Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Why do lasers cut? Is this a case of light acting as matter? All I found in Google was very broad. From a physics models perspective, why can photons emitted from a laser cut? Does this cut mean that the photons are acting like matter?
| Lasers "cut" as an interaction between matter and radiation (themselves different forms of natural substance)
A LASER (Light Amplification by Stimulated Emission of (Coherent) Radiation ) beam being coherent and high-energy (quantum-mechanically a-la Einstein-Planck relation, energy is directly analogous to laser frequency) when interacting with atoms and molecules, provides enough energy for electrons and molecules to move to higher energy-levels and underlying bonds become loose or break completely, resulting in the macroscopic result of "cutting".
As noted above only certain types of LASER "cut" certain types of matter compounds (related to frequency/energy of beam and the bonds base energy of the matter compound under study)
| {
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"timestamp": "2023-03-29T00:00:00",
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Galaxy rotation curve and dark matter I am reading "The Essential Cosmic Perspective" by Jeffrey O. Bennett, Megan O. Donahue, Nicholas Schneider, Mark Voit.
In Chapter 14, it is stated that an evidence of the presence of dark matter in our galaxy is that the rotational curve does not match that obtained from calculation.
In the calculation, the following formula is used
$$M_r=\frac{rv^2}{G}$$
where $M_r$ is the mass enclosed inside a radius $r$ from the galactic center.
My understanding is that the above equation is a result of the shell theorem. But we know that shell theorem applies to spherically symmetric mass distribution only, and most galaxies are disks. So why can we still do this?
| There are two points here. First, the mass is more spherically distributed than one would expect from images. There is something like 5 times as much dark matter as baryonic matter, so even if all the baryonic matter is in a disk, the dark matter halo will be spherical and this will dominate the potential.
Second, in practice we don't in fact make the assumption of sphericity to conclude there is missing mass. You can look at the stars in a galaxy and model its disk-shaped potential and then ask what velocities things should have. You get the wrong answer as a function of radius in the plane of the disk. Moreover, you find that orbital velocities are more or less spherically symmetric in the region around a galaxy, giving further evidence that the matter we see cannot be the whole story. For example, this was done for objects orbiting our galaxy in this paper.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why change in internal energy is zero in isothermal process In isothermal process $\Delta U =0$. But I am having trouble understanding it.
Say we have an ideal gas, and say my temperature is constant but I move the pressure, volume from $(P, V) \to (P-dP, V+dV) $. So the volume has expanded and system has done some work to the surrounding. So my work is non-zero.
So how come $\Delta U=0$? I am really confused here.
| See consider a cylinder with piston fixed, (i.e. it doesn't move) and the system is provided with a source with temperature $T$. Now as the piston does not move the volume is constant, so no work is done and internal energy is also constant and no heat is added since system and source are at the same temperature.
Now let us release the piston. Then the system does work using internal energy, but the change in internal energy is spontaneously overcome since system is copped to maintain at constant temperature. In other words we can say that system never uses internal energy since it is supplied with constant heat from a source.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does Minkowski space provide an accurate description of flat spacetime? What is the chain of reasoning (beginning, of course, from observations about the universe) that leads one to predict that Minkowski space provides an accurate description of space-time in the framework of special relativity?
That is, why should flat space-time have the Minkowski metric
$$\eta = \left(\eta_{\alpha\beta}\right) = \left(
\begin{matrix}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{matrix}\right)\,?$$
| Minkowski space provides an accurate description of flat space-time simply because it was derived from the equation for light traveling in Euclidean space. There is nothing unusual about the metric - Minkowski metric is just a way of presenting the good old Euclidean space. And as in Special Relativity there is no gravitation (acceleration) to curve this space-time, so it remains flat.
If you take a look at Einstein's "Relativity: The Special and General Theory", you will find in the Appendix I (just before equation (10)) that in order to derive the metric, Einstein simply started off with the Pythagorean Theorem in 3D, which he put like this:
$$r = \sqrt{x^2 + y^2 + z^2} = ct$$
So what he proposed (as well as Minkowski himself) is to show the vector of light traveling in a three dimensional space.
He then squared both sides and moved the $x^2 + y^2 + z^2$ to the right to obtain the equation $ct^2 - x^2 - y^2 - z^2 = 0$ (and later $0$ was replaced with $1$, which was to assure symmetry). Therefore we have the signature [$+, -, -, -$], which corresponds to the matrix you presented in your question.
So going back to our first equation we can see that it represents the movement of light in flat (Euclidean) space-time by definition. The basic equation that was used to derive the matrix refers to Euclidean space, and therefore the space-time it represents must be flat. And flat space-time is the framework of Special Relativity which pertains to inertial frames, i.e. non-accelerated ones (as opposed to General Relativity where gravity - i.e. acceleration - is the source of curvature).
| {
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Apparent dimensional mismatch after taking derivative Suppose I have a variable $x$ and a constant $a$, each having the dimension of length. That is $[x]=[a]=[L]$ where square brackets denote the dimension of the physical quantity contained within them.
Now, we wish to take the derivative of $u = log (\frac{x^2}{a^2})-log (\frac{a^2}{x^2})$. Here, we have taken the natural logarithm. It is clear that $u$ is a dimensionless function.
$$\frac{du}{dx} = \frac{a^2}{x^2}.\frac{2x}{a^2} - \frac{x^2}{a^2}.(-2a^2).\frac{2x}{x^3} \\
= \frac{1}{x} - 4. $$
Here, the dimensions of the two terms on the right do not match. The dimension of the first term is what I expected. Where am I going wrong?
|
Where am I going wrong?
Recall
$$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$$
with
$$f(\cdot) = \ln(\cdot) \rightarrow f'(\cdot) = \frac{1}{\cdot}$$
and
$$g(x) = \frac{a^2}{x^2} \rightarrow g'(x) = \frac{-2a^2}{x^3}$$
Thus
$$\frac{d}{dx}\ln\frac{a^2}{x^2} = \frac{1}{\frac{a^2}{x^2}}\frac{-2a^2}{x^3} = -\frac{2}{x}$$
An alternative approach is to recognize
$$\ln x^{-2} = -2\ln x$$
thus
$$ \ln\frac{a^2}{x^2} = \ln a^2 + \ln x^{-2} = \ln a^2 -2 \ln x$$
for which we can immediately write the derivative.
| {
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In the case of riding a bicycle, why can the system of the bicycle and rider be accelerated if no external net force? According to Newton, if no external net force acting on a system then the system will keep its initial condition whether at rest or moving uniformly in a straight line.
Let's consider a boy riding a bicycle. He exerts a force to the bicycle and the bicycle reacts with the same magnitude but opposite direction to the boy. From an observer outside the system of boy-bicycle, the net external force acting on the system is always zero so there will be no acceleration. However, in our daily life, we have seen that accelerating a bicycle is possible. It seems to me inconsistent, doesn't it?
What is wrong with my argumentation above?
| There are forces and torques in this system. Boy pushes on pedal and pedal pushes on boy. But for pedal to push on boy, bike frame pushes on pedal (at the axle of the pedals) and vice versa. Now there is a torque on the pedal which is transferred (through the chain) to the rear wheel. Rear wheel pushes on road, and road pushes on wheel (in the forward direction - because there is friction). Wheel pushes on frame. Frame pushes saddle. Saddle pushes boy. Boy moves with the whole bike.
I left out a few steps (like the fact that the torque is balanced by a different vertical force on rear wheel than on front wheel, or that angular momentum is conserved because of the force of the bicycle on the earth...) but I hope you get the idea.
Here is a little picture showing just some of these forces. In particular, the blue arrows correspond to forces from outside the "boy-bike" system: namely, the vertical force on the wheels (which stops the bike from disappearing into the earth) and the horizontal reaction force due to the force of the wheel on the earth (not shown) which in turn is a response to all the torques (described above - just some of them shown):
Source of bike cartoon
| {
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How to calculate the velocity of this body?
Graph of a train (the body) is provided which starts from rest. What is the velocity after the train has 10 metres displacement?
Thats the only things provided for the question, please help me out here, I find that more variables will be required but the book states that no more variables are required.
The answer key says that the answer is 10 m/s only! I am totally unable to understand how.
| According to the graph, accelration is linearly dependent on displacement. Now, assuming the motion is rectillinear(along a straight line, and quite justified for 30 m displacement of a train).
It is not very difficult to find the equation of this straight line (intercept form).
And then, you'll need some calculus to find velocity $\nu$ in terms of displacement $x$, which involves the following substitution : $$a=\nu\frac{d\nu}{dx}$$
Now transfer $dx$ to the other side and integrate. Try and see if it helps.
| {
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Is evaporative cooling more efficient with dry or moist air? I live in India, and in the summer season, the temperature can reach up to $45 \sideset{^\circ}{}{\mathrm{C}} .$ We use Split 1.5 Ton AC in our small office. The idea is to put an evaporative cooler on the inlet side of the heat exchanger of AC to give it more efficient cooling.
Will it help to increase efficiency? or COP? By how much?
| If you’re recirculating the office-side air, evaporative cooling is unlikely to help. You’d be adding water vapor to the air which the AC would eventually just have to use cooling power to remove.
Depending on humidity, though, it may be very effective to use evaporative cooling to reduce the temperature of the air reaching the outside (hot side) coils. That’s once-through air flow.
| {
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Why do turbine engines work? I know roughly how a turbine engine (let's say a gas turbine producing no jet thrust) is supposed to work:
The compressor forces fresh air into a combustion chamber, where it reacts with fuel to become hot exhaust gas. On its way out of the engine, the exhaust gas drives a turbine, and the turbine both makes the compressor go, and has enough leftover torque to do useful work.
However, how do the exhaust gases know they're supposed to push on the turbine blades to drive the shaft, rather than push back on the compressor blades to retard the drive shaft in equal measure?
In a piston engine there are valves that force things to flow in the correct direction at the right times. But with the turbine engine everything is openly connected all the time. Shouldn't that mean that the pressure differential the compressor must work against is exactly the same as that which is available to drive the turbine?
Something magical and irreversible seems to happen in the combustion chamber.
The descriptions I can find that go deeper than the three-step explanation above all seem to jump directly to a very detailed model with lots of thermodynamics and fluid dynamics that make my head spin. Is there an idealized system with fewer variables that I could think of to convince myself we're not getting something for nothing here (e.g., might the working fluid be incompressible, or massless, or have infinite heat capacity or whatever)?
| I have the same question myself but I'll raise two points:
Firstly, the compressor is not just a turbine, but also, in many cases, kind of a centrifugal pump. In such a case, I don't think the pressure in the combustion chamber can push the pump backwards.
Secondly, in a high speed air-flow situation, the dynamics is quite different from the static one. Just like the wing which makes air pressure under and above different, there must be structures to control the air pressure inside the combustion chamber, that make the air go where it is wanted.
| {
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Feynman's $i \epsilon$ prescription in loop expansion I have some questions about the $i\epsilon$ factor in Feynman diagrams. First, what is the physical meaning of $i\epsilon$ in loop amplitudes. Second, how does it ensures unitarity?
And third, Dyson series assume that incoming and outgoing particles are free, this can be implemented by assuming that the interaction Hamiltonian switches off adiabatically, $e^{-\eta\,t}H_{I}(t)$. Is this $\eta$ related with the $i\epsilon$?
| $\rm{i}\varepsilon$ is not a factor, but an addendum. It allows to integrate straight along the axis. It removes poles from axes in a "physical way" - by obtaining the right propagator. The right propagator does not need any adiabatic tricks.
Incoming and outgoing particles are not always free. In particular, the meaningful electron line has many soft photon lines emitted. It means the electron is coupled to the electromagnetic field even in the asymptotic regions. Only perturbatively (which is rather inexact) the lines may be "free".
| {
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Why Does Change of Magnetic Flux Induce an emf? Why does change in magnetic flux with time through a coil induce an emf across it? Please explain what happens to the charges in the coil when magnetic flux changes?
Also, why does a constant magnetic flux not induce an emf?
| It is an experimental fact that a changing magnetic field induces an electric field and a changing electric field induces a magnetic field. This has been mathematically described in the all inclusive Maxwell equations.
Conductivity appears with the motion of charge in a conductor.The difference between insulators and conductors is that in insulators the electrons around the atoms and molecules composing them are bound and cannot be detached from the potential that binds them to the nuclei of their atoms/molecules. In conductors some of the electrons are bound by a collective potential of the material, in bands where they can move large distances with respect to atomic distances, being shared with many atoms/molecules.
When an electric field is applied on a conductor the electrons are attracted to the poles and move in the direction of the electric field and a current appears. In a closed circuit conductor a changing magnetic field will be producing an electric field to which the electrons will respond by moving in the direction pointed by the field.
It is the changing magnetic field that induces the electric field. A steady one does nothing.
| {
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Is Newton's third law always correct? Newton's third law states that every force has an equal and opposite reaction. But this doesn't seem like the case in the following scenario:
For example, a person punches a wall and the wall breaks. The wall wasn't able to withstand the force, nor provide equal force in opposite direction to stop the punch.
If the force was indeed equal, wouldn't the punch not break the wall? I.e., like punching concrete, you'll just hurt your hand. Doesn't this mean Newton's third law is wrong in these cases?
| You can't apply to a wall the force bigger than it can withstand. If the wall breaks at some certain amount of newtons of the force applied then it follows that you applied this certain force. And of course the reaction of the wall equals the force applied.
| {
"language": "en",
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Chemical effect on gravitation? We know, that gravitation field of charged black hole is different than one of uncharged.
I this true only for objects with singularity or is true for all objects?
If true, then may we say, that gravitation field of chemical molecule has footprints of it's chemical composition and state and, in principle can be registered?
I know it is very small.
| Gravitational fields are not sensitive to chemical composition or state per se, except through the stress-energy tensor. The reason that a charged black hole has a different gravitational field is basically that the electric field has energy, and that energy is equivalent to some amount of mass, which is distributed over space rather than concentrated at the singularity. The stress-energy tensor isn't just a measure of mass-energy density; it also depends on the density of momentum and on the pressure. So chemical composition or the state of matter have no effect on the gravitational field unless they have an effect on the density of mass-energy, density of momentum, or pressure.
Note that electrical properties like the ones you've described may not even have any effect on the stress-energy tensor. For example, when you flip an electric dipole to point in the opposite direction, its stress-energy tensor remains the same, since the stress-energy tensor of the electric dipole field doesn't change under the flip. So not only is it impractical to use gravitational fields rather than electrical ones to find out about the dipole, the gravitational method gives less information even in principle.
| {
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Starting a nuclear reaction In Chemistry, an amount of energy has to be supplied for a reaction to occur. This energy, known as the "activation energy", breaks up the bonds between molecues in the substance. It is equivalent to the total bond energy of the reactants.
However, in high school I learnt that the energy required to start a nuclear reaction is the difference between the binding energy of the reactants and the binding energy of the products.
Why is it that the minimum required energy is not the binding energy of the reactants, similar to a chemical reaction?
|
In Chemistry, an amount of energy has to be supplied for a reaction to occur. This energy, known as the "activation energy", breaks up the bonds between molecules in the substance. It is equivalent to the total bond energy of the reactants.
In chemistry activation energy is not the energy required to break a bond. Instead, the activation energy is the difference in energy between a transition state and the reactants.
A product bond can begin to form before the reactant bond is completely broken.
A transition state may be stabilized by a catalyst, lowering the activation energy. This is critical to all life, as many biochemical reactions would not proceed at necessary rates without enzyme catalysis.
I learnt that the energy required to start a nuclear reaction is the difference between the binding energy of the reactants and the binding energy of the products.
As for nuclear reactions, the activation energy is not the difference between the binding energy of the reactants and the binding energy of the products. That would be the negative of the energy released by the reaction. For activation energy of a nuclear reaction, the energy difference between a transition state and the reactants should be considered.
| {
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Why is the ratio of velocity to the speed of light squared in the Lorentz factor? Why is the ratio of velocity to the speed of light squared in the Lorentz factor?
$${\left( {{v \over c}} \right)^2}$$
My only guess is the value must be positive.
| Consider a particle of light moving along a sick of length $c\Delta t$ from the frame of reference of a stationary observer. i.e. The stick itself has velocity zero relative to the stationary observer.
Then consider another observer moving with a velocity perpendicular to the sick with speed $v$. The stationary observer sees the light move straight up. While the moving observer sees the light moving along a diagonal path.
Since Maxwell made it clear that the speed of light remains the same in all reverence frames we need to make a correction to the classical transformations.
The height of the sick also remains the same in both frames of references, because the non-stationary observer is moving with velocity perpendicular to the stick.
This means the distance along the stick in the stationary frame divided by the time in the stationary frame must equal to the same length of the stick in the moving frame considering the speed of light remains the same.
Using the right triangle shown we get:
$$
(c\Delta t_p)^2 = (v\Delta t_p)^2 + (c\Delta t)^2
$$
$$
\Delta t^2_p(c^2-v^2)=c^2\Delta t^2
$$
$$
\Delta t^2_p = \frac{c^2\Delta t^2}{c^2-v^2} = \frac{\Delta t^2}{1-\frac{v^2}{c^2}}
$$
And hence the $\frac{v^2}{c^2}$ factor, Lorentz transformations, and the non-euclidean geometry of flat spacetime.
| {
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Temperature on the surface of the sun calculated with the Stefan-Boltzmann-rule In a German Wikipedia page, the following calculation for the temperature on the surface of the Sun is made:
$\sigma=5.67*10^{-8}\frac{W}{m^2K^4}$ (Stefan-Boltzmann constant)
$S = 1367\frac{W}{m^2}$ (solar constant)
$D = 1.496*10^{11} m$ (Earth-Sun average distance)
$R = 6.963*10^8 m$ (radius of the Sun)
$T = (\frac{P}{\sigma A})^\frac{1}{4} = (\frac{S4 \pi D^2}{\sigma 4\pi R^2})^\frac{1}{4}=(\frac{SD^2}{\sigma R^2})^\frac{1}{4} = 5775.8\ K$
(Wikipedia gives 5777K because the radius was rounded to $6.96*10^8m$)
This calculation is perfectly clear.
But in Gerthsen Kneser Vogel there is an exercise where Sherlock Holmes estimated the
temperature of the sun only knowing the root of the fraction of D and R.
Lets say, he estimated this fraction to 225, so the square root is about
15, how does he come to 6000 K ? The value $(\frac{S}{\sigma})^\frac{1}{4}$
has about the value 400. It cannot be the approximate average temperature
on earth, which is about 300K. What do I miss ?
| The relationship of temperature between a planet and a star based on a radiative energy balance is given by the following equation (from Wikipedia):
$T_p = temperature\ of\ the\ planet$
$T_s = temperature\ of\ the\ star$
$R_s = radius\ of\ the\ star$
$\alpha = albedo\ of\ the\ planet$
$\epsilon = average\ emissivity\ of\ the\ planet$
$D = distance\ between\ star\ and\ planet$
Therefore if Sherlock knows $\sqrt{\frac{R_s}{D}} = 0.06818$ and can estimate the Earth's temperature $T_p$ as well as $\alpha$ and $\epsilon$ then he can calculate the temperature on the surface of the sun which is the unknown variable $T_s$.
Both $\alpha$ and $\epsilon$ have true values between zero and one. Say Sherlock assumed $\alpha = 0.5$ and $\epsilon = 1$ (perfect blackbody). Estimating the temperature of the Earth $T_p$ to be 270 K and plugging in all the numbers we have:
Which is very near the true average temperature of the surface of the sun, 5870 K. Case closed!
| {
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Power fit to some experimental data I have to fit some data to a power law
$$ F=\alpha q^{\beta}$$
being $q$ and $F$ the experimental data points. What I usually do is taking logs so that
$$ \ln(F) = \beta \ln(q)+\ln(\alpha)$$
and apply least squares with uncertainties in both variables. How may I approach this in the case that some of the values of $q$ (and eventually $F$) are negative?. If it helps $\beta$ should be compatible with a value of $1$, but I want a general approach.
Edit:
The law to be fitted is Coulomb's law, $F$ is the force between to charged spheres. One of them has fixed charge, the other is a variable we call $q$. The proportionality constant $\alpha$ is related to $\varepsilon_0$ with a known expression in powers of $\beta=\frac{R}{d}$ where $d$ is the distance between the centers of the spheres and $R$ is their radius. So, I want to determine, from experiment, both $\alpha$ and $\beta$ and compare them with $\varepsilon_0$ and $1$.
The actual values are
$$
\begin{array}{|c|c|}
\hline
q / \mathrm{nC} & F / \mathrm{mN} \\ \hline
-26,8 \pm 0.8 & -1.5 \\ \hline
-18.8 \pm 0.5 & -1.1 \\ \hline
-9.9 \pm 0.2 & -0.6 \\ \hline
9.9 \pm 0.2 & 0.5 \\ \hline
18.8 \pm 0.5 & 1.0 \\ \hline
26.8 \pm 0.8 & 1.5 \\ \hline
35 \pm 2 & 2.1 \\ \hline
\end{array}
$$
with the uncertainty in the last decimal place for the forces.
| Taking the logarithms of negative numbers isn't a viable solution.
When you postulate $ F=\alpha q^{\beta}$ you are implicitly supposing that is F is a real number, q is positive.
Just make a least square fit using the absolute values of $F$, $\alpha$ and $q$:
If $ F=\alpha q^{\beta}$ then $ |F|=|\alpha| |q|^{\beta}$, so $ \ln(|F|) = \beta \ln(|q|)+\ln(|\alpha|)$.
That certainly will work.
Next study the sign: it is evident from data that with negative charge you get negative force. Then you can conclude that $\alpha$ is positive.
| {
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Lagrangian point or dark matter? We know that spiral galaxies spin in a way such that we have to assume that dark matter is responsible for the extra mass required to do so.
My question is, can Lagrangian points (L1 and L2) be used to describe a galaxy's rotation instead?
Can we explain that objects far away from the center of the galaxy have higher velocity because they are at the L2 Lagrangian point of a Lagrangian system which consists of a) the galaxy's super massive black hole at its center, b) a part of its spiral arm c) the far away object in question?
(I'm a computer engineer interested in physics. Please excuse my ignorance)
| The objections posted by the answerers to the original question really just point to the need for a solution to an unrestricted many-body problem to actually answer the question fairly. Since the asker is a computer engineer I would suggest he code a numerical simulation. My intuition is that, although the bodies involved are numerous and extended, that the gravitational fields will interact in interesting ways. I suspect the rough magnitudes involved are accounted for in existing models, but it would be interesting to see if the interacting fields have a significant effect on galactic motion.
| {
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Is it more efficient to stack two Peltier modules or to set them side by side? Is it more efficient to stack two Peltier modules or to set them side by side?
And why?
I have a small box that I want to cool down about 20 K below ambient -- cold, but not below freezing.
(I want to keep my camera cool, so I'm putting in this cool box.
The camera looks through a flat glass window on one side of the box).
The heatsink I have on hand is about twice as wide as the widest Peltier module I originally planned on using.
So there's room to put 2 Peltier modules side-by-side under the heatsink.
Or I could center a stack of 2 Peltier modules under the heatsink.
Which arrangement is more efficient?
I have to cut a bigger hole in the insulation for the side-by-side arrangement, so the unwanted heat that "back-flows" through the side-by-side arrangement is worse.
On the other hand, other effects are worse for the stacked arrangement.
(Is https://electronics.stackexchange.com/ a better place to post questions about Peltier coolers?)
| If efficiency is the issue, then definitely parallel TECs (or use a single unit rated for twice the power, same thing). The only reason for stacking TECs is to get a lower temperature. However that comes at great expense to efficiency and overall power consumption.
Another point is that paralleling TECs is actually more efficient overall. The reason is that TECs are more efficient at lower power. The cooling power is proportional to the current, but the internal heating is due to internal I2R losses, so goes with the square of the current. At low currents a small increment in current causes more cooling than additional internal heating. However, eventually the squared term wins and at the top of the parabola a small current increment causes the same heating as the additional current is able to cause the device to remove. That's the maximum cooling point, but is also the least efficient point along the normal operating range.
Two indetical TECs in parallel therefore need less than 1/2 the current each than a single one would need for the same cooling power. Another way to put this is that except for money and space constraints, you want to oversize the TEC for efficiency.
| {
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Electron indistintinguishability and handedness I just learned that right ad left handed electrons behave in a remarkably different way under the weak interaction.
Up till now I have been told that all the electrons are exact copies of one another and all this fermi-dirac statistics story.
Nonetheless, since right handed and left handed electrons DO have fundamental differences, shouldn't this remove their indistinguishability?
| A massive electron in its rest frame is a mixture of left- and right-handed components. All electrons mix the same way, and so they remain indistinguishable.
| {
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Light penetration in the ocean Is there an equation that describes the percentage of solar energy per wavelength that penetrates the ocean surface?
For example, this website states that only 44.5% of the surface light reaches a depth of 1 meter.
So, is there an equation that can be used to determine the percentage of solar energy per wavelength?
| The light intensity at a depth $d$ is given by the Beer-Lambert law:
$$ I = I_0 \exp\left(-\frac{d}{\ell}\right) $$
where $I_0$ is the intensity at the surface and $\ell$ is the depth at which the light intensity has fallen by 73%.
The depth $\ell$ is dependant on wavelength as seawater absorbs red light more strongly than blue light. I did a quick Google to see if I could find data on the absorptance of seawater, but while I found lots of articles I couldn't find any definitive figures. Presumably the absorptance of seawater is rather variable as different samples will contain different amounts of dissolved and suspended materials.
| {
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What does QM say about the past rather than the future? In QM, the wave function (in the Copenhagen interpretation) is not an actual physical wave but a device to derive probabilities about the outcomes of experiments. The wave function encodes all the information about the system we want to derive predictions for. Predictions are about future measurements. Once the measurement has been performed and the result is known, we adjust accordingly our expectation: the so-called collapse of the wavefunction just took place (let me add, in our minds). This subjective knowledge about the predictions of QM is crucial to avoid problem with causality in relativity when studying entangled systems. Fine.
What I am a bit confused about is what QM says about the past, rather than the future. What is the analog picture that QM gives about the state of a system in the past? What does QM say about the conditional probabilities of events? What does QM tell about, say, cosmology and the far past of the universe when e.g. string theory becomes relevant? I hope it is not a trivial, naive, question.
| On the fundamental level, QM is time-symmetric, hence it says the same about the past as about the future. The dynamics of the state is deterministic and given by the quantum Liouville equation (or, if you consider an isolated system in a pure state, by the Schroedinger equation). This state determines the probability distribution of measurable events at any time, the past as well as the future.
Accounting for new information about the past (or the future), as it becomes used by the observer to improve predictions about the past (or the future), is accounted for by projecting the state to the invariant subspace determined by the new information. This is the quantum analogue of taking conditional expectations in a classical stochastic model when new information about the past (or the future) becomes available.
| {
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What are the units for rate of movement through time? Thinking of time as the dimension orthogonal to depth, it seems logical that the rate at which a body moves through time could vary. Do units exist for rate of movement through time? The numerator seems clear (e.g. seconds), but what could the denominator be?
| When you are taking of RATE you already mean the flow of time.
So rate at which a body moves through time is not a physically correct statement.
According to the velocities of bodies the perception of time may be different but the rate of flow of events is TIME itself.
| {
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Why Microchannel Plates can be operated only in vacuum? Why it is said that the Microchannel Plates can be operated in vacuum? What is the maximum pressure in which it can be operated? Also, while it is not operating, should it be kept in vacuum?
Is this because the semiconducting property of the channels can be affected by the atmospheric pressure? Or while operating the gas molecules can be ionized and burn the detector?
| The electrons need to get from the top to the bottom without any interference from any gas molecules that might be in the channels. If nothing else, collisions with gas molecules will degrade performance. At atmospheric pressure, I don't think the device would work at all. You can blow a hole through an MCP with over-voltage, but I'm not sure how this relates to residual gas in the channels. Also, it is certainly possible to generate a discharge (spark) between the bottom of the MCP and whatever comes next, so you want to keep gases out of there.
You can store them in air, preferably in a dry box.
| {
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Is there any operator behind probability, in quantum mechanics? In Quantum mechanics, the probability of finding a particle at position $x$ is given by $|\psi(x)|^2$, where $\psi$ is the wave function. Wonder what is the operator which gives this probability? Is probability the result of any operator acting on $\psi$?
| The theory of quantum mechanics has been developed to explain observations, i.e. measurements. Without observations it is a floating mathematical construct.
One of the postulates to connect the mathematics with reality is:
To every observable there corresponds an operator which operating on the state function will give an eigenvalue. So the question becomes : is probability an observable? and then it becomes :what is an observable.
An observable in the framework of quantum mechanics is a variable of the system under consideration which a measurement can evaluate . The energy of a single photon. The momentum of a proton. The spin of an electron. We can always measure these variables on single particles with one observation, measurement. This is not possible with probability. It is an emergent value from a great number of measurements with the same boundary conditions: it is a normalized distribution, varying from 0 to 1, of the spread of the values found in the measurements.
So no, there exists no quantum mechanical operator for probability, as it is not an observable of a variable entering the quantum mechanical problem but an emergent quantity from a lot of measurements.
| {
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Why are the 'color-neutral' gluons confined? What makes the two 'color-neutral' gluons
$(r\bar r−b\bar b)/\sqrt2$ and
$(r\bar r+b\bar b −2g\bar g )/\sqrt6$ different from the pure $r\bar r +b\bar b +g\bar g $ ?
Why don't they result in long range (photon-like) interactions?
| One could state your problem as " why are there only eight gluons accepted as physically existing from the SU3 symmetry . This link gives the answer simply, and it says that if the ninth all color neutral gluon existed, then as you guess, it would act similarly to the photon with interactions between hadrons that have not been observed.
So the answer is: so that the SU(3) color model should fit the data , i.e. strong interactions between hadrons are confined because we have not observed colorless gluons in the interactions between hadrons.
In the matrix analysis the link goes on to show this :
But, no matter how we take complex linear combinations of trace-zero hermitian matrices, we cannot get
1 0 0
0 0 0
0 0 0
since this has trace 1. (I.e., the sum of the diagonal entries is 1: that's what the trace means, the sum of the diagonal entries.) So we cannot really get red anti-red. The closest we can get are things like
1 0 0
0 -1 0
0 0 0
or
1 0 0
0 0 0
0 0 -1
or
0 0 0
0 1 0
0 0 -1
But note, these three are not linearly independent: any one of them is a linear combination of the other two. So we can get stuff like
(red anti-red) − (blue anti-blue)
and so on, but not 3 linearly independent things of this sort, only 2: one less than you might expect.
It is thus fortunate that SU(3) color can be used to described what has been observed with hadrons, i.e. no neutral photon like gluons
| {
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Why does a ball bounce lower? If a ball hits the floor after an acceleration then why does it bounces lower? I mean the Energy is passed to the floor then why does the floor give back less Energy?
| At impact, most of the kinetic energy is transferred to elastic energy in the ball (by its deformation) and not to the floor. Some energy is also converted to other forms like heat and sound. These other forms of energy, are mostly losses and they are not recovered thus making the ball bounce back to a lower height.
Surely there is some energy passed to the floor, but that's not very relevant for the process. Only the energy stored as elastic potential in the ball produces the force against the floor to send back up.
| {
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Peaks in binding energy per nucleon Looking at the the binding energy per nucleon chart:
I observe peaks for N=4,8,12,16,20,24
while I expected to observe peaks for 2, 8, 20, 28, 50, 82, and 126 because I have heard that in correspondence of these numbers there is the completion of a nuclear orbital, so I expect a local maximum of the stability an so a peak of binding energy.
Is my reasoning wrong?
| Your peaks at $A=4n$ are the so-called "alpha-particle nuclei," which have even proton number, even neutron number, and $N=Z$. In a very hand-wavy way you can say that alpha-particle nuclei are especially tightly bound because the alpha particle is especially tightly bound.
Inside of a nucleus you have separate orbitals for neutrons and protons. The magic numbers you're thinking of, where a nucleon orbital closes, do correspond to especially tightly bound nuclei, but to see it you need to look at a full table of isotopes. For instance, tin (with $Z=50$) has ten or eleven stable isotopes, while indium and antimony ($Z=49,51$) have only two stable isotopes each. There are also six different elements with stable isotopes having $N=50$, which kind of stands out on the chart. There are some "doubly-magic" nuclei with closed shells in both $N$ and $Z$, such as calcium-40, calcium-48, which are both stable, and tin-100, tin-132, both near the drip lines.
| {
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Physical Interpretation of four velocity in GR I'm confused about the physical interpretation of the four-velocity $U^\mu=\frac{dx^\mu}{d\tau}$ in General Relativity. I know that it is a tangent vector to a particle's "worldline", but what does this mean more physically?
For example, I am comfortable with what $U^\mu$ means in Special Relativity. In your inertial frame, you cover a distance $\Delta x^\mu$ and your clock says time $\Delta \tau$ has passed, and by taking the limit as $\Delta \tau \to 0$ this defines your $U^\mu$.
But I'm unsure about what $U^\mu$ means in curved space, or even in an accelerated reference frame. In either case the frame is no longer an inertial frame, which makes it confusing to interpret $\tau$, because it's no longer the "proper time in a frame", there is no one frame we are working in.
| Accelerated particles do not have a "fixed" inertial reference frame. However, we can define infinitely many momentarily comoving reference frames (MCRFs) at each event. These are all related to each other through simple rotations. The four velocity at each event therefore is the time component of that event's MCRF's basis. As the particle is accelerated, so its MCRF will keep changing and so will its four-velocity.
| {
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Moon's pull causes tides on far side of Earth: why? I have always wondered and once I even got it, but then completely forgot. I understand that gravity causes high and low tides in oceans, but why does it occur on the other side of Earth?
| Imagine that we have a very massive object in space. At some distance away (call it ten units) we release three tennis balls in a row:
The tennis balls all fall towards the massive object. But because gravity goes like distance squared, the nearer balls feel a stronger attraction than the farther balls, and they move apart from each other:
You're riding on the middle tennis ball. You feel like you're in free fall, in a good inertial frame. You look towards the heavy object and you see the leading tennis ball moving away from you. You look away from the heavy object and you see the following tennis ball moving away from you. The heavy object is pulling the three tennis balls apart.
Likewise, if you had three objects at the same distance falling towards the massive object,
you'd see them converge as they all fell along slightly different rays towards the same center. This gives the tidal compression. You can imagine the process of launching a whole constellation of tennis balls, choosing the center one as your "rest frame," and having their motions approximate the arrow pattern in Joshua's figure.
The situation stays essentially the same if you add angular momentum, except that then your tennis ball constellation doesn't crash onto the massive object.
| {
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Placing two similarly charged particles in space Now, I will make a hypothetical situation. Assume that we place two similarly charged particles (lets take electrons) in space. Imagine that there is no other force acting on the particles except the repulsive force and the gravitational force of the particles. In other words, only these two electrons are present in the universe. So they are free from any outside interference. Now by nature, these electrons will start moving away from each other due to the repulsive force. Since there is nothing to stop them (gravitational force will only slow them down and not stop them as it is of a lesser magnitude than the repulsive force) they will keep moving and never stop. Over here we exclude expansion of space also for no complications. Now since the particles will keep moving as there is a constant repulsive force acting on them, they will do infinite work because $Work = Force * displacement$ and the displacement over here will keep increasing. Please tell me what is the problem in my thought experiment because it violates conservation of energy.
| Ok, so we all agree that $W = \vec F · \vec x$. If the force varies, then the total work on each electron is calculated using an integral:
$W = \int_{x_0}^\infty\vec F · d \vec x$
Here, $W$ is work, $\vec F$ is the electrical force, $\vec x$ is the distance of the charge from the center of your universe, and $x_0$ is the starting point from where you begin the experiment.
And the electrical force between the electrons is :
$F = q {e_1 e_2 \over (2x)^2}$
$q$ is the Coulomb constant, $e_1$ and $e_2$ are the charges (they are equal), and $2x$ is the distance between them. You know, one $x$ to the left and one $x$ to the right. And since the force and the displacement are colinear we can omit the arrows.
Now we can solve the integral:
$W = \int_{x_0}^\infty q {e_1 e_2 \over (2x)^2} dx$
$W = {q e_1 e_2 \over 4 }\int_{x_0}^\infty {1 \over x^2} dx$
$W = {q e_1 e_2 \over 4 } [{-1 \over x}]_{x_0}^\infty $
$W = {q e_1 e_2 \over 4 } {1 \over x_0}$
And that is basically the formula for the eletric potential energy!
| {
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Self-adjoint extensions with 'teletransporting' boundary conditions When choosing a self-adjoint extension of a Hamiltonian, in general one can obtain domains in which
(i) the probabilities teleport* between points on the boundary
and
(ii) boundary conditions locally conserve probabilities.
The ones which locally conserve probability currents somehow seems nicer to me. But this is not at all an argument especially since tunneling is allowed in quantum mechanics.
Is there any fundamental physical reasoning one can use to discard teleporting boundary conditions?
Thanks in advance for any useful inputs.
*I have used terminology from discussion about a related question :
Physical interpretation of different selfadjoint extensions
| Once you admit the tunnelling the "topology" of the problem changes to "particle in a circle with an infinite barrier". In fact, to avoid arguments with energy when crossing the barrier, it can be described as "particle in a circle with an infinite barrier in one point". So physically it can be argued to be a different problem.
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Why does a large train cause the ground to shake? I work in a 4 story building that is approx. 150 feet away from a set of train tracks. When a large (40+ car) freight train goes by, the shaking in the building is perceptible. As I've watched the train go by, there does not appear to be any side to side movement and the speed is constant. What exactly is causing the vibrations in the ground? Is it simply the train's traversal over each segment of track? Surely the train itself is not "bouncing" along the track with enough energy transfer to shake the ground, right?
| Since "enough" information is not available, one can only guess. My guess is that "most likely" resonance is what's causing the perception. The length, height, width, and composition of the building and its distance from the tracks, determines its "natural" oscillating frequency and the train's length and speed must create an oscillation that closely matches the building's frequency (or its harmonics), so that the small train vibrations get amplified to the point of being felt in the building. In order to find out if this is the case, there should be a train length that causes the "most noticeable" shaking. Longer or shorter than this, should produce less noticeable shaking.
| {
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Finding the components of the tensor for potential and kinetic energy I have a rather poor understanding of what a tensor is, but enough to apply it to the biggest part of the classical mechanics I'm studying.
However, I've run into a small problem while studying "Free vibrations of a linear triatomic molecule". (We're assuming all 3 atoms are in 1 straight line and the forces they exert on each other are represented by springs with constant values for $k$).
My potential energy is described as:
$$V=\frac{k}{2}(\eta_1^2+2\eta_2^2+\eta_3^2-2\eta_1\eta_2-2\eta_2\eta_3)$$
Where $\eta$ is a coordinate relative to the equilibrium position.
And 'hence' the tensor has the form:
$$\vec V=\begin{bmatrix} k & -k & 0\\ -k & 2k & -k\\ 0 & -k & k \end{bmatrix}$$
How do I go from 1 to the other and vice versa?
I've tried googling terms like "Equation to tensor", "Potential energy tensor", "Absolute value of tensor" etc. but they didn't yield anything usable for me.
| Your squared general coordinates correspond with the tensors diagonal axis. The others are the cross terms. Since we are dealing with commuting terms, it will always be of the form $2\eta_{i}\eta_{j}$ although it is truly $\eta_{i}\eta_{j}+\eta_{j}\eta_{i}$. Obviously, the $\frac{1}{2}$ is absorbed by the energy equation.
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How can $J_1^2, J_2^2, J_{1z}, J_{2z}$ commute mutually? I'm reading through J. J. Sakurai's Modern Quantum Mechanics book and currently looking at the "Angular-momentum addition" part.
Here, it says you have two options and that one option is to construct simultaneous eigenket $\vert j_1j_2;m_1m_2\rangle$ of $J_1^2, J_2^2, J_{1z}, J_{2z}$ since the four operators commute with each other. I understand that $J_1^2$ and $J_{1z}$ commute, but I'm not sure how $J_1^2$ and $J_{2z}$ can commute intuitively.
"commute" means that one can measure both at once right? But total angular momentum of spin 1 and angular momentum of spin 2 are independent.
Where am I wrong here?
| From your statement "measure both at once", it seems you've misunderstood what's meant by "simultaneous measurement". It does not mean that you can run a single experiment to measure both "independent" values together. Rather, it means that in principle, you can measure one quantity without "ruining" the results of measurement the other, so both quantities can be obtained.
| {
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Clocks in special relativity One book on special relativity says:
Any observer at rest relative to his own timepiece will see that other clocks moving with respect to him run fast - the greater their speed, the faster they are.
Other book says:
Observers measure any clock to run slow if it moves relative to them.
Don't they contradict each other? If yes - who's right? If no - why are they both right? I assume it is a very newbie question, but relativity is one of the topics where you recheck every statement again and again, so I want to be sure.
| The statement is
"Any observer at rest relative to his own timepiece will see that other clocks moving with respect to him run fast - the greater their speed, the faster they are".
According to one source, Don Koks (a physicist text book author)this statement is true..... on the condition A orbits B.
http://math.ucr.edu/home/baez/physics/Relativity/SR/movingClocks.html
Let A very closely orbit B. The orbit is so close that A is almost touching B; therefore time delays in signal exchanges can be neglected.
The faster A orbits B, the FASTER B's clock runs in A's view. Conversely, the SLOWER A's clock runs in B's view.
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Intuition behind mass corrections to massless fermions I'm trying to understand the intuition behind the mass correction to massless fermions. To be concrete lets consider a theory with a massless Weyl fermion ($\psi $), as well as two massive particles, a complex scalar ($\phi $) and another Weyl fermion ($ \psi ' $),
\begin{equation}
{\cal L} = {\cal L} _{ kin} - \frac{1}{2} m ^2 \left| \phi \right| ^2 - M \left(\psi '\psi ' + h.c. \right) - \frac{ \lambda }{ 4!} \left| \phi \right| ^4 - g \phi \psi \psi ' + h.c.
\end{equation}
This Lagrangian would be valid if for example we impose a $ U(1) $ symmetry such that,
\begin{align}
& \psi \rightarrow e ^{ i \alpha } \psi \\
& \phi \rightarrow e ^{ - i\alpha } \psi \\
& \psi ' \rightarrow \psi '
\end{align}
Now consider the lowest order mass correction the for massless fermion,
$\hspace{4cm}$
The mass correction can be calculated by setting external momenta to zero:
\begin{align}
i {\cal M} & \sim g ^2 \int \frac{ d^4 \ell }{ (2\pi)^4 } \frac{ 1 }{ \ell ^2 - m ^2 } \frac{ M }{ \ell ^2 - M ^2 } \\
& \sim \frac{g ^2}{16 \pi^2}M \left( \frac{1}{ \epsilon } + \log \left( \frac{ \mu ^2 }{ \Delta ( m , M ) } \right) \right)
\end{align}
where $ \Delta $ is a function of $m$ and $ M $. In $\overline{MS}$ we toss the infinity and the first order correction is
\begin{equation}
m _{ \psi } \sim g ^2 M \log \frac{ \mu ^2 }{ \Delta }
\end{equation}
This has three interesting features which I am trying to understand:
*
*It seems that the fermion doesn't decouple from the theory in the
limit that its mass goes to infinity. Why are we not justified in
integrating it out in this case?
*The calculation is very insensitive to the boson mass. Is this an
accident or has a deeper explanation?
*The calculation has large logs unless $ \mu \sim m, M $. It seems if you renormalize at this scale the log goes away and the physical mass goes to zero! So if I am understanding correctly the particle is massless when doing experiments at the scale of the other particles in the theory (at least at lowest order in perturbation theory) but heavy when its much below this scale. This seems very strange!
| I guess that the confusion here is that you mix Weyl notation and Feynman diagrams with Dirac propagators. As you notice yourself, the theory has a global (anomalous) U(1) symmetry that prohibits a mass term $m\,\psi\psi$ (in Weyl notation). If I try to draw your diagram (ab)using the usual Dirac propagator lines for Weyl fermions, I get
Here, the right (empty) vertex does not exist, as it violates U(1) charge conservation. On the other hand, in order to obtain the $M$ in the numerator of your expression, I need the mass insertion (indicated by the cross). (BTW, there is a typo in your transformation law $\phi\to e^{-\mathrm{i}\alpha}\psi$, it should be $\phi\to e^{-\mathrm{i}\alpha}\phi$). If you are unhappy with the Weyl argumentation, you may recast everything in Dirac notation, which yields the same result.
| {
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Why does my house seem to warm faster in summer than it cools in winter? In summers when we switch off the air conditioner, the room seems to instantly get hot again.
But in winter, when we switch off the heater the room seems to remain hot for some time. Why this difference?
| When your house heats up, it is receiving the contribution from the hot air outside, plus the sun's radiated heat. When cooling in the Winter, the factors are the cooler air and the radiation loss, which is not comparable to the sun's.
| {
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Why $SU(3)$ and not $U(3)$? Is there a good reason not to pick $U(3)$ as the colour group? Is there any experiment or intrinsic reason that would ruled out $U(3)$ as colour group instead?
| Suppose that $\text{U}(3)$ was the gauge group. We can decompose this as
$$\text{U}(3)=\text{U}(1)\times\text{SU}(3),$$
which implies that in addition to the $\text{SU}(3)$ that has eight generators corresponding to eight gluons, there would be an additional generator for $\text{U}(1)$. The latter in principle corresponds to an additional gauge boson, but a theory of the strong interactions containing such a particle is inconsistent with experiment.
| {
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How does the choice of a particular vacuum in a field theory problem decide the number of Goldstone bosons? How does the field expansion method (by this I mean expanding your fields about a chosen VEV and plugging into a given potential so that the masses of the fields are given by the coefficients in front) corrolate with the vacuum you choose. In SU(3), choosing v = (0,0,v) gives 5 NGB, while v = (v,0,0) gives 6 NGB (using the Gell-Mann matrices as a basis). However, when you do a field expansion and plug it into the potential, for either of the two vacua above, you get 5 massless fields and 1 massive field. How does this match up and where have I gone wrong?
| Where you have gone wrong is in accounting for the dimensionality of your
surviving subalgebra. In the first case, as you correctly see, the surviving SU(2) is, indeed, manifestly spanned by $\lambda_{1,2,3}$, so the remaining 5 generators are broken.
In the 2nd case, however, you should have noticed the less manifest SU(2) spanned by $\lambda_{6,7}$ and $(\sqrt{3} \lambda_8-\lambda_3)/2$, in the 2-3 subspace, now leaving the 1st component alone, and breaking the rest, including, of course, $(\sqrt{3} \lambda_3+\lambda_8)/2$.
Lie Algebras are linear structures.
| {
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Why can the entropy of an isolated system increase? From the second law of thermodynamics:
The second law of thermodynamics states that the entropy of an
isolated system never decreases, because isolated systems always
evolve toward thermodynamic equilibrium, a state with maximum entropy.
Now I understand why the entropy can't decrease, but I fail to understand why the entropy tends to increase as the system reach the thermodynamic equilibrium. Since an isolated system can't exchange work and heat with the external environment, and the entropy of a system is the difference of heat divided for the temperature, since the total heat of a system will always be the same for it doesn't receive heat from the external environment, it's natural for me to think that difference of entropy for an isolated system is always zero. Could someone explain me why I am wrong?
PS: There are many questions with a similar title, but they're not asking the same thing.
| While Bubble gave a nice example, let me try to explain this with "Clausius inequality". (You can read this on several sources, I like the explanation from Atkins' Physical Chemistry)
Let's start with the statement:
$$
|\delta w_{rev}| \geq |\delta w| \\
$$
Furthermore, for energy leaving the system as work, we can write
$$
\rightarrow \delta w - \delta w_{rev} \geq 0
$$
where $\delta w_{rev}$ is the reversible work. The first law states
$$
du = \delta q + \delta w = \delta q_{rev} + \delta w_{rev}
$$
since the internal energy $u$ is a state function, all paths between two states (reversible or irreversible) lead to the same change in $u$. Let's use the second equation in the first law:
$$
\delta w - \delta w_{rev} = \delta q_{rev} - \delta q \geq 0
$$
and therefore
$$
\frac{\delta q_{rev}}{T} \geq \frac{\delta q}{T}
$$
We know that the change in entropy is:
$$
ds = \frac{\delta q_{rev}}{T}
$$
We can use the latter equation to state:
$$
ds \geq \frac{\delta q}{T}
$$
There are alternative expressions for the latter equation. We can introduce a "entropy prodcution" term ($\sigma$).
$$
ds = \frac{\delta q_{rev}}{T} + \delta \sigma, ~~\delta \sigma \geq 0
$$
This production accounts for all irreversible changes taking place in our system.
For an isolated system, where $\delta q = 0$, it follows:
$$
ds \geq 0 \,.
$$
| {
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Time dilation at the Big Bang At the time the Big Bang happened the matter had enormous density. According the GR (I may be wrong here) such density dilates time.
If so, could it be that the time periods just after Big Bang which are usually considered happening in small part of a second (such as the Planck epoch), in reaity took billons of year (or may be, infinity) but due to time dilation appear to us as spanning only microscopic parts of a second? Could it be that the age of the universe is dramatically underestimated?
| Is the universe finite or infinite?
If the big bang had infinite mass at the start then time dilation due to gravity would be infinite and time would stand still. That's the problem with trying to conceive of the start of time because it took infinitely long to start.
An infinitesimal length of time under infinite gravity is infinitely long.
If the universe was finite then time dilation would have some maximum value, giving time a starting point.
| {
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Finding the electric field around an electric eel I'm having problems solving for the electric field and the current that an electric eel generates. Would I use Gauss's law and treat it as a long charged wire? How would I find the charge of the electric eel? I'm really confused about what I can do and where to start.
In the second part of the project I need to find the current and electric field that affects the prey of an electric field also. Would this calculation be like adding a (human) resistor in a circuit?
| Let's first make a 'textbook' calculation to get a feeling of the magnitude of the field. See below and sources for a more accurate description.
Suppose that the you can treat the eel as infinitely small. And take is length to be about 1 m long and can generate a voltage difference of 500 volts between its head and tail. Neglecting the directionality of the field, the magnitude of the field is
$$|\vec{E}| = \Delta V \cdot d = 500 V/m.$$
I based this value on the following
The higher intensity charges vary by the size of the eel. Smaller eels (about 10 cm in length) can produce charges of up to 100 V. Larger eels (over 1 m in length) can produce charges of 450 to 650 volts of electricity.Source
In fact an eel generates a electric dipole, i.e. it generates a positive pole at on location and a negative pole at a different location.
As you can see from this picture the electric field runs in curves lines between this poles.
A dipole field is characterized by a inverse cubic force
$$F \propto \dfrac{1}{r^3}. $$
This has a consequence that the electric field produced by the eel can only serve for communication over ranges of a few meters, for example it can only sens a prey using its field at about 1 m.
An old, quite funny question: Could we run an electric car on electric eels?
Source:
Physics in Biology and Medicine
By Paul Davidovits
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Why is the constant velocity model used in a projectile motion derivation? I was re-studying university physics last week, I'm now in the chapter about kinematics in 2 dimensions and specifically the one treating projectile motion. In page 86 of his book (Serway - Physics for scientists and engineers) he derives the equation of the range of the projectile motion to be: $$R=\frac{{v_i}^2\sin2\theta_i}{g}$$
But I don't know why he used one of his assumptions
$\color{red}{\bf Question1:}$ Why $v_{xi}=x_{x\rlap\bigcirc B}$? Where $\rlap\bigcirc {\,\sf B}$ is the time when the projectile stops.
$\color{darkorange}{\bf Question2:}$ Why did he use the particle under constant velocity model to derive that formula, whereas here we deal with a projectile under constant acceleration?
Any responses are welcome, I'm disappointed a lot about those matters!
| The force of gravity is in the y direction only.
There is no force on the particle in the x direction.
Therefore, the x-component of velocity is constant.
| {
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Find minimum distance between particles given initial position and velocity My friend gave me a question today:
We have a point $A$. At a distance of $x_0$ from the point. There is a particle $P_1$. There is another particle, $P_2$, at $A$. $P_1$ moves with velocity $u_1$ towards $A$. At the same instant, $P_2$ moves making an angle $\theta$ with $P_1A$ with a velocity $u_2$. What will be the minimum distance between the two particles when both of them start at the same time?
So can anybody help me with this problem? I tried decomposing it into vectors, used $\tan\theta$, but it did not give me a correct answer.
| You don't have enough information to properly answer the question.
If $u_1 \le u_2\ and\ \theta \ge \pi/2$, then the shortest distance occurs at the initial conditions.
When $u_1 > u_2$, then the shortest distance occurs when $\angle P_{1-initial}P_1P_2$ is $\pi/2$.
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Conservation of energy and the 'crazy ball' product Well I'm not sure how many people remember the crazy ball - a small ball made of rubber which bounced like crazy. What I noticed is that the ball seemed to bounce higher than the point from which it was actually dropped. How is this possible? Doesn't this violate the law of energy conservation?
| No, it doesn't violate energy conservation, and it doesn't bounce higher than the usual height, if you ''drop'' it (i.e. $v_{\rm initial} = 0$). If you add any force from your side, ''throwing'' it, then it will obviously bounce higher, but still won't violate energy conservation if you include the extra $\frac{1}{2} m v_{\rm initial}^2$ energy it carries at the time of throwing.
The way this process works is this:
The reason why usual balls rebound conspicuously lower than a crazy ball is because they lose energy on impact with the floor - one factor is friction, and another is the loud ''thud'' sound you hear when, let's say, a tennis ball hits the ground. Choice of material matters over here, if you choose a material with a higher coefficient of restitution, you can reduce the amount of energy lost and your ball bounces higher. Now, crazy balls are generally made of synthetic rubber (crudely; more specifically polybutadiene, but how does it matter!) which precisely achieves this. You may have noticed that the thud from a crazy ball is considerably weaker in intensity.
How all this came about is an interesting story in itself, which you can find here.
While the above reasoning was an intuitive way of stating this, one source where measurements of this effect were made some years back is this article (not sure if you would be able to access and it though, since I don't imagine your school being subscribed to this).
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Why is the Earth shaped like a sphere and not any other shape: cube, prism? Why is the earth shaped like a sphere and not any other shape: cube, prism?
| Feynman wants to say you something, listen to him!
What else can we understand when we understand gravity? Everyone knows the earth is round. Why is the earth round? That is easy; it is due to gravitation. The earth can be understood to be round merely because everything attracts everything else and so it has attracted itself together as far as it can! If we go even further, the earth is not exactly a sphere because it is rotating and this brings in centrifugal effects which tend to oppose gravity near equator. It turns out that earth should be elliptical, and we even get the right shape for the ellipse. We can thus deduce that the sun, the moon, and the earth should be (nearly) spheres, just from law of gravitation.$_1$
Credits: $_1$Feynman lectures on Physics-Page No.7-5. Page numbers are subjected to change depending on the editions.
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Why doesn't current flow when the wire is open? I realize this may be a hard to answer question but we are currently studying current in our school. One thing that struck me was why the heck doesn't it flow when the wire's closed?
When you connect the positive terminal of the battery (But do not connect the negative terminal) to a conductor, the loosely bound / free electrons in the conductor should be attracted to the battery's +ve terminal. This should create a chain reaction till every atom in the conductor has donated the maximum it can and no further flow is possible due to the atoms now strongly attracting their remaining electrons.
Why doesn't this happen? There should be momentary current and the wire should become positively charged. Why not? Why does it only flow when the circuit is closed?
| At first, try to understand the working principle of a battery. Here is a video explaining what positive terminal and negative terminal are. If you don't have the load connected there, you will not have infinite amount of depletion of electrons in the positive terminal as after some time no $H^+$ will be able to reach the positive terminal due to its electric field. similar statement for negative terminal.
so in the open circuit condition If you connect the conductor with the positive terminal only few of the electrons will go to that $PbO_2$ bar. Now the $H^+$
ions can again reach to the $PbO_2$, but it will sustain only for tiny amount of time as again $H^+$ will not be able to reach the positive terminal due to the electric filed of the terminal. Now this time $PbO_2$ bar+ the conductor becomes the positive terminal.
| {
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Bragg diffraction and lattice planes Crystalline substances show, for certain sharply defined wavelength and incident directions, very sharp peaks of scattered X-ray radiation.
From the illustration below we see that we get constructive interference when the path-length difference is a multiple of the wavelength $\lambda$.
In real crystalline materials we have a large amount of closely packed lattice planes. This large amount accounts for the sharp peaks for certain $\theta$. I do not understand how this follows from the Bragg reflection formula $$ n\lambda = 2d \sin \theta , $$ since $d$ is not constant anymore. I understand the model for two lattice planes as in the illustration.
Is it true that $d$ can only take on values of the seperation of lattice planes, so $d$ is defined to be the seperation of points in the reciprocal lattice, or in others words, is $d$ constrained to be the absolute values of some reciprocal lattice vector?
How does the Bragg condition account for very sharp peaks when we let $d$ run through all such absolute values?
| I think that the most appropriate way to think about the Bragg formula is in terms of a diffraction grating. In a diffraction grating one obtains sharp peaks because there are many slits with distance $d$ between them. The derivation of the intensity maximum for the diffraction grating case is similar to the Bragg case. To obtain a diffraction grating maximum, i.e. constructive interference, the path-length difference between each of the slits, $d\sin\theta$, must be an integer number of wavelengths, $n\lambda$.
Similarly, for the Bragg case, one must now have that the path-length difference between each of the lattice planes, $2d\sin\theta$ (because the light reflects into and out of the material), must be an integer number of wavelengths, $n\lambda$. This is the condition for constructive interference, and hence one gets sharp maxima just like in the diffraction grating case.
Hence, one may in some sense think of a perfect crystal as a diffraction grating for X-rays.
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Difference in decay for muon and anti muon In a couple weeks, I will conduct a lab experiment where I measure the lifetime of the muons from the secondary cosmic radiation. For that, we have two detectors above each other, one will give a start signal, the other will give a stop signal, assuming the muon came to rest in the second detector and decayed.
There is a preparation question that says:
Muons as well as anti muons arrive at the surface of earth. Which nuclear process is possible for muons but not for anti muons? How does this qualitatively affect the measured lifetime?
As far as I know, the decay of the muon goes like this:
$$ \mu^- \to \nu_\mu + \bar \nu_e + \mathrm e^-$$
And for the anti muon, it goes like this:
$$ \mu^+ \to \bar \nu_\mu + \nu_e + \mathrm e^+$$
The Wikipedia page says:
The mean lifetime of the (positive) muon is 2.1969811±0.0000022 µs.[1] The equality of the muon and antimuon lifetimes has been established to better than one part in 104.
What is that difference that they are asking about in the lab course manual?
| If you consider a muon decay in vacuum then there is no difference between the lifetimes of muons and antimuons.
However a muon can interact with a proton via the weak force to form a neutron, while an antimuon cannot. Since the air is full of protons this means the muon lifetime in air is slightly shorter than the antimuon lifetime. If the air were full of antiprotons instead of protons the effect would be the other way round.
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What makes nuclear binding energy so much stronger than chemical energy The strong force acting between quarks and responsible for holding protons together is 100 times stronger than the electromagnetic force. How come the nuclear binding energy derived from the strong force is millions time stronger than chemical energy. (rather than 100 times)
| Chemical forces are the result of the higher order moments, mainly of the electric field, of the neutral atoms. Look at these orbitals:
The five d orbitals in ψ(x, y, z)2 form, with a combination diagram showing how they fit together to fill space around an atomic nucleus.
These anisotropic shapes in space allow for spill over attractive and repulsive fields that are collectively called Wan der Waals forces. , spill over forces from the combined electric fields of the electrons and nucleus .
Correspondingly the nuclear force between protons and neutrons in the nucleus are spill overs of the strong force as it appears in the "geometry" of each nucleus. As the strong force is color confined , as rob describes in his answer, these forces do not exceed the fermi distances which characterize the size of the nucleons. They still are due to gluon exchanges and these are highly confining.
How come the nuclear binding energy derived from the strong force is millions time stronger than chemical energy. (rather than 100 times)
In this article the energy of intermolecular forces due to the VdWs forces is estimated as 1/r^6 , r distance between molecules. I do not have a functional estimate for the spill over forces of the strong interactions, but it is evident that the inter nucleon distance, of the order fermi, will play a corresponding role. The difference in the dimensions of the atomic orbitals and the nucleus dimensions allows for the large difference in the energies manifested as chemical and nuclear, overcoming factors coming from coupling constants just by "geometry".
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Normal ordering in curved spacetime In the flat spacetime, one can perform normal-ordering to set the energy of the vacuum state to zero. I read in some places that this procedure cannot be consistently performed in the curved spacetimes. I have not found any explanation of this fact in the literature.
Why is this a case?
| The normal ordering procedure is nothing but the a recurrent machinery to remove vacuum expectation values by means of the so-called Wick's theorem procedure. There, the vacuum state is the unique Poincaré invariant one, the Minkowski vacuum. In curved spacetime there is no such an invariant state. Therefore the normal ordering product must be defined referring to some other fixed states. These states (their two-point functions) should have a short distance behaviour similar to the one of Minkowski vacuum state. Technically they must be Hadamard states. Unfortunately there is no canonical (functorial) way to associate Hadamard states with (globally hyperbolic) spacetimes. This fact prevents one for defining a true generally covariant normal ordering procedure. A way out consists of removing not the expectation value of a state, but the "expectation value" of a local functional defined in terms of the local geometry only (the Hadamard parametrix). This procedure, in view of the fact that the functional is not a (weak) bi-solution of the field equations, has nonetheless annoying shortcomings, in particular the appearance of anomalies (the trace anomaly first of all). This machinery has been developed by various authors in the last 15 years (including myself).
For a recent overview see the PhD thesis http://arxiv.org/abs/arXiv:1008.1776 (I was one of the advisors, together with K. Fredenhagen and R.M. Wald).
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How is sweating a pipe an example of capillary action? I learned how to sweat a pipe today from my father. If you're not familiar with the process, this might help.
One thing that jumps out at me is this line (from the above link, as well as my father's explanation)
Solder, which melts at low temperatures, wicks into a joint by capillary action and bonds with copper at the molecular level.
This doesn't seem to be quite right to me. I always thought of capillary action as something like putting a piece of paper vertically into a dye and watching the dye rise up the paper. I also didn't think the solder bonded with the copper at the molecular level, just that it melted and filled in the gaps really well. Wikipedia seems to agree with my definition of capillary action
Capillary action ... is the ability of a liquid to flow in narrow spaces without the assistance of, and in opposition to, external forces like gravity.
When sweating a pipe you place the solder above the pipe, letting it drip down (ie with the assistance of gravity) which makes it not capillary action... right?
Is there something I'm just not understanding here?
| The basis for capillary action is attraction between the liquid and the solid, based upon intermolecular or interatomic forces.
When the liquid needs to rise against gravity to increase wetted surface area, an equilibrium is reached. However, capillary action does not need to be opposed to gravity. Capillary action still takes place without gravity and then is not limited by an opposing force.
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Why isn't invariant notation common? In principle, one can write quantities in a manifestly invariant - rather than covariant - fashion in e.g. special relativity. For example, rather than writing just $x^\mu$, we could write the basis explicitly, and ask that the basis transforms oppositely to the components,
$$
x = x^\mu e_\mu^{(i)},
$$
such that $x$ is invariant. Why isn't such an invariant notation more common in e.g. special and general relativity? Do mathematicians also work with covariant language? or the invariant language?
Is it because the basis disappears in contractions if the basis is orthonormal? e.g.
$$
x(y) = x^\mu e_\mu y_\nu e^\nu =
x^\mu y_\nu \delta^\nu_\mu = x^\mu y_\mu
$$
| A physicist would write your first equation $x^a = x^\mu e_\mu^a$. The notation $x^a$ is invariant in your terminology. The $a$ is an abstract index. It is ostensibly not supposed to be thought of as ranging over a set of numerical values, but is just a marker that indicates that $x$ is a vector (i.e., rank 1,0 tensor.) Similarly for each $\mu$, $e^a_\mu$ is a vector. You can read more about abstract index notation on wikipedia.
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Second fundamental form How do I calculate the integral of the trace of the second fundamental form on a surface?
The formula used in the Gibbons, Hawking, York paper Action integrals and partition functions in quantum gravity, how do I derive it? Is it a universal or does it have any assumptions about the kind of space time we are considering?
The formula is,
$$\int K d\Sigma = \frac{\partial}{\partial n}\int d\Sigma$$
$K$ is the trace of the second fundamental form, $d\Sigma$ is the area element and the derivative is along the outward normal to the surface.
| The boundary term of the Einstein-Hilbert action is given by,
$$S= \frac{1}{8\pi G}\int_{\partial M} \! \!\mathrm{d}^3x \, \sqrt{-h} \, K$$
where $h$ is the metric on the boundary of the manifold, i.e. $\partial M$, and $K$ is the trace of the extrinsic curvature. Specifically, we have,
$$K=\nabla_a n^a$$
where $n^a$ is a unit normal to the surface $\partial M$. The derivation follows by applying the Gauss-Codazzi formalism of differential geometry. Recall the variation of the Einstein-Hilbert action is,
$$\delta S_{EH}=-\frac{1}{16\pi G} \int_M \mathrm{d}^4x \, \sqrt{g} \, \delta g^{ab}G_{ab}-\frac{1}{16\pi G}\int_{\partial M} \mathrm{d}^3 x \, \sqrt{g} \left( \nabla_n \delta g -n_a\nabla_b g^{ab} \right)$$
The boundary is a submanifold, and a natural curvature scalar to select, rather than $R$, is the extrinsic curvature; the variation of $K$ is given by,
$$\delta K = \delta (\nabla_a n^a)=\nabla_a \delta n^a + \delta \Gamma^a_{ab}n^b$$
With some manipulation, which is found in virtually any general relativity textbook, it can be shown that the addition directly cancels the variation of the Einstein-Hilbert action, up to total derivatives.
As a practical example, consider the metric,
$$\mathrm{d}s^2 = \left( 1- \frac{2GM}{r}\right)\mathrm{d}\tau^2 + \left( 1-\frac{2GM}{r} \right)^{-1}\mathrm{d}r^2 + r^2 \mathrm{d}\theta^2 + r^2 \sin^2 \theta \mathrm{d}\phi^2$$
which is simply a Wick rotated Schwarzschild black hole, with periodic $\tau$ with period $\beta$. We may choose an inward pointing normal, correctly normalized such that $n_an^a = 1$, given by,
$$n^{a} = - \sqrt{1-\frac{2GM}{r}}\delta^a_r$$
Taking the divergence of the normal, we obtain,
$$K= -\left( 1-\frac{2GM}{r}\right)^{-1/2} \left( \frac{2}{r}-\frac{3GM}{r^2}\right)$$
The action is divergent, unless we introduce a radial cut-off, $R$. Note also we pick up a factor of $r^2$ from the determinant of the metric in the action; we find,
$$-8\pi G \, S= \int \mathrm{d}\tau \, \mathrm{d}\phi \, \mathrm{d}\theta \, \left( 2R -3GMR\right)=4\pi \beta \left( 2R -3GMR\right)$$
To regularize the integral, we must subtract off the contribution from flat space, with the same boundary.
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Is the magnetization direction of a permanent magnet fixed? I am now playing with a permanent magnet made of Neodymium.
It is impressively strong.
A question is, is the magnetization direction of the magnet fixed relative to its crystal structure?
| Practically yes, but the true answer is a little bit complexer.
Below the level of the crytals, there are the magnetic domains. In a domain have the atoms the same orientation (and thus, the same magnetic dipole moment).
In the case of a magnetized ferromagnetic material, these magnetic pole of these domains are directing to the same (or to the nearly same) direction.
To change the direction of the magnetic field of the ferromagnetic material, you need to apply so strong magnetic field, which is enough to "redirect" the orientation of the atoms. Of course, it is much easies, if there is an elevated temperature as well, because it also softens the metallic bonds between the atoms.
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Why $e$ in the formula for air density? I am reading a book that says that the density of air is approximately $D = 1.25 e^{(-0.0001h)}$, where h is the height in meters. Why is Euler's number $e$ used here? Was a differential equation used in deriving this formula?
| You can rearrange the terms to have any constant as the base of the exponent:
$D = 1.25 e^{(-0.0001h)}$
$= 1.25 (e^{0.0001})^{-h}$
$= 1.25 (2^{\frac{0.0001}{ln 2}})^{-h}$
$\approx 1.25 (2^{0.00014})^{-h}$
$= 1.25 \times 2^{(-0.00014h)}$
| {
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Why does angular momentum shorten the Schwarzschild Radius of a black hole? Angular momentum causes the event horizon of a black hole to recede. At maximum angular momentum, $J=GM^2/c$, the Schwarzschild radius is half of what it would be if the black hole wasn't spinning.
Can someone explain why angular momentum reduces the Schwarzschild radius?
| I will approach this question theoretically, although I feel the intuition follows nicely. If we talk about Kerr black holes - rotating black holes described by their mass and angular momentum, with no additional parameters such as charge etc. - then you can show that the radius of the event horizon is given by
$\boxed{r=M + \sqrt{M^2-a^2}}$
where $a=\frac{J}{M}$.
(This value of $r$ is found by finding where the Kerr metric blows up; hence event horizon. In fact, finding where the metric blows up involves solving a quadratic equation, so we get two values of $r$ and in Kerr black holes we therefore have two event horizons; unlike in Schwarzschild black holes.)
Regarding your first point about maximum angular momentum, if we set $G=1$ and $c=1$, the maximum angular momentum you stated is given by $a=M$ and if we plug this into our equation for $r$ above we see that we have
$r=M$.
We know that the radius of the event horizon in a Schwarzschild black hole (no rotation) is $r=2M$. So therefore we can see that at maximum angular momentum, the radius of the event horizon is half of what it would be if the black hole weren't spinning.
To this end, we can also see that at zero angular momentum, $a=0$, we have
$r=2M$
which is what we want as at zero angular momentum we of course should have the Schwarzschild radius.
Using the boxed equation for $r$ at the top, it's easy to test out different values of $a$ to see what happens to the event horizon. For example, this equation alone is sufficient to show that for $a>M$ we don't have an event horizon, in which case we have what is a called "Fast Kerr" which is just a singularity with no event horizon.
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Integrating the generator of the infinitesimal special conformal transformation (c.f Di Francesco, Conformal Field Theory chapters 2 and 4).
The expression for the full generator, $G_a$, of a transformation is $$iG_a \Phi = \frac{\delta x^{\mu}}{\delta \omega_{a}} \partial_{\mu} \Phi - \frac{\delta F}{\delta \omega_a}$$
For an infinitesimal special conformal transformation (SCT), the coordinates transform like $$x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2$$
If we now suppose the field transforms trivially under a SCT across the entire space, then $\delta F/\delta \omega_a = 0$.
Geometrically, a SCT comprises of a inversion, translation and then a further inversion.
An inversion of a point in space just looks like a translation of the point. So the constant vector $b^{\mu}$ parametrises the SCT. Then $$\frac{\delta x^{\mu}}{\delta b^{\nu}} = \frac{\delta x^{\mu}}{\delta (x^{\rho}b_{\rho})} \frac{\delta (x^{\gamma}b_{\gamma})}{\delta b^{\nu}} = 2 x^{\mu}x_{\nu} - x^2 \delta_{\nu}^{\mu}.$$
Now moving on to my question: Di Francesco makes a point of not showing how the finite transformation of the SCT comes from but just states it. $$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$$ I was wondering if somebody could point me to a link or explain the derivation. Is the reason for its non appearance due to complication or by being tedious?
I am also wondering how, from either of the infinitesimal or finite forms, we may express the SCT as $$\frac{x'^{\mu}}{x'^2} = \frac{x^{\mu}}{x^2} - b^{\mu},$$ which is to say the SCT is an inversion $(1/x^2)$ a translation $-b^{\mu}$ and then a further inversion $(1/x'^2)$ which then gives $x'^{\mu}$, i.e the transformed coordinate.
| The two questions posed above are exactly the same and one may prove both simply by substituting $x'_\mu$ and calculating. If
of not showing how the finite transformation of the SCT comes from but just states it. $$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$$
then
$$(x')^2 = \frac{(x_\mu-x^2 b_\mu)(x^\mu-x^2 b^\mu)}{(1-2xb+b^2x^2)^2}=\frac{x^2(1-2bx+x^2b^2) }{(1-2xb+b^2x^2)^2}=\frac{x^2}{1-2xb+b^2x^2}$$
and therefore
$$\frac{x'}{(x')^2} = \frac{x-x^2b}{x^2} $$
(some indices were omitted in a self-explanatory way) which is exactly the second equation. So the two equations you are asking about are exactly equivalent.
The special conformal transformations may be uniquely defined as the conformal transformations that map the infinity to a different point and that map a line of the multiples of $b^\mu$ onto itself. The inversion is really needed to get the point at infinity to a finite place, so that it may be moved elsewhere by the translation in the middle.
If you use a different definition of the special conformal transformation than the definition "inverse times translations time inversion" or the definition from the previous paragraph, you would have to specify what's your definition is and one could easily show the equivalence with the definitions above, too.
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Why does sand stick to my shoes? Well, that's easy: the sand is wet, and my shoes are wet, and hydrogen bonding adheres the wet sand to my wet feet and to my shoes.
But then I walk home, and my shoes dry, and the sand on them dries, and some of the sand falls off. But some does not. It's really stuck: even several days later I can turn the shoe upside-down and it won't fall off. What holds it on?
Sand sticks to my feet after my feet dries and the sand dries. Is this the same?
| I think this is because your shoes have tracks over the non-uniform surface to provide frictional force for movement,these holes allow the wet sands take longer time to evaporate, the outermost layer should get evaporate first, and fall off once the water is evaporated.If your shoe surface is facing directly downward, It would probably take a very long time for all the sand to fall off before you use the shoes again.
NB , the inner layer would take longer time to evaporate .
As the amount of sand slowly falls off, the reason why there are still some of them adhered to the shoes is because the tracks under the shoes are non-uniform and provide a normal reaction in a microscopic level, balancing the pull of gravity.
The reason why they remained stuck there is because the weight of these inner layers of sands are less "heavier" as there are no sands sitting above it, and the normal force provided by the tracks is enough to balance the gravity of each grain of sand By F = dm * g allowing them to stuck there ; dm is infinitesimally small mass.
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How much time passed for the passenger traveling with at speed-of-light spaceship? Let's suppose we have a spaceship with the exact speed of light. If a traveller takes this spaceship to go to proxima centauri (approximately 4 years light away from Earth) and come back, we (as observers on Earth) will see the ship coming back after approximately 8 years.
But how much time would have passed for the traveller on the ship? How can be this calculated with a formula?
| The quantity that tells you what time an observer travelling along a path $\gamma : [t_0,t_1] \rightarrow \mathbb{R}^4$ experiences is the proper time
$$ \tau = \int_\gamma \sqrt{\mathrm{d}x_\mu\mathrm{d}x^\mu}$$
Assuming flat spacetime, i.e Minkowski metric/special relativity, this reduces to
$$ \tau = \int_\gamma\sqrt{\mathrm{d}t^2 - \frac{1}{c^2}\mathrm{d}x^i\mathrm{d}x_i} =\int_{t_0}^{t_1} \sqrt{1 - \frac{\vec{v}(t)^2}{c^2}}\mathrm{d}t $$
For $\vec{v} = c$, i.e an object travelling with the speed of light, this is $\int_{t_0}^{t_1} 0 \mathrm{d}t = 0$, so anyone hypothetically travelling with the speed of light will indeed not experience any time at all.
By plugging in other values for the travelling speed $\vec{v}$, you are able to calculate the experienced time for arbitary travellers.
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How stellar aberration is measured? A simple calculation shows that stellar aberration due orbital motion of earth is roughly 20 arcseconds. My questions are:
*
*Practically how this small value is measured?
*Does this value is in the range of accuracy of a 11 inch reflective telescope with a camera?
*And how this measurement originally was done by Bradley in the 18th century?
| Here's a second, easier way. It has less precision, but it's more practical, at least if you want to simply see the basic effect.
Take a long-exposure telescopic photo of the celestial pole. Use the highest possible magnification. The star trails in the photo are centered on the pole, so you can infer the pole's position with respect to the stars.
You repeat this over the course of a year. The position of the pole will trace out a cycloid-like curve on the sky. The width of the cycloid is caused (mostly) by stellar aberration. The change in the center of the cycloid is caused by precession and nutation.
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Help explain how direction change relates to acceleration I was doing some simple harmonic motion problems and I came across this picture describing the position, velocity and acceleration of a linear oscillator. At the moment in time when v is 0 the linear oscillator should not be moving, only changing directions. I'm having a hard time understanding why the acceleration is the greatest at that time (according to these graphs), since there is no velocity change. Is it because acceleration is only the difference in velocity at two different points in time and not one? How exactly does the change in direction affect acceleration?
edit: I found another question that answered my question. haha.
|
Is it because acceleration is only the difference in velocity at two different points in time and not one?
I think you've basically hit on the answer to your question here. Acceleration is the derivative of velocity with respect to time, which means it is the instantaneous rate that the velocity is changing with time. Acceleration is a measure of how fast velocity is changing; it does not depend on the particular velocity at any one time. So even though the linear oscillator may not be moving at a particular time, it is undergoing a high acceleration as it switches directions.
Perhaps another easy way to recognize this in this specific case is by recalling Newton's second law: $\textbf{F}=ma$, where $\textbf{F}$ is the force applied to an object, $m$ is its mass, and $\textbf{a}$ is its acceleration. In the case of a block on a spring (a certain kind of linear oscillator), the spring will exert the most force on the block when the block is furthest away from equilibrium. This is also the point at which the block is motionless (i.e. its velocity is zero). Thus the highest acceleration will occur at zero velocity in the case of a linear oscillator.
Hope that cleared up your confusion!
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Can an LC oscillator be used to generate visible light? The LC oscillator is most commonly used to generate radio waves for practical use and the frequency $\omega$ of the LC oscillator equals that of the electromagnetic wave so produced. So, can they in principle be used to emit visible light?
The frequency of visible light is on the order of a few hundred terahertz, and the frequency of an LC oscillator is
$$\omega = \frac{1}{\sqrt{LC}}$$
I admit, the product $LC$ does become very small (on the order of $10^{-30}$) when the numbers are plugged in, but making an inductor and a capacitor with small values isn't difficult, is it?
| If, by LC oscillator, you mean a circuit composed of an 'ordinary' inductor and capacitor etc. then the answer is no.
The equations for the LC oscillator are derived within the context of ideal circuit theory which is the limit of a number of assumptions.
To apply the results from ideal circuit theory to physical systems, the physical systems must approximate the assumptions of ideal circuit theory.
One of these is that physical circuit elements can be represented with lumped element models.
And, for that to hold, the wavelengths of the signals of interest must be much larger than the dimensions of the physical circuit elements.
When this assumption does not hold, we must use the distributed element model such as in, for example, transmission line theory.
Another related consideration is that physical circuit elements have parasitic properties that cannot be avoided. Thus any physical system of conductors possess parasitic inductance and capacitance that become significant at high enough frequencies.
For visible light, the wavelengths are so small (hundreds of nanometers) that the assumptions of ideal circuit theory are not remotely valid so, unless one generalizes the notions of "inductor" and "capacitor" (and the other circuit elements) far outside the ordinary, the answer is no, an LC oscillator cannot produce visible light.
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Invariance of a tensor under coordinate transformation I know, that a tensor is a mathematically entity that is represented using a basis and tensor products, in the form of a matrix, and changing a representation doesn't change a tensor, is kind of obvious.
So does the invariance of a tensor under coordinate transformation mean what I stated above or does it mean that under a set of particular transformation the representation of a particular tensor also doesn't change.
Quoted from Wikipedia:
A vector is invariant under any change of basis, so if coordinates transform according to a transformation matrix $L$, the bases transform according to the matrix inverse $L^{−1}$, and conversely if the coordinates transform according to inverse $L^{−1}$, the bases transform according to the matrix $ L$.
Can someone please shed some light on this?
| Regarding what you quoted: a vector is represented by the sum of a set of basis vectors times the vector components. If the components transform according to $L$, then the bases will transform according to $L^{-1}$, which means that when you multiply the bases with the components (to make the vector), you will get the same result every time (since $L\cdot L^{-1}=I$). This is what is meant by invariance.
Invariance of a tensor means basically what you stated above- the tensor itself doesn't change under a change of coordinates (like I explained). However, the tensor's components can very well change.
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How much does the sound definition vary during an LP (Vinyl)? This question came to me when I realized how the linear speed varies while listening to a Vinyl LP.
The linear speed variation has to be compensated with a variation in the resolution of the grooves, that is, since the linear speed decreases, the groove resolution also has to decrease in some measure. What is this measure of reduction, or else, how much does the linear speed reduce? And how does that influence the sound definition?
| It should be obvious that the linear speed is inversely proportional to the distance from the center to that point on that groove.
Yes, speed varies over a record, so the wavelength of the wiggles in the groove gets shorter for the same frequency as you get further into the record. However, the master record was created with this same phenomenon in reverse, so it all works out.
The resolution that the vynil can be pressed with exceeds what is required even for encoding the high frequencies near the end of the track (closest to the center).
Since dust and dirt particles will be the same size regardless of where they land on the record, the noise from this dirt will have a overall lower spectrum at the end of the track than the beginning. Since the standard RIAA playback profile is heavily low pass filtered, this argues that there should be on average relatively more noise from dirt at the end of a track. There are other sources of noise too that don't scale the same way.
Overall, the fact that linear speed changes over 2:1 from the beginning to the end of a track is one of the engineering tradeoffs that went into records. This is also true of floppy disks and hard drives. It is interesting to note that it is not true for CDs. The CD standard specifies constant linear track speed for the same data rate. CD drives (and their decendents like DVD drives) have variable speed motors that compensate for the diameter effect to keep linear track speed constant.
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How can such a wave exist at the surface of the sun? Recently, I came across the following picture from NASA's SOHO observatory:
It seems evident that this is a transverse wave (mind the
ring which is bright and dark). But how can this be the case if the
sun consists of gaseous matter?
| I agree it looks like a transverse wave - like the ripples on a pond. But I believe you are fooled by a simple thing: the waves you are looking at look like "illuminated ripples" but are in reality just changes in temperature (changes in brightness of the sun's surface).
If you have a shock wave traveling out across the surface of the sun, what happens? The medium (solar plasma) will alternately be compressed and expanded. That is likely to affect the temperature - and I think that's what you are looking at here.
Now if you could somehow believe that there really were "surface ripples" on the sun's surface, that is really no different than the ripples in water on a pond: while the sun is a plasma (not a gas - not at that temperature), its particles are subject to considerable gravity. And the laws of motion for a wave traveling along the surface of a medium subjected to gravity really don't care about the state of matter of the medium - just the differential forces generated by a slope on the surface (which leads to wave propagation).
Either way there is no contradiction - but I'm pretty sure you are looking at changes in intensity/density, not a "transverse wave".
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The Einstein-Hilbert Action On-Shell If one consider the Maxwell action as
$$S=-\int \mathrm{d^{4}}x\! \ \frac{1}{4}F_{ab}F^{ab} \,$$
one find the usual Maxwell equation
$$\partial_{a}F^{ab}=0$$
Then one can simply arrive the following the Maxwell on-shell action
$$-\int \mathrm{d^{4}}x\! \ \frac{1}{2}\partial_{a}(A_{b}F^{ab}) \,$$
Now my question is for Einstein Hilbert action. What is the expression of on-shell Einstein Hilbert action
$$S=\int \mathrm{d^{4}}x\! \ R \,$$
I know how to find Einstein equation from variational principle, which is given as
$$R_{ab}-\frac{1}{2}g_{ab}R=0$$
How to write on-shell Einstein Hilbert action with above equation?
| The action you're considering yields Einstein's equations in vacuum, so $R=0$ (this follows immediately from contracting Einstein's equations). Therefore the action vanishes on shell.
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Could Legolas actually see that far? The video “How Far Can Legolas See?” by MinutePhysics recently went viral. The video states that although Legolas would in principle be able to count $105$ horsemen $24\text{ km}$ away, he shouldn't have been able to tell that their leader was very tall.
I understand that the main goal of MinutePhysics is mostly educational, and for that reason it assumes a simplified model for seeing. But if we consider a more detailed model for vision, it appears to me that even with human-size eyeballs and pupils$^\dagger$, one might actually be able to (in principle) distinguish smaller angles than the well known angular resolution:
$$\theta \approx 1.22 \frac \lambda D$$
So here's my question—using the facts that:
*
*Elves have two eyes (which might be useful as in e.g. the Very Large Array).
*Eyes can dynamically move and change the size of their pupils.
And assuming that:
*
*Legolas could do intensive image processing.
*The density of photoreceptor cells in Legolas's retina is not a limiting factor here.
*Elves are pretty much limited to visible light just as humans are.
*They had the cleanest air possible on Earth on that day.
How well could Legolas see those horsemen?
$^\dagger$ I'm not sure if this is an accurate description of elves in Tolkien's fantasy
| One thing that you failed to take into account. The curve of the planet (Middle Earth is similar in size and curvature to Earth). You can only see 3 miles to the horizon of the ocean at 6 feet tall. To see 24 km, you would need to be almost 100m above the objects being viewed. So unless Legolas was atop a very (very) tall hill or mountain, he would not have been able to see 24km in the first place due to the curvature of the planet.
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Role of physics in the zeta function $\zeta$ and the Riemann hypothesis Hilbert and Polya suggested a physical way to verify the Riemann hypotesis about $\zeta(x)$. If the Riemann hypotesis is true, we can state all eigenvalues of physical problems are real. What is the connection between the eigenvalues and the $\zeta$ function?
| In http://arxiv.org/abs/1608.03679 , the authors consider a Hamiltonian
$\hat{H}=\frac{1}{I-\exp(-i\hat{p})}(\hat{x}\hat{p}+\hat{p}\hat{x})(I-\exp(-i\hat{p}))$,
where $I$ is the identity matrix (I suppose - I cannot type the symbol they use), and claim that "a formal calculation of the eigenstates $\{\psi_n\}$ and
eigenvalues $\{E_n\}$ of $\hat{H}$ shows that with the boundary condition $\psi_n(0)=0$ for all $n$ the eigenvalues satisfy the property that $\{\frac{1}{2}(1-iE_n)\}$ are the nontrivial zeros of the Riemann zeta function. This non-Hermitian form of the Hilbert-Polya conjecture thus confirms the Berry-
Keating conjecture."
| {
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The fundamental equations of electromagnetism I'd like to know what are the basic equations of electromagnetism, that can be used to formulate all the other laws and equations.
Those basic equations I can think of are Maxwell equations, Lorentz force equation and Coulomb's law. What are the other fundamental laws and equation?
| The basics about the direction of force and field comes from the "Fleming's Left Hand Rule" and the "Maxwell's Corkscrew Rule". In addition to these the Lorentz force law, i.e. F=q[E+(vxB)] gives the force on a charge moving through a magnetic and electric field [Neglect E if electric field is absent.] The Biot- Savart Law gives the relation between current and magnetic field.
The Ampere's Circuital Law also plays a pivotal role in finding out magnetic field due to current through an enclosed area.
Coming to electric field due to magnetism[commonly called, Electromagnetic Induction], there exists the Lenz's Law. The concepts of motional emf and Eddy currents are essential for understanding the topic.
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If the energy of an ant is smaller than an elephant, does that mean the ant is more stable than the elephant? I know that when a system is in its lowest level of energy, it is most stable. However, what if system 1 has lower energy than system 2, does it keep meaning so? Or do we need to examine their binding energies of them? If the systems have only two bodies, easy: the one which has lower binding energy is stabler because their bodies don't need so much energy to combine them ($E_{binding}=E_{total}-E_A-E_B$). But what about many-body systems? If the binding is only meaningful in the context of two, then what is the subtraction $E_{total}-E_A-\sum_i{E_i}$?
By saying "more stable", I mean that it is "easier to combine", not "more energetic".
| I guess that you should not compare them with each other.
in fact stability is not a real thing.
we know that when a system has potential energy it has ability to do work and turn them to kinetic energy so that its velocity will be increased and as we know velocity(not important velocity of particles or bigger masses)means that system is not stable(because of motion).
so if you want to compare elephant and ant you have to comare their velocity after their potential energies turned to kinetic energy
| {
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Weighing head by angular momentum A popular Phys.S.E question asks how can I measure the weight of my head. One of the answers suggests measuring the moment of inertia.
My suggestion was to construct an apparatus that places the subject on a translatable carrier located on top of a rotating table. Sensors would allow the change of angular velocity to be measured as the carrier is translated relative to the centre of rotation. The subject would begin with their head at the centre of rotation and end with their feet there, many measurements being taken in between.
It would (equivalently) rotate at constant speed and measure the forces necessary to hold the translating carrier in place.
I hoped that this (with x & y translation, if desired) would give enough information to computationally determine a mass density map of the subject.
Q) Are these two approaches basically equivalent?
Q) Would these approaches actually work? A comment on the moment of inertia answer states:
"Isn't the moment of inertia of a rigid body a tensor? More specifically, it's a 3×3 matrix, so all your measurements above can be calculated knowing just nine numbers."
Is this the case, and does this mean that a density image could not be generated?
| Regarding the second question:
The parallel axis theorem states, that if you have rotation around another axis than a body's center of mass, you only have to add the inertia tensor of a point mass at the location of the body's center of mass. And because of your setup, you will probably not be able to measure more than one component of the person's inertia tensor.
Regarding the first question:
As I understand it, both measurements will yield the same information (one component of the inertia tensor). As mentioned in the linked answer, you will need a model of a human to get the head's weight, which means additional assumptions about anatomy etc.
| {
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Why ONLY Maxwell's equations are the basic equations of electromagnetism? In electromagnetism we say that all the electromagnetic interactions are governed by the 4 golden rules of Maxwell. But I want to know: is this(to assume that there is no requirement of any other rule)only an assumption, a practical observation, or is there a deeper theoretical point behind it? Could there be a deeper theory behind assuming that there is not requirement of rules other than Maxwell's equations?
| The Maxwell equations only approximately describe electromagnetism, even in a pure vacuum. This is a consequence of quantum electrodynamics. One can derive corrections to the Maxwell equations; this was first done by Heisenberg an Euler in the regime where the fields only change appreciably over distances much larger than the electron Compton wavelength, see here.
| {
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Physical Role of Batter in Baseball Physically, what is the role of a batter in baseball?
My question is inspired by How does the speed of an incoming pitch affect the speed of a baseball after it's hit?
The answer to that question, that a faster pitch results in a farther hit, surprised me. I had thought that a batter was solely applying impulse to the ball (mass * velocity of bat = Force * time on ball). Clearly the batter is doing more.
Specifically, I'm interested in what information we'd need to know and what math we'd use to calculate the final velocity of the ball.
| It helps to think of the extreme cases here. If the bat was not moving (like a bunt), the resulting bounce does depend on incoming speed. Now, make the incoming ball really slow and the batter hitting fast. The result does not depend on incoming speed. The true answer is the superposition of these two cases.
| {
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Classic home experiments for an 8-year-old child My 8-year-old daughter's school report says that she's good at understanding the basic science she's doing, but she's having trouble seeing how experimental results lead to conclusions. Specifically, it says she struggles to appreciate how changing parameters in an experiment can be used to prove or rule out a hypothesis.
I'm a biologist/chemist by degree, and I found it hard to think of any parameter-based experiments in my field that we could do at home that wouldn't bore a child. They like creepy-crawlies well enough, but biology experiments tend to need time and repeat observations to get a result. Chemistry tends to need more specialist equipment and reagents.
I've got a couple of books of home science, and I scoured the net as well, but most of the living-room science experiments I found that were suitable for an 8-year-old were more demonstrations than actual experiments. There were few parameters: nothing you could really vary.
So can any of your learned people suggest some simple, parameter-based physics experiments we could do at home or in the garden, without specialist equipment, which involve science at a level an eight-year-old could engage with?
| How big is an atom?
Fill a sink with water. Find a chemical which, when dropped into water, forms a contiguous floating disk. Drop one drop of this chemical into water. Measure volume of drop and area of floating disk. This provides an upper bound on the size of an atom.
Falling speed versus mass/shape
Drop a book and a piece of paper at the same time. Book hits floor first. Now place paper on top of book and drop book. They fall together. This suggests that air resistance is the reason the paper normally falls slower, and that it has nothing to do with mass.
You can also crumple the paper and then drop it with the book to get the same effect.
| {
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How does the Alcubierre drive warp space in the vicinity of it's destination?
I don't understand how this could work. So lets say the drive is set to go 30,000,000 meters in a straight line to get to a distant planet. Now lets say the ship size is 20 meters. Now let's make the assumption that the warp field in the photo is 30 meters (10 meters bigger than the ship). So the warp bubble is applied. Now the 30 meters of space the ship occupied is cut. But it only could apply the warp bubble to this confined region which means 30 meters are cut instead of 30 million and now 29,999,970 meters are left and the ship is not even close to the planet it wants to go to (barely moves). Are there any solutions to this problem?
| The idea is that the object generating the "pit" in the front is in the center of the flat region in the middle. What happens is that the object in the middle begins to "fall" into the "pit" in front of it, due to gravitational attraction. The "pit", however, moves forward because it is a fixed distance away from the object in the middle. Basically, as the object moves forward, the "pit" moves forward with it.
Imagine a dog with a treat dangling in front of its nose. The treat is hanging from a harness attached to the dog. As the dog moves forward to try to reach the treat, the treat moves forward the same amount, preserving the arrangement of the system, but causing an increase in velocity. The Alcubierre drive works in an analogous manner.
As to how the drive would warp spacetime, I'd consult Wikipedia, which does a much better job explaining this than I could.
| {
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Why do we deal only with large scale magnetic fields in astrophysics, and not electric fields? In astrophysics there is a lot going on about strong, large scale magnetic fields: in stars (prominences), magnetic dynamos, compact accretors collimating jets, etc. There's even a special computational formalism called magnetohydrodynamics (MHD), which allows to deal with space plasma.
Yet I've never read about large scale electric fields. I know that most of the matter we model in astrophysics is plasma but, naively, one would assume that this introduces both $\mathbf{E}$ and $\mathbf{B}$ fields on an equal footing.
So where does this asymmetry come from?
| The lack of the electric field in modeling plasmas stems from the Lorentz force,
$$
\mathbf F=q\mathbf E+q\boldsymbol\beta\times\mathbf B
$$
where $\boldsymbol\beta=\mathbf v/c$. For most astrophysical plasmas, the force is zero, so we have that
$$
\mathbf E=-\boldsymbol\beta\times\mathbf B
$$
So any time we see an electric field, we can simply replace it with the above cross product. This happens in Faraday's law:
$$
\frac{\partial\mathbf B}{\partial t}=-\nabla\times\mathbf E=-\nabla\times\left(-\boldsymbol\beta\times\mathbf B\right)=\nabla\times\boldsymbol\beta\times\mathbf B
$$
The justification for $\mathbf F=0$ is as Chris White says: we assume perfect conductivity.
| {
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Uncertainty principle with two photons Imagine an experimental setup in which you have to measure the momentum and location of a particle. To measure it we know we will have to affect it, and the uncertainty principle would come into the picture, but I have a different setup. The classical setup is that you fire a photon to measure the location of the particle, but the particle will change its momentum due to the collision with the photon.
I decided to take two photons. I will shoot one photon from either side of the particle, so the effects of the two photons cancel each other, giving an accurate measurement. To understand this, see the picture below.
*
*The classic experiment
*My thought experiment
In the second experiment, we shoot a photon of the same energy as the first one and counteract the effect of the first photon, so the electron would continue on its original path. Please tell me where I am wrong.
EDIT
We will have to take multiple photons but equal from both sides and in opposite directions.
| Your classic experiment 1) is quantum mechanical. Electrons and photons are elementary particles and interact individually as quantum mechanical entities.
The plot shows the elastic scattering of photon on electron, a computable process quantum mechanically thus the angular distribution is known. As we are talking quantum mechanics there is a probability for each angle where the electron may scatter that is given by the Klein-Nishina formula. So the individual photon-electron pair does not have a fixed angle which you could use in your thought experiment to counteract the effect on the electron's direction. ( always presuming you could time the photons to hit at the same delta(t) of Heisenberg's uncertainty principle) .
The thought set up 2) is impossible.
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Why does wavelength affect diffraction? I have seen many questions of this type but I could nowhere find the answer to "why". I know this is a phenomenon which has been seen and discovered and we know it happens and how it happens. But my question is why would wavelength affect the amount of diffraction? I am looking for a very simple logical explanation rather than a complex mathematical answer. Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray? I need an answer that will answer "why" does diffraction depend on wavelength of light.
Image sources: http://www.olympusmicro.com/primer/java/diffraction/index.html
| The greater the wavelength the heavier the wave. If you think of it visually, the heavier the wave the more energy needed to move the wave in a different direction. As a result, the greater the wavelength, less diffraction.
| {
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Natural entanglement system I'm a beginner and amateur interested in quantum physics.
I would like to know if entangled systems of natural states exist or whether such systems require human intervention?
Is it possible? Either no or yes, Why?
| The answer is definitely yes. The ground states (and low-lying eigenstates) of many-body systems are generically entangled. Examples include the ground states of local quantum field theories (which describe the fundamental particles and forces of nature) and ground states of fermionic lattice models (which describe much of the solid matter we see around us). The reason for this is simple: these systems are composed of many interacting constituents. Take for example the magnetic dipoles associated with the spins of atoms in a lattice. These spins are quantum mechanical, which means the orientations of their dipole moments fluctuate even at low temperatures, due to the uncertainty principle. In addition, the dipoles interact with each other because of the magnetic fields that they produce. Therefore these intrinsic quantum fluctuations become correlated with each other, via the mutual magnetic interaction of the spins. The appearance of local quantum fluctuations that are globally correlated is one of the essential features of entangled states.
The entanglement of many-body ground states is highly relevant, since many systems are nearly in their quantum ground states even at everyday temperatures. Examples include the electrons in a typical metal, or the electromagnetic radiation field at optical frequencies. This is true whenever the typical energy scales of the system are large compared to room temperature (i.e. the Fermi energy for a metal, or the frequency of optical radiation).
Interestingly, the principle of locality places quite severe restrictions on the kinds of entangled ground states that actually appear in nature. Typically, these states obey area laws. That is, the amount of entanglement between a sub-region of the system and the rest scales with the area of the boundary of the sub-region, rather than its volume. For more information on area laws, see J. Eisert et al., Rev. Mod. Phys. 82, 277 (2010).
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Difference between single mode and multi mode optical fibres? What is the difference between single mode and multi mode optical fibres? First off, I guess that by modes we mean the spatial modes of the electric (or magnetic?) field right?
Now: what makes a fibre able to support more than a single mode? I mean, what aspect of its structure corresponds to which mode(s) can be transmitted?
| The V-number defines the mode of any fiber. It depends on wavelength, diameter of fiber and refractive index.
The v number is
\begin{align}
\ V^2 &= \frac{2}{} (_1^2−_2^2)\\\\
\end{align}
Each mode has an effective index that can be defined by:
\begin{align}
\ _ = _ \frac{2}{}\\\\
\end{align}
The effective index tells you how tightly the mode is confined to the waveguide core
*
*Guided mode _ < _ < _
*Tightly confined to the core _ ~ _
*Weakly confined to the core _ ~ _
*Unguided or radiating modes _<_
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Relative wind velocity explanation I am having trouble understanding the reasoning behind the solution in this Irodov General Physics problem:
problem 1.6.) A ship moves along the equator to the east with velocity
$v_0=30\;km/h$. The southeastern wind blows at an angle $\theta=60°$ to the equator with velocity $v=15\;km/h$. Find the wind velocity $v'$ relative to the ship and the angle $\theta'$ between the equator and the wind direction in the reference frame fixed to the ship.
My calculations:
After drawing the initial diagram I consider that in order for the wind direction to be relative to the boat, I need to subtract the boat vector from the wind vector.
In my diagram I have repositioned the boat vector to be subtracted from the wind vector and using trigonometry I solve as follows:
First the cosine law to determine the magnitude
The magnitude = $\sqrt{30^2 + 15^2 - 2\cdot 15\cdot 30\cdot \cos{60}}$
= $\sqrt{675} \approx 25.98$
Then for the angle I can use sine law
$\frac{sin(\theta)}{30\frac {km}{h}} = \frac{sin(60)}{\sqrt{675}}$
$\theta= \arcsin{\frac{30\frac {km}{h}\cdot sin(60)}{\sqrt{675}}}$
$\theta = 90°$
The answer I get is therefore that the wind, in reference to the boat (as felt by the people standing on the boat), is approx $26\:km/h$ at an angle of $60°$ South of West (in relation to the equator).
However the answer given in the solutions is always that the approximate speed of the wind is $40\:km/h$ and the angle is approx $19°$.
I have looked at the solutions and I think the best explanation is given in the following diagrams obtained from link #3:
I do not understand, however, why he is using $\cos{(\pi +\theta)}$. I want to understand why my solution is incorrect and yields a different answer. What is my mistake and what is the proper intuition for understanding this problem if I have done so incorrectly?
Solutions I have viewed:
*
*This Youtube Video
*This online blog
*This online blog
| I think you are referring to the I.E.IRODOV (Problems in General Physics) question number 1.6 (KINEMATICS)
South-Eastern winds don't go towards South-East, in fact it's the complete opposite, they originate from South-East and go towards North-West.
So, if you were to reverse the direction of the wind vector, the solution would be correct and so will be the answer obtained by you.
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Calculate electricity generation A friend of mine stays near a water stream and wanted help to use its water. The height up to which the water could be raised using the water flow is 12ft. How much electricity can it produce?
Is it at least enough to power half the house?
| The efficiency of large hydroelectric generators can be very high - up to 95% in ideal cases - however the efficiency of small installations is a lot lower and in particular it's hard to get efficient electricity generation if the flow rate is low, which is likely to be the case for your friend.
But lets see what the potential is. You don't say what the water flow rate is, so let's just call it $F$ kg/sec. 12 feet is about 3.7m so the power is:
$$ P = Fgh \approx 36F \space\text{Watts} $$
Let's guesstimate the efficiency of electricity generation at 10%, then the electrical power your friend can generate is:
$$ P \approx 3.6F \space\text{Watts} $$
I would guess most houses use a few kW of electricity. If we take 1kW and work out what flow rate is needed to produce this we get:
$$ F \approx \frac{P}{3.6} \approx 280 \text{kg}/\text{sec} \approx 0.28 \text{m}^3/\text{sec} $$
I would guess this flow rate is more than the typical stream, but it's not all that great.
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Thermal expansion coefficient times temperature: under which condition is it unity? In Landau's Fluid Mechanics p. 8 (2nd edition) he writes for the thermal expansion coefficient $\beta = (1/V) (\partial V/\partial T)_P$:
For a column of gas in equilibrium which can be taken as a thermodynamically
perfect gas, $\beta T = 1$ [...]
This is not not obvious to me. Can somebody provide me with a derivation or an argument under which circumstances I can approximate this expression like that?
Or is it as easy as taking Clapeyron's equation for $V$, i.e. $PV = NT \implies V = NT/P$, so that we get
$$
\beta T = \frac{TN}{VP} = 1
$$
This leaves me somehow unsatisfied; is there maybe a more physical argument for this?
| What he writes down is the relative change in volume with temperature, so it is true by definition. What may be confusing is that sometimes β denotes the thermal expansion coefficient in two dimensions, whereas the three-dimensional coefficient is meant here.
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Why does the Walecka model not include pions? The Walecka or $\sigma$/$\omega$-model is an effective theory describing nucleon-nucleon interaction by an exchange of $\sigma$/$\omega$-mesons. Why does it not include interactions by pions?
| I want to cite the original paper of Walecka where he awnsers your question:
From Annals of Physics 83/2 (1974) p. 491 "A theory of highly condensed matter
"The reader might object to the fact that there is no one-pion exchange tail in this interaction; however, the strong spin and isospin dependence of the potential arising from the exchange of an isovector, pseudoscalar pion implies that the contribution of the one-pion exchange potential to the bulk proerties of nuclear matter largely averages to zero"
Hope this helps
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Magnetic field and Newton's third law If a magnet exerts force on a iron block (opposite and EQUAL), does a iron block also exerts force on magnet (via Newton's third law)? If yes then what magnetic property does it has to produce equal and opposite force on magnet considering that its not a ideal environment? If no then is it not the violation of newton's third law?
| It's not a violation of the Newtons law. (As AcuriousMind answers on the comments)
It's the very same magnetic force which prevents one solid the penetrate another solid. The only difference is that this force is weaker and works only over very small distance. If you have a iron block and a rubber pushed to gether ie. by gravity. Doesn't the Iron block have more force similarly as the magnet? No, the both produce the same force, but in the rubber, there is just more deformation needed to produce this same force. Even the pulling force works; ie. with cold welding you can produce a similar connection like the two magnets has. Principally it's the very same. The magnetic fields get too close each others, and so they get connected. If this magnetic force becomes the weak enough, the solid becomes a fluid, and this penetration becomes possible.
Is this then a violation of the newtons 3rd law?
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
At least the fluid dynamics can't be derived from the newtons laws according to the mainstream physics. And you definetly become problem simply by defining the "body" of the fluid and how the forces are exerted. (I see this possible, and this idea even fits to the QED.)
This can been put more simple; even in fluids it's not a violation of newtons laws, the hydrostatic pressure is clear implication for this force. And if this force is missing in some direction, then the flow accelerates until there is no pressure left. Exactly as said by the Newton's first law. This video might provide the best explanation to your question; Feynman, Fun to Imagine 4, Magnets
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Singularity in Newton's gravitational law If $r=0$ in the well know equation $F= G\dfrac{m_1\cdot m_2}{r^2}$, it will not follow that the force will be infinite?
May someone please clarify it to me?
| Assuming that $m_1$ and $m_2$ take up a finite amount of space (e.g., two spheres of mass with radius $r_0$), that equation isn't even valid for $r < r_0$, so there's no inconsistency.
The derivation follows from Gauss' law; it is analogous to the application of Gauss' law in electrostatics; the $m_1$ and $m_2$ are the mass enclosed at some distance $r$.
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Gradient is covariant or contravariant? I read somewhere people write gradient in covariant form because of their proposes.
I think gradient expanded in covariant basis $i$, $j$, $k$, so by invariance nature of vectors, component of gradient must be in contravariant form. However we know by transformation properties and chain rule we find it is a covariant vector. What is wrong with my reasoning?
My second question is: if gradient has been written in covariant form, what is the contravariant form of gradient?
| Most of the answers posted here are incorrect. The Wikipedia page for the gradient says
The gradient of $f$ is defined as the unique vector field whose dot product with any vector $v$ at each point $x$ is the directional derivative of $f$ along $v$.
A look at Theodore Frankel's The Geometry of Physics confirms this. Other posters have said that the components of the gradient of $f$ are given by $\partial_i f$; these are in fact the components of the differential of $f$, which is a covector. The gradient is this with the index raised.
Let's now calculate both sides of the expression from Wikipedia. The inner product of $\mathrm{grad}(f)$ with a vector $v$ is
$$
\mathrm{grad}(f)^{i} g_{i j} v^j =\mathrm{grad}(f)^i v_i.
$$
The directional derivative of $f$ along $v$ is
$$
D_v f = v^i \partial_i f = g^{ij} \partial_i f ~v_j.
$$
We can clearly identify
$$
\mathrm{grad}(f)^{i} = g^{ij} \partial_j f
$$
or
$$
\mathrm{grad}f = g^{-1} \mathrm{d} f.
$$
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Is there a difference between "average acceleration" and centripetal acceleration? Question adapted from Examkrackers MCAT prep book:
A particle moves along a half circle (diameter=$10\text{ m}$) at a constant speed of $1\text{ m/s}$. What is the average acceleration of the particle as it moves from one side of the half circle to the other side?
A. $0$
B. $0.2/\pi$
C. $0.4/\pi$
D. $1$
The book says C is correct. Acceleration is change in velocity divided by time. Initial velocity is $1\text{ m/s}$ up; final velocity is $1\text{ m/s}$ down. The change in velocity is therefore $2\text{ m/s}$. The time is found from speed equals distance divided by time. Distance is $2\pi r/2$. Thus
$$a= \frac{2}{(2\pi(5)/2)/1} = \frac{2}{5\pi} = 0.4/\pi$$
I thought all of the answers were wrong because I thought they should've used the centripetal acceleration equation: $a= v^2/r$
SO my question like the title: Is there a difference between "average acceleration" and centripetal acceleration?
I searched for a couple hours and couldn't find this issue directly addressed.
|
Is there a difference between "average acceleration" and centripetal acceleration?
Yes, in fact they're almost completely unrelated.
The average acceleration is defined as
$$\vec a_\text{avg} = \frac{\Delta\vec v}{\Delta t}$$
It is one quantity that partially describes the motion of a particle over an extended time. In other words, average acceleration encapsulates the fact that a particle started with some velocity at time A and ended with some velocity at time B, but completely ignores what the particle did between A and B. This is by design.
Centripetal acceleration, on the other hand, is an instantaneous quantity: it's the radial component of acceleration. (This requires that you have chosen some point to be the center of a polar coordinate system.) It partially describes the motion of a particle at one moment, not over an extended time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/126970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Wrong calculation of work done on a spring, how is it wrong? So I would have thought that this would be how you derive the work on a spring: basically the same way you do with gravity and other contexts, use $$W=\vec{F}\cdot \vec{x}.$$ If you displace a spring by $x$, then it exerts a force $-k x$, so $F=-kx$, since the displacement is $x$.
So $$W=-kx^2.\qquad \leftarrow\text{ (however, apparently wrong!)}$$
I've seen the correct derivation of work in a spring (with an extra half) and don't doubt that it's correct, but also don't see where my logic fails in this alternate derivation.
| The factor $\frac{1}{2}$ is due to the integral.
The wrong sign of yours is due to the fact that you have to counter the force of the spring.
So the Force if the Spring is $-kx$, but you have to pull in the direction it is extended, so apply the force $kx$, therefore the energy is positive $W=\frac 1 2 k^2 x$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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In what sense is a quantum field an infinite set of harmonic oscillators? In what sense is a quantum field an infinite set of harmonic oscillators, one at each space-time point?
When is it useful to think of a quantum field this way?
The book I'm reading now, QFT by Klauber, claims its not true, which is it?
I would like to understand this analogy a little better.
| For a free theory, say for one scalar field for simplicity, which gives a a linear differential equation for the field $\phi$, one can cast the Hamiltonian
$$
H=\frac{1}{2}\int d^3x \dot\phi^2+(\partial_i \phi)^2+ m^2\phi^2
$$
in this form (basically by taking a Fourier Transform):
$$
H=\mathrm{const}+\int \frac{d^3 k}{(2\pi)^3} \omega(k)a^\dagger(k)a(k)\,,\qquad \omega(k)^2=k^2+m^2
$$
where $[a(k),a^\dagger(p)]=(2\pi)^3\delta^3(p-k)$ and $[a(k),a(p)]=0$. You see that this is the Hamiltonian for infinitely many harmonic oscillators, one in every point of momentum space. Since the energy levels for a harmonic oscillator are evenly separated for every $k$, you get the particle interpretation of the free field theory given that the state of the system is identified by the occupation numbers of any such harmonic oscillator.
If the theory is weakly coupled, that is a small perturbation of the above system, you can keep thinking of the system as a collection of harmonic oscillators, although now coupled, with new phenomena such as non-trivial scattering (elastic or inelastic with particle production...) because the time evolution is deformed by one of the free oscillators/particles that we have prepared at infinity where the interactions can be neglected.
Vice versa, several QFTs, especially the strongly coupled ones, do not admit such a particle/oscillators interpretation and you should not think of them as a collection of harmonic oscillators.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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"answer_id": 2
} |
Why does blowing on hot coffee cool it down? And will it cool off faster if you blow across the top of the cup or directly into the coffee?
Does it have to do with the fact that when you blow across the top of the cup the velocity of the air increases which causes an area of low pressure above the cup, resulting in steam from the coffee to be forced up and out of the cup? Or is that not it?
| When we blow air it's temp is about 98.5°F the cup of coffee also heat when more heat a given to coffee its evaporation rate increases and gets cool down quickly
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Distance from redshift I am looking for a exact derivation of a relation between redshift $z$ and distance $d$.
What I know is the definition
$$z=\frac{\lambda_{\text{observed}}}{\lambda_{\text{unshifted}}}-1=\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}-1$$
and that the Hubble constant $H$ as a function of $z$ is:
$$H^2=H_0^2\left(\Omega_m\left(1+z\right)^3+\Omega_{\Lambda}\right)$$
How can I use this to derive the distance?
| Depending on the shape of the universe the luminosity distance is given by :
\begin{equation}
d_L(z) = \left\{ \begin{array}{rl}
\frac{(1 + z) c}{H_0 \sqrt{|\Omega_k|}} \sin \left[ \sqrt{|\Omega_k|} \int _0 ^z \frac{dz'}{H(z')/H_0} \right] & \mbox{for $k = 1$} \\
\frac{(1 + z) c}{H_0} \int _0 ^z \frac{dz'}{H(z')/H_0} & \mbox{for $k = 0$} \\
\frac{(1 + z) c}{H_0 \sqrt{|\Omega_k|}} \sinh \left[ \sqrt{|\Omega_k|} \int _0 ^z \frac{dz'}{H(z')/H_0} \right] & \mbox{for $k = -1$}
\end{array}
\right.
\end{equation}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What will be the effect of placing a light source very close to a photodiode? What will be the effect of placing a photodiode really close to a laser source and what should be the appropriate distance between a light source and photodiode to get maximum output current?
| I assume that the laser is emitting a wavelength that the photodiode is sensitive to. If the laser has too much power, you would burn out the semiconductor. Since a laser is a collimate light source, the power lose is minimal for any distance unless there is fog.
What type of photodiode do you have? Some are like transistors with low gain and others are FETs or even Darlington transistors (very high gain). What is the purpose of the light signal? Is it just an indicator, or does the signal depend on the intensity?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Correlation in electron gas In the textbooks that I read (namely Ashcroft/Mermin , Marder, etc.) it seems that a distinction is made between the correlations in electron gas and a Couloumb interaction between the electrons. What is exactly meant by the concept of correlations? How is that connected to the interactions in electron gas, and how does the screening enters the picture?
| It would be useful to have the exact reference in the text where "correlation" is mentioned, but I would argue that correlation is a word which characterizes the collective behavior of electrons in certain circumstances. Such circumstances are varied, but imply that their behavior (and that may be spatial dynamics, but also spin dynamics or other quantities) is not independent. Examples include Coulomb interactions: such interaction mar for instance prevent double occupation of a site in a solid if it is strong enough. The Pauli exclusion principle also imposes constraints on the spin arrangements, thereby inducing correlations between the spins of electrons of a lattice. Other spin spin interactions such may induce ferromagnetism or antiferromagnetism on a lattice, which is another form of correlation.
As a conclusion, there is indeed a difference between correlation, which is a general description of the behavior of electrons, and the interaction or other phenomenon (not necessarily an interaction) which generates it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why do bubbles make a sound? I have an understanding of how bubbles work. They encapsulate air (or other fluids) in a membrane caused by surface tension.
When they pop, there is often a sound. Sound is a type of energy, kinetic to be precise, that usually occurs from collisions. When a bubble pops I would assume that a sound implies that air rushes out due to a pressure change. Why is there a pressure change? I wouldn't expect the bubble to exert enough pressure to compress air. If the sound is caused by the air now being able to move into the rest of the room due to Brownian motion, then why wouldn't I hear air moving in a still room?
| Cavitation is the formation of bubbles in a liquid when a sufficiently strong negative pressure is applied. A point in the liquid experiences a “negative pressure” if the local pressure goes below the average pressure in the liquid. This can happen when water in a pipe has a very abrupt turn, near the propellers of ships and submarines, in presence of a high-intensity sound (you can imagine sound as an oscillation of pressure).
In the Cavitator, sound is pulling the water apart, leaving a bubble of gas which is due to the excess pressure inside
which is why smaller bubbles tend to pop with a greater sound (more pressure)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 3
} |
Is a falling, perfect sheet of fluid possible to create? This is a bit of an abstract question so I'll try explain this as best I can from the bottom up.
I would like to know if it is possible to observe a sheet of fluid, much like a sheet of glass, falling completely flat and level.
Imagine a volume of water that is at rest on some sort of very, very fast trap door, which once opened, would allow the volume of water to retain its shape and fall through the trap door.
Is this possible if it were done in a room containing a near-perfect vacuum to discount drag effects? would things like surface tension stop this happening? perhaps using a super-fluid?
Thanks!
| EDIT: replaced "fluid" with "liquid", thanks to Kyle.
I am not aware of any material with a liquid phase in near-vacuum. Probably, the liquid would evaporate and maybe a part of it freezes solid due to evaporation cooling.
EDIT: NeuroFuzzy pointed to a youtube video containing an ionic liquid, which is able to retain liquidity in very near vacuum. What would happen, if you prepared such a liquid in sheet form and dropped it in vacuum? I think, that the strong cohesive forces would pull the liquid together, so that the sheet is transformed into a droplet. However, this would take some time, and it may be that the shape, which splashes on the ground, very well resembles the original sheet.
If you supply enough pressure for a liquid phase to exist, there will be two forces:
*
*Drag from the surrounding medium
*Cohesive force between fluid molecules
Together they will probably effect formation of isolated droplets.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Pool in a submarine A common theme in aquatic science fiction is the submarine pool/access to the ocean. That terrible TV show Seaquest had it, The Deep & Deep Blue Sea (Samuel L Jackson is standing in front of it when the shark chomps him). My question is how this could possibly work? From what little knowledge I have, I'd say the cabin where the pool resides would have to be pressurized to the water at that depth. The implications are that you'd have to pass through an airlock to get to the room, and that it would only work to a certain depth.
Is this correct, or it too far to the "fiction" side of the science fiction axes?
|
The implications are that you'd have to pass through an airlock to get to the room, and that it would only work to a certain depth.
Correct x 2.
You have 2 choices with an underwater habitat - build it really strong to take the pressure, or just pressurize it and you can make the whole thing out of plastic.
Submarines chose the former, as they may need to surface quickly and don't want the crew exploding when they open the hatch. Exiting a sub below a certain depth is not possible, and a small hole can create a steel-cutting jet inside. Descending below crush depth is ..... bad.
If you want to put something on the bottom with a convenient door you just raise the internal pressure of the whole thing to match the external pressure. Done for underwater construction all the time. Mixing the gases is important but also not complicated, the crew goes to the surface in a pressurized diving bell and spends a lot of time decompressing. Accidents do happen, and are rather spectacular. Look up "Byford Dolphin" in the usual place for the gory details. In theory this will work all the way down to the Mariana Trench, if you can get a powerful enough air compressor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
Which ball touches the ground first? This is a very well known problem, but I can't find an answer in the specific case I'm looking for.
Let's consider two balls :
*
*Ball 1 weighs 10 kg
*Ball 2 weighs 1 kg
*Balls have identical volumes (so Ball 1 is much more dense)
*Balls have identical shapes (perfect spheres)
Let's drop them from a rather important height, on earth, WITH air. (That's the important thing, because all the proofs that I browse take place in a vaccum).
I am arguing with a colleague. He thinks that ball 1 will fall faster in air, and that the two balls will fall at the same speed in a vacuum. I think that the identical shapes and volumes make air friction identical too and that the vaccum has no importance here. Could someone tell who's right and provide a small proof?
| I am not satisfied with the way @Bernhard answered, since it just shows the maximum velocity, thus only answering partially the question.
The air resistance can be written as :
$$
R = \frac{1}{2}\,C_x\, \rho\, S\, v^2
$$
Note : The mass of the object is not in this equation. This is very important.
Applying Newton's law to one of the object gives at any moment of the fall:
$$
a = g - \frac{1}{2m}\,C_x\, \rho\, S\, v^2
$$
As you can see the acceleration is function of the mass of the object $m$. A heavier object will accelerate more than a lighter one, therefore, will go faster during the whole fall. Both objects will at one point reach the maximum velocity that is explained well in @Bernhard answer.
So at any point of the fall, your heavier object will be faster than the lighter one.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 6,
"answer_id": 2
} |
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