:
*
*The pulley is moving towards the right.
*All blocks have different masses. (The pulley and the strings are massless.)
What I don't understand:
*
*Is the tension the same for both A and B?
*If the tension is the same, then both blocks should have different accelerations, but this is not true?
|
Is the tension the same for both A and B?
Yes. A massless string needs to have constant tension, otherwise it would feel a net force and have infinite acceleration.
If the tension is the same, then both blocks should have different
accelerations
Yes.
but this is not true?
Why do you think the boxes should have the same acceleration? I can't help find the flaw in your reasoning if you don't say why you think that.
Incidentally, the details of the situation are rather unclear. It is not obvious which way gravity points or what the precise nature of the box is. However, I can tell you that the detail that the pulley is moving to the right is immaterial, due to the principle of relativity.
| {
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What does imaginary number maps to physically? I am taking undergraduate quantum mechanics currently, and the concept of an imaginary number had always troubled me. I always feel that complex numbers are more of a mathematical convenience, but apparently this is not true, it has occurred in way too many of my classes, Circuits, Control Theory and now Quantum Mechanics, and it seems that I always understand the math, but fail to grasp the concept in terms of its physical mapping. Hence my question, what does imaginary number maps to physically?
Any help would be much appreciated
| A rotation by a right angle. So complex numbers come up whenever you have periodicity/oscillations/phases etc.
For a simple example, consider the physical intuition for Euler's formula -- if you push something with force $F=kx$, you get exponential motion, but $F=-kx$ is periodic motion. The latter, where you have periodic, trigonometric functions, is where there are complex numbers/imaginary exponents.
Similarly when describing light, the polarisation of a light wave behaves as a phase. The same idea carries over to quantum mechanics.
| {
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Similarity Transformation How can I find the similarity transformation $S$ between gamma matrices in the Dirac representation $\gamma_D$ and Majorana representation $\gamma_M$ in 4 dimensions theory?
The relation is $\gamma_M = S \gamma_D S^{-1}$
Actually, my question is about the general method of finding the similarity transformation between 2 give gamma matrices in different representation, and the given problem above is just for the sake of demonstration.
| If you know the change of the (vector) basis, the answer is straightforward.
If you don't know the change of the (vector) basis, but only want some particular representation for the gamma matrices (for instance you want only real matrices, or only imaginary matrices), you may try for $S$ :
$$S=\frac{1}{\sqrt{2}} \begin{pmatrix} A&B\\-\epsilon B&\epsilon A \end{pmatrix}, S^{-1}=\frac{1}{\sqrt{2}}\begin{pmatrix} A&-\epsilon B\\ B&\epsilon A \end{pmatrix}$$
where $\epsilon = \pm1$, $A, B$ are $2*2$ matrices, such as $A^2= B^2=1$, and $[A, B]=0$.
For instance, you may take one of the matrix equals to $\pm \mathbb{Id}$, and the other being a Pauli matrix $\pm \sigma_i$.
For obtaining Majorana representation from Dirac representation, we may use : $\epsilon = -1, A = \mathbb{Id}, B = \sigma_y$
| {
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What is the definition of a quantum integrable model? What is the definition of a quantum integrable model?
To be specific: given a quantum Hamiltonian, what makes it integrable?
| Quantum integrability basically means that the model is Bethe Ansatz solvable. This means that we can, using the Yang-Baxter relation, get a so-called "transfer matrix" which can be used to generate an infinite set of conserved quantities, including the Hamiltonian of the system, which, in turn, commute with the Hamiltonian. In other words, if we can find a transfer matrix which satisfies the Yang-Baxter relation and also generates the Hamiltonian of the model, then the model is integrable.
Please note that, oddly enough, a solvable system is not the same thing as an integrable system. For instance, the generalized quantum Rabi model is not integrable, but is solvable (see e.g. D. Braak, Integrability of the Rabi Model, Phys. Rev. Lett. 107 no. 10, 100401 (2011), arXiv:1103.2461).
A nice introduction to integrability and the algebraic Bethe Ansatz is this set of lectures by Faddeev in Algebraic aspects of the Bethe Ansatz (Int. J. Mod. Phys. A 10 no. 13 (1995) pp. 1845-1878, arXiv:hep-th/9404013)
| {
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How do I find the amount of atoms in a pure gas in a confined area? Say I have Argon gas in an area such as a cube with pressure around pressure between 10-5 and 10-3 Torr (pressure) at 25 degrees at 320 Kelvins . How do I find the amount of atoms in that cube? Any formulas?
| If you assume it is an ideal gas and you know the volume of the cube, the temperature of the gas, and the pressure, you can calculate the number of atoms using the ideal gas law:
$PV=nRT$
Where P is pressure, V is volume, n is the nuber of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
| {
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Where to place my second image charge? (Spherical ungrounded conductor) I am trying to find the potential $V$ inside a sphere using the method of image charges.
I have a conducting spherical shell. The charge $q$ is inside the sphere. The sphere is ungrounded and is an equipotential because it is a conductor.
If I place an image charge $q'$ outside the sphere, I can make it equipotential if it is grounded i.e. potential$ V=0$ on the sphere. But since it is not grounded, there is some potential $V_0$ on the sphere.
To make the equipotential $V_0$ on the sphere, I can put another image charge $q''$ at the centre except that that is not allowed because I can't put image charges inside the space that I'm investigating.
Where else can I put the image charge? No matter where I put it outside the sphere it will not give me the $V_0$ equipotential on the sphere. How do I solve this?
| Your first image charge made the spherical shell an equipotential (@ V=0). So, by superposition, you can now add a uniform surface charge density to the shell, and it will stay an equipotential, but no longer at zero...
| {
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Deriving the Lorentz force from velocity dependent potential We can achieve a simplified version of the Lorentz force by
$$F=q\bigg[-\nabla(\phi-\mathbf{A}\cdot\mathbf{v})-\frac{d\mathbf{A}}{dt}\bigg],$$
where $\mathbf{A}$ is the magnetic vector potential and the scalar $\phi$ the electrostatic potential.
How is this derivable from a velocity-dependent potential
$$U=q\phi-q\mathbf{A}\cdot\mathbf{v}?$$
I fail to see how the total derivative of $\mathbf{A}$ can be disposed of and the signs partially reversed. I'm obviously missing something.
| Hints: Use
$$\frac{\partial U}{\partial {\bf v}}= -q{\bf A}, $$
and the defining property of a velocity-dependent potential:
$${\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.$$
See e.g. Herbert Goldstein, Classical Mechanics and Wikipedia for more details.
| {
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Why should nature of light(or any quantum object) depend on observation? We know that, in the double slit experiment, observation changes the behavior of a quantum object, that it behaves like a particle when observed and a wave when not observed. But why should its nature depend on observation? What if we didn't exist and hence no observation...? The nature of the quantum objects must remain same, right? Why is it based on observation? Am I trying to understand the wave-particle duality in a wrong way?
| Your confusion is perfectly justified. Congrats, you just discovered what is called the measurement problem. There have been several attempts to resolve this problem, most of them based on introducing interpretations of quantum mechanics other than the Copenhagen interpretation (or more concisely standard or orthodox interpretation; the original Copenhagen interpretation is considered unbearable these days). An important role here is played by decoherence which is in part capable to explain the collapse of the quantum state. However, this whole problem is still largely unresolved and there is a lot of research on it right now although the problem has essentially already been raised by Einstein.
The standard interpretation that is taught in schools and universities does not really explain what exactly is to be considered an observation. In fact, Mermin once summarized it with the words
Shut up and calculate
as it is more like a working theory. It helps us make accurate predictions on the behavior of a quantum state without having to consider the interaction of the quantum system with the measurement apparatus which is thought to be the source of the 'collapse'.
If you are interested in the role of decoherence in the solution of the measurement problem and some of the more recent concepts regarding it, I recommend this paper by Schlosshauer.
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Surface gravity of Kerr black hole I'm going through Kerr metric, and following the 'Relativist's toolkit' derivation of the surface gravity, I've come to a part that I don't understand.
Firstly, the metric is given by
$$\mathrm{d}s^2=\left(\frac{\Sigma}{\rho^2}\sin^2\theta\omega^2-\frac{\rho^2\Delta}{\Sigma}\right)\mathrm{d}t^2-2\frac{\Sigma}{\rho^2}\sin^2\theta\omega \mathrm{d}\phi \mathrm{d}t+\frac{\Sigma}{\rho^2}\sin^2\theta \mathrm{d}\phi^2+\frac{\rho^2}{\Delta}\mathrm{d}r^2+\rho^2 \mathrm{d}\theta^2$$
With
$$\rho^2=r^2+a^2\cos^2\theta,\quad \Delta=r^2-2Mr+a^2,$$
$$\Sigma=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=\frac{2Mar}{\Sigma}$$
The Killing vector that is null at the event horizon is
$$\chi^\mu=\partial_t+\Omega_H\partial_\phi$$
where $\Omega_H$ is angular velocity at the horizon.
Now I got the same norm of the Killing vector
$$\chi^\mu\chi_\mu=g_{\mu\nu}\chi^\mu\chi^\nu=\frac{\Sigma}{\rho^2}\sin^2\theta(\Omega_H-\omega)^2-\frac{\rho^2\Delta}{\Sigma}$$
And now I should use this equation
$$\nabla_\nu(-\chi^\mu\chi_\mu)=2\kappa\chi_\nu$$
And I need to look at the horizon. Now, on the horizon $\omega=\Omega_H$ so my first term in the norm is zero, but, on the horizon $\Delta=0$ too, so how are they deriving that side, and how did they get
$$\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2}{\Sigma}\nabla_\nu\Delta$$
if the $\Delta=0$ on the horizon? Since $\rho$ and $\Sigma$ both depend on $r$, and even if I evaluate them at $r_+=M+\sqrt{M^2-a^2}$ they don't cancel each other.
How do they get to the end result of $\kappa$?
| Ok, every book I looked has this solved by looking at four velocity and four acceleration of a free particle at the horizon, so that must be it :\ Altho I'm sure there's a way to do it via Killing vector $\chi^\mu=\partial_t+\Omega_H\partial_r$.
So I'll just go through this derivation with acceleration...
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Riemann tensor in 2d and 3d Ok so I seem to be missing something here.
I know that the number of independent coefficients of the Riemann tensor is $\frac{1}{12} n^2 (n^2-1)$, which means in 2d it's 1 (i.e. Riemann tensor given by Ricci Scalar) and in 3d it's 6 (i.e. Riemann Tensor given by Ricci tensor).
But why does that constrain the Riemann tensor to only be a function of the metric? Why not a tensorial combination of derivatives of the metric?
What I mean is why is the Riemann tensor in 2D of the form
\begin{align}
R_{abcd} = \frac{R}{2}(g_{ac}g_{bd} - g_{a d}g_{b c})
\end{align}
and in 3D,
\begin{align}
R_{abcd} = f(R_{ac})g_{bd} - f(R_{ad})g_{bc} + f(R_{bd})g_{a c} - f(R_{bc})g_{ad}
\end{align}
where $f(R_{ab}) = R_{ab} - \frac{1}{4}R g_{ab}$?
Wikipedia says something about the Bianchi identities but I can't work it out. A hint I got (for the 2d case at least) was to consider the RHS (the terms in parenthesis) and show that it satisfies all the required properties of the Riemann tensor (sraightforward) and proceed from there - but, I have not been able to come up with any argument as to why there must be a unique tensor satisfying those properties.
Of course I could brute force it by computing $R_{abcd}$ from the Christoffel symbols etc., but surely there must be a more elegant method to prove the statements above.
Help, anyone? I haven't been able to find any proofs online - maybe my Googling skills suck.
| The simplicity of geometry in lower dimensions is because the Riemann curvature tensor could be expressed in terms of simpler tensor object: scalar curvature and metric (in 2d) or Ricci tensor and metric (in 3d). That fact, of course, does not alter the possibility to write Riemann tensor (as well as Ricci tensor and scalar curvature) as a combination of metric derivatives. But each term in such a combination is not a tensor - only the whole object.
Now, let us recapture, why in lower dimensions we are able to reduce the Riemann tensor to a combination of lower rank tensor objects.
For 3d case, definition of Ricci tensor:
$$ R_{ab} = R_{abcd}g^{ac}$$
contains 6 independent components, exactly the number of independent components in Riemann tensor. So, this equation could be reversed, thus expressing $R_{abcd}$ in terms of $R_{ab}$ and $g_{ab}$.
In 2d case we could similarly start with definition of Ricci scalar:
$$ R = R_{ab} g^{ab} ,$$
and reverse it expressing $R_{ab}$ through $g_{ab}$ and $R$. The next step would be to express Riemann tensor with $g_{ab}$ and $R_{ab}$ (and thus through scalar $R$ only).
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Small oscillations of the double pendulum From the Lagrangian I've got the following equations of motion for the double pendulum in 2D. (The masses are different but the lengths of the two pendula are equal.) Let $m_2$ be the lowest-hanging mass.
$$(m_1+m_2)\ddot{\theta_1}+2m_2\ddot\theta_2\cos(\theta_2-\theta_1)=\\ -2m_2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2)-(m_1+m_2)g/l\sin(\theta_1)$$
and
$$m_2\ddot{\theta_1}+2m_2\ddot\theta_2\cos(\theta_2-\theta_1)=\\ 2m_2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2)-m_2g/l\sin(\theta_1)$$
In the small angle approximation these become, respectively
$$(m_1+m_2)\ddot{\theta_1}+2m_2\ddot\theta_2= -2m_2\dot\theta_1\dot\theta_2(\theta_1-\theta_2)-\theta_1(m_1+m_2)g/l$$
and
$$m_2\ddot{\theta_1}+2m_2\ddot\theta_2= 2m_2\dot\theta_1\dot\theta_2(\theta_1-\theta_2)-\theta_1m_2g/l$$.
Most sources don't have the terms of order $\dot\theta$. This is because they apply the small angle approximation to the Lagrangian before taking the derivatives, thereby ignoring terms of order $\theta.$ What justification do we have for getting rid of these terms?
| I think the issue here is that you need to keep a consistent level of approximation in your "small angle approximation." By small angles, we typically mean $\theta_1$ and $\theta_2$ are both of order $\epsilon$, where $\epsilon \ll 1$. Then the question is - to what order in $\epsilon$ do you want to write down the equations of motion?
When you neglect the term $\frac{3}{2} \dot\theta_1 \dot \theta_2 (\theta_1 - \theta_2)^2$ in the Lagrangian, you are saying that terms of size $\epsilon^4$ are small compared to terms like $\dot \theta_1^2$, which is of size $\epsilon^2$. In the equation of motion, you get terms that are $\dot \theta_1 \dot \theta_2 (\theta_1 - \theta_2)$, which are of size $\epsilon^3$, compared to $\theta_1$, which is size $\epsilon$.
So neglecting the additional term in the Lagrangian gets you the same equation of motion as keeping the whole Lagrangian, and then dropping terms that are of size $\epsilon^3$.
This kind of argument is a little handwavy, and (in principle) could blow up if the time derivatives of $\theta_{1,2}$ were large - at some point, you might want to check out some books on perturbation theory in a more formal sense.
| {
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Quantum entanglement as practical method of superluminal communication As I understand it (from a lay physics perspective), quantum entanglement has been experimentally demonstrated - it is a reality. As I understand it, you can measure something like the spin of an electron and know that its entangled pair will, in that same instant, no matter where in the universe it is, have the opposite spin.
This would not seem to have any utility as the foundation of a superluminal communications device. Is this true, or has it been established that is there some aspect of quantum entanglement that can ultimately lead to the development of such a device.
In other words: is superluminal communication via quantum entanglement an open scientific question, has it been settled as an impossibility, or is it currently more of an engineering problem than a scientific one?
| This 'spooky action at a distance' will not and can not lead to communications technologies. The point is that the correlation between the two states cannot be used for information transmission. The two observers can influence each others' observations, but they can never communicate their own observations to the other superluminally, and thus will have no way of checking who influenced who (before waiting for subluminal information transmission devices), rendering the correlation between the states useless for all communications purposes.
| {
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Is there an easy way to get water at roughly 70°C in our kitchen? Some green tea requires to pour water at 70°C. I have no thermal sensor or kettle with adjustable temperature with me.
Do you know a way to get water at roughly 70°C like “boil water and wait for x minutes” or “mix x part of boiling water with 1-x part of fresh water” ?
| You can start with water at $0^{\circ} C$, as suggested in Vibert's answer, measure the height of water in the container, and start heating the water. The density of water at $0^{\circ} C$ is $1 g \cdot cm^{-3}$, at $65^{\circ} C$ it is $0.981 g \cdot cm^{-3}$, at $70^{\circ} C$ it is $0.978 g \cdot cm^{-3}$. So if at $0^{\circ} C$ the height of water in the container is 10 cm, then the difference in heights of water between $65^{\circ} C$ and $70^{\circ} C$ will be approximately $0.3 mm$ and can be detected using a ruler (or a beam compass). Thermal expansion of a solid container is typically much less than that of water. I would think that not much water will evaporate when the water is heated from $0^{\circ} C$ to $70^{\circ} C$, but maybe this issue should be studied in more detail.
One could say that what I suggest uses the principle of an ordinary thermometer, but I am not sure this is worse than the calorimetric principle.
| {
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What are the assumptions of the Navier-Stokes equations? I wanted to model a real life problem using the Navier-Stokes equations and was wondering what the assumptions made by the same are so that I could better relate my entities with a 'fluid' and make or set assumptions on them likewise. For example one of the assumptions of a Newtonian fluid is that the viscosity does not depend on the shear rate. Similarly what are the assumptions that are made on a fluid or how does the Navier-Stokes equations define a fluid for which the equation is applicable?
| The Navier-Stokes equations assume (assuming we are looking at a vector conservative form):
*
*The continuum hypothesis, which is applicable for Knudsen numbers of much less than unity.
The Navier-Stokes equations must specify a form for the diffusive fluxes (e.g. otherwise you would have the Cauchy momentum equation not the Navier-Stokes momentum equation), e.g.
*
*Newtonian fluid for stress tensor or Cauchy's 2nd law, conservation of angular momentum
*Definition of the transport coefficients (e.g. viscosity)
The solution of the Navier-Stokes equations involves additional assumptions, (but this is separate from the equations themselves) e.g.
*
*An equation of state for closure (e.g. thermally perfect gas, calorically perfect gas)
*Stokes' assumption for zero bulk viscosity
| {
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Confusion With How Dimensions Work Form what I understand if you have an equation such as:
$$v = v_0 + at$$
then the dimensions must match on both sides i.e. $L/T = L/T$ (which is true in this case), but I have seen equations such as 'position as a function of time' $x(t) = 1 + t^2$, and obviously time is in $T$, but apparently the function gives you position which is $L$... so what happens to $T$ and where does the $L$ come from? I thought dimensions must always match...
Also, let us say that you know the time to reach a destination is proportional to distance i.e. double the distance and you get double the time, now this makes sense to me, but as I said earlier I thought that dimensions must always be consistent or else you can not make comparisons in physics, so if you are giving me $L$ (the distance), how can that become $T$ (time) all of a sudden?
| Units must always be consistent, that is correct. So using your example of:
$$ x(t) = 1 + t^2 $$
where the left hand side has units of $L$ (distance). This means the constant $1$ on the right side has implied units of $L$ while the coefficient in front of $t^2$ (which has the value of $1$) has implied units of $L/T^2$.
In other words, the units do match but they get attached to the constants multiplying each term.
| {
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Distribution of point charges on a line of finite length How will $N$ freely moving charges confined to a line with length $L$ be distributed? What are their equilibrium positions?
| Aside from the special cases of $N = 1,2, 3$, in general the solution will be a quite involved system of simultaneous equations. I don't know if there's any way to not do the algebra at all and simply rely on physical arguments... but the full solution is:
Let $q = 1, \epsilon_0 = 1, L =1$ i.e. set all constants $=1$ for simplicity. Let there be $N$ particles, and let the particles be indexed by $i$ which runs from $0$ to $N-1$, where $i$ is ordered such that it describes particles going from left to right along the line.
Obviously the $0$th particle will be at the left most end, the $N-1$th particle at the right most end.
Let $x_i$ = distance from the $i-1$th particle to the $i$th particle. Here $i$ runs from $1$ to $N-1$.
Then the length constraint is
\begin{align}
\sum_{i=1}^{N-1} x_i = 1,
\end{align}
and the $N-2$ force equations are
\begin{align}
\frac{1}{x_1^2} &= \frac{1}{x_2^2}+\frac{1}{(x_2+x_3)^2}+\cdots+\frac{1}{(x_2+\cdots+x_{N-1})^2} \nonumber \\
\frac{1}{(x_1+x_2)^2}+\frac{1}{x_2^2}&=\frac{1}{x_3^2}+\frac{1}{(x_3+x_4)^2}+\cdots+\frac{1}{(x_3+\cdots+x_{N-1})^2} \nonumber \\
\frac{1}{(x_1+x_2+x_3)^2}+\frac{1}{(x_1+x_2)^2}+\frac{1}{x_3^2}&=\frac{1}{x_4^2}+\frac{1}{(x_4+x_5)^2}+\cdots+\frac{1}{(x_4+\cdots+x_{N-1})^2} \nonumber \\
&\vdots
\end{align}
giving $N-1$ independent equations for $N-1$ variables. Solve.
The system will be symmetric about the mid-point, so you can reduce some of the equations, but I don't think you can just 'see' the solution off-hand.
| {
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How is relativity equation approximated What is the technique with which I can approximate the equation
$$\frac{mc^2}{\sqrt{1-(\frac{v}{c})^2}}-mc^2$$
when $v\ll c$?
Any hint would be much appreciated
| When $v\ll c$, the ratio $\beta = v/c$ is small, so we perform a Taylor expansion about $\beta = 0$;
\begin{align}
\frac{1}{\sqrt{1-\beta^2}} = (1-\beta^2)^{-1/2} = 1+\frac{1}{2}\beta^2+\frac{3}{8}\beta^4+\cdots
\end{align}
Now plug this into your expression and simplify.
| {
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Lightcone singularity of a 3 point function in CFT I had a quick question regarding the title of the question. In e.g. 2D CFT (for simplicity), the three point function of three operators with conformal dimension $a$, $b$ and $c$ are given as
$$
\langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\mathcal{O}_3(x_3)\rangle~=~\frac{c_{123}}{(x_1-x_2)^{a+b-c}(x_2-x_3)^{-a+b+c}(x_1-x_3)^{a-b+c}}
$$
Now, I will expect that this correlator shows UV divergences when any two of these operators are coincident. But from the RHS, it seems that e.g. for $c>a+b$, there's no light cone singularity for $x_1=x_2$. What am I missing here?
| You need to refine your intuition a little bit. If you bring two operators together, you indeed get singular behaviour, which is taken into account by the OPE. The unit operator is the most singular, and the higher the scaling dimension of an operator $\phi$ in the $O_1 \times O_2$ OPE, the less singular its contribution will be. (In your case, the contribution of $O_3$ is regular if $c > a+b$).
In the limit $x_1 \rightarrow x_2$, any 3-pt function
$$ < O_1(x_1) O_2(x_2) \phi(x_3) >$$
will measure the overlap of the $O_1 \times O_2$ OPE with $\phi$ inserted 'far away' at $x_3$. According to the above paragraph, to see singular behaviour, you need $\phi$ to be a light operator. If $\phi$ is too high in the spectrum (as in your case), you will only measure a term that lives in the tail of the OPE so you'll get a small, regular result.
I hope that this is more or less what you expected - otherwise I can add more details.
EDIT to answer the remaining questions. You have hopefully learned that two-point functions are diagonal, so
$$ < \phi(x) \phi'(y) > = 0$$
unless $\phi = \phi'.$ So the 3-point function with $O_3$ can only measure the contribution of $O_3$ in the OPE - that's the overlap. It's blind to the presence of any other operators.
To see how singular the contribution of some operator $\phi$ is, just write the leading OPE term explicitly:
$$O_1(x_1) O_2(x_2) \sim \frac{1}{|x_1 - x_2|^z} \left\{\phi(x_2) + \text{descendants} \right\}$$
and we want to determine $z$. But you can just apply a dilatation $x \mapsto \lambda x$ and compare the left and right hand sides, and you'll see that $z = a+b-\Delta$ where $\Delta$ is the dimension of $\phi$. The unit operator has dimension zero, all other operators (in a unitary theory) have positive scaling dimensions.
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What happens if an already excited electron gets hit by another photon (lasers)? Slightly vague title but it is the best I can do. My question stems from some interest in how 3 stage lasers function. A photon from the pumping source comes in and strikes a atom in the active medium. The photon is absorbed, an electron is excited, and the energy instantly falls to a metastable state. My question is what happens when I try to excite that electron again with another pump photon. Does the electron jump to an even higher state if one is available (the pump photon provides the right dE again to excite the electron to a valid higher energy state)? Or do we choose our active medium properly such that no "double-pumped" state exists and the excited electrons become transparent to the pumping photons?
| The answer is yes. If there would be a state available so that an electron in an excited state could be further excited (by the laser wavelength), this would also happen and reduce the efficiency of stimulated emission.
Hence, active media are chosen in such a way that this can not happen, or is at least at reduced probability.
| {
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How do you determine the value of the degeneracy factor in the partition function? In the partition function, expressed as $$Z = \sum_j g_je^{-\beta E_j}$$ I'm wondering what determines the $g_j$ factor. I've been trying to look around the internet for an explanation of it but I can't find one. I guess it is the number of degenerate states in a given energy level? How do you determine how many degenerate states there are? A simple example involving how to determine the $g_j$ factor would be great. Thanks for the help!
| Consider a quantum system with state (Hilbert) space $\mathcal H$. For simplicity, let the Hamiltonian $H$ of the system have discrete spectrum so that there exists a basis $|n\rangle$ with $n=0,1,2,\dots$ for the state space consisting of eigenvectors of the Hamiltonian. Let $\epsilon_n$ denote the energy corresponding to each eigenvector $|n\rangle$, namely
\begin{align}
H|n\rangle = \epsilon_n|n\rangle
\end{align}
Now, it may happen that one or more of the energy eigenvectors $|n\rangle$ have the same energy. In this case, we say that their corresponding shared energy eigenvalue is degenerate. It is therefore often convenient to have a the concept of the energy levels $E_j$ of the system which are simply defined as the sequence of distinct energy eigenvalues in the spectrum of the Hamiltonian. So, whereas one can have $\epsilon_n = \epsilon_m$ if $n\neq m$, one cannot have $E_n = E_m$ if $n\neq m$. Moreover, it is often convenient to label the energy levels in increasing index order so that $E_m < E_n$ whenever $m<n$.
The degeneracy $g_n$ of the energy level $E_n$ is defined as the number of distinct energy eigenvalues $\epsilon_m$ for which $\epsilon_m=E_n$. For simplicity, we assume that none of the levels is infinitely degenerate so that $g_n\geq 1$ is integer for all $n$.
The partition function of a system in the canonical ensemble is given by
\begin{align}
Z = \sum_n e^{-\beta\epsilon_n}
\end{align}
In other words, the sum is over the state labels, not over the energy levels. However, noting that whenever there is degeneracy, sum of the terms in the sum will be the same, we can rewrite the partition function as a sum over levels
\begin{align}
Z = \sum_{n} g_n e^{-\beta E_n}
\end{align}
The degeneracy factor is precisely what counts the number of terms in the sum that have the same energy.
As for a simple example, consider a system consisting of two, noninteracting one-dimensional quantum harmonic oscillators. The eigenstates of this system are $|n_1, n_2\rangle$ where $n_1,n_2 = 0, 1, 2, \dots$ and the corresponding energies are
\begin{align}
\epsilon_{n_1,n_2} = (n_1+n_2+1)\hbar\omega.
\end{align}
The canonical partition function is given by
\begin{align}
Z = \sum_{n_1,n_2=0}^\infty e^{-\beta \epsilon_{n_1, n_2}} = \underbrace{e^{-\beta\hbar\omega}}_{n_1=0,n_2=0} + \underbrace{e^{-\beta(2\hbar\omega)}}_{n_1=1,n_2=0} + \underbrace{e^{-\beta(2\hbar\omega)}}_{n_1=0,n_2=1} + \cdots
\end{align}
If you think about it for a moment, you'll notice that, in fact, the energy levels of this composite system are
\begin{align}
E_n = n\hbar\omega, \qquad n_1+n_2 = n
\end{align}
and that the degeneracy of the $n^\mathrm{th}$ energy level is
\begin{align}
g_n = n
\end{align}
so that the partition function can also be written in the form that uses energy levels and degeneracies as follows:
\begin{align}
Z = \sum_{n=1}^\infty g_n e^{-\beta E_n} = \underbrace{e^{-\beta\hbar\omega}}_{n=1} + \underbrace{2e^{-\beta(2\hbar\omega)}}_{n=2} + \underbrace{3e^{-\beta(3\hbar\omega)}}_{n=3} + \cdots
\end{align}
| {
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How does humidity affect the path of a bullet? Background
Last night, I was reading the FM 23-10 (The U.S. army official field manual for sniper training), and I've noticed that they're potentially teaching snipers incorrect information.
Generally speaking, when we say "impact goes up" it means that the bullet was either somehow made faster or its path was easier, therefore the curve in its ballistic trajectory is smoother. Thus, it will hit higher. When we say impact goes down, we mean the opposite.
For example, atmospheric heat will, loosely speaking, make the air "thinner" and therefore the impact will be higher. Cold weather will do the opposite. This part is correct.
What about humidity?
The FM 23-10 says:
The sniper can encounter problems if drastic humidity changes occur in
his area of operation. Remember, if humidity goes up, impact goes
down; if humidity goes down, impact goes up.
They're basically saying that when humidity goes up, then the bullet's travel will be more difficult-> steeper trajectory curve -> lower point of impact.
However, as far as I know, dry air is denser than humid air because air has higher molecular mass than water vapour. In humid air water vapour replaces other gases, thus bringing the whole density down. So, the point of impact should be higher with higher humidity.
So my question is:
All other factors being equal, does humid air pose less resistance to the bullet making the point of impact higher than in dry air?
| Could be that the expanding gasses behind the projectile diminish faster as the outside air influenced the outcome ,herefore decreasing ramp time(bullet arc)wich lead to bullet drop on its intended path!pesonal experience in sub tempreatures using a 45acp had the same outcome at 15metres.2seprate 1911 pistols were fired and both litterally experienced bullet drop using the same reloaded ammo(185gr swc).normally we used the very same range and ammo when climate is humid_hot and bullets seemed to print on sight of aim or slightly higher 2-5cm.
| {
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Entropy inequality Assume that you have two bipartite systems $\rho_1^{AB},\rho_2^{AB}$ then I would like to prove the following:
$$S(\frac{1}{2}( \rho_1^{AB}+I^A\otimes\rho_2^B))+S(\frac{1}{2}(\rho_2^{AB}+I^A\otimes\rho_1^B)) \geq S(\frac{1}{2}(\rho_1^{AB}+I^A\otimes\rho_1^B))+S(\frac{1}{2}(\rho_2^{AB}+I^A\otimes\rho_2^B))$$
where $S$ is the von Neumann entropy, $\rho_1^B=tr_A(\rho_1^{AB}),\rho_2^B=tr_A(\rho_2^{AB})$ and $I^A$ is the maximally mixed state on $A$. It looks like it should pass with some monotony property, any hints or counterexample are welcome.
| (very) partial answer:
In a very particular case, this is true.
Let's have $\rho_1^{AB} = I^A \otimes \rho_1^{B}$ and $\rho_2^{AB} = I^A \otimes \rho_2^{B}$
The left hand side is $L = 2 S( \frac{1}{2}(\rho_1^{AB}+\rho_2^{AB})) $
The right hand side is $R = S(\rho_1^{AB}) + S(\rho_2^{AB})$
By the concavity of the Von Neumann entropy, we have $L \ge R$
| {
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Quantum tunneling effect in a potential of the kind $V(x)=A\frac{x^2}{1+x^4}$ Given a potential: $$V(x)=A\frac{x^2}{1+x^4}$$ with $A\gt 1$ and a quantum particle inside the well around the point $x=0$. I'm stuck on the calculation of the transmission and reflection coefficients for this particle vs. its energy.
| In order to compute the transmission coefficient, we can use the first correction in the WKB approximation. Ignoring constants that we can pull outside the integral, we essentially are faced with the integration problem,
$$I(x) = \int \mathrm dx \, \sqrt{V(x)-E}.$$
In the case of your potential, we thus have,
$$I(x) = \sqrt{E}\int \mathrm dx \, \sqrt{\frac{Cx^2}{1+x^4}-1} $$
where $C := A/E$. We can now employ the generalised binomial theorem to expand the square root, using the Pochhammer symbol ${}_rP_k$, obtaining,
$$\sqrt{\frac{Cx^2}{1+x^4}-1} = \sum_{k=0}^\infty \frac{(-1)^k {}_{1/2}P_k}{k!}C^{1/2-k} \left(\frac{x^2}{1+x^4} \right)^{1/2-k}.$$
We can now integrate a general term over $x$, which yields a hypergeometric function. There are further simplifications for the cases $x>0$ and $x<0$. These lead to,
$$I(x) = \mathrm{sgn}(-x)\frac{\sqrt{E}}{4}\sum_{k=0}^\infty i^{k+1}\frac{(-1)^k {}_{1/2}P_k}{k!}C^{1/2-k} B \left(-x^4; \frac{1-k}{2},\frac{1+k}{2} \right)$$
where $B(z;a,b)$ is the incomplete beta function. If $x_1$ and $x_2$ denote the two classical turning points, we then have that,
$$T = \exp \left[ -2 \frac{\sqrt{2m}}{\hbar}[I(x_2)-I(x_1)]\right] \left( 1 + \frac14 \exp \left[ -2 \frac{\sqrt{2m}}{\hbar}[I(x_2)-I(x_1)]\right]\right)^{-2}.$$
| {
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Gravitational acceleration on the Moon and Mars There are plenty of formulas that use gravity acceleration of Earth. This is represented with the symbol $g$. In my school work (I am a high school student) we usually take it as $g= 9,8 \,\text m/\text s^2$.
This thing is obviously a number that is only usable on Earth. What I want to know is that, what if I want to make my calculations according to another planet? How the number is going to change?
| Let's see how the acceleration due to gravity is obtained for any planet, and then we can apply this to Earth or the Moon or whatever we want.
Newton's Law of gravitation tells us that the magnitude of the gravitational force between to objects of masses $m_1$ and $m_2$ is given by
\begin{align}
F = G\frac{m_1m_2}{r^2},
\end{align}
where $r$ is the distance between their centers of mass. Now, suppose that object 1 is a planet of mass $m_1 = M$ and radius $R$, and object 2 is a much smaller object of mass $m_2 = m$ located at a height $h$ above the surface of the planet that is small compared to the radius of the planet. The magnitude of the gravitational force between the two objects will be
\begin{align}
F = G\frac{Mm}{(R+h)^2}
\end{align}
on the other hand, Newton's Second Law tells us that the acceleration of object 2 will satisfy
\begin{align}
F = ma
\end{align}
Combining these facts, namely setting the right hand sides equal, causes the mass $m$ to drop out of the equations, and the acceleration due to gravity of the object of mass $m$ becomes
\begin{align}
a = \frac{GM}{(R+h)^2} = \frac{GM}{R^2}\left(1-2\frac{h}{R}+\cdots\right)
\end{align}
where in the second equality, I have performed a Taylor expansion of the answer in terms of the small number $h/R$. Notice that to zeroth order, namely the dominant contribution when object 2 is close to the surface of the planet, is some constant that is independent of the height and depends only on the mass and radius of the planet;
\begin{align}
a_0 = \frac{GM}{R^2}
\end{align}
This is precisely what we usually call the acceleration due to gravity near the surface of a planet. If you plug the numbers in for Earth, you will get
\begin{align}
a_0^\mathrm{Earth} \approx 9.8\,\mathrm{m}/\mathrm{s}^2
\end{align}
and I'll leave it to you to determine the number for other planets. The important property of this acceleration due to gravity is that it scales linearly with the mass $M$ of the planet, and it scales like the negative second power of the radius of the planet.
| {
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"timestamp": "2023-03-29T00:00:00",
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If I am travelling on a car at around 60 km/h, and I shine a light, does that mean that the light is travelling faster than the speed of light? The title says it all.
If I was on a bus at 60 km/h, and I started walking on the bus at a steady pace of 5 km/h, then I'd technically be moving at 65 km/h, right?
So my son posed me an interesting question today: since light travels as fast as anything can go, what if I shined light when moving in a car?
How should I answer his question?
|
If I was on a bus at 60 km/h, and I started walking on the bus at a
steady pace of 5 km/h, then I'd technically be moving at 65 km/h,
right?
Not exactly right. You would be correct if the Galilean transformation correctly described the relationship between moving frames of reference but, it doesn't.
Instead, the empirical evidence is that the Lorentz transformation must be used and, by that transformation, your speed with respect to the ground would be slightly less than 65 km/h. According to the Lorentz velocity addition formula, your speed with respect to the ground is given by:
$$\dfrac{60 + 5}{1 + \dfrac{60 \cdot 5}{c^2}} = \dfrac{65}{1 + 3.333 \cdot 10^{-15}} \text{km}/\text h \approx 64.9999999999998\ \text{km}/\text h$$
Sure, that's only very slightly less than 65 km/h but this is important to your main question because, when we calculate the speed of the light relative to the ground we get:
$$\dfrac{60 + c}{1 + \dfrac{60 \cdot c}{c^2}} = c$$
The speed of light, relative to the ground remains c!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/79331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the importance of state functions in physics? I'm currently reading about the Carnot cycle and its significance on the formulation of entropy (because I want to try to understand the concept better), but I can't seem to answer the following question:
*
*Why are state functions important in physics? What does it mean, physically, that a certain variable is a state function? I know the definition of state function, but I can't figure out why it would be important at all.
In the aforementioned Carnot cycle, some variables that are state are internal energy and entropy. Apparently Rudolf Clausius felt the need to define a certain state function he discovered to be 'entropy', so it must have a reason.
| state functions are important because you can analyse the system at a specific moment of time if you know its configuration at that moment, it doesn't matter how the system got to the state it is in now, they also describe the equilibrium state of the system. its similar to the importance of conservative forces in classical mechanics.
| {
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"url": "https://physics.stackexchange.com/questions/79490",
"timestamp": "2023-03-29T00:00:00",
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How to establish relation between flow rate and height of the water column of the tank? Suppose a water tank has 1" diameter drain at the bottom and is filled with water up to one meter height above the drain. What time it will take the tank to drain out completely. Now say, the tank is filled up to two meter height above the drain then what time it will take the tank to drain out? Will it be double or less than it? Can we establish a relation between flow rate for the given height of water column?
| This is a very simple problem. The relation is given by using
$$A \frac{dh}{dt} = \frac{-\pi D^2}{4}v;$$
where $D$, $h$, $A$ are the diameter, height and area of the tank and v is velocity.
I will leave the derivation up to you
| {
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"timestamp": "2023-03-29T00:00:00",
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Nature of frictional force I was thinking about a situation where a person in standing on the ground with some friction. The frictional force is directly proportional to the normal reaction acting on him by the ground. Assume that he leaned forward i.e his center of mass is not vertically up the point of contact of him with the ground.Then, does the frictional force change?
My thoughts:Although the person has leaned forward the force $mg$ acting on him will be vertically downward. Since normal reaction is always normal to the surface of contact its magnitude will remain same and thus frictional force will remain same.But, I have seen athletes starting the running race with a leaning forward position which would be mostly for increasing the friction between their shoes and the track. So, I'm in a dilemma. Please help.
| It's not correct that runners lean forward to begin a race in order to increase friction. They lean forward because otherwise, they would experience no propulsion whatsoever because static friction is zero when the runner is completely upright.
When the runner leans forward and flexes his leg muscles, he exerts a horizontal force on the track in the backward direction. The track responds by exerting an equal and opposite frictional force (unless there is slipping) on the runner in the forward direction that propels him forward.
Generally speaking, the more a runner leans forward at the start, the larger the horizontal component of the force exerted by his legs against the ground, and the larger the frictional force he will experience. As a result, his initial acceleration will be greater.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is it possible for the entropy in an isolated system to decrease? As far as I can tell, the concept of entropy is a purely statistical one. In my engineering thermodynamics course we were told that the second law of Thermodynamics states that "the entropy of an isolated system never decreases". However, this doesn't make much sense to me.
By counter-example: Consider a gas-filled isolated system where the gas has maximum entropy (it is at equilibrium). Since the molecular motion is considered to be random, at some point in the future there will be a pressure gradient formed by pure chance. At this point in time, entropy has decreased.
According to Wikipedia, the second law purely states that systems tend toward thermodynamic equilibrium which makes sense. I then ask a) is the second law as we were taught it wrong (in general), and b) what is the use of entropy (as a mathematical value) if it's effectively an arbitrary definition (i.e. what implications can we draw from knowing the change in entropy of a system)?
Thanks in advance for your help.
| As you said, entropy is a statistical phenomenon. As with everything statistical, the parameters of your sampling effect the quality of your conclusions.
In your given example - a random molecular gas spontaneously forming a pressure gradient - you examine a brief timescale during which the entropy of the system is less than a previous value. Given that we are considering a process that evolves over time (and assuming that the particle number is "large enough" from a statistical standpoint) "brief period" indicates a small sample. Perhaps a fitting analogy would be demographics: the population of a college town changes dramatically in a week's time several times a year, but a longer study may reveal a much more consistent trend.
Likewise, in the given example, the pressure gradient described can only last a brief time. A statistically sound examination of the system will show that the system is indeed in thermal equilibrium, and at a higher entropy than similar large-sample studies performed at earlier times in the evolution of the system from an initial non-equilibrium state.
| {
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Do orbitals overlap? Yes, as the title states: Do orbitals overlap ?
I mean, if I take a look at this figure...
I see the distribution in different orbitals. So if for example I take the S orbitals, they are all just a sphere. So wont the 2S orbital overlap with the 1S overlap, making the electrons in each orbital "meet" at some point?
Or have I misunderstood something?
| Yes, orbitals do overlap. However, these orbitals do not necessarily mean that is where the electron is. The orbitals are only the probable locations that an electron would be. In fact, electrons are not even necessarily in those orbitals; they could be anywhere. This is able to be shown mathematically, but it is relatively dense stuff. If you are curious and want a detailed explanation, I would be happy to edit the answer to show all of that.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/79985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to understand wavefunction in quantum mechanics in math I am reading some introduction on quantum mechanics. I don't understand all but I get the point that the wavefunction tells some probability aspects. In one book, they show one example of the wavefunction $f(x)$ in position space as a complex function, so they said the probability of finding the particle is $f^*(x) f(x) = |f(x)|^2$. In other book, the same example shown but in so-called bra and ket vector form, I know if I calculate the absolute square, I should get the same answer. But I am still learning the bra, ket notation, so I wonder if $\langle f(x)|f(x)\rangle$ or $|\langle f(x)|f(x)\rangle|^2$ gives the probability? If the last one gives the probability, what is $\langle f(x)|f(x)\rangle$? Is $\langle f(x)|f(x)\rangle = f^*(x)\cdot f(x)$ ?
| "$| f(x) \rangle$" does not mean anything and is not proper bra-ket notation. For translating back and forth beteween wavefunction and bra-ket notation, here is the #1 thing to keep in mind:
$$
f(x) = \langle x \mid f \rangle
$$
So, the probability density to find the particle at $x$ is
$$
\left|f(x)\right|^2 = \left| \langle x \mid f \rangle \right|^2
$$
Since $\langle a \mid b \rangle = \langle b \mid a \rangle^*$, this can also be written
$$
|f(x)|^2 = \langle f \mid x \rangle \langle x \mid f \rangle
$$
Remember, this represents a probability density in $x$. What this means is that
$$
\int dx\, A(x) \left| f(x) \right|^2 = \left< f \right| \left\{ \int dx \, A(x) \left| x \rangle \langle x \right| \right\} \left| f \right>
$$
should be the expected value of the function A(x). The quantity in the brackets is an operator:
$$
\hat A = \int dx \, A(x) \left| x \rangle \langle x \right|
$$
(Edit: As pointed out by Trimok, the above is not true for most operators. It is only true for any operator that is diagonal in the x basis, or equivalently that can be written as a function of the operator $\hat x$. These are the only kind of operator for which expectation value and higher moments can be computed using $|f(x)|^2$ as a probability density function.)
The expectation value of this operator is
$$
\langle A \rangle = \left< f \right| \hat A \left| f \right>
$$
| {
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"url": "https://physics.stackexchange.com/questions/80107",
"timestamp": "2023-03-29T00:00:00",
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Why are we not affected by the radiation of the radioactive decay going inside the Earth? I was reading the question Why has Earth's core not become solid?, and one of the answers says that
The core is heated by radioactive decays of Uranium-238, Uranium-235, Thorium-232, and Potassium-40
Why are we not affected by the radioactive emission of the condition below? Is this due to the fact that there is a very thick layer of mantle and crust between us and the core? Or I am wrong and we suffer from it's radiation in everyday life up to some extent?
| You are always subjected to a low background of ionizing radiation from a number of natural and artificial sources, which include cosmic rays, trace amounts of radioactive nuclei in the air and in food, and indeed from the ground. A good place to read up on this is the corresponding Wikipedia article.
The radiation from the core, however, has no chance of making it to the surface. Gamma radiation is typically stopped by a few to maybe 20 or 30 cm of rock or soil. (There is also the danger that material used to shield against gamma radiation becomes radioactive itself, but of course this is hardly an issue with the Earth's core.) Alpha and beta rays are even easier to stop. Trace amounts of radioactive nuclei, though, can be present in the soil and buildings around you and will then expose you to a small amount of radiation.
It's important to note that this is natural and nothing to worry about. The most significant contribution to background radiation, away from zones like Hiroshima, Chernobyl or Fukushima, is the trace amount of radon in the air you breathe.
| {
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"url": "https://physics.stackexchange.com/questions/80171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Why can't Iron fusion occur in stars? It is said that iron fusion is endothermic and star can't sustain this kind of fusion (not until it goes supernova). However star is constantly releasing energy from fusion of elements like Hydrogen and Helium. So, can't that energy be used for fusion of Iron nuclei?
| As you correctly stated in normal situation the star cannot sustain the process. This doesn't mean that there are no such reactions going on in the core. The difference is that during the pre-supernova phase of the star the production of iron is negligible compared to the star. When it goes supernova, it produces a comparable amount of iron.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Myopia / Hypermetropia eye glasses inverting image on retina As far as I'm aware, the eye acts like a pinhole camera in that it inverts the image on the rentina. This makes sense as the rays converge and form a focal point that is upside down.
Myopia (shortsightedness) is described as the rays focusing before the retina, resulting in a blurred distance image. This would still be the right side up. However with Hypermetropia (farsightedness) the focal would be behind the retina, so the rays should project a blurry but right side up (but interpreted by the brain as upside down) image.
I know this is not the case, by why?
| The eye does not act as a pinhole camera. It is a multi element optical system with the cornea and inter-ocular lens doing most of the work. An image is inverted because the light entering the eye from above is headed DOWN, below the optical axis and will therefore image on the lower half of the retina. The light coming from below is has a positive slope and will image on the top half of the retina. This has nothing to do with focus. Regardless of where your image plane is (retina), the converging cones of light are still headed up or down depending on where they came from in object space. While walking on the sidewalk, the light from the gum you're about to step in is headed UP towards your eye and will image on the upper half of the retina. The light from the bird nest that is about to hit you is coming down to you from above and will continue that trajectory even as it passes through the lens in your eye. This has nothing to do with focus.
| {
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"timestamp": "2023-03-29T00:00:00",
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Any quadrupole approximation? Any example? In atomic and molecular physics we quite often encounter with electric dipole approximation. The dipole approximation we do when the wave-length of the type of electromagnetic radiation which induces, or is emitted during, transitions between different atomic energy levels is much larger than the typical size of a light atom. This is mostly the case. I have two questions regarding this:
1) Is there any case where we use quadrupole approximation or higher?
2) In the case of transition in molecules (for eg. large organic molecules or polymers) the size of the molecule is larger than the EM radiation. This case how we choose the approximation?
|
The tides are essentially caused by the quadrupole component of the moon's gravitational field, as shown in this picture.
If you think about it, the classic Stern-Gerlach experiment also depends on a quadrupole field. It's a basic fact of magnetostatics that you can't have a field gradient in the x direction without a complimentary gradient in the y direction. This has consequences: most importantly, it means the usual description of the experiment, with "two dots on the screen", is nonsense. You can't split a ray of silver atoms into two paths, up-and-down, without also splitting them into two paths, left-and-right. Because there's just as much field gradient in the y direction as the x direction.
I discuss this in my blogpost here:http://marty-green.blogspot.ca/2011/12/quantization-of-spin-revisited.html . The actual pattern you would get for a ray of silver atoms would be a donut shape, not two dots. And here is the pattern you would get on the screen for a polarized ray, if you could create a ray with all spins up:
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is moment dependent on the distance from the point of rotation to the force? The formula for moment is:
$$M = Fd$$
Where F is the force applied on the object and d is the perpendicular distance from the point of rotation to the line of action of the force.
Why? Intuitively, it makes sense that moment is dependent on force since the force "increases the intensity". But why distance? Why does the distance from the line of action of the force to the point of intensity affect the moment?
I am NOT looking for a derivation of the above formula from the cross product formula, I am looking for intuition. I understand how when I am turning a wrench, if the wrench is shorter its harder to turn it but I don't understand WHY.
Thanks.
| Understand that there is a tradeoff. You can apply a smaller force with a larger wrench, but you have to move it through a larger distance (arc length) to accomplish the same amount of work (force x distance).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/80538",
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Can I make a rod in the vertical plane move with its one end on the ground in a slanting position? Consider a rod kept vertically on the ground. I keeps the rod in a slanting position making some angle with the horizontal. Can I now move this rod along the horizontal plane by applying a force at its bottom? The torque due to my force at the bottom will act against that due to gravity keeping the rod at equilibrium in that slanting position and I could move it.
I tried it experimentally using one of my books instead of rod and it kind of worked.
But, when I thought about the situation in different ways I get opposite answers. Consider the rod at rest at that slanting position. If I leave it it will fall by rotating about the point of contact. When I'm moving the rod I'm applying force on the bottom point. As my force passes through the axis of rotation torque due to it would be zero and I cannot move it the way I want.
Another way of my thinking-my force at the bottom of the rod would actually provide a torque along the center of mass of the rod which will act against the torque due to gravity and in this way my experiment would work the way I thought.
Why am I getting different results if I think in different ways? Please help.
| The situation in your question is not clear. Without a sketch, all I can offer is the following. The governing equations of a sliding, slanted rod driven by a force $F$ at the bottom are:
$$ \ddot{\theta} = \frac{ \frac{\ell}{2} \left( m \cos\theta \left( g - \frac{\ell}{2} \cos\theta\,\dot{\theta}^2\right) + F\,\sin\theta\right)}{I_C + m \frac{\ell^2}{4} \cos\theta\sin\theta} $$
$$ N = m g - m \frac{\ell}{2} \left(\sin\theta\,\ddot{\theta} + \cos\theta\,\dot{\theta}^2\right) \ge 0 $$
$$ \begin{aligned} \ddot{x}_C & = \frac{F}{m} \\ \ddot{y}_C &= \frac{N}{m} - g \end{aligned} $$
So to keep the slant angle $\theta$, we solve with $\dot\theta = \ddot\theta = 0 $ yielding
$$ F = -m g \cot \theta \\ N = m g \\ \ddot{x}_C = \frac{F}{m} \\ \ddot{y}_C =0 $$
| {
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