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Plotting hydrogen wave functions This may sound a bit dumb but how do I plot the hydrogen wave functions? For example, what is exactly being represented in this image? Is it just the norm-squared of the wave function and is the z-axis sticking out of the page?
I'm not sure how to use any other application but I'm using the Mac grapher tool. Say I wanted to plot $\psi_{200}$. The combined radial and angular equation for this state is $$\psi_{200}=\frac{1}{4\sqrt{2\pi}a_0 ^\tfrac{3}{2}} \left(2-\frac{r}{a_0}\right)e^{-\tfrac{r}{2a_0}}$$ where $a_0$ is the Bohr radius. As I said, i'm pretty clueless. I'm not sure how to plot this in spherical coordinates so I just converted $r=\sqrt{x^2+y^2+z^2}$. Basically, I plotted $$z=\frac{1}{4\sqrt{2\pi}a_0 ^\tfrac{3}{2}} \left(2-\frac{\sqrt{x^2+y^2+z^2}}{a_0}\right)e^{-\tfrac{\sqrt{x^2+y^2+z^2}}{2a_0}} $$When I do plot this, I get a flat plane with a sort of half-sphere in the centre, which is not what the image I linked above shows. I also tried graphing $\psi^2$ but I still did not get it. I feel like i'm missing something big.
| For Mac there still is Atom in a Box. It can display hydrogen wave functions in different ways. I found the view mode with "phase as color" very helpful for my understanding. The wave functions are displayed in 3D and animated in time. One can also display superpositions of different eigenfunctions (for example the hybridizations from chemistry).
| {
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Two Rolling logs Suppose we have two logs rolling down a hill, one of gold and the other of wood; the acceleration for both will be equal, something which is unclear to me; I get that this may be due to their form, which is the same, but how come the mass of the objects doesn't matter?
| All objects are accelerated by the same value in a gravitational potential in accordance with newton's laws of gravitation and motion
$F = mGM/r^2 = mg$ where G is newtons constant and M is mass of the earth.
Applying the equivalence of gravitational and inertial mass we have
$F = mg = ma => g = a$ (independent of mass)
where $a=9.8 m/s^2$ is the acceleration due to the earth.
For an inclined place the acceleration is reduced by the sin of the angle ($\theta$) the plane makes with the ground. I.e. $a = g*sin(\theta)$
Since mass cancels out of this equation then you will find that the 2 logs accelerate equally.
The deep concept here is that inertial and gravitational mass are equivalent.
Astronauts performed this type of experiment on the moon by dropping a feather and a er at roughly the same time and found they both landed at roughly the same time
Hope that helps!
:D
| {
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Why do we not use the SI system for distance in space? One of the closest stars to Sol is Alpha Centauri at 4.367 Ly according to wikipedia. Why do we not say that it is 41.343 Peta-meters rather? (4.367 Ly = 41.343 Pm)
Why does Light-years or Parsecs seem to be the standard rather than SI?
| Light years and parsecs have been used since long before SI existed, so a lot of it is tradition.
But using light years also makes it very obvious how long the light has traveled to get here, and thus which era of the universe we are seeing the object in. Something that is 11 billion light years away dates from the era of early galaxies, for example. If you gave the distance as 100 yottameters instead, it would be far from obvious unless your listener was particularly familiar with distances on that scale.
For closer objects, parsecs are useful because they directly relate to the amount of parallax shift seen between opposite sides of the Earth's orbit. This is not so relevant now, but in the days when parallax was a standard method of measuring distance to astronomical objects, it was very convenient to be able to convert a parallax directly to a distance. (The name "parsec" actually comes from "parallax arcsecond".)
| {
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Is there a prohibited region in $P-V$ plane? Polytropic process generalize the particular thermodynamic processes with
$$P V^{n}= \mathrm{constant}$$
Where, if $n$ changes, the curve on $P-V$ plane changes, as shown in the diagram.
The orange region is not touched by any curve, so there is no value of $n$ for which the gas goes directly in to the orange region.
Why is that? I do not see any particular reason why there should not exist a process to make the gas go into the orange part.
| Assuming there is only one molecular in this box and assuming it is a closed system with initial P-V state is defined. The question becomes: can the system move to anywhere on the PV diagram?
Well we can adjust volume to any number. Then the question becomes: can pressure reaches to any values on PV diagram?
Pressure relates to impacting intensity and frequency. By increasing its kinetic energy or its speed, both intensity and frequency increase. So it seems that we can control pressure using temperature or heating/cooling until the limit of molecular is reached. At this time, you can say there is a prohibited region but not as the orange zone in your plot.
| {
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Why don't high pressure gases stored in containers lose energy? Containers holding gas at a high pressure don't slowly lose the internal energy of the gas. It seems like the high speed particles would collide with the metal walls and slowly transfer their energy to the slower particles outside the container.
Even if the pressure is from more particles in the container, they can do work when released so they have energy. Shouldn't that energy dissipate over time?
|
It seems like the high speed particles would collide with the metal walls and slowly transfer their energy to the slower particles outside the container.
The mechanism you describe is correct, but you have to keep in mind that average kinetic energy, $\langle K \rangle$, is only proportional to temperature:
$$\langle K \rangle = \frac 3 2 N k T$$
So (since the volume is fixed and therefore macroscopic work is ruled out) there will be a net energy transfer only if the temperature inside the vessel is different from the temperature of the outside environment and if heat transfer is allowed (i.e. if the walls of the containers are not adiabatic).
| {
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Why Do Glueballs Have Mass, When Individual Gluons Are Massless? From Wikipedia Glueballs
Glueballs are predicted by quantum chromodynamics to be massive, notwithstanding the fact that gluons themselves have zero rest mass in the Standard Model. Glueballs with all four possible combinations of quantum numbers P (parity) and C (c-parity) for every possible total angular momentum have been considered, producing at least fifteen possible glueball states including excited glueball states that share the same quantum numbers but have differing masses with the lightest states having masses as low as 1.4 GeV/c2 (for a glueball with quantum numbers J=0, P=+, C=+), and the heaviest states having masses as great as almost 5 GeV/c2 (for a glueball with quantum numbers J=0, P=+, C=-).
Rather than going through a list of possible mechanisms that unfortunately I know next to nothing about, such as can the mass be attributed to virtual quarks, or binding energy between the gluons, I would rather leave the question as in the title to find out as much as I can.
Also, although the SM is firmly established, would the discovery of Glueballs buttress it further?
My apologies for not knowing more about the interior of hadron like particles or if the answer is readily available (or worse, blindingly obvious).
| Because in relativity the mass of a collection of particles is not necessarily the sum of the masses.
Even two photons (treated as a unit) can have mass. Consider the total four-vector of a system with component four-vectors $(E,\hat{z}E/c)$ and $(E,-\hat{z}E/c)$. It has mass $(mc^2)^2 = (2E)^2$.
| {
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How to show mathematically that the electric field inside a conductor is zero? The electric field is characterized by the equations
$$\nabla\cdot \mathbf{E} = \dfrac{\rho}{\epsilon_0}$$
$$\nabla \times \mathbf{E} = 0$$
Or equivalently, $\nabla^2 V = -\rho/\epsilon_0$ and then $\mathbf{E} = -\nabla V$. The boundary conditions which should be used are the discontinuity of the normal component of $\mathbf{E}$ when crossing a charged surface and the continiuty of the tangential component. That is:
$$\mathbf{n}\cdot (\mathbf{E}_2-\mathbf{E}_1)=\sigma/\epsilon_0,$$
$$\mathbf{n}\times(\mathbf{E}_2-\mathbf{E}_1)=0.$$
I'm trying to show mathematically, just with the equations, that the electric field inside a conductor is zero. I've seen many "conceptual" arguments that if there was a field the charges would move and produce a field canceling this one out.
That's fine, but still I wanted to see a more concrete proof of this. I believe that it is a matter of picking $\rho$ correctly and using the correct boundary conditions. In truth I believe it all boils down to: how do we model a conductor? The concept is simple, but I mean, how the equations take form for a conductor and how using them we can show that $\mathbf{E}=0$ inside a conductor?
| You need to use Ohms law: $J = \sigma E$ which has to be added to Maxwell's equations as a bulk observation, as explained by this answer.
You can then conclude that the electric field is zero in a conductor for:
*
*perfect conductor where $\rho = 1/\sigma = 0$ and $J$ is finite
*static case where $J = 0$ and $\sigma$ is finite
| {
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If change in position over time is average velocity, why doesn't change in position over time squared equal average acceleration? For example, let's say a car is experiencing an acceleration of $1$m/s$^2$, for $6$ seconds so it goes $18$m. Now the average velocity is found through dividing $18$m by $6$s which is in line with the formula $v_\text{avg} = \frac{\Delta x}{\Delta t}$. And indeed, the average velocity is $3$m/s.
Acceleration has units of distance divided by time squared, however the average acceleration is not $18/6^2 = 0.5$m/s$^2$, the average acceleration is $1$m/s! So I have two questions from this:
*
*What exactly is that $.5m/s$ signifying? I know the kinematic equations including $\Delta x = v_0t+\frac{1}{2}at^2$ and this would allow us to find acceleration.
*Aren't the units a bit deceiving on acceleration? Maybe I'm just not super comfortable visualizing second derivatives yet but if I have an $m/s^2$ I feel like I should be able to plug in meters and seconds and get the average acceleration. And I feel like it's part how we define the units also. Because:
$$v = \frac{\Delta x}{\Delta t}$$
$$a = \frac{\Delta v}{\Delta t}$$
Substitution then gives us:
$$a = \frac{\Delta\frac{\Delta x}{\Delta t}}{\Delta t}$$
Which is different than the units seem to imply, from my perception.
| Why your computed average acceleration is wrong?
the average acceleration is defined as:
$\overline a=(v_2-v_1)/(t_2-t_1)$
where the $v$'s are instantaneous speeds. If you start with zero initial speed you can simplify it to
$\overline a= v /t $
$v$ is still the instantaneous speed at $t$. For a constant acceleration $v=at$ so in this case you recover the fact that $\overline a=a $, as expected.
But if instead of using the instant velocity $v$ you now use the average velocity $\overline v$, then you will get the wrong result:
$\overline a_{new} = \overline v /t=(x/t)/t=x/t^2 $, where again, $x$ is the instantaneous position. For constant acceleration $x=at^2/2$, and so you get $\overline a_{new} =a/2$.
So the short answer is that by calculating average acceleration as $x/t^2$ you are not really using the correct definition of average acceleration because at some point you replaced an instantaneous speed by an average speed.
| {
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Adjoint of the gauge covariant derivative Suppose $A=A_1dx_1+A_2dx_2$ is a 1-form connection in $\mathbb{R}^2$ and $D_A \phi=d\phi-iA\phi$ is the gauge covariant derivative with $\phi=\phi_1+i\phi_2$ is a complex scalar field. May I ask what the adjoint $D^*_A$ of the gauge covariant derivative? Thank you so much.
| For $D=d+A$,with respect to the usual inner product on $\mathbb{R}^2$ and the ones induced by it on differential forms, one has $D^{*}_{A}=-*D_{A} *$ where $*$ stands for the hodge star operator. For example, $$D^{*}_A (f_1dx_1+f_2dx_2)=-*D_{A} *(f_1dx_1+f_2dx_2)=-*D_{A} (f_1dx_2-f_2dx_1)=-*(\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}+A_1f_1+A_2f_2)dx_1\wedge dx_2=-(\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}+A_1f_1+A_2f_2).$$
To check, one considers turning off $A$, and making the test one form exact, i.e. $f_i=\frac{\partial \phi}{\partial x_i}$, and the above gives $d^{*}d$ to be the usual Laplacian operator on $\mathbb{R}^2$ but with a minus sign, which is indeed what one expects from general computations.
| {
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How exactly does string theory make general relativity and quantum mechanics compatible? Correct me if I'm wrong, but the reason that quantum mechanics and general relativity are incompatible is because the quantum foam at Planck scales renders space-time discontinuous and doesn't allow Lorentz transformations to occur. See also this Phys.SE post. How does string theory solve this problem?
| String theory does not say that GR or quantum field theory hold at those scales. It posits strings, and gets to the Planck scale and predicts what it might look like, foam or stringy things arising and changing and so on. At lower energies it is consistent with quantum field theory and GR. So, GR is a low energy description, and does not worry about the Planck scale. String theory also has a model for the graviton as some kind of string vibration or twist.
Unfortunately there are few exactly solvable models in string theory, and so a lot of the calculations are approximations and perturbations. Also there are multiple versions with a very large number of parameters so nobody knows which it exactly predicts. Also, some of the most reasonable models are supersymmetric, but supersymmetric particles have still not been detected and there is some concern that it might mean the lightest ones, which should exist, should have already been detected. This is true for supersymmetry also without string theory, though string theory may have versions where this non-detection is still ok. All these uncertainties and unresolved issues in string theory goes towards making it difficult for some people to take it as a theory. Still, it is still one of two or three (one is loop quantum gravity with its own problems) theories of everything (string and more generally M theory) or gravity
| {
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Does the speed of light in vacuum define the universal speed limit?
*
*Is light the thing causing the universal speed limit to be $299\,792\,458\,\mathrm{m/s}$? So the universal speed limit would be different if light travelled faster or slower?
*Or, is $299\,792\,458\,\mathrm{m/s}$ the universal speed limit anyway and light just goes that fast? Light is just something we commonly associate with it because it goes super fast.
| The numerical value of $c$ does not have any fundamental significance. Rather it is the number we get based on the experimental fact (according to the number & unit system employed) . If some alien civilization ended with some different value of $c$ compared to us. Even that is not a problem. They will reach the conclusion that this is upper bound of the speed limit for any object, provided both civilizations governed by the same set of laws of physics. In that sense the speed of light is universal.
| {
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Conceptual doubt in Tension force I recently studied that Tension in a string is a kind of force originated from electrostatic attraction between the atoms of the string in which the force is originating. My doubt was that:
Assume that I am pulling a rope with a force $F$, and the rope will develop a tension $T$ in itself and will pull me with $T$ but what about the force $F$ with which I started pulling the rope? Where would the reaction pair of this force would be felt? I know that this force is not included in the free body diagram of me (the one who is pulling the rope) as FBD only incorporates the forces acting on a body and not the ones exerted by it and hence I am kind of confused where I can find the reaction pair of $F$ being felt.
| To simplify this answer assume that the string is made up of a line of molecules so each molecule bar the end ones have only two nearest neighbours.
When there is no tension force in a string then on average the molecules which make up the string are at their equilibrium spacing and have no net force acting on them.
Imaging that you apply a force $F$ on the first molecule in the string.
That molecule will exert an equal but opposite force on you - Newton's third law.
That first molecule is attached via a bond (electrostatic interaction) to its neighbouring molecule.
The spacing between the molecules increases and the neighbouring molecule exerts a force on the first molecule.
In turn the first molecule exerts a force on the second molecule - Newton's third law..
Those forces between the molecules we call the tension in the string.
You can liken the situation as having a line of point masses initially being connected by unextended springs.
Pull the end mass and the springs extend thus producing forces between the masses.
If the string is held at a fixed point at the other end and nothing is moving (static equilibrium has been established) then all of the molecules are separated by a distance greater than their equilibrium separation and so have forces on them due to their nearest neighbours but the net force on each of those molecules is zero.
The first molecule which you are pulling with a force $F$ has a force of the same magnitude but opposite in direction acting on you which you call the tension $T$.
The first molecule has a force $F$ acting on it due to you and a force $T(=F)$ acting on it due to the second molecule.
| {
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What would occur between the parallel plates of the capacitor? In capacitors dielectric materials are sometimes inserted between the parallel plates. What would happen if a diamagnetic, or a paramagnetic material is inserted between the parallel plates of a capacitor?
| As long as neither diamagnetic or paramagnetic material acts as a conductor (no flow of charge between plates below breakdown voltage), it will still behave as a capacitor.
The dielectric polarisability of the material will however have an influence on the total storage of charge and charging profile of the capacitor, as the material can store additional charge through polarisation under the applied field. The degree and nature of this influence will depend on the specific nature of the material.
| {
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Multiplicity Identity in Kittel's Thermal Physics On page 25 of Kittel's Thermal Physics, the author derives the multiplicity of $N$ harmonic oscillators with total quanta of energy $n$, $g(N,n)$.
He writes
\begin{align}
g(N,n) &= \lim_{t\rightarrow 0} \frac{1}{n!}\left( \frac{d}{dt}\right)^n \sum_{s=0}^{\infty}g(N,s)t^s\\
&= \lim_{t\rightarrow 0}\frac{1}{n!}\left(\frac{d}{dt}\right)^n(1-t)^{-N}\\
&=\frac{N(N+1)(N+2)\cdots(N+n-1)}{n!}.
\end{align}
I understand everything after the first equation but I fail to see where the first equation comes from. I've tried expanding out the derivatives and summation but I still can't get it. How can I derive the first equation?
| I figured it out. If you pull out the summation out front, everything except the $s=n$ term vanishes. The terms with a higher power than $n$ vanish when taking the limit while the terms with a lower power than $n$ vanish when taking the $n$th derivative.
However, it would be great if someone can come up with a better, more constructive way of deriving that formula.
| {
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Notation about basis of gamma matrices in $4d$ In Quantum Field theories, we encounter gamma matrices a lot.
Reading from various textbook, i encountered some textbook use different basis for their gamma matrices.
Gamma matrices are defined such that $\gamma^{a}\gamma^{b}+\gamma^{b}\gamma^{a}=2\eta^{ab}$. Multiplying them in all possible way furnish the following list
\begin{align}
\{ \Gamma^A \} = \{1, \gamma^{a_1}, \gamma^{a_1 a_2}, \cdots \gamma^{a_1 \cdots a_d} \}
\end{align}
with $a_{1}<a_{2}<a_{3}\cdots<a_{d}$.
where $d$ is the dimension of spacetime for given gamma matrices.
Applying above it for $4d$ i have
\begin{align}
\{ \Gamma^A \} = \{ 1, \gamma^{a_1}, \gamma^{a_1 a_2}, \gamma^{a_1 a_2 a_3}, \gamma^{a_1 a_2 a_3 a_4} \}
\end{align}
In usual qft textbook, writes
\begin{align}
\{ \Gamma^A \} = \{1, \gamma_5, \gamma^{a_1}, \gamma_5 \gamma_{a_1}, \gamma_{a_1 a_2} \}
\end{align}
I know they are equivalent, $i.e$,
\begin{align}
&\gamma_5 \propto \gamma^1 \gamma^2 \gamma^3\gamma^4 \propto \textrm{four product of gamma}\\
& \gamma_5 \gamma_{a_1} \propto \textrm{three products of gamma}
\end{align}
What i am interested is instead of writing the first one modern qft textbook prefers to write the second form. Is there any reason for that?
I think it might be just a matter of convention, like eastern or western approach of metric $(-1, 1, 1, \cdots, 1)$, $(1, -1, -1, \cdots -1)$, etc
| I'd argue that the root of this is that different Gamma matrix bases give you different "good" features, and while the choice is equivalent (and amounts to a choice of basis for your Dirac spinor), which choice is "right" depends on what features you want to be obvious/trivial -- particle/antiparticle, left/right handedness, ease of performing Legendre transforms, etc. Which of these features you want to expose depends on what type of research you want to do, so different textbooks will likely reflect choices that optimize study in the hot research subfield at the time of writing.
Ultimately, however, the only thing that is physical is that the Gamma matrices obey the Clifford algebra.
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How do I calculate a upward-pulling force on a swinging pendulum? I'm trying to implement a simple pendulum using a 2d physics system that can model rigid bodies with gravity. The problem is that I don't know how to calculate the upward-pulling force of the rope, as in this image:
I've only found equations for calculating velocity of the pendulum, as shown on wikipedia, but the problem is that I can't change the velocity directly, as I need to be applying a force.
The problem is probably just the magnitude of the force vector. Since in the idle position the upward force is equal to negative gravity, I thought I could simply take something like $cos(\theta) \cdot v \cdot G$ where $v$ is the pendulum vector and G is gravity magnitude, but that doesn't work in my simulation.
| A properly drawn free-body diagram will have a tension force vector acting along the line of the rope, toward the pivot point and a gravity vector acting straight down. If you establish a coordinate system which is instantaneously parallel and perpendicular to the rope, you then will decompose the gravity vector into two components ( $mg$ times trig functions of the angular position, $\theta$). Usually, $\theta$ is measured with respect to the vertical.
The vector sum of the components parallel to the rope must equal $m\frac{v^2}{l}$ where $l$ is the length of the pendulum.
$$ F_{tension}-mg\cdot\text{trig_function}(\theta)=m\frac{v^2}{l}
$$
The gravity component perpendicular to the rope must be $ml\alpha$ where $\alpha$ is the instantaneous angular acceleration, $\alpha= \dfrac{d^2\theta}{dt^2}$.
| {
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Temperature distribution in a column of air Say there is a sealed cylinder of air that has a height $\mathrm{h}$ and area $\mathrm{A}$ on the ends. The initial temperature throughout the column is $T_0$ and has a uniform initial density $\rho_0$. If the bottom of the cylinder is at sea level, what is the temperature at the top of the cylinder when the system reaches equilibrium?
As we know, hot air rises and cold air sinks. So it stands to reason that the bottom will be cooler then the top in a very tall air column. However is there a formula?
| CAVEAT - I am giving a possible calculation, but I believe the answer may be off by a factor 2x (compared to the lapse rate observed in the atmosphere). I am leaving it here for you to ponder. Perhaps it can inspire you to find the correct solution yourself. Or perhaps the difference is due to the fact that this calculation doesn't assume convection - so that the adiabatic expansion terms in the derivation of lapse rate don't apply.
If the column of air is sealed we should probably assume there is not much air flow. In the steady state, we can use conservation of energy to solve this.
First - assume that the sum of (mean) kinetic and potential energy of the molecules is constant, independent of height. We know that at a given temperature the kinetic energy of a light diatomic gas (like most of the components of air) is
$$KE = \frac52 k_B T$$
Where the factor 5 comes from 5 degrees of freedom (3 translation, 2 rotation). Now the potential energy for a molecule of mass $m$ is $PE = m~g~h$. For air we will use an "average" mass of 28.8 amu (20% oxygen, 80% nitrogen; ignoring CO2, water, argon, ...). If the sum of $KE+PE$ is constant, then
$$m~g~h + \frac52 k_B T = \rm{C}$$
This means that there will be a linear change in temperature with height:
$$T(h) = T(0) - \frac{2mgh}{5 k_B}$$
Putting in numbers, we get
$$T(h) = T(0) - 0.014 h$$
which results in a temperature change of 1 K for every 70 meters. In reality, the slope in the atmosphere (according to this NASA page) is about 0.00649 K/m
That's about a factor 2x off from my calculation. I don't know what simplifying assumption I am making (or whether there is simply an arithmetic error in my work).
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Transforming Qubits Into Bits From what I understand, a qubit exists in a superposition of states and once it has been measured, it must fall into one of the two possible states. Now, I have been told that once a qubit is measured, it is no longer proper to call it a qubit but a bit since it no longer exists in a superposition of states. Is this correct?
Along the same lines, if a photon with unknown polarization (the polarization state can be our qubit) hits a polarizing beam splitter, then its no longer exists in a superposition of states but must be either horizontally or vertically polarized. So would this mean that the polarization no longer is a qubit, but a bit, since it can only exist in one of two states? This would not make sense because many regimes for experimentally realizing quantum logic gates involve polarizing beam splitters. So if my reasoning is correct, that would mean that in the gate itself the qubit actually is no longer a qubit, but a bit.
One final thing, since measuring a qubit is inherent to a functional quantum computer, does this mean that quantum computers actually use bits as well as qubits?
|
One final thing, since measuring a qubit is inherent to a functional quantum computer, does this mean that quantum computers actually use bits as well as qubits?
The entire advantage of a quantum computer lies in the use of qubits. I've implemented a quantum computing algorithm(Grover's Search) on a classical computer before, and it was incredibly slow compared to classical alternatives.
It isn't accurate to think of a qubit as a bit while it is in a classical state. Even in a classical state, the physical implementation of a qubit is far more complex than that of a bit, and it can be impelled into a superposition at any time using a Hadamard gate. So no, quantum computers do not use bits. They may interact with a classical computer that uses bits, but that is a different story.
| {
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Distance between adjacent planes in a crystal This question has been asked before, but there doesn't seem to be a decent answer.
Many sources state that " For cubic crystals with lattice constant a, the spacing d between adjacent (ℓmn) lattice planes is:
$$ {\displaystyle d_{\ell mn}={\frac {a}{\sqrt {\ell ^{2}+m^{2}+n^{2}}}}}$$
"
https://en.wikipedia.org/wiki/Crystal_structure
Could someone please explain what "adjacent" means in this case (Is it planes that share the same side, is it parallel planes, are these panes in the same unit cell or neighbouring cells etc)? Better yet, does anyone know of a sketch explaining this ?
I am really at a loss here and this has been driving me nuts the whole day
|
I could only find this poor-quality picture. It should give you an idea, anyway. For example, consider the first picture in the first row: $(l,m,n)=(1,0,0)$ in that case, and it is easy to verify that the distance between the grey planes is
$$d=a$$
In the second case, $(l,m,n)=(1,1,0)$, and you can see that
$$d=\frac a {\sqrt 2}$$
etc.
| {
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Does the universe expand in every direction evenly? I've heard that the universe is expanding constantly and that galaxies are moving further and further away from each other because of this. However, does the universe expand in every direction evenly or does it expand in one direction more than another direction?
| Short answer: Yes
Explanation: The answer to this question is something well documented in astrophysics. The "Size" of a universe is modeled by metaphorical expanding fluids known as the Freidman Equations. These equations say that from a singular point, the universe will expand at rates according to the travel of its components: energy and matter, for example, moving at different speeds. But each of the uniform types of matter expands with equal speed, so it becomes a matter of proportions to figure out the speed.
| {
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How we chose the height while calculating potential energy? I'm really confused how to chose height when calculating potential energy. If an object is right above a desk, suppose the will we take height from desk?
If we take it from a height $x$ from the desk and the level of desk is $y$ from the ground and we change the position of the object such that now the object is directly above the floor. So will the potential energy change as height changes and if so, we know that potential energy is stored energy.
Will the stored energy increase?
| Potential energy stored in a body is relative. We have to first choose the potential at a finite point or infinity. In the case given above, we take potential energy to be 0 at the centre of the Earth. So according to the relation, $PE = mgh$ where $h$ is the height from the centre of the Earth. Generally, we take height from the surface of the Earth and take $9.8 m/s^2$ as the acceleration due to gravity (the $g$ at the surface). Therefore, in your case Potential energy will be $PE = mg(x+y)$.
And since height is directly proportional to the potential energy, as height increases, potential energy increases.
| {
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Are black holes perfect spheroids? What I know about black holes (correct me if I'm wrong) is that they are the most compact objects in the universe that have been discovered. Due to all that gravity, wouldn't black holes be a perfect spheroid, sort of like planets are spheroids (due to centrifugal forces)? Can you measure the geometry of a black hole due to its power of warping space-time itself?
| The shape of a black hole's event horizon depends on who is asking. Observers who are moving quickly towards a hole, for example, will see a different shape compared to those who are not.
Per @benrg the event horizon of a static black hole is not observer dependent, similarly to how the shape of an expanding flash of light is not.
In the coordinates appropriate to very distant "inertial" observers, the event horizon of a nonspinning uncharged black hole in equilibrium is spherical. If the hole is rotating, the event horizon will bulge out along the equator like any other rotating object.
| {
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Can we exit the event horizon of merging black holes? I have an intuitive scenario. Consider we have a spaceship just below the event horizon of a BH, which is merging with another black hole.
Finally, the singularities merge and we have a single black hole again.
But, in the transient stage, it is unclear to me if a timelike world-line would exist to leave the system.
I suspect, the metric is probably far too complex for an analytical solution, but in the worst case, it could be maybe solved numerically.
As far I know, black hole merges are examined mainly in an inspiral scenario. I suspect, maybe the escape is possible only if they have a hyperbolic-like orbit (i.e. there is no inspiral, but they simply collide).
Is it possible?
| Classically speaking, since total BH evaporation takes finite time, yet being at the event horizon stops all time (in your IRF), the black hole will evaporate before you can get to the event horizon.
| {
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What frequency is the scratching of finger nails on a blackboard? This is the frequency/intensity that sets my teeth on edge.
Does anybody know what frequency (roughly) it is? I am guessing it is near the top of normal human hearing, 20kHZ, but I'm not sure if that's why it affects me.
I am sure the same frequency is played on some of the music I listen to, but somehow, it does not make me wince.
There is a related question here, with no answer Scratching on a Blackboard, but I just want a frequency value.
| In addition to the other answer by Cort Ammon, I have heard of other psychophysical/evolutionary explanation:
The frequency distribution of that sound closely corresponds to the frequency of a crying baby, which has been shown to drive people crazy when exposed to it for a short amount of time (we are genetically predisposed to get distressed by such a call). In this case the specific frequency distribution of nails in a chalkboard is better as stimulating those neural pathways than the stimulation produced by the original sound.
| {
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How does the LHC explore extra dimensions? The Large Hadron Collider (LHC) has been smashing particles for a long time and sometimes people say that it has found new dimensions. How is it even possible for a particle accelerator to find new dimensions?
| It is important to understand how experiments work. With very few very very basic exceptions, all experiments and their measurements involve a theoretical framework.
Fact is, we almost never measure things explicitly. For crude examples consider:
*
*Temperature: a mercury thermometer measures length (that of the column of mercury). An electric thermometer measures either voltage, current, or resistance.
*Speed: we usually measure position and time, and calculate the speed. The speedometer in your car very likely measures current.
With modern cutting-edge experiments, there is a lot more theory behind. The measures themselves are just numbers that the machines output (and again, the actual sensors are probably just measuring electric current, magnetic charge, etc., and for this they use the accepted electro-magnetism theories, say, which are very well established but they are theories, not facts). This data is processed by software designed following the principles of the accepted theory. Said like this, it doesn't sound very exciting, but it definitely has its merits.
In any case, experiments either confirm the numbers predicted by the theory, in which case they contribute to the theory's standing, or they contradict the theory, and then the theorists need to work on understanding what is wrong with the theory.
The bottom line is that experimental gadgets almost never measure directly the effects they intend to measure. They will measure some consequences of the effects, which are then studied according the accepted theories.
| {
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Is a light wave's amplitude stretched, along with the "red shift" stretch - making it brighter? When light waves are stretched and "red-shifted", is the amplitude of the light wave stretched as well, affecting the intensity/brightness of the light wave?
| Light waves are not necessarily stretched. The formula
(frequency measured by the observer) = (speed of the light relative to the observer)/(wavelength)
suggests that the frequency shift (blueshift or redshift) can be due to the variation of the speed of the light relative to the observer, not to wavelength change. When the initially stationary observer starts moving towards the light source, this is obviously the case:
http://www.einstein-online.info/images/spotlights/doppler/doppler_static.gif (stationary observer)
http://www.einstein-online.info/images/spotlights/doppler/doppler_detector_blue.gif (moving observer)
I believe that the frequency shift is ALWAYS due to the variation of the speed of the light relative to the observer (the wavelength of the traveling light never changes).
| {
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Largest Mass Diffraction I have read "Matter-wave interference with particles selected from a molecular library with masses exceeding 10000 amu" which claims to observe diffraction patterns in objects of around 10'000 amu. What is the largest mass objects shown to have diffraction patterns and show wave-particle duality? I have heard a claims of this type have been shown for small amino acids, and possibly protein strands or even small viruses, but have struggled to find any references.
| Yes, up to now, that is the paper showing the largest-mass particle interference experiment. The possibility of using larger particles, like small viruses and other kinds of biomaterials are discussed in this paper:
http://iopscience.iop.org/article/10.1088/0031-8949/91/6/063007/meta
This is the most up to date reference I can find on the topic.
There is also a recent blog post here that is relevant and might be of interest to you:
https://thiscondensedlife.wordpress.com/2016/06/24/schrodingers-cat-and-macroscopic-quantum-mechanics/
| {
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Satellite revolving problem gives two different answer Assume there's a satellite revolving about the Earth. If I would like to decrease its radius, should I increase or decrease its velocity?
I know the answer apparently should be decreasing its speed, but the following two formulas give different answers. Can someone explain why two formulas give two different answers?
$r = mv² / F$, where r and v are directly proportional
$v = √(GM / r)$, where v and r are inversely proportional
| Your question is puzzling but I think i have understood the answer.
When the satellite undergoes slow down (due to friction or something) it comes closer to the earth and gain additional velocity. This velocity is same as in your equation and as in @sammy gerbil 's answer.
$v^2=GM/r$
Friction reduces the energy of the satellite and the energy of the satellite is
E=KE+PE or
$E=\frac{1}{2}mv^2-\frac{GM}{r}= -\frac{GM}{2r}$
Hence after slow down the satellite comes closer to the earth and gain velocity, although the final equilibrium velocity (after the slow down) is higher than the initial velocity (before the slow down) the final energy is lower than the initial energy. I think this will solve the contradiction.
| {
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Using fermion-based analog computers to solve NP-hard problems in polynomial time If the fermion sign problem is an NP-hard problem as it seems to be proved by this work, is it possible to take an NP-hard problem, convert it into an equivalent fermionic state evolution problem, prepare the system physically, let it evolve, average over many experiments, and expect the result to converge to the solution of the original problem in a reasonable time? (i.e: less than exponential time)
| Scott Aarson gives examples of problems that might look like enabling such a reduction procedure, but fail in the paper 'NP-complete Problems and Physical Reality'.
Can NP-complete problems be solved efficiently in the physical universe? I survey proposals
including soap bubbles, protein folding, quantum computing, quantum advice, quantum adiabatic
algorithms, quantum-mechanical nonlinearities, hidden variables, relativistic time dilation,
analog computing, (...)
On classical approaches:
(...) There are other proposed methods for solving NP-complete problems that involve relaxation to
a minimum-energy state, such as spin glasses and protein folding . All of these methods are subject
to the same pitfalls of local optima and potentially long relaxation times.
On quantum computing:
(...) In other words,
there is no “brute-force” quantum algorithm to solve NP-complete problems in polynomial time,
just as there is no brute-force classical algorithm.
http://www.scottaaronson.com/papers/npcomplete.pdf
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Is this constraint holonomic or non-holonomic? $$f(q,q^\prime, t) = 0, ~\mathrm df = \frac{\partial f}{\partial q}~\mathrm dq + \frac{\partial f}{\partial q^\prime}~\mathrm dq^\prime+ \frac{\partial f}{\partial t}~\mathrm dt = 0$$
I really want to know whether this constraint is holonomic or non-holonomic.
(As far as I know, Non-holonomic constraint has a term of velocity and do non-integrable. But this formula does not dependent on a path, because it is a total differential form.)
*
*prime is a time derivative.
| *
*Firstly, recall that virtual displacements don't change time $\delta t=0$. Time is fixed, say $t=t_0$.
*Secondly, let $M$ be the position space, often call the configuration space, with generalized position coordinates $q^i$.
*OP is right that the constraint $\delta f(q,v,t) \approx 0$ doesn't depend on virtual displacements $(\delta q,\delta v)$ in the tangent bundle $TM$ (with appropriate boundary conditions imposed). However, a velocity-dependent constraint $f(q,v,t) \approx 0$ is not well-defined on the position manifold $M$ itself. And it is down in the base manifold $M$ where the applications, such as, e.g., d'Alembert's principle take place.
*TL;DR: A holonomic constraint has by definition no velocity dependence.
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Why can any general motion of a rigid body be represented as translation + rotation about center of mass?
*
*Why can any general motion of a rigid body be represented as translation + rotation about center of mass?
*I am beginning to read rotational dynamics and my textbook states this fact without proof. I am wondering - Is this fact only true for a center of mass?
*Then - The phrase "rotation about center of mass" strikes as vague to me. Rotation about which axis?
| If you want to describe the position of a rigid object in space, it is clearly not sufficient to give just the position of the center of mass - you also need to specify the orientation.
That orientation can be reached with a rotation about the center of mass - but you need to figure out what the direction of the rotation axis has to be, and what the angle is through which you rotate.
You need pick any three points in the object (not on the same line) in order to describe how it was rotated; that is the necessary and sufficient number of parameters. You can then write three equations in three unknowns, and solve for three parameters - exactly the number of parameters needed to describe an axis of rotation (2 parameters) and angle of rotation (third parameter).
You could pick a rotation about a different point than the center of mass - and while it is possible to describe motion with any 3+3 parameters, it is MUCH harder when the axis is not going through the center of mass (because, for example, there will be an apparent centrifugal force in the rotating frame of reference because the center of mass is not on the axis of rotation).
| {
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Ground state of local parent Hamiltonians and invariance under local unitaries Assume that a finite-dimensional pure state $|\psi\rangle\in \mathcal{H}\simeq \mathbb{C}^m$, $m<\infty$, is the (unique) frustration-free ground state of a local parent Hamiltonian and suppose that the locality notion is given in terms of a connected set of neighbourhoods $\{\mathcal{N}_k\}$. My question is the following one: Is it true that any unitary $U$ satisfying $$U|\psi\rangle\langle \psi|U^\dagger=|\psi\rangle\langle \psi|$$ can be decomposed into a finite product of invariance-satisfying unitaries acting only on the neighbourhoods $\{\mathcal{N}_k\}$, that is $U$ can be written as $U=\prod_{i=1}^N U_{\mathcal{N}_{k_i}}$, where every $U_{\mathcal{N}_{k_i}}$ acts only on the neighbourhood $\mathcal{N}_{k_i}$ and it is such that $U_{\mathcal{N}_{k_i}}|\psi\rangle\langle \psi|U_{\mathcal{N}_{k_i}}^\dagger=|\psi\rangle\langle \psi|$
?
Any (partial) answer/comment/reference is very welcome.
Thanks in advance.
| Consider the toric code Hamiltonian situated on a spherical geometry. This has a unique ground state. Expand the sphere to an infinite radius. Consider a string excitation of the ground state, and loop the string around the sphere (an infinite number of local operations) such that it meets itself, returning our system to its ground state. We know by analogy to the topologically protected ground states of the toric code on an infinite toroidal geometry that such an evolution is not possible with a finite number of local operations. Thus we have described a unitary evolution of the system, with the ground state as its eigenstate, which can not be expressed as a finite decomposition of local operations.
That being said, I may just be cheating by how I am taking the thermodynamic limit, and you may have to clarify such considerations in your question.
| {
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What is the Continuity Equation in QM? I have an exercise for my homework that mentions the "continuity equation". Don't tell me how to solve it please, just tell me what the continuity equation is. I tried googling it but I couldn't find anything exact. Just stuff mentioning the probability current. The question is:
Show that the following expression satisfies continuity equation:
$$\xi=\psi^*\psi+\frac{\partial}{\partial x}(\psi^*\psi). $$
| So the continuity equation is usually written as
$$\frac{\partial\rho}{\partial t} + \nabla\cdot \mathbf{j} = 0$$
where $\rho \equiv \rho(\mathbf{r},t) = |\psi|^2$ is the standard probability density and $\mathbf{j} = \frac{\hbar}{2mi}\left(\bar{\psi}\nabla\psi - \psi\nabla\bar{\psi}\right)$ is called the probability current. Just wanted to write this out so you can better understand what you are reading if you look online.
Now in your problem they have defined $\xi \equiv |\psi|^2 + \frac{\partial}{\partial x}(\bar{\psi}\psi)$ which is very close to $\rho + \mathbf{j}$ but not quite the same. What they want you to do is show $\partial_t\xi + \partial_x\xi = 0$
| {
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To effectively suppress normal audible sound, how wide and how absolute would a vacuum space have to be? I don't know if this question is too specific or simple, but: to effectively suppress (say 99%) of normal audible sound (say 20-20kHz @ 100dB), how wide (mm?) and how absolute (torr?) would a vacuum space have to be?
| This is really a nice question. After reading your question I have searched the web with the idea that sound is a pressure wave and in order to hear them it must transfer energy to your ears. I have found this article which is really nice. It says that (and so was my intuition) the energy density of the sound wave is directly proportional to the density of the medium. Hence if you decrease the density of the medium the energy stored in the medium decreases and less energy is transferred from that medium. Hence if you want to decrease the sound intensity by a factor of ~100 decrease the pressure to ~10 mbar.
Now if you have a vacuum curtain (i.e. suddenly the pressure of a portion of the medium is decreased) then the sound will not propagate. Now question is how thin we can make this curtain. I believe that the sound wave setup an oscillation in the gas molecules and the extent of this oscillation is of the order of wavelength. Hence if you want to effectively suppress the sound wave then the thickness of the vacuum curtain must be larger than the sound wavelength (or else the molecules will overshoot the curtain and transfer energy at other side). It may be noted that this analogy of free vacuum curtain is adopted to avoid any other type of effects i.e. sound propagation through the metal enclosures (I can hear sound of motors from inside the vacuum chambers) or scattering of sound waves from such enclosures.
| {
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Why aren't trigonometric functions dimensionless regardless of the argument? Consider this equation :-
$$y = a\sin kt$$
where $a$ is amplitude, $y$ is displacement, $t$ is time and $k$ is some dimensionless constant.
My instructor said this equation is dimensionally incorrect because the dimension of $[kt] = [\text{T}^1]$ and since $\text{angles}$ are dimensionless, we can conclude that it is dimensionally incorrect.
I don't understand why it is so. Why do we need to check the dimension homogeneity of the term inside the $\sin$ to conclude whether the equation is dimensionally correct or not?
Why isn't the whole sine function is dimensionless $(\sin kt = \text{[T}^0]) $ regardless of the dimension of the argument inside as the range of sine function is $[-1, 1]$.
| The quantity $kt$ is dimensionless.
If $t$ has the dimensions of time then the dimensions of $k$ must be $\text{time}^{-1}$ so your series expansion works.
You will meet this idea again and again in Physics and checking the dimensions often is a good way of checking a derivation.
The charging and discharging of a capacitor $C$ through a resistor $R$ has a term $e ^{-\frac {t}{RC}}$ and so if $t$ is a time then $RC$ will also have the dimensions of time so that $\frac{t}{RC}$ will be dimensionless which would not be case if the derived expression contained $\frac{tC}{R}$.
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Why is the surface of a liquid slanted when it is accelerated? Consider a uniformly horizontally accelerated tube of water. I know that the fluid experiences a pseudo force in addition to its own weight, so that it reaches equilibrium in the below diagram.
But why can't the water also exert a force like this, so it can be in equilibrium horizontally?
| The reason that the second diagram you drew cannot represent what is happening is that it will not satisfy Newton's 2nd law for all parcels of fluid in the tank. Imagine that you had a tank like the one shown in the diagrams and, rather than accelerating it, you just tilt it at an angle so that base is no longer horizontal. Basically, what you ave done is change the direction of gravity relative to the sides of the container. Would you expect the water surface to remain parallel to the base of the container, or would you expect it to be horizontal again (but tilted relative to the base). What you have done in the acceleration experiments is to add a pseudo-gravitational force component in the direction opposite to the acceleration. So now, the effective gravity is no longer pointing in the vertical direction. Thus, the surface of the fluid must readjust to again be perpendicular to the new effective gravitational direction (which is not vertical).
If you do a force balance on a small parcel of fluid within the system having sides dx, dy, and dz, the force balance in the y (vertical) direction reduces to:$$\frac{\partial p}{\partial y}=-\rho g$$The force balance in the x (horizontal) direction reduces to:$$\frac{\partial p}{\partial x}=-\rho a$$where a is the acceleration. The variation of pressure with position is given by:
$$dp=\frac{\partial p}{\partial x}dx+\frac{\partial p}{\partial y}dy=-\rho adx-\rho g dy$$ It follows from this that the surfaces of constant pressure are given by:$$\frac{dy}{dx}=-\frac{a}{g}$$ The free surface is a contour of constant pressure.
| {
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How is potential energy lost when a water droplet is dropping down slowly on a wall? When a water droplet is on a vertical wall, it usually drops slowly, which is different from free falling. As the dropping speed is slower than free drawing, so I guess some energy must be lost.
I guess it is lost as internal energy, but if it is true, how a water droplet gains internal energy when dropping slowly in microscope view?
| Energy lost because it create a layer of water as it slides down, and layer created due to the viscous force between the liquid - liquid layer and liquid - container layer.
What happens is the fluid experience, s a Shear force as it slides (tangential) , and this Shear force is cause of energy. Loss, as Water flows down drop become small and energy lost will continue till the size of drop become negligible.
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Superpotential is Supergauge Invariant? I was studying by the X. Tata & Bauer and I'm stuck with something:
In Chapter 6 - Supersymmetric Gauge Theories, it states that the superpotential is already invariant by a supergauge transformation, and explains that is because "it's polynomial in the chiral superfields".
I'm thinking about it, but I guess i'm too tired or something, but I can't prove that every polynomial function of the chiral superfields will be supergauge invariant....
Could someone give me a tip?
| A chiral superfield with variable change $x_\mu\rightarrow x_\mu+ \frac{1}{2}\bar\theta \gamma_5\gamma_{\mu}\theta$ will have similar gauge transformation as it's components and if the action does not contain derivative or complex conjugate of the chiral superfield and contains a polynomial part then under a local gauge transformation involving $exp(it_A\Omega_A(x))$ action will be invariant if it is invariant under global transformation with $\Omega$ independent of $x_\mu$.
| {
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What is optical density? I'm a zoology minor and we are doing protein estimation by colorimetric method. I have stumbled upon a term 'Optical density'. I don't understand the term well. Is it a measure of the extent of light that can pass through a particular object?
I've checked a related question of this community and it doesn't solve my question completely.
| For all intents and purposes, OD is the negative of the order of magnitude the factor by which the intensity of the light is reduced by the attenuating element with said OD.
In other words: OD = 6, means that the intensity will be reduced by a factor of 10 to the power of -6, a.k.a by a factor of a million.
| {
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Spring force on both sides of spring I am a little confused about springs. I just wanted to know that if I pull an ideal spring of spring constant $k$ such that the spring has been symmetrically pulled and its elongation (total) comes out to be $x$ then would the force on one side by $$F=kx$$ or $$F=kx/2$$ I am a little bit confused and hence I resorted to ask it here.
| When you state $F = k x$ the variable $x$ represents the extension of the spring and not the position of one of the ends.
In a symmetric pull each end moves by $ \frac{\delta}{2} $ then the total displacement is $\delta$. The spring force is then $F = k \delta$ on each end of the spring.
| {
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Why magnetic component of light is not shown in polarization diagram? Does polarization eliminate magnetic field? So far I knew about light-polarization is like this ...
in an event of a plane-polarization, the polarizer-crystal does NOT separate the electric-component and magnetic component of the light. The magnetic component after polarization is just Not drawn to avoid some (?) complications. So ,
Fig 1. the vertical arrows is just an abbreviation showing the vertical electric field, and not showing the horizontal magnetic fields. (in fig. 1 the dot-sign indicates we're viewing the ray coming towards our eye). (In the right-side, realistic image, i did not shown double-headed arrow b'coz i've shown condition at one moment)
So, if we could draw the plane-polarization event including the magnetic field, it should look like this.
Fig 2.
If we could show the magnetic field, it should look like this. Fig 2a when we would see from side, fig 2 b when we would see from front. (in fig 2b i didn't used double-headed arrow because i shown condition of one moment.)
I can't recall exact source from my memory... it would be my college chemistry classes.
Now a geology college-student is telling the above-thing is completely wrong. According to them,
The polarizing-crystal actually filters-out the magnetic-component and allows only the electric-component only, like this-
Fig 3.
Only electric-component coming-in.
Both of us searched internet for hours make the dispute clear, but I could not display them any diagram that displays that , after being filtered through a polarizing crystal, the wave retains both the components. So they didn't believed this and stood on the same point (fig. 3)
So, My question is, which-one of the concept of linear-polarization (from above 2) is correct? If both is wrong, then what would be the correct concept?
| You are (mostly) right, your geology friend is wrong.
As stated in the comments, an electromagnetic wave can not consist of only an electric nor of only a magnetic field. They go hand in hand.
| {
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Does empty space have energy? My physics friend suggested that "the answer to why matter exists in the universe" is because all massive particles are just the fabric of space excited into little packets. To illustrate, imagine a blanket on the ground. Then, pinch a small bit of the blanket and twist it. This is a particle that has mass.
It was intriguing to hear this (he's only studied up through Freshman year of college physics), but there are clear flaws (i.e. angular momentum of a "particle" tied to a "blanket"??).
Regardless, it made me wonder about vacuums. Is there any theory that suggests that a vacuum actually has energy in some form or another?
| Yes empty space does have energy.When you apply quantum mechanics and special relativity, empty space inevitably has energy. The problem is, way too much energy. It has 120 orders of magnitude more energy than is contained in everything we see!
| {
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Finding an equivalent system of an infinitely long cylinder with polarization vector $\vec P = P\hat y$ I am given an infinitely long cylinder, with the axis of symmetry on the $z$ axis, $y$ is right to left, and $x$ towards us.
It is polarized with $\vec P = P\hat y$.
My solution manual states that an equivalent system is two cylinders, one with density $\sigma_B=P $ and one $\sigma_B=-P$.
I don't understand how exactly they arranged the two cylinders.
One on top of the other? Or are they next to each other? And what is that based on?
And if they are next to each other, which one do I assign the positive or negative charge density
| The polarization will cause a displacement of all the positive charge relative to the negative charge in the y-direction, so that a top view of the cylinders will look like this:
Since the polarization vector points in the positive y direction, the electric dipole moment vector will point the same way, which tells you that the negatively charges cylinder is to the left and the positively charged one to the right ($\vec{p}$ points from negative to positive charge)
Now, all you have to do is find the potential using the Legendre polynomials (at least that is how Griffiths solves it), then take the gradient to find the electric field both outside and inside the cylinder.
Side note:
However, there is a neat trick that you can use to find the electric field due to the cylinder (I find it is relevant to your question, and I want to put it out there) Note, however that this can only be used when the polarization is constant in one direction, as in your case. You can exploit this constant polarization that effectively displaces all negative and positive charges by the same distance relative to each other. Then, you treat the situation as that of two separate cylinders with a constant volume charge density ($\rho_0$) Let the radius of the cylinder be R
Inside the cylinder
$\vec{E}$=$\frac{\rho_0 r}{2\epsilon_0}$ (General formula for E-field inside uniformly charged cylinder)
So, from the two cylinders,(refer to figure)
$\vec{E_{net}}$=$\frac{\rho_0 \vec{r}}{2\epsilon_0}-\frac{\rho_0 (\vec{r}-\delta \vec{r})}{2\epsilon_0}
=\frac{\rho_0 \vec{\delta r}}{2\epsilon_0}=-\frac{\vec{P}}{2\epsilon_0}$
Outside the cylinder
$\vec{E}$=$\frac{\rho_0 R^2}{2\epsilon_0 r}=\frac{Q}{2 \pi \epsilon_0 hr}$ (General formula for E-field outside uniformly charged cylinder)
So, from the two cylinders,(refer to figure)
$\vec{E_{net}}$=$\frac{1}{2 \pi \epsilon_0 h}(\frac{Q}{r}-\frac{Q}{r-\delta r})\hat{r}=\frac{1}{2 \pi \epsilon_0 h}\frac{\vec{p}}{r^2}\hat{r}$, where $\vec{p}$ is the net dipole moment in the volume $(\vec{p}=\vec{P}V)$
$\vec{E_{net}}$=$\frac{\vec{P} R^2}{2 \epsilon_0 r^2}$
I hope I answered your question!
| {
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What is the difference between these two ways to calculate average velocity? Average velocity:
$$v_{\rm avg,1}=\frac{v_{\rm final}+v_{\rm initial}}{2}$$
and average velocity:
$$v_{\rm avg,2} =\frac{\rm total\;displacement}{\rm time \;taken}=\frac{\Delta x}{\Delta t} $$
What is the difference between them and when do we use them?
| The correct equation for the average velocity is the second equation of the both equations you gave. The first one is correct only under given condition that the acceleration of the body is constant. Second equation even holds for variable acceleration.
| {
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Can gravitational wave create anti-gravity, i.e. repulsive gravity? A very layman question as in title. Like every wave having a negative side, can a gravitational wave have anti-gravity.
To put it in different words, a gravitational wave passing through a complete vacuum, if in positive cycle, can create a denser space-time, in it's negative cycle, create a rarer space-time?
| Gravitational waves, though transverse, can be thought of as similar to sound waves:
A sound wave, as it moves through a medium the sound wave creates alternating volumes of greater and lesser particle density.
Gravitational waves do something similar, except the medium is spacetime itself. The result is that as a gravitational wave passes through a region of space, at one crest the spacetime is "stretched" in one direction and contracted in the perpendicular direction, like when you stretch a rubber band and it gets narrower. At the trough of the wave, the same thing happens, except the direction that was contracted is now stretched and the direction that was stretch is now contracted. This is why the good ol' perpendicular lasers and mirrors trick worked for detecting them.
| {
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Different net external forces If I apply an external force to an object, it gets transmitted internally and cancels out according to Newton's 3rd law. This leaves a net external force, causing the object to accelerate.
But when I attach two objects of different masses by an inextensible string and pull one object with a force, say 50 N, why does it not get transmitted throughout equally, like the internal forces, but instead shares itself between the two objects, say 40 N and 10 N, such that the acceleration is the same?
| Realistically, the redistribution of the force between the two masses is the same idea as the force being transmitted internally.
Within a single object, each individual "particle" will receive an appropriate force to cause it to undergo an acceleration that is equal to the rest of the particles within the object. Particle here referring to small parts of the object, be that molecules or atoms, the effect is the same.
As the two masses are joined by a string, so too must the total force be transmitted to all the particles in the second mass such that the entire system undergoes the same acceleration. The 10N and 40N are just the totals of the forces on each particle in each separate block, in just the same was as 50N is the total of the entire system.
On another note, if the acceleration of two masses were not the same, then either the string would need to extend, which you have stated is not possible, or the string would lose all tension and become slack, preventing one mass from accelerating until the slack was taken up and the tension restored.
| {
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Why doesn't Helium freeze at 0K? I have read that Helium does not freeze at absolute zero under normal pressures.
How could this be possible given that the absolute zero is the lowest attainable temperature and at that temperature, all random movements of the atom stop?
Shouldn't the atoms just stop vibrating and solidify instantly? Why do they possess kinetic energy at absolute zero?
| At $0K$ there is still zero point energy. As He is very light and inert the associated zero point motion this is enough to prevent solidification.
| {
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Does a higher water volume increase pressure? I am constructing a gravity flow water system. I have 100ft point where I can put my tank. My question is does the size of my tank matter? I am using a 1" pipe. Will I get more pressure if I use a bigger tank? For example what is the difference in pressure if I use a 10 gallon tank or a 50 gallon tank?
| Gravity flow pressure can be figured, or measured, simply by two methods:
1) divide the drop in elevation (in feet) by 2.31 OR........2.31 feet of drop = 1# of pressure
2) if you have a system already established, screw an inexpensive water pressure gauge on a hydrant, or hose bibb and read the pressure
NOTE: the volume of water in a tank does not increase the water pressure. Height, or amount of drop, is what creates water pressure. CAVEAT: I am a lay person, not an engineer, so can only explain this in simple terms. And, being a lay person, I have found that it is difficult to wrap your head around the fact that the volume of a tank does not impact water pressure. Only the height of the tank plays into the amount of water pressure you'll have.
Friction will reduce water pressure. The more elbows and the distance/length of the pipe will change the pressure.
| {
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Use of angular diameter to determine distance in special relativity Summary: at relativistic speeds, if you compute a planet's relative distance using angular diameter (roughly proportional to 1/angular diameter), will that computed distance increase or decrease linearly, assuming you are traveling directly towards or directly away from the planet? Example:
A ship starts at Earth, accelerates to 0.8c, and travels to a planet P 10 light years away.
The ship pilot knows the diameter of P, and uses the angular diameter to compute his distance from P.
When the ship accelerates from 0 to 0.8c, P's angular diameter does not change (is that correct?), so the pilot defines that diameter to correspond to 10ly (he's using the reference frame of the Earth/planet for this definition only).
The journey takes 7.5 years ship clock time; as the ship approaches t=7.5, P's angular diameter tells the pilot his distance is approaching 0.
Question: does the distance as computed from the angular diameter (which is roughly the reciprocal of the angular diameter) decrease linearly during the pilot's journey? The angular diameter is perpendicular to direction of flight, so foreshortening shouldn't be an issue, but I could be wrong about this.
| Acceleration changes your speed which does change the angular diameter. You'll find this in the relativity textbooks under 'stellar aberration'.
Consider a photon from the rim of the planet. It has a transverse momentum $p_T=p \sin\theta$ and a longitudinal momentum $p_L=p \cos \theta$.
When you increase your speed, in the new frame the transverse momentum is the same, and the longitudinal momentum increases to $p_L'=\gamma(p_L+v E/c^2)$. ($E$ is just $c\sqrt{p_T^2+p_L^2}$)
So the ratio $\tan \theta'=p_T'/p_L'<p_T/p_L$. The disc gets smaller.
This is confirmed by the shot in Star Wars (the first one) when the Millenium Falcon accelerates - it's right at the end of this clip https://www.youtube.com/watch?v=vNoCDvJpmPU - and the stars appear to move inwards, making smaller angles with the axis of the direction of travel.
| {
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What is a phase arrow? Feynman say's that a photon takes every path while reflecting off a mirror when going form A to B, but we only see the middle one(where incident angle = reflected angle) because all the others are cancelled out as they have longer routes and while following them their phase arrows cancel out each other.
If you can, please explain it in general terms without mathematics, as I am studying it the same way.
| Feynman's understanding of this effect probably came from optics where it is known as Fermat's principle. The mechanism behind this principle is based on the notion of stationary phase, which is the phenomenon that light will go in that direction where the phase coming from different paths varies slowly.
I'll try to explain this without mathematics. The scenario is one where there is some field that propagates linearly through space or through some linear system, as in the case where we have linear optical components. At the output, one would then find locations where the field has a large amplitude and other locations where the amplitude is small or even zero.
If one looks at a particular output point, the field amplitude at that point is the sum of all the bits of field that arrived there via different paths. The phase of these bits is determine by the length of the path that they followed to get to the output point. If this phase value changes rapidly as a function of the varying path, then the different values will tend to cancel each other. You can think of it as a bunch of arrows pointing in all directions and you are adding these arrows to find the combined arrow. For the rapidly varying case the amplitudes (lengths of the arrows) will vary slow compared to the phase (directions of the arrows). Arrows that point in opposite directions will cancel each other. So the net result is a very small amplitude.
If, on the other hand, the bits of the fields arriving at the output point, have a slow varying phase, then the arrows would all point in more or less the same direction. So they would add up to give a larger amplitude (longer arrow).
| {
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How does an off balance Load affect Direct Drive Dc Motors on a robotic "vehicle" I write the autonomous program for my local robotics team. The robot has 4 wheels directly fixed to an Andy Mark motor. there is no leveling or balancing system. due to placement of different modules with in the robot each side is under a different amount of stress due to weight. The off balanced weight causes the robot to drive slightly to the left when it should be driving straight. I believe this can be fixed by Throttling the power(thus Throttling the RPM). so I plan to place a scale under each of the motors (4 at one time) to show the weight that each motor must pull. I believe if I convert that to a percent of the total weight and subtract it from 25
(percent of load - 25(expected percent) =(percent to compensate for)
x = percent to compensate for
with this the equation would look like
Front Left Power = ((desired power value)+x)
Hopefully I was clear enough but if there is anything I'm missing or anyone has any suggestions I would appreciate it greatly. Thanks!
| The 'power proportional to weight' algorithm is unlikely to be completely correct. If your wheels deform under load, that deformation may be changing the effective
radius of a driven wheel, not only the retarding force. That would
call for load compensation of velocity, not power.
In forward acceleration, we can see from F= mA, the
power to left wheels and right wheels ought to be corrected for the
left/right mass distribution (so as not to apply torque around the
center of mass). Both left wheels can be treated equally,
and both right wheels likewise; front-back weight distribution is not relevant,
for this correction.
A differential gear, as in an automobile, is a good solution for two driven wheels and one motor. Do you need four motors for some reason?
| {
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Deceptively simple mass-spring problem? This question is inspired by two other, similar, so far unanswered questions (posed by different OPs).
Mass $m_2$ sits on a incline with angle $\theta$ that provides just enough friction for it not to start sliding down. It is connected by a massless string $S$ and perfect spring (with Hookean spring constant $k$) to mass $m_1$. Pulley $P$ is frictionless and massless. At $t=0$ the spring is not extended at all. Then $m_1$ is released.
Question:
What is the minimum $m_1$ to cause movement of $m_2$ up the incline?
Attempt:
Ignore the spring.
Determine static coefficient $\mu$ first.
\begin{align}m_2g\sin \theta &=\mu m_2g\cos \theta\\ \implies \mu &=\tan \theta\end{align}
To overcome the $m_2g$ component parallel to the inclined and the friction:
\begin{align}m_1g &\gt\mu m_2g\cos \theta+m_2g\sin \theta\\ \implies m_1 &\gt 2m_2\sin \theta\end{align}
But apparently this overestimates $m_1$. It has to be taken into account that $m_1$ starts accelerating before $m_2$ starts moving, because of the spring.
But how? Like several other members I can't see how the work done on the spring affects the minimum $m_1$. Conservation of energy?
| You have derived an equation which predicts the minimum force $F_{\text{min}}$ required to get the block $m_2$ moving up the slope.
$$F_{\text{min}} = \mu m_2g\cos \theta+m_2g\sin \theta$$
Until the tension in the string is equal to that value of force the block $m_2$ will not move so the spring mass $m_1$ system can be thought of a a standard spring-mass system with block $m_2$ being the rigid support.
The static equilibrium position of the spring-mass $m_1$ system is when $m_1 g - k l =0$ where $k$ is the spring constant and $l$ is the static extension of the spring.
When mass $m_1$ is released the magnitude of the force at both ends of the spring increases as the spring stretches.
However the mass $m_1$ overshoots the static extension position and continues on downwards until the extension of the spring is a maximum, $= 2l$, and hence the force on each end of the spring is $2k_l = 2 m_1g$ where the mass $m_1$ stops.
This is the maximum force exerted by the spring on block $m_2$ via the string and is your required $F_{\text{min}}$.
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What's the difference between Quark Colors and Quark Flavours?
Each of the six "flavors" of quarks can have three different "colors". The quark forces are attractive only in "colorless" combinations of three quarks (baryons), quark-antiquark pairs (mesons) and possibly larger combinations such as the pentaquark that could also meet the colorless condition. Quarks undergo transformations by the exchange of W bosons, and those transformations determine the rate and nature of the decay of hadrons by the weak interaction.
What's the difference between Quark Colors and Flavors, I've heard them used in the same way before. So what exactly is the difference between the three colors and 6 flavours?
| One is talking quantum mechanics and attributed quantum numbers to elementary particles.
A simple quantum number is charge and it it assigned to quarks ( and antiquarks) as +/-1/3 or +/-2/3 as in the table
Charge is connected with the electromagnetic force.
Flavor is assigned as a quantum number to each quark, and it is connected with the weak interaction.
Each quark at the same time is connected with the strong color force of quantum chromo dynamics. So it can also come in the three color quantum numbers, for identification called red blue and green ( analogous to the weak Strange Charm Bottom Top).
The identifications are not random, they are within the SU(3)xSU(2)xU(1) group representations and algebra of the standard model of particle physics.
The quark forces are attractive only in "colorless" combinations .
Attractive is a wrong attribute. Color is always attractive, but it can be "nullified" in certain color combinations so that stable bound states of quarks appear, as with the rest of the quote.
Flavor characterizes the weak interactions of the quarks. Color the strong ones.
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What exactly are "primordial fluctuations"? Are "primordial fluctuations" essentially the same as "virtual particles" and "quantum fluctuations" that created the universe from nothing like what is featured in the Lawrence Krauss book, A Universe from Nothing?
| I haven't read Krauss' book, but it seems the "quantum fluctuations" referred to is something invoked to explain the origin of the Universe itself; i.e. why is there something rather than nothing.
This is not what the term "primordial fluctuations" refers to. Although this is also quantum fluctuations, in the sense that their origin is described by quantum mechanics, their existence assumes the existence of a Universe in the first place. Quantum mechanics' uncertainty principle implies that the Universe — however it came into existence and not really considering the cause — was clumpy on very small scales. When the Universe was $10^{-36}$–$10^{-33}$ seconds old, during the epoch called inflation, these clumps grew in size to cosmological sizes, later collapsing under the force of gravity to the structures we see today as clusters, galaxies, and stars.
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Finding out how long a light bulb will light A cylindrical wire used to form a light bulb filament has radius 3.7 micrometers and length 1.7 cm. The resistivity of the wire is 5.25 * 10^-5 ohm meters. The light bulb is connected to a 12V battery.
With the given information; resistance equals 20.75 ohm meters. Current will be 0.58 amps and power will be 6.94 W.
The question I do not understand the meaning of is, if the battery has a total stored charge of 0.5 A hr, and produces a constant potential difference until discharged, how long will the light bulb light?
Can anyone explain to me what the question means?
| You've done all the hard work and now there's one little part left.
You're being asked, given a certain battery capacity in amp hours and a certain current in amps flowing out of the battery, for how many hours (or seconds) will the current flow? The assumption is that as long as the current is flowing the light will be on and as soon as the capacity of the battery is drained the light will go off.
Explanation
You know the power the bulb will use. You know the voltage of the battery, you know the total charge store in the battery in the form "amp hours" (A hr). Amp hours are an engineering term for charge and are equivalent to coulombs.
1 amp flowing for 1 hour = 1 coulomb / second flowing for 3600 seconds = 3600 C
A battery capacity of 1 amp hour means that:
*
*a current of 1 amp can flow for 1 hour
*a current of 2 amps can flow for 0.5 hours
*a current of 0.25 amps can flow for 4 hours
Notice that in each of these the current (amps) * time (hours) = capacity (amp hours): 1 * 1 = 2 * 0.5 = 0.25 * 4 = 1
A constant potential difference across the light means that the current will also be constant. In the real world as batteries discharge the voltage across their terminals decreases. This drop in voltage is due to internal reistance of the battery increasing as chemical reactions take place inside the battery but that's outside the scope of this question. If the driving voltage wasn't contant the current would also change which makes the problem harder.
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Can someone please explain what happens on microscopic scale when an image becomes unfocused on a screen from a projector lens? My questions is basically asking when you move a projector back farther from a screen the image tends to blur unless you focus it. Logically I would think that every point(ray of light) of the image would expand proportionally allowing just a larger clear image. Instead you get a blurred image that is larger and requires focusing. My question is what exactly is happening to the light when you create a larger distance between the projector and the screen. Why can't the image just get larger and stay clear without the need to focus. Is every light ray independent? So each ray of light expands and over laps the other when the image is not focused? If so what exactly is a single ray of light and how thick is it? Am I thinking too deep about this ? Why does such a simple concept seem impossible to explain?
| When a sharp image is formed, every point on the object is reproduced in the image, and all the points around that point on the object are reproduced in the same relative positions of the image.
So, a diverging cone of rays from a point on the object hits the lens and the lens then refracts those rays so that they all pass though a point. That point is the image of the point on the screen.
Now suppose you increase the distance between the lens and the screen. Those rays from the object after refraction still converge at the same point as before and then continue onwards as a diverging cone of light that produces a circle of illumination on the screen rather than the point when the image was in focus.
This happens to rays from each point on the object and so these circles of illumination overlap one another and form a blurred image.
| {
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Does gravitational time dilation happen due to height or difference in the strength of the field? The reason why you have to tune differently the atomic clocks in GPS is because the GPS is higher or because there is less gravity there, or both? In other words in a constant gravitational field which doesn't differ with height, will time dilation still occur?
They say that the reason why people on the first floor age slower than people on higher floors is because as you get further from the earth gravity weakens. Is that true? Does the difference in the field cause that or just the distance?
If pure distance doesn't matter then why do we say that for a spaceship accelerating forward clocks at the front tick faster than clocks at the back since both are accelerating at the same rate?
| Gravitational time dilation is caused purely by "resisting" the gravitational field, in other words by accelerating in order to maintain a fixed position within the potential well. There is no gravitational time dilation in free fall.
| {
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What is the physical meaning of the Schwarzschild radius for objects that aren't black holes? Earth has a Schwarzschild radius of a little less than a centimeter. What does this mean for the matter of Earth's core that is within this radius?
A related question comes up for what happens when an almost black hole accretes matter and slowly becomes a black hole. Prior to the moment of the Schwarzschild radius crossing the boundary of the object, what does the matter within the radius experience?
|
What does this mean for the matter of Earth's core that is within this radius?
Nothing, since the mass outside the radius does not contribute to the force (see Newton's shell theorem).
A related question comes up for what happens when an almost black hole accretes matter and slowly becomes a black hole.
In the system of the coordinate bookkeeper the velocity of the infalling matter not only slows down but converges to zero when it approaches the Schwarzschild radius because of the gravitational time dilation. So there is never enough mass inside the Schwarzschild radius to form a true horizon in a finite coordinate time. The combined mass of the initial body and the infalling material will therefore be larger than before, but so is the volume over which the total mass is spread out. Therefore in the system of the coordinate bookkeeper the radius that contains the mass will always be larger than the Schwarzschild radius of the mass.
Prior to the moment of the Schwarzschild radius crossing the boundary of the object, what does the matter within the radius experience?
The observer that crossed the horizon in a finite proper time took an infinite amount of coordinate time to even reach the horizon, so there is no longer a connection to the outside world. He will also experience spaghettification before he inevitably ends in the singularity.
| {
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How is heat represented on a quantum level? Heat is just a form of kinetic energy for molecules, because as temperature rises, the heated molecules are "shake" and "vibrate" more and more. But how does that show up on a quantum scale? What element actually carries the kinetic energy: the heated molecule as a whole, its atoms, the nuclei, or the electrons' orbits? (Maybe even the quarks found in the nuclei?). Or is it that the shaking described is only an analogy for a notion of energy that is more difficult to grasp as their is no real physical movement in the heated object?
| The equipartition theorem says that all modes of excitation carry heat. There may be some modes which are too energetic to be excited at a given temperature, but the remaining modes are all excited. In overly simple terms, everything that can shake will shake.
| {
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At what distance we see actual size of object? At different distances object is seen of different sizes but there is some distance at which actual size of object is seen through eyes. Is it 25cm?
| The "size" of an object that you see is related to the angle subtended by the object at your eye.
The angle is called the visual angle.
This will determine the size of the image of the object on your retina.
The closer the object is to your eye the bigger the image on your retina and so the bigger the object appears to be.
However there is a minimum distance that the object can be from your eye and with the object closer than the "least distance of distinct vision" the image on your retina is no longer in focus.
The position of the image at the least distance of distinct vision is called the near point.
As a a rule of thumb for the "average" eye the least distance of distinct vision is taken to be 25 cm and when an object is placed at that distance from the eye a focussed image has its maximum size.
| {
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Why is a $5-60 mph$ time slower than a $0-60 mph$ time for some automobiles? This doesn't make a lot of sense to me, from a physics 101 point of view. I've read a few blog entries on why this is, but none of them explain it well or are convincing. "something-something launch control. something-something computers." Nothing in physics terms or equations.
For instance, Car and Driver magazine tested the Porsche Macan GTS. The $x-60$ times are:
*
*Rolling start, $5-60\; \mathrm{mph}: 5.4\;\mathrm{ s}$
*$0-60\;\mathrm{mph}: 4.4\;\mathrm{s}$
That's a whole second - about $20$% faster from a dead stop than with some momentum - which seems rather huge.
edit:
here is the article for this particular example. But I've noticed this with many cars that are tested for $0-60$ and $5-60$ times.
Here is another example - an SUV.
Another example.
And finally, interesting, even for the Tesla Model S (EV) where power doesn't depend on engine RPM, $0-60$ is still slightly faster than $5-60.$
| In the rolling start there is no tire slip or revving on the engine and so the run starts at low rpm where the engine is making less power.
A rolling start might have the engine at 2000rpm making for example 200 lb-ft (or 76hp) resulting in 0.45g of acceleration at 5mph (this example yields the acceleration to be 0.002253 times the torque produced).
With a launch from zero the engine is revved first then its kinetic energy transferred to the car yielding the first 5 mph almost instantaneously. At this point the clutch is either still slipping, or the tires spinning allowing the engine to be at about 4500rpm. The higher engine speed and the slightly higher torque (like 220 lb-ft) results in significantly higher engine power at about 188 hp (Power = Torque × RPM/5250). Some of this power is lost due to the clutch/tire slipping so the wheels see like 50%-60% of it, or 113 hp. At the same 5 mph this power at the wheels means about 0.67g of acceleration (or 0.0030 times torque produced) or 35% more.
In summary,
*
*Rolling start: Engine bogs down and it takes time to get up to the "power band". Peak acceleration decided by engine torque only.
*Launch: Keep engine spinning in the "mid range" and slip clutch or spin tires enough to match available traction.
| {
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How does curved spacetime affect gravitational waves? How differently will a LIGO detector detect a gravitational wave which came directly to it with a detector which happened to have a black hole between it and the source?
| A weak gravitational wave with $\square \hat h_{++}~=~0$ and $\square \hat h_{\times\times}~=~0$ is similar to an electromagnetic wave. We could think of the weak graviton as a sort of "diphoton," so the gravitational wave is similar to an electromagnetic wave with two polarization directions. It is also much more weakly interacting.
The gravitational wave is then a massless wave with a certain mass-energy content and thus will behave in a stationary gravity field as would any other massless field or wave. This is assured by the Einstein equivalence principle(s). This would then mean a gravitational wave would be lensed around a stationary gravity field, such as a black hole or an elliptical galaxy.
It would require more than three LIGO style detectors to measure this. With only three this would not show up. It is possible this could be measured if the source of a gravitational wave is associated with a distant galaxy or quasar in a gravitational lens observed optically.
| {
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Solving first order perturbation exactly in this situation I have this homework problem in QM Perturbation Theory
The Hamiltonian of a system is given by $$H_0 = A L^2 + B L_z$$ where
$A$ and $B$ are constants. If a perturbation $V = C L_y$ is added to
the system (where $C \ll A, B$ , find the lowest order correction to
the energy. Also solve the problem exactly
The lowest order correction to energy is
$$\langle l,m \rvert C L_y \lvert l, m \rangle = 0$$
as $L_y$ will either give me $\lvert l, m-1 \rangle$ or $\lvert l, m+1 \rangle$
But how do I solve it exactly? If I fix $l$, then $L_y$ is a $(2l+1) \times (2l+1)$, which is still more tedious than this assignment is supposed to be. Please help.
| As Valter Moretti answered in the comments, adding this here for archive purposes.
Define $N=B^2+C^2$. Using a unitary transformation $U$ corresponding to a certain rotation, you have $U(AL^2+BL_z+CL_y)U^∗=AL^2+NL_z$. Unitary transformations do not change the eigenvalues. Therefore the exact eigevalues are $Al(l+1)+Nm$: the same eigenvalues as for $H_0$ with $B$ replaced for $N$.
| {
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D3-brane in AdS/CFT correspondance I was reading a paper by Veronika Hubeny The AdS/CFT correspondence 1. Maldacena chose a D3-brane system to derive his conjecture. So I was wondering, why "D3-brane"? In other words, I need to know the importance of D3-brane system, so that is used in the AdS/CFT correspondence. It would be nice if a reference is recommended if the answer isn't so straightforward. Thanks in advance.
| $3$-branes are special in the following sense: only for $p=3$, the black p-brane solution admits a constant dilaton, while it is running for $p\not=3$. In particular, the dilaton $\phi$ diverges at the horizon of extremal $p$-brane solutions for $p\not=3$, which means that the string coupling $g_s=e^\phi$ cannot be kept small. In the Maldacena decoupling argument, two limits are taken:
*
*A near-horizon limit in the $p$-brane background
*A supergravity limit in which string loop corrections are supressed, i.e. $g_s\ll 1$
For $p\not=3$, the second limit cannot be taken near the horizon since the string coupling diverges. Only for $p=3$ the constant dilaton can also be taken to be small near the horizon.
In summary, only 3-brane solutions admit a simultaneous near-horizon and supergravity limit!
For all the mathematical details, you can refer to pages 16 - 19 in the MAGOO review.
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Weight factor in Path Integral Formalism In Quantum Mechanics, transition amplitude between two states in given by (path integral approach):
$$
\left\langle q';t'|q;t\right\rangle= \int[\mathrm dq] \exp \left(i \int L(q,\dot{q})~\mathrm d\tau\right)
$$
This tells that contribution of the paths to the amplitude is given by the weight factor : "i times the action".
Can anybody explain "intuitively" why this should be the weight factor?
| The intuitive explanation is really not there. I have a very fuzzy argument in the favor calling this "the weight factor". Let's see if that helps you at all.
The action is given by $S = \int L(q,\dot{q}) dt$. Now classically the path of the particle is determined by minimizing the action. i.e. the particle will follow the path for which the action is minimized(or in general extremized). This tells you that action somehow gives you the idea of how long the path is. Just like Fermat's Principle in optics.
Now come to the quantum mechanics using path integrals. Here we assume that the particle travels through all possible paths. But we know that the paths are not of same length(here length stands for the fact that action is different along different paths). So when the particle reaches to same final point and starts from same initial point but through two different paths, we expect the wave-function will catch a relative phase and its intuitively obvious that this phase must be proportional to the length of the path i.e. action. That's why you can call it "the weight factor" associated with the path.
| {
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Number of degrees of freedom of the coupled pendulum problem In Chapter 4 from the book Theoretical Mechanics of Particles and Continua by A. L. Fetter and J. D. Walecka, it is solved the problem of a coupled pendulum system while considering small oscillations.
There, they say the number of degrees of freedom needed to describe the Lagrangian, are the infinitesimal displacements from equilibrium $\eta_1$ and $\eta_2$, corresponding to each pendulum mass.
My question is: why there are needed two degrees of freedom? Isn't the spring that is attached to both masses a constraint of motion that reduces the degrees of freedom to only one?
Actually, they explicitly write the following equation:
$d-d_{0}=\eta_{2}-\eta_{1}$
which is the equation of the change in length of the spring.
Thank you for any answers or suggestions!
| There are naively three degrees of freedom: the length $d$ of the spring, and the angles $\theta_1$ and $\theta_2$. We also have the constraint
$$d - d_0 = l(\theta_1 - \theta_2).$$
You can use this constraint to eliminate any one of these variables, leaving two independent degrees of freedom.
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Why does increasing resistance decrease the heat produced in an electric circuit? If $H=\frac{V^2}{R}{t}$ ,then increasing resistance means decreasing the heat produced.
But, isnt it that the heat in a circuit is produced due to the presence of resistors? Moreover metals with high resistances are used as heating elements ,like Nichrome?
Why does the equation state that the heat produced is inversely proportional to Resistance
| Agreed it is confusing, but consider two ohmic resistors, where resistor 1 is half of the value of resistor 2. And let us say that the same voltage is applied across both resistors. Now since the current through each resistor is inversely proportional to the resistance, then current in resistor 1 is twice the current in resistor 2, and the heat dissipated in resistor 1 is twice the heat dissipated in resistor 2 since heat dissipated, $H=IVt$. Same conclusion is reached by using $H=I^2Rt$
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Quantification of entropy mathematically when $T$ and $V$ both change
$$ln\frac{W_f}{W_i}=N ln\frac{V_f}{V_i}=n N_a ln\frac{V_f}{V_i}$$
$$\Delta S=nRln\frac{V_f}{V_i}$$
$$ln\frac{V_f}{V_i}=\frac{1}{n N_a}ln\frac{W_f}{W_i}$$
$$\Delta S=\frac{R}{N_a}ln\frac{W_f}{W_i}=kln\frac{W_f}{W_i}=klnW_f-klnW_i$$
hence $$S=klnW$$
For T change taken from Atkins physical chemistry:
Above micro entropy S=klnW and macro entropy $\Delta S=\frac{\Delta Q}{T}$ are united isolated for T and V change. Entropy being a state function my problem is how one can understand that this is the $quantification$ that works for paths in change of both V and T? How does T and V quantify against each other? Can this be justified or explained?
| It seems to me the first derivation is simply assuming that S = k ln W, because it looks like it was inserted along the way and then pulled out at the end as if it was a conclusion of the logic. What's more, by considering only position and not momentum, it seems to me that derivation is assuming the particles are not changing kinetic energy. So that's why it ends up looking like there's no explicit mention of heat input or dQ/T, because to change V but not change kinetic energy, there would need to be heat input that is implicit. The second derivation shows more clearly where the connection between dQ/T and k ln W comes from, but it does it only for constant volume. But a reversible change in V does not change the entropy, so as long as everything is done reversibly, it doesn't matter if V changes or not, and the second derivation is less general than it could be.
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Entropy and reversible paths I'm a little bit confused about calculating entropy changes along irreversible paths by integrating over a reversible path. When using the central equation I can understand the argument, entropy and all the quantities we use to calculate the entropy change are state functions that have well defined end points. So the process between these two states doesn't matter, we will get the same entropy change. I find it difficult to grasp how we can use dS=dQ/T to calculate entropy changes in irreversible processes. Since Q is path dependent I don't see how the original argument applies. We can take the free expansion as an example, if we try to use dS=dQ/T we will get zero entropy change as no heat flows into the system. But using the central equation where all quantities are state functions we get a entropy change which is larger that zero as expected.
| There is an implicit but crucial assumption in thermodynamics (your calculations are based on it): Any irreversible process can be closed by a reversible process to become a cycle. If it is false, thermodynamics would collapse:
http://philsci-archive.pitt.edu/archive/00000313/
Jos Uffink, Bluff your Way in the Second Law of Thermodynamics, p. 39: "A more important objection, it seems to me, is that Clausius bases his conclusion that the entropy increases in a nicht umkehrbar [irreversible] process on the assumption that such a process can be closed by an umkehrbar [reversible] process to become a cycle. This is essential for the definition of the entropy difference between the initial and final states. But the assumption is far from obvious for a system more complex than an ideal gas, or for states far from equilibrium, or for processes other than the simple exchange of heat and work. Thus, the generalisation to all transformations occurring in Nature is somewhat rash."
| {
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Partial derivative of probability density (squared modulus of the wave function) wrt to position (1D)? Here's a snippet from Introduction to Quantum Mechanics by David Griffiths (Sec 1.5):
I understand how we used the Schrodinger equation to go from a partial in time to a double partial in position. But why is there a $-$ sign in the parentheses? Shouldn't there be a $+$ instead, since
$\Large \frac{\partial |\Psi|^2}{\partial x} = \frac{\partial (\Psi^*\Psi)}{\partial x} = \Psi^*\frac{\partial\Psi}{\partial x}\ +\ \frac{\partial\Psi^*}{\partial x}\Psi $
| You should try solving this on a piece of paper. I don't think that you understand correctly how the author passes from $\partial/\partial t$ to $\partial^2/\partial^2 x$. You can not use Schrodinger's equation for $|\Psi|^2$, it is only valid for $\Psi(x, t)$:
$$ i \hbar \frac{\partial}{\partial t} \Psi(x, t) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} \Psi(x, t). $$
For $\Psi^*$ another equation holds, which can be obtained by conjugating both sides of the Schrodinger's equation:
$$ - i \hbar \frac{\partial}{\partial t} \Psi^*(x, t) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} \Psi^*(x, t). $$
Note the minus sign on the l.h.s. It comes from conjugating an imaginary unit ($i^* = -i$). This minus sign is responsible for the minus sign in your answer.
Now I trust you to carefully expand both parts of your equation and see for yourself that they are indeed equal.
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Does Gravity Depend on Spatial Dimension? Consider a line containing two point masses, $m$ and $M$. The line is a $1D$ space.
What's the gravitational force between the two masses?
Newton's formula for the gravitational force $F$ between two masses $m$ and $M$ in 3D space is
$$F=\frac{G M m}{r^2}$$
where $G$ is a constant and $r$ is the distance between the two masses.
The $r$ term is good in a $3D$ space, but in general it's $r^{n-1}$ where $n$ is the dimension of the space. So putting $n=1$ for $1D$ space we get
$$r^{1-1}=r^0=1 \Rightarrow F=GMm \, ,$$
Which means $F$ is independent of distance. Gravity has the same strength no matter how far apart the two objects are!
Of course, this calculation uses Newton's theory of gravity. Perhaps General Relativity would give a different result.
| There is nothing in Newton's laws that restricts the force of gravity to behave as $\propto r^{1-D}$.
Yes, if you imagine the gravitational vector field to be like the velocity field of flowing water, then it should have zero divergence, and that indeed implies $\propto r^{1-D}$ behavior, including the constant $D=1$ behavior that you mentioned.
So $r^{1-D}$ behavior implies zero divergence and vice versa, but there's no way in Newtonian mechanics to prove either from Newton's laws. That's why Newton had to use experimental data to get the correct force law! If it was discovered that the three dimensional force law was like $e^{-r}$, Newtonian mechanics would chug along just fine!
Quantum field theory and general relativity change the picture, giving strict conditions on the force behavior as consequences.
| {
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Boltzmann statistics and density matrix for a simple spin 1/2 system. (NMR) I have a system with spin 1/2 particles (protons) in a magnetic field. The Zeeman effect leads to two energy levels (Spin up, spin down). Thus I have a state
$$
\left|\Psi\right> = a\left|\uparrow\right> + b\left|\downarrow\right>
$$
At room temperature, the probability to find a spin in one of the two states is determined by the Boltzmann distribution (high temp limit of Fermi). Hence, the probabilities are given by $p_\uparrow = e^{-E_\uparrow/k_B T}$ und $p_\downarrow = e^{-E_\downarrow/k_B T}$
Now my question: How does $a,b$ relate to $p_\uparrow$ and $p_\downarrow$. I have heard that you simply need to use density matrices to understand it, but I don't get it actually...
| There is no relation. $|\Psi\rangle$ is a pure state, while the thermal state $\rho$ is a statistical mixture. Since they describe different states, there is no relation.
| {
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Collapse of wave function Suppose a quantum system is initially at a state $\psi_0$ and that a measurement of an observable $f$ is performed. Immediately after the measurement, the system will be in a state that is an eigenvector of the operator $\hat f$ associated to $f$, the eigenvalue being the result of the measurement.
My question is the following: What if the candidate for this eigenvector does not represent a valid state? For example, the space of states of a 1D-system is $L^2(\mathbb{R})$ and there are operators acting on the space of all functions on $\mathbb{R}$ whose eigenvectors may not belong to $L^2(\mathbb{R})$. How does the wave function collapse to such an eigenvector?
| Hermitian operators corresponding to physical observables act on the Hilbert space of physically valid states. It's clear from the definition of an eigenvector that for any vector space $\mathcal{H}$ and linear map $f: \mathcal{H} \to \mathcal{H}$, the eigenvectors must lie in $\mathcal{H}$. Therefore, the eigenvectors for any physical observable will be physically valid states, and your issue can't come up.
For example, since position eigenkets $| x \rangle$ and momentum eigenkets $| p \rangle$ do not lie in the $L^2(\mathbb{R})$ Hilbert space (although they do lie in a more general "rigged Hilbert space"), the position and momentum operators technically aren't physical observables - only operators that are slightly smeared in position or momentum space are. Strange but true. They're still extremely useful mathematical idealizations though. Physically, this just means that no real measurement could ever have infinite precision.
In practice, this is almost never an issue, because all the usual formulas of quantum mechanics are true "in the distributional sense" - they're true if you multiply both sides by a smooth "envelope" function and then integrate. Or you can often discretize your Hilbert space into a large but finite set of points, in which case everything's well-behaved (this is what's almost always done in computational physics).
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Super-renormalizable theory and $\beta$-function There is the statement that $\beta$-function vanishes for super-renormalizable theories. In $D=2$, scalar field has mass dimension zero. So any polynomial interaction is super-renormalizable. Then shouldn't all of them have vanishing $\beta$-functions? But there are many theories (e.g, sine-Gordon) in $2D$ which have nontrivial $\beta$-function. I must be missing something very basic here.
| In a qft, it may be possible to redefine other parameters than coupling to absorb the infinities coming from higher order corrections. In this way, coupling constant does not get renormalized and hence the beta function vanishes. It is a possibility in super-renormalizable theory as fewer diagrams are divergent and the condition may be satisfied.
As an example for the Sine- Gordon model, the action is
$$\mathcal{S}(\theta)=\int d^2x [\frac{1}{2}(\partial_\mu\theta(x))^2-\frac{m^2}{k^2}cosk\theta(x)]$$
Redefining, $\theta=k\theta$ gives
$$\mathcal{S}(\theta)=\frac{1}{t}\int d^2x [\frac{1}{2}(\partial_\mu\theta(x))^2-m^2cos\theta(x)]$$ with $t=k^2$.
Perturbative expansion in the power of k only modifies the $cos\theta$ term as a self interaction and the divergences arising can be absorbed by a redefinition of m. In this way, coupling constant does not get renormalized and hence beta function vanishes.
This property is not true in general as the vanishing of beta function to all orders implies a finite theory ($\mathcal{N}=4$ SYM) which is a result need to be obtained from a non-perturbative analysis unless it is trivially true as in the former case. Most qft exist perturbatively and the existence of fixed points is not known non-perturbatively. A super-renormalizable theory does not have a vanishing beta function generally as can be seen from $\phi^3$ theory beta function which in d dimension reads,
$\beta(g)=(d/2-3)g-\frac{3g^3}{256\pi^3}+O(g^5)$ ( Collins "Renormalization," eqn. 7.3.7)
$\phi^3$ theory is super-renormalizable for $d<6$ but $\beta$ function is not zero. It however shows asymptotic freedom which is a property of super-renormalizable theories ( I am not aware of the proof though).
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How do electrons move at an atomic level? This was meant to be a sub question in the comments of my last question but I think it is big enough to have its own post.
I know that electrons move because of the potential difference across the wire.
But do the electrons jump from the valence shell of 1 atom to the valence shell of another atom or do they flow in a swarm without interacting with any atoms? Here I am talking about electron flow in a metallic conductor.
And what about insulators? Let's say we have an insulator in the middle of a conducting circuit. How will the it prevent electrons from going through?
| It does not make sense to talk about how an electron moves in an atom. The standard orbitals, shown below, are amplitudes for the occurrence of the electron. For very high Rydberg atoms with the electron put in extremely high orbitals the wave packet can then localize and orbit around the nucleus. One has to apply a weak magnetic field to define the plane of this orbit.
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Is speed of Hot air rising gravity dependent? Would say a heated air rise twice as fast in 2G than in the environment with standard Earth gravity?
| Initially, on an air parcel, buoyancy force would be twice as much when gravity doubles, which will impart higher initial acceleration, helping air parcel reach a higher speed before being counterbalanced by viscous forces. Of course there is much complication here, because an air parcel loses its momentum also by mixing, if the flow is turbulent. If somehow the parcel is able to maintain its identity by not mixing (very unlikely), then its final speed will be twice as much only if viscous force acting on it is a linear function of speed, but this is not the case. It may be simpler to think of a balloon rising in air. If Reynolds number is high for flow around balloon to be turbulent, then viscous force will increase quite rapidly with balloon speed, something like $\sim v^2$, so balloon's speed will be increased about $\sqrt{2}$ times when gravity is doubled.
| {
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Why does this planet (J1407 b) and Saturn's ring center on its equator? Why is the material in J1407 b's or Saturn's rings stay in a disk not scattered? Is it the gravity and/or magnetic field that causes this? Does it differ with other large bodies that might not have a electromagnetic field?
| The reason it's a disk and not spread out is actually simpler than you might imagine.
Consider all of the mass in the system as a whole. It will have a certain centre of mass, angular momentum, etc, which much be conserved no matter how the distribution changes. That defines a preferred axis of rotation, and perpendicular to it, a plane that is the natural direction for a ring.
Now consider any mass that is not, initially, in that plane. Say a rock that is travelling around the centre of mass "top to bottom", in what would be a polar orbit around a planet. By definition, there is some other mass with the opposite direction on average. So if you wait a couple of hundred million years, the objects that were orbiting top-to-bottom will eventually meet one travelling bottom-to-top, their non-plane momentum will be cancelled out, and any remaining momentum will be in the plane.
That gets you something like a disk. After that other issues come into play that can shepherd or randomize the distribution. In the case of a ring around a large planet, like Saturn, the equatorial bulge helps out.
It is entirely possible to construct rings that don't spin the same way as the planet, but they will tend to scatter. There has been some work on using such a system, with artificial shepherds, as a way to build a skyhook.
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Discontinuity of the geometric phase Does the geometric phase accumulated along a closed trajectory (in some parameter space) has to be continuous?
| The geometric phase can experience discontinuous dependence on the parameter space in the vicinity of diabolical points (i.e., points where the Hamiltonian eigenvalues are degenerate but the eigenvectors are distinct) and exceptional points where both eigenvalues and eigenvectors are degenerate. Exceptional points appear in non-Hermitian Hamiltonians used to model open quantum systems.
Please see the following article by: Nesterov and de la Cruz for a detailed exposition.
The Berry curvature has a monopole singularity at the diabolical points. Diabolical points corresponding to higher classes of degeneracies can give rise to higher Chern numbers, please see the following article by: Garg.
This subject is still under very active research, please see for example the following work by Viennot
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In what coordinates is the following Kerr metric writing in? In this book the kerr metric was given by
I am confused because of the appearance of the $2d\nu dr$ term because in the standard Kerr metric I know this term doesn't appear. Does anyone know what coordinates is this metric given by and the relationship between these coordinates and the standard Kerr coordinates?
Thanks.
| Sorry but have you read that book/section you referenced? The line element you gave is given in eq. (19.45) and in the text directly below it says
"The coordinates $(\nu,r,\theta,\bar \phi)$ are the Kerr coordinates."
The section 19.4.1 in your reference discusses in detail how to get from Boyer-Lindquist coordinates to this so called Kerr coordinates. The term you are confused about come from the $-dt^2$ and the $dr^2$ terms in Boyer-Lindquist coordinates, using the coordinate change $d\nu=dt +\frac{r^2+a^2}{\Delta}dr$, which is described in eq. (19.42).
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Heisenberg uncertainty and Lorentz contraction Consider a particle in a frame moving with speed $v$ relative to the lab frame. By Lorentz contraction, the width of the wavefunction will be smaller in the lab frame, resulting in smaller $\Delta x$. If $v$ is high enough, then the uncertainty principle $\Delta x \Delta p \ge \hbar/2$ will be violated in the lab frame.
What's wrong here? Does $\Delta p$ increase somehow? This seems unlikely, since simply translating momentum distribution by a constant should not alter the standard deviation.
| You're trying to apply relativity (Lorentz contraction) to a result from nonrelativistic quantum mechanics (Heisenberg uncertainty), so of course you get a contradiction.
In nonrelativistic quantum mechanics, the effect of a boost is given by the Galilean transformation
$$\psi(x) \to \exp((im/\hbar) (vx + v^2t/2))\, \psi(x-vt)$$
as explained in more detail here. You can verify by explicit calculation that this shifts position eigenvalues as
$$x \to x-vt$$
and shifts momentum eigenvalues by
$$p \to p-mv$$
as expected classically. Since both position and momentum are simply shifted, $\Delta x \Delta p$ stays the same, and the uncertainty principle is preserved.
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Why was an 8 TeV collider needed to find a 125 GeV Higgs? This might be very naive, but why wouldn't a (say) 209 GeV LEP do the job?
| You not only need enough kinetic energy to create the new particle, but enough to create the new particle after losing kinetic energy to all the shrapnel particles. At the very least, collisions produce a lot of photons that carry away energy due to the electromagnetic interactions between the colliding particles. Also, head-on collisions are extremely rare. Glancing collisions are much more common, so not all the energy in the beam goes into the collision. The original beam particles merely scatter off at glancing angles, taking away energy from the collision. Increasing the energy way beyond the amount necessary to produce a new particle increases the chance that even glancing collisions will have enough energy to produce the new particle.
In particle physics, it's not a matter of getting one collision to produce the new particle. Every collision produces a gigantic spray of particles that has to be sifted through for actually relevant data. We need to create the new particle many times to build up enough statistics to be confident that we're actually seeing something real. Higher energies allow us to produce more samples of a rare event to make it easier to study.
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Question about tensor form of Maxwell equation By variating the Maxwell Lagrangian we get the equation of motion. The remaining two Maxwell equations can be written as
$$\epsilon_{\mu\nu\rho\sigma}\partial^{\rho} F^{\mu\nu} = 0.$$
I have also seen it written as the Bianchi identity: $$\partial_{[\lambda}F_{\mu\nu]} = 0.$$
Why are these two forms equivalent?
| It's basically just a duality relation analogous to the cross product in three dimensions. But if you want to do some work to show the equivalence, then:
Going from the second equation to the first is easy, just hit it with $\epsilon_{\mu\nu\rho\sigma}$.
Going from the first to the second equation, is a little trickier and relies on knowing how to evaluate the products of Levi-Civita symbols.
The basic idea is that you should contact the first equation with $\epsilon^{\mu'\nu'\lambda'\sigma}$ and compare the resulting antisymmetric combination of $\delta$s with the antisymmetrization of the indices in the second equation.
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Lagrangian and finding equations of motion I am given the following lagrangian:
$L=-\frac{1}{2}\phi\Box\phi\color{red}{ +} \frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4$
and the questions asks:
*
*How many constants c can you find for which $\phi(x)=c$ is a solution to the equations of motion? Which solution has the lowest energy (ground state)?
*My attempt:
since lagrangian is second order we have the following for the equations of motion:
$$\frac{\partial L}{\partial \phi}-\frac{\partial}{\partial x_\mu}\frac{\partial L}{\partial(\partial^\mu \phi)}+\frac{\partial^2}{\partial x_\mu \partial x_\nu}\frac{\partial^2 L}{\partial(\partial^\mu \phi)\partial(\partial^\nu \phi)}=0 $$
then the second term is zero since lagrangian is independent of the fist order derivative. so we will end up with:
$$\frac{\partial L}{\partial \phi}=-\frac{1}{2} \Box \phi+m^2\phi-\frac{\lambda}{3!}\phi^3$$
and:$$\frac{\partial^2}{\partial x_\mu \partial x_\nu}\frac{\partial^2 L}{\partial(\partial^\mu \phi)\partial(\partial^\nu \phi)}=-\frac{1}{2}\Box\phi$$
so altogether we have for the equations of motion:
$$-\frac{1}{2}\Box\phi+m^2\phi-\frac{\lambda}{6}\phi^3-\frac{1}{2}\Box\phi=0$$
and if $\phi=c$ where "c" is a constant then $\Box\phi=0$ and then the equation reduces to $$m^2\phi-\frac{\lambda}{6}\phi^3=0$$ which for $\phi=c$ gives us 3 solutions:$$c=-m\sqrt{\frac{6}{\lambda}}\\c=0\\c=m\sqrt{\frac{6}{\lambda}}$$
My question is is my method and calculations right and how do I see which one has the lowest energy (ground state)? so I find the Hamiltonin for that?
| Thanks to all you guys I have found that my mistake was at confusing the kinetic and interaction terms. so here is my answer to this question:
this problem is basically finding the values for $\phi$ that minimizes the effective potential and I have found them above named $c_1$,$c_2$ and $c_3$ considering those are correct, I have for my effective potential now:
$$V(\phi)=-\frac{1}{2}m^2 \phi^2+\frac{\lambda}{4!}\phi^4$$since $L=KE-V$ then my Hamiltonian will be $$H=-\frac{1}{2}\phi\Box\phi -\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ for c=0 its just gonna give me zero but for $c=\pm \sqrt{\frac{6m^2}{\lambda}}$ now substituting this into the hamiltonain:$$<H>=E=0-\frac{1}{2}m^2(\sqrt{\frac{6m^2}{\lambda}})^2+\frac{\lambda}{4!}(\sqrt{\frac{6m^2}{\lambda}})^4\\E=\frac{m^4}{\lambda}(-\frac{6}{2}+\frac{36}{4!})\\E=\frac{m^4}{\lambda}(-\frac{6}{2}+\frac{3}{2})\\E=-\frac{3}{2}\frac{m^4}{\lambda}$$ so there are two solutions that have the lowest energy which is $c=\phi=\pm\sqrt{\frac{6m^2}{\lambda}}$.
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Neutral points in a system of charges on the vertices of a square There are four positive charges of equal magnitude placed at the four vertices of a square. Is there any point where the electric field vanishes (neutral point) within the square and in its plane, other than its center?
| No.
Think of this Geometrically.
if each charge generates a field: $ \vec{E}_i = {kq \over {|\vec{r}-\vec{r}_i|}^2} \hat{({\vec{r}-\vec{r}_i})} $ one can see that the center of the square is the only point where $\Sigma \vec{E}_i =0$. This is because of the fact that any point, to the left or to the right, up or down, from the center, has a nonzero field. Now, the direction of the field would depend on where that point is.
Perhaps if you drew some field lines it would help you understand, finding a field plotting software would be most helpful here, but I know not of one which i could recommend.
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Space bends relative to what? We all are aware of gravitational waves, as it bends space and time, black hole squeeze space, but the squeezing, bending, expanding happens reference to what? Since the observable universe is the universe existing within itself, so it bends in reference to whose perspective?
| You do not need to have an "external reference" to see consequences of the bending of spacetime. For instance, light travels on a straight line if spacetime is flat. If spacetime is locally bended by a very massive object, a ray of light will follow a curved path when travelling close to this object.
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Understanding tension based on assumptions of pulley system If we consider a simple pulley system with two masses hanging on each end of a MASSLESS and INEXTENSIBLE string around a MASSLESS and FRICTIONLESS pulley, how then can one reason that the tension at each end of the string must be the same?
My own reasoning:
MASSLESS ROPE means that for any segment of the rope with tension $T_1$ and $T_2$ we have that $\sum F = T_ 2 - T_1 = 0$ (since $m = 0$) and thus the tensions must be the same, on a non curved rope at least!
INEXTENSIBLE means that no energy can be stored in the string, however I fail to see how this is a neccesary condition (for equal tension)
MASSLESS PULLEY means that no rotational inertia exists, and thus no force can alter the tension of the string (?)
FRICTIONLESS PULLEY is hard for me to figure.
Needless to say, I feel quite at a loss conceptually!
| In general case such system moves with acceleration...
If rope is extensible we cannot assume that the magnitudes of accelerations of the two masses are the same( they will NOT move "jointly"). If the pulley has a mass then tensions on left and right of the pulley must differ to ensure the pulley's rotational acceleration. If pulley is not frictionless the friction force must be accounted for when writing 2-nd law of Newton.
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Can the carbon nanotubes in Vantablack absorb sound? Say you wanted to create a sensory deprivation room surrounded with the Vantablack material. As well as absorbing light, would the carbon nanotubes also be able to absorb sound, eliminating the shape of the walls and floors of an anechoic chamber?
| https://www.acoustics.asn.au/conference_proceedings/INTERNOISE2014/papers/p124.pdf This research paper here has done the experiments with the CNT that you were curious about. Along with the performance of the CNT they've also compared it to normal materials used nowadays. The shape will definitely play an important role. The CNT used here(from my preliminary reading of the article) is multi-walled CNT nano-forest. The comparison is with glass-wool and melamine foam.
They've also reported the comparison for different frequencies of sound.
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Finding conserved quantities I have the 1-form $$\alpha = pdq-\frac{(p^2+q^2)}{2}dt$$
and the vector field
$$v=p\frac{\partial}{\partial q} - q\frac{\partial}{\partial p}.$$
How would I find conserved quantities using Noether's Theorem?
Additionally, is there a difference between finding conserved charges and conserved quantities?
I'm a math student who hasn't taken many physics classes so I am a bit lost with this.
| Well, the system is secretly on Hamiltonian form, meaning that there is lurking a symplectic structure $$\omega ~=~ dp\wedge dq$$ in the background, and there is a Hamiltonian $$H~=~\frac{p^2}{2}+\frac{q^2}{2}.$$ Here $$\alpha~=~\mathbb{L}_H~=~pdq-Hdt$$ is the Hamiltonian Lagrangian one-form. If $\gamma: [t_i,t_f]\to \mathbb{R}^2$ denotes a path in phase space, then the corresponding action functional reads $$S[\gamma]~=~\int_{t_i}^{t_f} \! \gamma^{\ast}\mathbb{L}_H.$$ The symmetry-generating vector field $$v~=~-X_H~=~-\{H,\cdot\}_{PB}$$ is the Hamiltonian vector field generated by $-H$.
This means that we can use a Hamiltonian version of Noether's theorem, cf. this Phys.SE post. We leave the details to the reader, but the main answer is that the Hamiltonian $H$ itself is the sought-for conserved charge/quantity.
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Help needed with applying pseudo force I have read in my book that to apply a pseudo force,
We make sure that the object ( our reference frame ) is accelerating and then we add the negative of it's acceleration vector to the object that we are trying to apply the pseudo force at.
Now take a look at this picture where B and A are fixed points on a circular plate.
I want to apply a pseudo force on A w.r.t. B . I see, that B has a centripetal force $F_{c2}$ and it's direction is clearly along $BO$.
Obviously, according to B, A will be at rest. And B knows that on A there is a centripetal force $F_{c1}$ and thus there must be some force on A ( the pseudo one ) in the $O$ to $A$ direction which should be equal to $F_{c1}$ to make A look stationary.
Question- Why doesn't B apply a pseudo force on A which is equal to [ $-$ $F_{c2}$ ] ( according to the definition of pseudo force in my book ) but it applies a force of [$-$ $F_{c1}$] on A ?
| Your statement about pseudo force is not entirely correct. For example, a rotating observer at the center of a merry go round is not accelerating, but still you need to add centrifugal forces.
Does your observer at $B$ self-rotate as well? If he is not self-rotating, then observed by him, $A$ will be rotating. If he is self-rotating so that he always faces $O$, $A$ will be observed to be at rest, but then you need to add another pseudo force, the centrifugal force, which together with the pseudo force you have taken into account, will exactly balance the centripetal force.
Let the position vector of $A$ and $B$ measured by $O$ be $\vec{r}_A$ and $\vec{r}_B$, respectively. Then the acceleration of $B$ itself is $$-\omega^2 \vec{r}_B$$
So one need to first add a pseudo force
$$m_A \omega^2 \vec{r}_B$$
to $A$.
Then if $B$ is self-rotating with the same $\omega$ so that it always faces $O$, then a centrifugal force
$$m_A\omega^2(\vec{r}_A - \vec{r}_B)$$
is needed as well.
So you can see that the sum is
$$m_A \omega^2 \vec{r}_A$$
which just balances the real centripetal force.
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Orbital maintenance of a Mars cycler How much orbital maintenance (i.e. corrections), if any, would a Mars cycler need.
As far as I understand it, a cycler needs no corrections to its orbit unless something extraordinary happens like gravitational interference from a passing comet or something of that sort.
I'm having trouble finding anything about the topic with my very limited 'knowledge' of orbital mechanics.
For those unfamiliar with cyclers: http://en.wikipedia.org/wiki/Mars_cycler
|
As far as I understand it, a cycler needs no corrections to its orbit unless something extraordinary happens like gravitational interference from a passing comet or something of that sort.
Buzz Aldrin's Road map to Mars
The paper from which this table is taken is only 10 pages long, and most of it is spent discussing the various types of orbit available, with surprisingly little attention paid to orbital corrections, other than the one listed below. The link to the paper is reference 1 at the bottom of the Mars Cycler Wikipedia article.
It seems from the last two columns in the above table that $\Delta v $, supplied by rocket power, would be required to "complete the turn" around Earth in more than 70% of the cycles.
It's not a free lunch, but it seems very efficient.
When two bodies orbit a third body in different orbits, and thus different orbital periods, their respective synodic period can be found. If the orbital periods of the two bodies around the third are called $P_1$ and $P_2$, so that $P_1 < P_2$, their synodic period is given by:
$${\displaystyle {\frac {1}{P_{\mathrm {syn} }}}={\frac {1}{P_{1}}}-{\frac {1}{P_{2}}}}$$
Mars has a synodic period of 2.135 relative to Earth.
One feature of this approach is that the cycler system cannot make use of Hohmann Transfer orbits (shown below).
Finally, to address Jon Custer's valid point about having to take into account various gravitational sources, I think, from reading Aldrin's science fiction stories (which feature this idea extensively) and his biography, his idea was to adjust each flyby to allow for these influences as much as possible.
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Can a lens focus all types of em waves? A glass lens is usually used to focus visible light, but is this glass lens capable of focusing other types of electromagnetic waves? If not, are there lenses made of different materials that can be used for em waves of different wave lengths (radio waves, micro-waves, x-rays, ...)?
Logically I don't see why a glass lens could not focus em waves with different wave lengths. Visible light is bent because of the refraction that occurs between the two mediums, so would changing the wave length of this wave now stop the refraction from occurring?
| This question is really about how the refractive index of glass (and other materials) varies with frequency.
Here the term refractive index being used in its broadest sense it that it has a real part which is a measure of the speed of light through the material and an imaginary part which is called the attenuation coefficient.
For light the frequency dependence of refractive index can be seen with the dispersion of white light and absorption being frequency dependent with the absorption of ultra violet by glass.
I do not think that there is a material which is transparent to electromagnetic radiation at all frequencies and so it would seem that you cannot have a "one material fits all frequencies" lens.
So that is the answer to your question but for an explanation there are two very good answers related to this question in this forum but for a fuller explanation of the topic you could refer to Feynman I-31 and Feynman I-32.
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Average Velocity: $( v_1+v_2)/2$ While searching for the answer regarding, why acceleration is needed to be constant for using the formula $(v_1+v_2)/2$ , I found many simple and easy proofs regarding this, here in this Physics.SE website, one of which is ,
But can anyone come up with a daily life simple explanation for understanding why acceleration is needed to be constant for using the formula $(v_1+v_2)/2$ , for a freshman student in physics like me.
| A visually geometric answer is that the average velocity with respect to time over an interval is the area under a velocity-time curve (assuming rectillinear motion, of course), divided by the length of the time interval.
This area is only equal to the mean of the end velocities if the curve is a straight line, in which case the said mean is simply using the formula for the area of a trapezoid.
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Schwarzschild metric in expanding Universe In Schwarzschild coordinates the line element of the Schwarzschild metric is given by:
$$ds^2=\Big(1-\frac{r_s}{r}\Big)\ c^2dt^2-\Big(1-\frac{r_s}{r}\Big)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta\ d\phi^2).$$
In the asymptotic limit where $r>>r_s$ the Schwartzschild metric becomes:
$$ds^2=c^2dt^2-dr^2-r^2(d\theta^2+\sin^2\theta\ d\phi^2),$$
which is the Minkowski metric of flat spacetime.
But observations show that real astronomical objects are embedded in an expanding spatially flat FRW metric given in polar co-ordinates by:
$$ds^2=c^2dt^2-a^2(t)\ dr^2-a^2(t)\ r^2(d\theta^2+\sin^2\theta\ d\phi^2).$$
Therefore maybe the Schwarzschild metric should be given by:
$$ds^2=\Big(1-\frac{r_s}{r}\Big)\ c^2dt^2-a^2(t)\Big(1-\frac{r_s}{r}\Big)^{-1}dr^2-a^2(t)\ r^2(d\theta^2+\sin^2\theta\ d\phi^2).$$
Perhaps this metric would only be useful to describe a gravitational system whose size is comparable to the Universe itself?
| Matching the Schwarzschild metric onto the metric of an expanding universe is not trivial. Einstein and Straus tried it in the 1940s but their paper, as I recall, has a mistake. A solution was given in 1956 by C. Gilbert in the MNRAS:
http://adsabs.harvard.edu/full/1956MNRAS.116..678G
As I said, the solution is not trivial.
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Could the universe be shrinking? It is thought the universe is expanding because of the red shift of most galaxies but if all the matter in the universe was actually falling into a massive black hole wouldn't most galaxies still be red shifted because the black hole would accelerate every galaxy and every galaxy that was nearer the black hole would be accelerating away from us and we would be accelerating away from any galaxy that was further from the black hole than us. There must be something wrong with this thought can you tell me what it is?
| The universe as a black hole would have observable consequences. The Schwarzschild metric
$$
ds^2~=~c^2(1~-~2m/r)dt^2~-~(1~-~2m/r)^{-1}dr^2~-~r^2d\Omega^2
$$
defines the curvature of spacetime. Here we have $m~=~GM/c^2$. The Riemann curvature tensor components to $O(c^2)$ are
$$
R_{trtr}~=~-\frac{2mc^2}{r^3},~R_{t\theta t\theta}~=~\frac{mc^2(r~-~2m)}{r^2},~R_{t\phi t\phi}~=~\frac{mc^2(r~-~2m)sin^2\theta}{r^2}.
$$
There are the components $R_{r\theta r\theta}$, $R_{r\phi r\phi}$ and $R_{\theta\phi\theta\phi}$, but these are less than the three above by a factor $1/c^2$. These enter into the geodesic deviation equation
$$
\frac{d^2x^\mu}{ds^2}~+~{R^\mu}_{\alpha\nu\beta}U^\alpha x^\nu U^\beta~=~0
$$
which determine the separation rate of two test masses separated by the vector $\bf x$ with components $x^\nu$. For test masses separated by a radial distance $x^r~=~r$. Some thought illustrates that the radial distance between the two test masses increases. Similarly the distance defined by the angles $\theta$ and $\phi$ decreases.
From a more physical perspective this is the tidal force. The Riemann tensor above determines the tidal force on an extended mass, or the acceleration between test masses. Below a diagram illustrates the motion of a spherical or ellipsoidal shell of test masses as it falls radially to a black hole. In the extreme situation this leads to the so called spagettification of any extended system of masses.
If the universe were a black hole the optical signature of this would be that a set of galaxies at antipodal regions of the sky would be redshifted. They would be accelerating away for an observer. There would also be a set of galaxies arrayed on a plane or annular region of the sky that would be more blue shifted and approaching any observer. Very clearly we do not observe the universe to be of this character.
| {
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How would you include gravity in a momentum problem?
Say you have a big ball of mass $m_1$ and a little ball on top of that of mass $m_2$ (assume they are a small distance apart, like $1~\mathrm{mm}$). Now lets drop these from a height of $h$ so that the big ball will bounce off the ground and collide into the little ball in an elastic collision.
Now I know gravity would play a key role in this example but how would one perform calculations with it? I know $F=p/t$ and momentum will not be conserved since there is an external force (gravity). So, knowing this how can one determine the height each ball will rise after the collision?
| Use kinematics to find the velocity of both balls, so that tells you their momentum, except the lower one has its momentum reversed when it bounces. Then conserve that total momentum, and kinetic energy, during the collision at the ground. Then solve the kinematics of what follows. (By kinematics, I mean the fomulae that involve a known acceleration of g, like v^2 = 2gh.) In short, I think the key answer to your question is that conservation of energy is global and can be used at any location, but conservation of momentum is only useful in the instant of a collision.
| {
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Equations of motion for a free particle on a sphere I derived the equations of motion for a particle constrained on the surface of a sphere Parametrizing the trajectory as a function of time through the usual $\theta$ and $\phi$ angles, these equations read:
$$ \ddot{\theta} = \dot{\phi}^2 \sin \theta \cos \theta $$
$$ \ddot{\phi} = - 2 \dot{\phi} \dot{\theta} \frac{1}{\tan \theta} $$
I've obtained them starting from the Lagrangian of the system and using the Euler-Lagrange equations.
My question is simple: is there a way (a clever substitution, maybe), to go on and solve the differential equations? I would be interested even in a simpler, partially integrated solution. Or is a numerical solution the only way?
| I checked, and these equations of motions correspond to motion in of an otherwise free Lagrangian. Something that will make your life easier in solving this is to recognize that angular momentum is conserved here. Because velocity is guaranteed to be perpendicular to the radius:
$$\begin{align} |\mathbf{L}| &= mvr \\
& = mr \left(\dot{\theta} + \dot{\phi}\sin\theta\right) \\
|\dot{\mathbf{L}}| & = mr\left(\ddot{\theta} + \ddot{\phi}\sin\theta + \dot{\phi}\dot{\theta}\cos\theta\right) = 0.\end{align}$$
You should be able to combine the equations of motion you have to show that last line. You also have that $|L| = I\omega$ and, because $I$ is fixed, that means that the rate of change of some angular variable is constant. The final hint I give is that you should look at how rotations can be defined as rotations around an axis by an angle.
| {
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"timestamp": "2023-03-29T00:00:00",
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Motivation behind the theory of cosmological perturbations What is the main motivation behind the theory of cosmological perturbations? Is it the observational data, observational hints, or perhaps the theory of inflation?
| None of those. If you mean by motivation what it means in English, what was the reason for doing perturbations, it's more or less straightforward.
It was to try to find approximate solutions to how deviations from an ideal homogeneous and isotropic universe evolves. Since people figured that probably gravity had something to do with it (not a very hard thing to think), and in the large going back to the past the universe still looks homogeneous and isotropic, but we also do see very strong inhomogeneities in the 'small', like stars, galaxies and clusters, they went to earlier cosmological times and put in small perturbations and saw how they grew.
Even before cosmology there were perturbation calculations in Newtonian gravity to model and help determine how stars and galaxies formed.
So perturbations is just a normal way of doing the physics of changes in some semi-stable solution, and see how other things happen, particularly where exact solutions with those kinds of irregular distributions are just not possible. It was not just Einstein, there is also the raw numbers of particles and bodies, and you need to also include the nuclear
and electromagnetic interactions.
The only reason inflation enters in is because it is a way to get to an initially mostly homogeneous, isotropic, and flat universe. The the question is what are the initial sources of the perturbations, and at those energies and temperatures it came down to quantum fluctuations of the fields, with sizes approximately the Planck size.
From that point on you can treat cosmology without reference to inflation. Some people simply call that post-inflation the beginning of the Big Bang.
| {
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Classically, why is the gyromagnetic ratio always $q/2m$? For uniform charge and mass distribution for a rigid body rotating with a uniform angular velocity about its centre of mass, I want to know if it is just a mathematical artifact of integration which cancels out the factors of moment of inertia (angular momentum) and the magnetic dipole moment or if there is something much deeper.
I realize the spin magnetic moment $g$-factor is 2 quantum mechanically, which is another reason why i want to know if there is some kind of emergent property which nicely cancels everything and brings the final result as $q/2m$ in classical physics
| Although you question is not particularly precise, here is my honest guess to what might give you an answer:
The classcial gyromagnetic factor is obtained by the following straightforward calculation. Consider a single particle of mass $m_e$ and charge $e$ orbitting on circle of radius $r$ with velocity $v$. Then its magnetic moment is
$\vec{m} = I \cdot \vec{A} = (ev/2\pi r)\cdot (\pi r^2 \vec{e_z})= e/2 \cdot vr\,\vec{e_z}$.
Whereas his angular momentum is equal to
$\vec{l} = J\cdot \omega = (mr^2) \cdot (v/r\,\vec{e_z}) = m \cdot vr\,\vec{e_z}$
Therefore is is easy to see that $\vec{m} = (e/2m)\vec{l}$.
This was the classical calculation. The quantum mechanical gyromagnetic factor is usually seen by invastigating the Dirac equation but may already be obtained by linearizing the Pauli-equation (see [Greiner, Quantum Mechanics]).
Nevertheless, the two notions describe utterly different gyromagnetic ratios, since clasically we desribe extrinsic angular momentum (relative to the origin $r=0$) but quantum-mechanically it is the intrinsic angular momentum (i.e. spin) we bother about. Actually it is also a very pleasing example of why thinking of an electron as a small rotating sphere (why even a sphere?) is just leading you onto the wrong track!
| {
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Fusion boosted fission reactor With weapons, fusion boosted fission seems to already be a thing. But I'm not interested in weapons.
Are there any types of fission reactor (other than a nuclear explosive) that would make more efficient use of fissile material by mixing in fusionable material?
Obviously the average temperature of a fission reactor isn't high enough for fusion, but the fission products will be in the right energy range for a couple of collisions while they are in the process of cooling down.
| Something very close to this is the fusion-fission hybrid idea where fusion does not have to be at the point of self-sustaining burning (so it is below the ignition threshold) but rather driven by an external power source. Still, such a driven fusion reactor can produce lots of neutrons that can be used for driving fission burn of surrounding fissionable material. For example this could be used to extract energy from (and dispose of) nuclear waste that is not rich enough to go into self-sustaining fission reaction but with external neutron radiation one can drive fission reaction in it. Here is the Wikipedia article with detailed explanations https://en.wikipedia.org/wiki/Nuclear_fusion-fission_hybrid
| {
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Why is it that the change in internal energy always uses the formula with Cv in regards to pressure/volume/temperature changes on a gas? Normally I would associate the use of $C_v$ with finding the energy taken into or leaving a system when the volume is kept constant. However, the formula to find $\triangle E_i$ (change in internal energy) is $nC_v \triangle T$. Why $C_v$? Also, does this apply to pretty much anything? Or are there limitations?
| First, for arbitrary process, this formula applies only to ideal gas. For the special case of constant-volume process, this applies to all gases, by definition of $C_v$.
Now internal energy $U$ is a state function. So if you know temperature in initial and final states for an ideal gas, change in $U$ is thereby completely determined, no matter which process the system went through in between.
| {
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"url": "https://physics.stackexchange.com/questions/283054",
"timestamp": "2023-03-29T00:00:00",
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