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The torque created using an oval chain ring I have seen the explanation on "How bicycle gear works?" but this seem to be using a standard round chain ring. I would like to know if there is any difference when an oval chain ring is used. As I can clearly see, one can have an oval chain ring on a fixed/single gear bike. There fore there is no slaking or tightening of the chain. So I would say that there is no difference in gear ratio during the complete rotation of the oval-chain-ring. So the torque will remain the same.
Is there an answer to my problem? and if there is could you proved any equations
| The principal benefit of elliptical chainrings is to reduce the resisting torque when the pedals are oriented vertically (one foot higher than the other) and increase the generated torque when the pedals are oriented horizontally (both feet at the same height). This is because a person can produce much more torque in the horizontal case by shifting weight to the forward foot.
Elliptical chainrings are not as particularly common because they are more complicated to produce and complicate the design of chain tensioners.
| {
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How do we know the number of photons in a decay? How can we determine the exact number of photons produced in a decay or other event? This has puzzled me because photons can have arbitrarily low energy and momentum, so how do we tell if two photons are produced or three photons one of which has arbitrarily low energy? (or infinitely many soft photons for that matter)
For instance we hear the singlet state of positronium decays mostly to two photons but the triplet has to decay to three photons or higher odd numbers due to charge parity or the Landau Yang theorem.
What if one of those odd photons had arbitrarily small energy, wouldn't that look like a decay to even photons? Don't we expect infinitely many soft photons in any process in any case?
|
how do experimentalists distinguish between the number of photons produced in a decay or other event?
By designing an experiment that can detect photon direction and energies of the photon , and using previously established conservation laws , energy, momentum, and quantum numbers to interpret the data.
because photons can have arbitrarily low energy and momentum,
Not if coming from a specific quantum mechanical state, as is electron positron annihilation. Energy must be conserved, so the two photons detected must have at least the energy of the mass of two electrons.
This is used in astrophysics, as an example:
The production of positrons and their annihilation in the galactic interstellar medium (ISM) is one of the pioneering topics of γ-ray astronomy. Since the detection of the 0.511-MeV line
With high enough energies electrons and positrons can annihilate to many other particles, the whole LEP experiment studied these interactions.
Every photon added in a decay reduces the probability of happening because the feynman diagrams will be depressed by 1/137, the electromagnetic coupling constant. There are some publications on this.
The paper presents results of experimental imaging of point-like sources using the 3-photon annihilations registered by a system of three high energy resolution detectors in coincidence. After filtering out the irrelevant random coincidences images of activity distributions are reconstructed. The positions of the sources are reproduced with good accuracy. The influence of random triple coincidences arising from the predominant 2-gamma annihilations, which may contribute to image noise is discussed. The analysis of experimental results is reinforced by computer simulations.
| {
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Induced electric field in circular wire around solenoid I have a stationary solenoid of radius $a$ and length $L$ with $n$ windings per unit length. There is a time varying current in the wire $I = kt$, with $k$ a constant. A conducting wire with radius $r$, mass $m$ and resistance $R$ is placed around the wire centred around the solenoid and is free to move.
I am at first interested in the induced electric field. However I am a bit confused.
I know that $$\oint_C \mathbf{E} \cdot d \mathbf{l} = - \frac{d \phi}{dt}$$
For my integral, we are used to choosing circular path, but I am unsure about what radius to choose for this? Shall I choose a radius $r$ or radius $a$? Similarly, what radius do I choose to use for the area that comes into play with the $\phi$?
And what is the difference between the cases $r \geq a$ and $r<a$?
Thanks :)
| Inside the solenoid:
The field is $$B(t)=\mu_0ni(t)$$
So for $r<a$ the flux is $$\phi_B=\mu_0ni(t)*\pi r^2$$
$$\oint E\,ds=\frac{d\phi_B}{dt}$$
$$E2\pi r=\mu_0nk\pi r^2$$
$$E=\frac{\mu_0nkr}{2}$$
Outside the solenoid:
$$B=0$$
So the magnetic flux is only due to the field inside the solenoid:
$$\phi_B=\mu_0ni(t)*\pi a^2$$
$$\oint E\,ds=\frac{d\phi_B}{dt}$$
$$E2\pi r=\mu_0nk\pi a^2$$
$$E=\frac{\mu_0nk a^2}{2r}$$
| {
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Low power loss in electricity transmission lines To reduce the heat lost during transmission of electricity, we say we increase the voltage of transmission, taking the formula $I^2R$ in consideration. Couldn't I consider $V^2/R$? If I consider the second form, increasing voltage will increase the power dissipated. No?
| $P = I^2R$ gives the power consumed by the transmission lines if $I$ is the current through the transmission line and $R$ is the resistance of the transmission line.
$P = \frac{V^2}{R}$ gives the power consumed by the transmission lines if $V$ is the voltage across the transmission line only and $R$ is the resistance of the transmission line.
We aren't increasing the voltage across the transmission line. We are increasing the voltage of the source. The following image should clear your doubts.
The above diagram is an oversimplification of real world transmission lines and load. However, the above diagram is adequate enough to show where the question asker has made a mistake without complicating the answer.
Let the resistance (or impedance) of the transmission line be $R_t$
Let the resistance (or impedance) of the load (the devices used by industries, homes, etc) be $R_l$
Let the R.M.S voltage drop (potential difference) across the transmission line be $V_t$
Let the R.M.S voltage of the source be $V$
Let the R.M.S current through the transmission line be $I$
From Ohm's law, we have:
$$I = \frac{V}{R_t + R_l}$$
The total power consumed by all processes is given by
$$P_{total} = VI$$
The voltage across the transmission line is given by:
$$V_{t} = IR_{t} \tag{1}$$
The power dissipated by the transmission line is given by:
$$P_t = \frac{V_{t}^2}{R_{t}}$$
Since $I$ is reduced, the voltage across the transmission line is reduced (deducible from equation $(1)$). Therefore, $\frac{V_{t}^2}{R_{t}}$ also reduces. Thus, we are saving power.
| {
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What does it mean by "an atom at rest"? I was reading a document when an article about atomic clock passed by. There is statement that I don't understand
The second is the duration of 9 192 631 770 periods of the radiation
corresponding to the transition between the two hyperfine levels of the
ground state of the cesium 133 atom.
"This definition refers to a cesium atom at rest at a temperature of 0 K"
Could I ask what does it mean by an atom at rest ? Does it mean that the atom is not moving ? If it means "not moving"; how can it be possible?
| In this context, "at rest" means in a reference frame centered at the atom.
In practice, you can't get any system to exactly 0 K, nor can you make an observation from directly on the atom.
But, you can plot the periods of radiation relative to temperatures that are very low and to make any necessary special relativistic tweaks (which would be tiny, but we are talking about ten significant digit precision here, so even a tiny tweak might be relevant) to the observations based upon average velocity at the temperature at which the a cesium atom would be moving at that temperature. You can then extrapolate from this data to the value it would have in theory if the data were analytically extrapolated to O K and the rest frame of the atom based upon your measurements at various very low temperatures.
| {
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Can a neutrino antenna be made one day? I am wondering if it is theoretically possible that some kind of device/material could absorb neutrinos much better than everyday materials (preferably non-thermal absorption). This could enable a sort of neutrino antenna.
| How does an electromagnetic antenna work at the photon, quantum mechanical level? By inducing varying electric fields which induce accelerated charges which radiate coherently photons which build the classical electromagnetic wave.
The neutrino interacts with the weak interaction , is a fermion and can in no way be in a way similar to a photon, so cannot be regimented as the photon can.
The neutrino is produced in weak interactions, beta decays naturally, and in interactions in accelerators and cosmic rays. Let us see whether a neutrino beam that could be modulated is possible.
As in radioactivity they participate in three body decays there is no way to control their energy and direction so radioactively produced they cannot be coherent in any sense.
This leaves accelerators and cosmic rays. No way to control cosmic ray neutrinos of course.
If muon beams are produced in accelerators from pion decays, at the same time there will be a neutrino beam from the two body decay. If the production of the muon beam is modulated, the production of neutrinos will be modulated The problem will be on the detection side, since there are no receiving antennas other than huge detectors, like the OPERA.
What makes neutrinos attractive for gathering signals from the sun and stars, the weak interaction, prohibits any practical uses for the beam.
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Why doesn't saturation current in the photoelectric effect depend on the frequency of light absorbed by the metal emitter? If current $I$ is given by $I = nAev$, where $n$ is the number of electrons per unit volume, $A$ is the area, $e$ is the charge of an electron and $v$ is the velocity of the electron, it must mean that the current increases with increase in velocity of the electron which increases with the frequency of light incident on the metal emitter. Why doesn't then saturation current increase with increase in frequency?
| Well in the equation $I=neAV$ with the increase in $V$, $I$ increases only if n, e, A does not decrease . Now just think if you increase the drift velocity of the electron then they would move from their position and the no of electrons per unit volume (n) should decrease. Thus the resultant effect is that I does not change.
Also you want to increase the velocity of the electron keeping $n$ constant so that would mean that the electrons are randomly moving in a given region only and hence their contribution to the net current is zero.
| {
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Halbach array and levitation Full disclosure: I am a total novice when it comes to physics apart from being very curious and basic physics classes at school level (long long ago).
Using 16 mm diameter disc magnets. I am attempting to make a magnetic track and levitating vehicle. First tries quickly taught me that the influence of the "other side" of the magnet, plays havoc in trying to keep my vehicle levitated. I eventually got a basic levitation in a confined block (track), but it was VERY unstable. The slightest movement would collapse the levitation attracting my repelling magnets to the attracting corner of my track's repelling magnets.
So the vehicle would fall on one side, and get stuck to the side of my track magnets, overcoming the repelling force on that one corner.
This eventually led me to discover the "Halbach array". Problem solved it seemed, and after 3 prototypes I successfully build a nice magnet block consisting of 5 x 5 disc magnet stacks arranged in the Halbach array formation.
I have a successful "weak" and "strong" side but upon testing, I discovered that the strong side had a mix of positive and negative flux(es)(?)
This poses a problem for my levitation plans, as in my head I was under the impression that the strong side would be strong in one type of field (+ or -), and not a mix of the two.
Researching magnetic shielding is what eventually got me learning about the Halbach array, but evidently, that won't be my solution. Is there a different configuration/technique to get a stronger flux on one pole and a weaker (much weaker) flux on the other?
Or did I mess up the Halbach configuration?
| The only stable ferromagnetic system I know is the "Levitron", which levitates a spinning magnet. That might not be suitable for your project. So instead you can use rollers or air bearings to the sides of your vehicle to provide stability. Those rollers/bearings would not need to support the weight of the vehicle; would only need to provide a restoring force when the vehicle wanders sideways.
| {
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How does one measure the curvature $k$ in FLRW metric? How does one measure the curvature parameter $k$ in the FLRW metric? $$ds^2=-c^2dt^2+a^2(t)[\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2]$$ In particular, what is the convenient equation (involving $k$) that is/can be used to measure $k$?
EDIT I'm looking for an answer that will explain the measurement of curvature $k$ with the same clarity as the measurement of spring constant $\kappa$ from the one-dimensional simple harmonic equation $F=-\kappa x$ i.e., having measured the applied force $F$ (can be done with a spring balance may be) and the corresponding displacement $x$ (by a meter rule), one can measure $\kappa$. Similarly, if the equation involving curvature $k$ contains non-trivial physical quantities (such as the components of Riemann curvature tensor etc), I would like to know how each of them is measured.
| You simply measure the ratio of the circumference of a circle to its radius.
Take a spatial submanifold, and for convenience we'll take $a=1$ (the units of the radial distance can always be chosen to make $a=1$ as any chosen time). Then the spatial metric becomes:
$$ d\ell^2=\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2 $$
Draw a circle with yourself at the origin. To get the circumference of the circle we integrate around the equatorial angle $\phi$ while keeping $r$ fixed and $\theta$ fixed at $\pi/2$. Since $dr = d\theta = 0$ the metric becomes:
$$ d\ell^2=r^2\sin^2\theta d\phi^2 $$
The circumference is then:
$$ C = \int_0^{2\pi}\,rd\phi = 2\pi r $$
which shouldn't surprise us unduly :-)
Now we take a measuring tape and measure out the distance to the circle. In this process we are keeping $\theta$ and $\phi$ fixed so $d\theta = d\phi = 0$ so our metric becomes:
$$ d\ell^2=\frac{dr^2}{1-kr^2} $$
So the distance we measure is:
$$ R = \int_0^r\,\frac{dr}{\sqrt{1-kr^2}} $$
The integral depends on the sign of $k$. For positive $k$ (closed universe) we get:
$$ R = \frac{\sin^{-1}(\sqrt{k}\,r)}{\sqrt{k}} $$
and for negative $k$ (open universe) we get:
$$ R = \frac{\sinh^{-1}(\sqrt{|k|}\,r)}{\sqrt{|k|}} $$
To find $k$ simply substitute $r = C/2\pi$, where $C$ is our experimentally measured circumference, and solve the resulting equation for $k$.
| {
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How is entropy a state function? Is there only one reversible way to move from one state to another?
If we consider two states $A$ and $B$ on an isotherm and we move from $A$ to $B$ by first reversible isochoric process and then reversible isobaric process. Now the path followed should be reversible since both the processes were reversible. But what about simply following the reversible isothermal process?
According to me both processes should be reversible. Now entropy is the heat added reversibly to move from one state to another divided by the temperature at which it is added. But we know that the heat added to the system is different in both the cases. Then how is entropy a state function?
| Your question goes right in the kernel of the meaning of the term state function.
A state function is a function defined over all possible states of the system such that its value for every state does not depend on how the system reached the state. Each state has a definite and unique value for the given state function.
The state $A$ has a definite value for the state function entropy, $S(A)$. The same for the state $B$, which gives $S(B)$. Thus the difference in entropy between the states $A$ and $B$ is simply $\Delta S=S(B)-S(A)$ and this value does not depend on the process that takes $A$ to $B$. The difference $\Delta S$ between $A$ and $B$ exists even for irreversible paths and it has always the same value.
In the case of entropy, there is some subtlety though. The way we calculate the difference $\Delta S$ is always
$$\Delta S=\int_{\mathrm{rev}}\frac{dQ}{T},$$
where the integral has to be computed through a reversible process. There is a plenty of reversible process from $A$ to $B$ but we just choose the simplest one for calculations.
| {
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How a moving car becomes electrically charged? Car has been electrically charged as it travels along the road.how is this possible?
| Look up Triboelectricity.
When objects come into contact with one another, temporary chemical bonds form between the touching objects. When these are afterwards broken mechanically, they can leave the two objects with an surfeit / dearth of electrons on either side of the broken bond.
This process happens continuously for moving objects like a car and particularly for aeroplanes. Air and road continually make and break contact with the vehicle through the latter's friction with the former.
| {
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Misconceptions on Electronics and circuits So i had a question regarding basic circuits
*
*What do negative currents and voltages mean.? This really stumped me for a while as negative current and voltage seemed odd to me if someone can clarify much appreciated.
p.s Sorry didn't realise there was too many questions asked before that were too broad, so i just edited it to include just this one as its the one that stumped me the most in class.
| *
*Power dissipated in a resistor $R$ is $I^2R$
*Positive charges can also contribute to the current in gasses. Only electrons can move in a solid conductor.
*What do you mean by 'just a wire'. Does your wire have resistance?
*Yes, A is at a higher electric potential than B. It's the result of chemical reactions inside the Battery.
Batteries. Batteries use a chemical reaction to do work on charge and produce a voltage between their output terminals. The basic element is called an electrochemical cell and makes use of an oxidation/reduction reaction. An electrochemical cell which produces an external current is called a voltaic cell.
Batteries - Hyperphysics
*Power is Joules/sec. $VI$ is indeed $Js^{-1}$.
*Directions of currents and voltage differences are sign sensitive. If $V_A-V_B is +10V$ then $V_B-V_A = -10V$
| {
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Rotation by a given distance in given direction on a Bloch sphere I'm using the following matrix to rotate a states by a distance given by $\theta$ in a direction given by $\phi$:
$$
U = \left(\begin{array}{cc} \cos{\theta\over{2}} & -e^{- i\phi}\sin{\theta\over{2}} \\ e^{i\phi}\sin{\theta\over{2}} & \cos{\theta\over{2}} \end{array}\right)
$$
It works well only in the case, when my initial state is a state $|0\rangle$ or $|1\rangle$. E.g. when I get new state by:
$$|\psi\rangle = U_{\theta,\phi}|0\rangle$$
distance between $|0\rangle$ and $|\psi\rangle$ is equal $\theta$ but when I do it for any other state, e.g. $|+\rangle$ - distance between two states after that operation isn't always equal to $\theta$ and depends on $\phi$ parameter.
Am I doing something wrong or it's normal and expected result? If so, is there any matrix which will allow me to get the results that I expect?
|
when I do it for any other state, e.g. $|+\rangle$ - distance between two states after that operation isn't always equal to $\theta$ and depends on $\phi$ parameter.
That's the expected behaviour. Every rotation of the Bloch sphere has stationary points (or, in Hilbert-space language, every unitary has a basis of eigenvectors), otherwise known as the rotation axis, which means that the Bloch-sphere distance between $|\psi⟩$ and $U|\psi⟩$ can always be zero. Thus, the Bloch-sphere angle between $|\psi⟩$ and its image under a rotation by angle $\theta$ can be any number between zero and $\theta$, depending on where $|\psi⟩$ is with respect to the axis of the rotation.
If what you want is a transformation from the Bloch sphere into itself such that every point ends up at a fixed distance from itself, then no, I don't think this is possible - but, more importantly, it does not represent a rotation in the Bloch sphere nor a unitary on Hilbert space, so even if it exists it is not particularly useful.
| {
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Why does a laboratory centrifuge cause heavier particles to go to the bottom of the tube? From what I have read online, I can understand that there is a "centrifugal force" that pushes the heavier objects away from the center, but I cannot understand why.
I asked my teacher and she told me that I have to think about what changes between a particle with larger mass vs. smaller mass. I had thought about this, and decided that it must be the radius, since the particles with lower mass are pulled towards the center, and must therefore have a smaller radius of rotation than the higher mass particles. I then tried to plug in numbers into equations in order to see if I was correct. I used the net force equation I was taught in class
$$
\sum \mathbf F = 4\pi^2 r f^2 m
$$
which I rearranged for $r$
$$
r = \frac{\sum \mathbf F}{4\pi^2 f^2 m}\,.
$$
I solved for $r$ with one mass (say 10 kg) and then a larger mass (say 15 kg), keeping all other variables constant. To my surprise, the radius of the larger mass was smaller than the one with the smaller mass. This would mean that the smaller particles go to the bottom of the tube, not the heavier ones. I assume that I did something wrong since this is not what happens in real life. I think it might have something to do with the variables that I kept constant actually changing when the mass does. In any case after much thought I was not able to come up with a reasonable explanation for how the apparatus works. What did I do wrong? What is the correct reasoning and explanation?
| It is not mass but density which is the important parameter.
What you have is a "local" value of $g$, the "gravitational field strength" which is $R \omega^2$ where $R$ is the radius of the orbit and $\omega$ is the angular speed - this provides your centrifugal force which is the weight of a mass in this local gravitational field.
The all you need to do is to use Archimedes principle to find the upthrust on a particle which is equal to the weight of fluid displaced.
The denser material will "sink" ie move towards the outside of the rotation.
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Photoelectric effect:- Reduction of wavelength increases current? I did a question in which, the intensity of the incident radiation on a metal surface was kept constant but the wavelength of the photons has been reduced. The question inquired what will be the effect on the maximum photoelectric current? The initial wavelength was smaller than threshold wavelength of the metal surface.
My thinking was since the intensity remains constant, thus the number of photons emitted from the source remains constant and thus the number of electrons emitted from the metal surface. And since number of electrons per unit time isn't changed, the current will remain the same.
However, the answer key stated "Fewer photons (per unit time) so (maximum) current is smaller"
How does decreasing wavelength (equivalent to increasing the energy of photons) result in a fewer photon emission?
| Intensity of light means total energy per unit time per unit area.As in later case wavelenght of light was decreased so the energy of individual photon will increases but the intensity was kept constant so there should Be less number of photon falling per unit time per unit area.Just remember intensity of light is not determined by either the energy of each photon or the number of photon but as combination of both because total energy depend upon total number of photon and energy of each photon.
| {
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Is the existence of a photon relative? If an observer passes an electron, in such a way that the observer is accelerating, the observer would see photons because accelerating charges induce electromagnetic waves.
But from point of view of the electron or an inertial observer there is no magnetic field nor an acceleration which could 'produce' an electromagnetic wave.
So for the first observer there exists a photon but not for the second observer. How is this possible, are photons relative?
| The short answer is yes. When one tries to generalise the theory of quantum fields in Minkowski spacetime to more general spacetimes, one finds that several familiar features of the theory are absent or ambiguous. For instance, it is not generally possible to unambiguously define a vacuum state when studying QFT in a curved spacetime. That is to say, a state which one observer sees as a vacuum, another may see as a thermal bath of particles.
Of course, the spacetime of concern here is just Minkowski spacetime, not a curved spacetime. However, the same problems one faces with QFT in curved spacetime appear also when studying QFT in flat spacetime in some complicated coordinate system, such as accelerating coordinates. Minkowski spacetime in accelerating coordinates is sometimes called Rindler spacetime (although it is the same spacetime), and a lengthy QFT calculation reveals that if our system is in the vacuum state according to an inertial observer, then our accelerating observer will see a background of particles with a perfect black body energy spectrum, with temperature
$$ T = \frac{\hbar A}{2 \pi c k_B} \,,$$
where $A$ is the magnitude of the proper acceleration. This is known as the Unruh effect.
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Why is the mol a fundamental physical quantity? I am starting to study physics in detail and as I read about physical quantities, I was puzzled why mol (amount of substance) is taken as a physical quantity.
A physical quantity is any quantity which we can measure and has a unit associated with it. But a mol represents the amount of substance by telling the number of particles (atoms, molecules, ions, etc.) present. So it is a pure number and numbers are dimensionless. So mol should not be considered a physical quantity.
Also, fundamental physical quantities should be independent of each other. I am wondering whether mass and mol are independent. This is so as they surely affect each other as we can evidently see while calculating the number of moles and using the mass of that sample for calculation.
So how is the mol a fundamental physical quantity and independent of mass?
| Is it a fundamental number in nature? It's (currently) a number resulting from atomic structure (fundamentally defined by the masses of quarks, Planck's constant and the way quantum mechanics works) and our definition of the gram, which is based on the international Kg prototype. Avogadro's constant is currently defined by experiment, and therefore has no absolute "right" number, just an agreed working definition.
This is a messy way to define things though, and there are many arguing that the Kg should be defined in terms of a particular element and Avogadro's number, which would put it on a more "fundamental" level in my book. (See https://en.wikipedia.org/wiki/Kilogram#Avogadro_project )
This would mean "fixing" Avogadro's constant by simply picking a number, then defining the Kg in terms of this, in the same way the second was "fixed" in terms of the so-many-oscillations of a particular frequency of light, rather than being a 60th of a 60th of a 24th of one rotation of Earth (a messy, variable number).
| {
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Adiabatic free expansion of real (Van der Waal's model) gas below/at/above inversion temperature For an adiabatic free expansion, $W = 0$ and $Q=0$. Therefore, by the first law of thermodynamics, $\Delta U = Q -W = 0$.
For a Van der Waals model of a real gas, $\Delta U = n C_v \Delta T - a\,n^2 \left(\frac{1}{V_2}-\frac{1}{V_1} \right)$. This means that $n C_v \Delta T - a\,n^2 \left(\frac{1}{V_2} -\frac{1}{V_1} \right) = 0$, or $n C_v \Delta T = a\,n^2 \left(\frac{1}{V_2} -\frac{1}{V_1} \right)$. Now, since the gas is expanding, $V_2 > V_1$. So, $a \, n^2 \left(\frac{1}{V_2}-\frac{1}{V_1} \right) < 0$. Therefore, $n C_v \Delta T<0$. This means that the temperature of the gas decreases in an adiabatic free expansion.
Will the temperature always decrease? I mean, what if the gas is above its inversion temperature or at it? Does inversion temperature play no role in adiabatic free expansion?
| Your calculation indeed shows that the temperature of the van der Waals gas always decreases in adiabatic free expansion. It is only when the gas flows between two regions of different pressures that the sign of the variation of temperature depends on the temperature. Enthalpy, rather than energy, is conserved in this case. Using the expression of pressure for the van der Waals gas
$$P={Nk_BT\over V-Nb}-{N^2a\over V^2}$$
enthalpy reads
$$\eqalign{
H=U+PV&\simeq {3\over 2}Nk_BT-{N^2a\over V}
+\left({Nk_BT\over V-Nb}-{N^2a\over V^2}\right)V\cr
&={3\over 2}Nk_BT-{2N^2a\over V}
+Nk_BT\left(1-{Nb\over V}\right)^{-1}\cr
&\simeq {3\over 2}Nk_BT-{2N^2a\over V}
+Nk_BT\left(1+{Nb\over V}\right)^{-1}\cr
&={5\over 2}Nk_BT-{2N^2a\over V}+{N^2bk_BT\over V}\cr
&={5\over 2}Nk_BT+{N^2\over V}\big(bk_BT-2a\big)\cr
}$$
which can be rewritten as
$$H=C_pT+{N^2bk_B\over V}(T-T_i)$$
where the inversion temperature is
$$T_i={2a\over bk_B}$$
For $T>T_i$, the expansion leads to an increase of the temperature.
| {
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Uncertainty: $f(n) = K \log n$? I've been searching for the derivation of Shannon's info theory derivation and I landed upon this page from Stanford:
http://micro.stanford.edu/~caiwei/me334/Chap7_Entropy_v04.pdf
They've repeated referenced formulas (page 9) like $f(n) = K \log n$, $f(mn) = f(m)+f(n)$ where $f$ seems to represent uncertainty.
I'd really like some idea as to how this formula came about?
Thank you!
| These aren't referenced equations. These are equations which are logically derived from the above text.
(The rest of the answer references the text -- I will put another link here for convenience: http://micro.stanford.edu/~caiwei/me334/Chap7_Entropy_v04.pdf.)
Let's try to go through the logic. Equation (9) simply defines the function $f(N)$ as being the uncertainty $S(1/N,\ldots,1/N)$ of picking one object out of $N$ objects with uniform probability. This function is assumed to be monotonically increasing (the more objects, the higher the uncertainty). Simple enough.
Next, let's break the group of $N$ objects into $m$ groups of $n_k$ objects with $k=1,\ldots, m$. We then break the experiment of picking one of the $N$ objects into two steps: picking the group then picking the element of the group. Picking group $k$ comes with probability $p_k\equiv n_k/N$. Thus, if $A$ is the process of picking the group, $S(A)=S(p_1,\ldots,p_m)$. If $B$ is the process of picking the element of the group, $S(B|A)=f(n_k)$ after we picked the group. All in all, using hypothesis 3) from page 8, we have that the uncertainty of the whole experiment is
$$f(N)=S(AB)=S(A)+\sum_{k=1}^{m}p_kS(B|A)=S(p_1,\ldots,p_m)+\sum_{k=1}^{m}p_kf(n_k)$$
This equation determines the form of $f$ almost uniquely. In particular, take the case when $n_k=n=N/m$ for all $k$ (uniform groups), so that $p_k=1/m$. Then this relation becomes
$$f(N)=S(1/m,\ldots,1/m)+\sum_{k=1}^{m}\frac{1}{m}f(n)=f(m)+f(n)$$
Which must now hold for all $m$ and $n$. The only continuous increasing functions which satisfiy $f(mn)=f(m)+f(n)$ for all $m$ and $n$ are precisely
$$f(x)=K\log{x}$$
For some constant $K$. This was not pulled out of thin air -- it was logically deduced.
The rest of the proof is straightforward. Go back to the general case with arbitrary $n_k$. Then
$$K\log{N}=S(p_1,\ldots,p_m)+K\sum_{k=1}^{m}p_k\log{n_k}$$
Since $\sum_{k}p_k=1$, we can rewrite this as
$$S(p_1,\ldots,p_k)=K\sum_{k=1}^{m}p_k\left(\log{N}-\log{n_k}\right)=-K\sum_{k=1}^{m}p_k\log{p_k}$$
And Eurika! We have derived Shannon entropy from first principles!
I hope this helped clear up some confusion!
| {
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Miller indices vs Laue indices? I am looking at the difference between the Miller indices and Laue indices in crystal structures. I will denote the former by $hkl$ and the latter by $HKL$.
I understand that $hkl$ must be coprime integers, however, I have also heard that $hkl$ represent families of lattice planes. For the case of e.g. the bcc lattice these seem contradictory since $(200)$ rather then $(100)$ forms faimiles of lattice planes.
For $HKL$ it appears that they to must be integers (but not necessarily coprime). These are the indices used to determine if we have a systematic absence or not. But this seems to be assuming we only get diffraction peaks from families of planes which are families of lattice planes - yet as far as I can tell you will get a diffraction peak from the planes $(0.5 \ \ 0.5\ \ 0.5)$ .
Please can someone therefore explain to me the difference, what values they can take and how they can be defined?
| The distinction between the two indices is quite subtle.Perhaps it is better to start from the laue condition. Let's say I have rows of atoms with spacing $a$. We are in the fraunhofer limit so when two x-ray beams come in and scatter off neighboring atoms they come in parallel and go out parallel. We can then consider the path length difference between the incoming beams and out going beams i.e $ \Delta^1 = \Delta_1 - \Delta_2= a \cos \alpha_1 - a \cos \alpha_2 $ where $\alpha_1$ is the angle between the incoming beam and the row of atoms and $\alpha_2$ is the angle between the outgoing beam and the row of atoms.The condition for constructive interference is that $\Delta^1 = h \lambda$ where $\lambda$ is the wavelength and h is an integer. This analysis can be done in all three directions and then the condition has to be true in all three directions giving $\Delta^2 = k\lambda \text{ and } \Delta^3 = l\lambda $.(There will be corresponding angles for the other path length differences).
So we can define incoming unit vectors $s_1 = (\cos \alpha_1,\cos \beta_1, \cos \gamma_1)$ and outgoing $s_2 = (\cos \alpha_2, \cos \beta_2, \cos \gamma_2)$. From this we consider the scatter wave $s_1-s_2= G \lambda $ with $G = \frac{1}{a}(h,k,l)$ since $s_1 ,s_2$ are unit vectors $G$ will be perpendicular to the vector that bisects the angle between $s_1 $
and $s_2$.
Now comes the first crucial point, draw planes that are parallel to the plane that bisects the angle between the incoming and outgoing beams. These are the lattice planes and they define the miller indices.Laue indices will be defined by another set of planes. Back to the lattice planes , call the distance between each successive plane $d_{hkl}$
this is the distance that appears in Bragg's law.When we pick a primitive cell then (hkl)are our miller indices.
What about laue indices? Well we look at the bragg's law $n \lambda = 2 d_{hkl} \sin \theta_n $ .
Now comes the second crucial point, note that the $m^{th}$ order reflection off the (hkl) plane can be regarded as $n^{th}$ order if I $ \textit{define}$ new planes at distance $\frac{d_{hkl}n}{m}$ i.e $ m \frac{n}{m}\lambda=2 (d_{hkl}\frac{n}{m})\sin \theta_m $ This family of planes is then given by the notation $ \frac{m h}{n} \frac{m k}{n} \frac{m l}{n}$ without parantheses. These are the laue indices, note how they can have common factors even if we started off with a primitive cell.
| {
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if white things reflect light and mirror reflect light why don't they look the same We learned at school that white object reflects all the light that falls on it. We also learned that a mirror reflects all light as well. However, we cannot see ourselves in a white object while we can see ourselves in a mirror.
What makes a mirror different from a white surface?
If both white surface and mirror reflect all the light that fall on them, then why don't they look the same?
| A white object only appears white if white light is striking it. If only red light is striking it, it appears red. A mirror has less distortion than other surfaces so it reflects light in a straight line. You don't see the surface of the mirror but rather the objects from which the light originates.
NB, when I speak about white or red light, it's important to remember that light itself has no colour. Colours are merely how our brains interpret different wavelengths.
| {
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Kinetic, potential and total orbital energy in General Relativity In Schwarzschild geodesics the total orbital energy $E$ is
$$E = \dot{t} \left( 1 - \frac{r_{\rm s}}{r} \right) m \, c^2$$
with the time dilation factor $\dot{t}$ in dependence of the local velcity $v$
$$\dot{t} = \frac{1}{\sqrt{ \left( 1-\frac{r_{\rm s}}{r} \right) \left( 1-\frac{v^2}{c^2} \right)}}$$
so plugged into the equation for $E$ we get
$$E = \frac{m \ c^2 \ (r-r_{\rm s})}{\sqrt{r \ (r_{\rm s}-r)(v^2/c^2-1)}}$$
which seems to be
$$E = m \ c^2 + E_{\rm \ kin} + E_{\rm \ pot}$$
But how would one factor out the kinetic and the potential component of the total Energy in terms of the coordinate derivatives $\dot{r}, \dot{\phi}, \dot{t}$ or in terms of $v^2=v_{\perp}^2+v_{\parallel}^2$ (radial and transverse components)?
The other constant of motion, the angular momentum, is easy to get because with
$$\dot{r} = v_{\parallel} \sqrt{\frac{1-2 M/r}{1-v^2}} \ , \ \dot{\phi} = \frac{ v_{\perp}}{r \sqrt{1-v^2}}$$
we get
$$L = m \ \dot{\phi} \ r^2 =\frac{m \ v_{\perp} \ r}{\sqrt{1-v^2}}$$
but what about $E_{\rm \ kin}$ and $E_{\rm \ pot}$? Those seem to be very different than with Newton or Special Relativity, at least one of them since the sum does not match up. I only managed to calculate to total energy but failed to split it into it's components.
| Your expression for the total energy can be written as:
$$E=mc^2\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2}}}$$
This is slightly wrong because in general relativity under spherical symmetry you have:
$v_{light}=c(1-\frac{2GM}{rc^2})$
in the radial direction and:
$v_{light}=c\sqrt{1-\frac{2GM}{rc^2}}$
in the pure non-radial direction.
You really need to compensate the Lorentz factor for this, and get:
$$E=mc^2\left(\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})^2(\hat{r}\cdot\hat{v})^2+(1-\frac{2GM}{rc^2})|\hat{r}\times\hat{v}|^2\right)}}}\right).$$
This can be re-written as:
$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})(\hat{r}\cdot\hat{v})^2+|\hat{r}\times\hat{v}|^2\right)}}}\right).$$
Written in this form, the Schwarzshild radius, the innermost stable circular orbit and the photon radius is not so hard to find. You can still find the classical kinetic and potential energy in the weak field limit.
| {
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For a tall skyscraper, what are the requirements for young/shear modulus? I am teaching physics 101 for the first time and am discussing stress/strain. As an illustrative example to students, I'd like to make the broad statement, "The primary supports in skyscrapers need a high Young's modulus (to support the structure's own weight without deformation) but a low shear modulus (to allow for sway in case of wind/earthquake/etc)." However, I'm neither a structural engineer nor have I taught this material before so it's somewhat fresh to me. In a general sense, are those statements true?
| You are confusing the modulus of a material, with the stiffness of an object made from that material.
Let's assume the skyscraper has a steel frame. Steel is isotropic, so like all isotropic materials, there is an equation linking its Youngs modulus, shear modulus, and Poisson's ratio: $$E = 2G(1 + \nu).$$
Since for most structural metals Poisson's ratio is about $0.3$, $G$ is about $2.6$ times smaller than $E$.
On the other hand, the vertical members of the skyscraper frame might be considered as a uniform square bar with side $a$ and length $L$. (This is greatly over simplified, but it illustrates the point). We can then find the cross section area of the bar, $A = a^2$, and its second moment of area $I = a^4/12$.
The stiffness of each bar under a vertical compressive load (i.e. the weight of the building) is $$k_{\text{axial}} = \frac{EA}{L} = \frac{Ea^2}{L}$$
and the bending stiffness for a wind load is $$k_{\text{bending}} = \frac{3EI}{L^3} = \frac{Ea^4}{4L^3}.$$
The ratio of the stiffnesses is $$\frac{k_\text{axial}}{k_{\text{bending}}} = \frac{4L^2}{a^2}.$$
So the huge difference in the two stiffness (or flexibility) terms actually has nothing to do with the material itself, but with the geometry of the bar.
In a real design, you would change the geometry to increase the bending stiffness, by replacing a single thick solid bar with an arrangement of thinner bars, held apart by a weaker (but cheaper and lighter) material such as concrete. This would make no difference to the axial stiffness, but the second moment of area would then depend on the distance separating the thinner bars, not on the thickness of each individual bar.
Incidentally, note that the formula for the bending stiffness is written in terms of $E$, not $G$. In fact, there is an additional stiffness term which does depend on $G$, but unless $L < 10a$, it is too small to bother about. The only "elementary" stiffness formulas that I can think of which directly involves $G$ are for the torsional stiffness of an object, where the deformation of the material is mainly in shear.
These formulas and basic ideas in this answer will be in any "strength of materials" textbook or website.
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Why do all electromagnetic waves travel at the same speed when travelling through vacuum? What does my teacher mean when he says that all electromagnetic waves travel at the same speed when travelling through a vacuum? If you may, please answer as simple as possible.
| Assume you are walking down the road. You carry a little stick with you. Just for fun you decide to wiggle the stick rhythmically up and down at the rate of one up/down wiggle per second (you are a bit of an olympic expert at stick wiggling so it's very accurate and reliable). Your stick wiggling is at 1Hz. The speed at which you walk down the road is not related to the rate at which you wiggle the stick. You can walk down the road at any speed you like and still wiggle at 1Hz.
Have you got it now ?
The speed of transmission of an e-m wave is not related to the frequency/wavelength.
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Multiple star system, stable orbits? Inspired by worldbuilding SE, I know that there are relatively stable star systems with two or three suns, but any more than that and they start to become very unstable (e.g. trapezium systems), but I'm more interested in the concept of >3 stars, each of similar mass. How could they be arranged in a stable (for a few billion years), non hierarchical manner? I tried sketching out a few possibilities but lack the understanding of how suns interact with each other (heat and pressure being the foggiest elements).
Are there any stable >3 star, star orbits, and if so, what do they look like?
edit to clarify: I am looking for an answer within a single solar system, not a star cluster / galaxy (which would fit the question)
| If space is not a concern (so no perturbations from the outside) then there should be no limit on the number of stars. Because you can always take a duplicate of your current system place them in orbit of each other far enough apart from each other such that the tidal perturbations can be neglected.
Instead of doubling the number of stars you might also be able to tipple the number of stars, for example by placing the copies into a figure eight pattern.
| {
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Does Earth experience any significant, measurable time dilation at perihelion? Is there any measurable time dilation when Earth reaches perihelion? Can we measure such a phenomena relative to the motion of the outer planets?
| The GPS depends on corrections to the timing from General and Special relativity because the satellites are in a smaller gravitational field and they are running with a high enough velocity.
Because an observer on the ground sees the satellites in motion relative to them, Special Relativity predicts that we should see their clocks ticking more slowly (see the Special Relativity lecture). Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion .
Further, the satellites are in orbits high above the Earth, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away (see the Black Holes lecture). As such, when viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground. A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.
The combination of these two relativitic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)! This sounds small, but the high-precision required of the GPS system requires nanosecond accuracy, and 38 microseconds is 38,000 nanoseconds. If these effects were not properly taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and errors in global positions would continue to accumulate at a rate of about 10 kilometers each day! The whole system would be utterly worthless for navigation in a very short time.
The Time dilation - Earth & Jupiter , should tell you that corresponding differences will be found between the perihelion and aphelion of the earth.
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Why does the proton have a parity $P=1$ and an anti-symmetric wave function? The overall parity of a proton is 1 because the parity of a quark is 1. How does this go together with the proton's wave function being anti-symmetric?
Is the reason for the proton's wave function's anti-symmetry the fact that in $SU(3)_C$ you consider the $u,d,s$ quark flavors to be identical for the strong interaction?
| The spin-flavor part of the proton wavefunction,
$$
|p_\uparrow\rangle= \frac{1}{\sqrt {18}} [ 2| u_\uparrow d_\downarrow u_\uparrow \rangle + 2| u_\uparrow u_\uparrow d_\downarrow \rangle +2| d_\downarrow u_\uparrow u_\uparrow \rangle \\ - | u_\uparrow u_\downarrow d_\uparrow\rangle -| u_\uparrow d_\uparrow u_\downarrow\rangle -| u_\downarrow d_\uparrow u_\uparrow\rangle
-| d_\uparrow u_\downarrow u_\uparrow\rangle -| d_\uparrow u_\uparrow u_\downarrow\rangle -| u_\downarrow u_\uparrow d_\uparrow\rangle ]
$$
is completely symmetric under interchange of any two quarks with each other; even though, individually, flavor and spin interchanges are of mixed symmetry: this is the cornerstone of the model, really.
The position (orbital angular momentum) wavefunction is then an S-wave, symmetric, so, given the total antisymmetrization by the color, normally omitted, the three fermion constituent quarks are in a fully antisymmetric state, as they should be.
Consequently, since S contributes + to the overall parity, this parity then turns out to be ++++ $\leadsto$ + for the overall parity of the proton.
The idea linking generalized Pauli antisymmetrization of fermions to parity amounts to appreciating that there is no P wave spatial component, so the parity of the nucleon is the product of the parities of the three constituents.
| {
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Is the speed of light the only speed that exists? Well, it seems to me that if I move faster in space I move slower in the dimension of time which is orthogonal to the dimension of space.
All speeds are then equal. Is this statement correct?
| My answer incorporates features of the earlier answers, but tries to clearly disentangle the implied uses of "speed" in space and in spacetime [when a phrase like 'slower in the dimension of time' is used].
"Massive objects (like a basketball) all move at constant 'speed' c in spacetime"
really means that
*
*its 4-momentum vector can be normalized to a "unit" 4-velocity vector
with Minkowski-norm c.
*It has a spatial speed (slope) v < c because
the spatial component of its 4-momentum has an absolute-size that is
smaller than that of the time-component of its 4-momentum.
"Massless objects (like a light-signal) all move at constant 'speed' zero in spacetime"
really means that
*
*its 4-momentum vector has Minkowski-norm zero [and thus can't be normalized].
*It has a spatial speed (slope) c because the spatial component of its 4-momentum has the same absolute-size as that of the time-component of its 4-momentum.
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Can plasma retain a magnetic field? even for the tiniest fraction of a second? If I subject plasma to a magnetic field, can it retain that field after the magnetic field suddenly ceases(the magnetic field disappears is a planck second)
for even the tiniest fraction of a second is the plasma "ferromagnetic"?
| The magnetic field in a conductor (such as a plasma) obeys a version of the diffusion equation:
$$
\frac{1}{\mu_0 \sigma} \nabla^2 \vec{B} = \frac{\partial \vec{B}}{\partial t}
$$
where we have assumed negligible magnetic susceptibility ($\mu \approx \mu_0$) and uniform conductivity $\sigma$. This means that if we create a magnetic field in some region of a plasma, it will eventually decay away, just like a solute concentrated in a small part of a fluid will eventually spread out and reduce its concentration to practically zero.
This does not happen instantaneously, though; the time scale on which this occurs is on the order of
$$
\tau \approx \mu_0 \sigma L^2,
$$
where $L$ is the approximate size of the region of magnetic field non-uniformity. We can see that the larger the conductivity is, or the larger the region of non-uniform magnetic field, the longer the field will last; but eventually it will decay away.
Finally: while this time scale for demagnetization can be quite long, and so the plasma can behave a little like a ferromagnet in that regard, it is not what I (or most physicists, I think) would refer to as proper ferromagnetism. In particular, the behavior of a magnetic field in a plasma can be described entirely classically, is largely due to the large-scale motion of charges, and does not change radically when plasma gets too hot. None of the above statements are true of ferromagnets.
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Why does the Redfield equation model thermal relaxation while the Lindblad equation does not? In open quantum systems, we model a process known as thermal relaxation. What is this process, and why is it that only the Redfield equation models this process, and the Lindblad equation doesn't?
| In open quantum systems, the evolution of the total System is unitary. The freedom degrees of the environment is large, so we always think the environment will not evolve and that is the Born approximation.$$\rho(t) = \rho_{s}(t)\otimes\rho_{E}$$ The relaxation time means the time what the environment need back to equilibrium again. Redfield equation is always the Non-Markovian master equation, so we have to consider the relaxation time, the Lindblad equation is always Markovian, so we don't consider the relaxation time here.
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Is a mirror-less telescope possible? I was reading about telescopes and the Hubble Telescope for example has a 2.4m mirror which reflects lights to a sensor. Other type of telescopes use lenses to focus light to the imaging sensor.
I was wondering, is it possible to have a telescope without a mirror or lens?
So in Hubble's case, instead of having a 2.4m mirror reflecting light to a small sensor, why not have a big imaging sensor (same size as the mirror - 2.4m). Would this type of telescope have similar capabilities as the mirror one?
I know that we use mirrors because it's way cheaper and easier than building large sensors but I'm curious if a mirror-less telescope would be better / worse or just the same.
| A phase(d) array antenna is an imaging device without either a mirror or a lens. Of course, it needs a reference oscillator that is phase coherent with the incoming wave so when the two are mixed in the array, sensor element by sensor element, only the I and Q ("inphase" and "quadrature") components of the incoming wave are processed. I admit, this is conceptually easier for a radar antenna that generates its own illumination but I can imagine that, for example, by some magic if you could delay the incoming wave illuminating the array but ahead capture its carrier to phase lock your own oscillator then you could play the same game locally...
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Goldstone bosons, quark and gluon masses counting in color-flavor locking QCD Consider QCD, with three flavors of massless quarks, we like to focus on the possible Cooper paired phases.
For 3 quarks $(u,c,d)$ and 3 colors $(r,g,b)$, the Cooper pairs cannot be flavor singlets, and both color and flavor symmetries are broken. The attractive channel favored by
1-gluon exchange is known as “color-flavor locking.” A condensate involving left-handed quarks alone locks $SU(3)_L$ flavor rotations to $SU(3)_{color}$, in the sense that the condensate is not symmetric under either alone, but is symmetric under the simultaneous $SU(3)_{L+color}$ rotations. A condensate involving right-handed quarks alone locks $SU(3)_R$ flavor rotations to
$SU(3)_{color}$. Because color is vectorial, the result is to breaking chiral symmetry. Thus, in quark matter with three massless quarks, the $SU(3)_{color} \times SU(3)_L \times SU(3)_R \times U(1)_B$ (the last one is baryon) symmetry is broken down to the global diagonal $SU(3)_{color+L+R}$ group.
question:
1) How many quarks among nine ($(u,c,d) \times (r,g,b)$) have a dynamical energy gap? What are they?
2) How many among the eight gluons get a mass? What are they?
3) How many massless Nambu-Goldstone bosons there are? What are they? How to describe them?
| These questions are answered in the original literature:
1) All quarks are gapped. The nine quarks arrange themselves into an octet with gap $\Delta$ and a singlet with gap $2\Delta$.
2) All gluons are gapped.
3) There is an octet of Goldstone bosons related to chiral symmetry breaking, and a singlet associated with $U(1)$ breaking.
Postscript:
i) When pair condensates form there is a gap in the excitation spectrum of single quarks (this is just regular BCS). However, the gapped excitations may be linear combinations of the microscopic quark fields. In the present case the nine types of quark fields ($N_c\times N_f=9$), form an octet and a singlet of an unbroken $SU(3)$ color-flavor symmetry.
ii) Pair condensation and the formation of a gap take place near the Fermi surface. There is no Fermi surface for anti-quarks (if $\mu$ is positive and large), and therefore no pairing and no gaps.
iii) There is both a $U(1)$ GB (associated with the broken $U(1)_B$) and a masssless $U(1)$ gauge boson (associated with the $U(1)_{Q}$ gauge symmetry that is not Higgsed).
iv) The [8] GB correspond to spontaneous breaking of chiral symmetry. In ordinary QCD these would be quark-anti-quark states, but at high density anti-quarks decouple. A detailed analysis shows that the GBs are predominantly 2-particle-2-hole states, $(qq)(\bar{q}\bar{q})$.
| {
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Zero momentum frame in special relativity I am considering a case of collision in special relativity in which there is a change in the number of particles; for example, two particles colliding and sticking together to form a single particle. I was thinking about approaching a problem like this from the zero momentum frame, but then I realised that I did not know whether to consider the initial scenario (two particles approaching each othet) or the final scenario (a single particle travelling with speed v) to obtain the velocity of my zero momentum frame.
Something tells me that it should not matter which I choose, as if it did I could tell whether I am moving relative to some other frame, which would violate one of the principles of relativity. However mathematically I cannot see why it would be the case that these two considerations would lead to the same result for the velocity of the ZM frame.
| Like @Ofek Gillon said, it is about conservation of momentum. In any frame it is conserved, so in the ZM frame before it is 0, after the collision in that same frame it is also zero. Both ZM frames are the same.
A lot of collision physics experiments are analyzed this way because it can be simpler, and after if one needs to one can Lorentz transform to any other frame.
| {
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What determines how much power goes into each diffraction order? Imagine a grating with infinite number of slits, and the spacing D between slits is larger than the wavelength so that there are high order diffractions. In each of the diffraction directions the waves constructively interfere, but what decides the percentage of power that goes into each order?
My thinking is that each slit is a Huygens source, radiating cylindrical waves homogeneously in every direction, but due to interference, only those with constructive interference can exist. I guess the energy going into each order should be equal, which is not the case. So I am confused on how the light will distribute its energy to different orders. Thank you.
| The physical setup is:
$$
\text{light source}
\quad
\overset{\mathcal{F}}{\underset{\text{(far field)}}{\Longrightarrow}}
\quad
\text{grating}
\quad
\overset{\mathcal{F}}{\underset{\text{(far field)}}{\Longrightarrow}}
\quad
\text{observation screen}
$$
where $\mathcal{F}$ denotes Fourier transform, as known in Fourier optics.
Let a single slit be $a(x)$, and denote $\delta_D(x)$ as the Dirac comb of spacing $D$. Denote $\star$ as convolution, the grating is:
$$
g(x) = a(x) \star \delta_D(x)
$$
Assume we illuminate the grating with a point light source $\delta(x)$, it becomes plane wave when hitting the grating:
$$
\delta(x)
\quad
\overset{\mathcal{F}}{\underset{\text{(far field)}}{\Longrightarrow}}
\quad
1 \cdot g(x)
\quad
\overset{\mathcal{F}}{\underset{\text{(far field)}}{\Longrightarrow}}
\quad
\text{sinc}(\pi D_1 x) \cdot \delta_{1/D}(x)
$$
where $D_1$ ($< D$) is the width of a single slit.
We identify the intensity be $|\text{sinc}(\pi D_1 x)|^2 \cdot \delta_{1/D}(x)$, i.e. discretized $|\text{sinc}(\pi D_1 x)|^2$ sampled at $D$ period, which is also known as the diffraction orders.
Let fill-factor be defined as $D_1/D$.
*
*If $D_1 = D$, i.e. 100 % fill-factor, there will be no higher-order diffractions but only $0^{\text{th}}$.
*If $D_1 < D$, usually around 90 % fill factor, the intensity would contain difraction orders like this:
So the energy depends on the fill-factor $D_1/D$ of your grating. The larger $D_1/D$, more energy at $0^{\text{th}}$-order.
| {
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If pressure is proportional to temperature, why aren't compressed gases always hot? When you compress a gas, say within a deodorant can, it temporarily heats up, but then cools to room temperature whilst in the can. It will then cool down to below room temperature once decompressed.
It seems like the gas equation doesn't apply when the gas is cooling down within the can and the pressure remains the same. Does the equation only apply at the instant of compression? It does not apply to liquids, but exactly why does it apply to gases? Also, I have learnt that gases are harder to increase in temperature (require more heat) at high pressures. I can't fully understand why.
Also, can't this concept be used to create a perfect stirling engine? I mean if the gas within the can cools back down to exactly room temperature?
So the question in summary:
1) when and where does this gas law apply and show a proof explaining why
2) why are gases harder to increase in temperature at high pressure, does the same work vice versa?
3) could this in principal make a 100% efficient engine?
| When you compress a gas into a deodorant can it does indeed heat up. Then if you let the can cool the pressure falls again. The pressure and temperature will remain related by (approximately) the ideal gas law:
$$ P = \frac{nR}{V}\,T $$
You say in your question:
It seems like the gas equation doesn't apply when the gas is cooling down within the can and the pressure remains the same. (my emphasis)
But the pressure doesn't remain the same while the can is cooling.
For completeness we should note that in real deodorant cans the gas used liquifies under pressure so the behaviour is more complicated than a simple ideal gas.
| {
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Why during annihilation of an electron and positron 2 gamma rays are produced instead of 1? $1 \gamma \rightarrow 1 e^- + 1 e^+$ (pair production)
Then why
$1 e^- + 1 e^+ \rightarrow 2 \gamma$ (annihilation of matter)
instead of
$1 e^- + 1 e^+ -> 1 \gamma$ ?
| This has a simple answer: the process $e^++e^-\to\gamma$ cannot satisfy both momentum and energy conservation at the same time. To see, this let's choose a reference frame in which the total momentum of the system is zero (that is, the electron has the opposite momentum to the positron). This reference frame is always possible to be chosen by a simple Lorentz transformation.
Now, the photon produced must have zero total momentum. However, this simply isn't possible, since photons must always travel at the speed of light. More importantly, if the produced photon has frequency $\omega$, then the momentum must satisfy $|\textbf{k}|=\hbar\omega/c$. Since $\hbar\omega$ is the energy of the photon, $\hbar\omega\geq 2m_e$ by energy conservation, and so $|\textbf{k}|>0$.
The $e^++e^-\to\gamma+\gamma$ process is perfectly allowed since, if the outgoing photons have opposite momenta, the center-of-mass frame can still be perfectly satisfied.
I hope this helped!
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Motion of bodies connected by springs Two blocks $A$ and $B$ of masses $m$ are connected by a spring of length $L$ and spring constant $k$. They rest on the frictionless floor. Another body of mass m moving with velocity $v$ collides elastically with $A$. The spring compresses and at maximum compression velocity of both $A$ and $B$ are $v/2$ each. Why did these bodies get equal velocities?
| Assuming that the collision takes place over a period of time much shorter than the period of oscillation of the two mass & spring system the collision can be treated as the moving mass, velocity $v$ mass $m$, hitting head on a stationary mass of equal mass.
This results in the originally moving mass stopping and the originally stationary mass moving off with velocity $v$.
This can be shown by using the conservation of linear momentum (no external forces acting) and the conservation of kinetic energy (elastic collision).
The two mass & spring system has momentum $mv$ and so the velocity of their centre of mass must be $\frac v 2$ and it will stay that value because there are no external forces.
Because the total momentum in the centre of mass frame must be zero the velocities of the two masses in the centre of mass frame must always be equal in magnitude but opposite in direction.
When the spring has a maximum compression the two masses must be at rest in the centre of mass frame so they must be moving at the speed of the centre of mass, $\frac v2$, relative to the ground.
| {
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Is the diffraction pattern of a vertical slit horizontal? I am familiar with the mathematical aspects of single slit diffraction pattern, at the undergraduate level.
Consider the following pictorial representation from the book Optics, by Hecht:
The fact that I find puzzling here is - even though the slit is shown vertical, the pattern on the screen is shown horizontal. Is this correct? Why so?
My logic:-
The reason why I find this strange is because of a translational symmetry argument. Any two points vertically separated by some distance have the same horizontal attributes. So, one expects the pattern also to have this sort of vertical symmetry, irrespective of what happens along the horizontal axis.
Am I mistaken? If yes, can someone please point out why is the vertical slit producing a horizontal pattern here?
| To observe the diffraction from a slit assume that the vertical dimension of the slit is much larger than its horizontal dimension. To illuminate it you need a line source not a point source otherwise you will get 2-dimensional diffraction pattern not a simple 1-dimensional sinc thing.
The diagram shows that the rays break at $L_1$ and $L_2$, resp., implying a pair focusing (collimating) lenses at $L_1$ and $L_2$ distance. The one at $L_1$ converts the point source into a line source parallel with the long (here vertical) dimension of the slit. The one at $L_2$ collimates the emerging rays from the slit and project them at the screen for observation. The latter lens at $L_2$ focuses the rays that approximately homogeneous in the vertical dimension emerging from the slit into a vertically narrow and mostly horizontally distributed diffraction pattern on the screen for observation. The rays that emerge from the slit are diffracted horizontally but their distribution is nearly homogeneous in the vertical dimension because of the narrowness of the slit.
If there were no lens at $L_2$ you would see a mess on the screen, but if you looked straight into the slit the lens in your eye will collimate and you would see a diffraction pattern but do not do this or you will get badly burned.
| {
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Solar neutrino momentum flux through Earth According to wikipedia, the sun emits enough neutrinos that the number passing through a square meter of area oriented perpendicular to the sun at Earth distance is around $6.5 \times 10^{14}$ per second.
What is the momentum flux of these neutrinos? If you counted up the momentum of all the solar neutrinos passing through that square meter of area at $1 \cdot au$ from the sun, what's the order of magnitude? For example, if you made an impossible sci-fi material that was opaque to neutrinos, could the "lift" generated by the neutrino "wind" through a $1\cdot m^2$ "sail" overcome Earth gravity?
| You really only need one more piece of data to finish the problem: the typical solar neutrino has a momentum of a few $\mathrm{MeV}/c \approx 1.5 \times 10^{-21} \,\mathrm{kg \cdot m / s}$.
Multiplying that by the flux you list above gives about $P_\nu \approx 1 \times 10^{-6} \,\mathrm{Pa}$ for the pressure on a neutrino absorber at about 1 AU.
For reference this should be compared to the effectiveness of a solar sail which is around
\begin{align*}
P_\gamma &= 2 \left(\frac{\text{solar constant}}{c}\right) \\
&= 2 \left(\frac{1400\,\mathrm{W/m^2}}{3 \times 10^8\,\mathrm{m/s}}\right) \\
&\approx 1 \times 10^{-5} \,\mathrm{Pa}
\end{align*}
or an order of magnitude larger.
| {
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Is the Charge on proton is $3.2\times10^{-19}$ greater than that of charge on electron? As the charge on electron is $$e^-=-1.6\times10^{-19}C$$ and charge on proton is $$p^+=+1.6\times10^{-19}C$$
Does this mean that the charge on electron is $3.2\times10^{-19}C$ less than that of charge on proton?
| Negative doesn't necessarily mean less, especially in Physics. In most cases, negatives just represent the opposite direction. Or negatives represent something which has the opposite effect of the positive quantity. People often consider that negatives are less than 0. Yes, we write $-5<0$. But this 'less' is not the the same 'less' that we use in everyday world. It just represents a sense of order. $0$ represents nothing and anything can't possibly be below nothing.
Ask yourself this: Is 5 units North 'greater than' 3 units South. No, they're different things. But tracking South by a minus sign helps us in determining what is the end result or the effect. For example, go 5 units North, then 3 units South, you end up at $5-3=2units$ North. If the result is a negative number, then you end up at South. Even North could be represented by a '-' to keep track of sign.
But, if by greater, you mean 'more positive', then yes, the charge on a proton is $3.2\times10^{-19}$ greater than the charge on an electron only due to the sign convention (It could have easily been the other way around). Now, this 'greater' is not the same as 'greater' in everyday language. This 'greater' doesn't represent any 'improvement' in properties.
Another great example is work-done: You might ask: One can either do some work or no work. How can one do less than no work? This is not at all a 'mind-babbling' question because negative work is also some work. Here too, negatives are just keeping track of the direction. Just like going to South doesn't mean you've gone to a place less than nothing.
| {
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Why can we only "see" reflected light? This is a question thats been bothering me a while. I don't even know if it makes sense or not (like if it is a physics question or becoming a philosophical one). But here it goes. The crux of my question basically is that we all know that we can't see light (like in its photon or electromagnetic wave form) directly when it is traveling past us. However, we also know that the way we see objects is by light reflecting off them. This then means that we are "seeing" the light reflecting from the object which then sends the signal to our brain saying that we are seeing a particular object. We know that both light traveling past us and light reflected from objects are made of photons (so they are the same kind)? So then my question is that what is happening to the photon of a light after it is reflected from the objects, that causes us to see it or the object, but on the other hand we can't see light as it is directly traveling past us.
| Well done. You grasp a concept which many can't. The simplest explanation is that we don't see light, we feel light. By light I mean photons, not brightness. We see brightness because it is a visual sensation created by our brain. When our retina detects a photon it sends a message to the brain and the brain interprets this message as an image. Everything you see is created by the brain. A 3 dimensional visual representation of our surroundings.
Seeing something means detecting the shape, size, location and what wavelength of light it reflects (colour), without touching it. We only detect light that strikes our retina. Therefore we feel light, we don't see light. Feeling light is part of the process we call seeing.
| {
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Is there a Birkhoff-like theorem for stationary axisymmetric metrics? I know about the theorem by Robinson and Carter about the uniqueness of the Kerr metric in the case of stationary axisymmetric (SA) black holes. Are there any uniqueness theorems like Birkhoff's theorem for stationary axisymmetric metrics?
| *
*Note first of all that $U(1)$ axial symmetry is much smaller than $SO(3)$ spherical symmetry.
*(Let us put the cosmological constant $\Lambda=0$ to zero.) Where as spherically symmetric vacuum solutions are static and there are no spherically symmetric gravitational waves, the axisymmetric vacuum solutions are not necessarily stationary and there are axisymmetric gravitational waves. Even if the axisymmetric vacuum solution is additionally assumed to be stationary or static, there is still too much freedom left. Hence there is no axisymmetric version of Birkhoff's theorem.
*The following electrostatic analogy with 3D Poisson's equation is telling: Spherically symmetric solutions $\phi$ to Laplace equation are restricted to just $\phi= Ar^2+B/r$. On the other hand, for axisymmetric solutions to Laplace equation in cylindrical coordinates, we have no control over the $z$-dependence.
| {
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Circuit with three capacitors and a switch I have a homework problem where I have a battery connected to a capacitor and a switch that connects the capacitor to 2 others in series/disconnects the circuit from the battery.
The question is:
S is initially closed to the left until c1 is completely charged. Once charged, it closes to the right and remains closed there until it reaches equilibrium. Calculate the difference in potential of c1.
I'm stumped because I don't know how the charges distribute between the three capacitors once the circuit is closed, and so I'm not able to calculate the final voltage. I probably didn't do too good of a job translating the text so if you don't understand ask. Thank you.
| The total charge on $C_1$ ends up distributed across the three capacitors, so that the final voltage on the three of them will be the same.
Now you know that $Q=CV$. When the switch is thrown, the total charge $Q$ is the same, but the capacitance $C$ becomes 5 times greater (10+20+20).
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Opposite of particle decay I have read about particle decay, a process in which one particle becomes several other particles. However, I have not been able to find much information about its opposite: several particles combining into one particle. Is such a process possible, and if so, under what conditions? For example, a free neutron may decay into a proton, electron, and electron antineutrino. Could a proton, electron, and electron antineutrino somehow be joined into a neutron?
Edit: Everyone, thank you for your help, but let me try to make what I'm looking for clearer. I want to know whether several particles can join into ONE particle, in an exact reverse of that one particle decaying into several particles. As far as I know, I don't think an atomic nucleus counts as one particle. Please correct me if I'm wrong.
| Yes, the reverse of the "particle decay" process is possible. The proof of this is the existence of the universe (and us in it). If it were not possible, there would not be molecules, elements, compounds, etc.. The best example of this, is hydrogen (the most abundant element), which is formed by "uniting" protons and electrons,etc.. The conditions required, obviously, are the ones that existed right after the Big Bang.
The reason there is more "literature" on the decay process, is that it is a lot easier and less expensive to observe and measure the decay processes, than to combine particles. It takes a lot of energy and money to perform the combination process.
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How can length be a vector? Length and current both are not vectors. Then how can we assign the vector $l$ to the length of a wire carrying current while calculating for a current carrying conductor in a magnetic field. Also why in Biot—Savart law do we take small length element $dl$ as a vector?
Why is length sometimes a vector, sometimes not, whereas current always is a scalar?
| *
*$\vec l$ is displacement (or position), and $l$ is it's magnitude, called distance or length.
You might hear people call displacement for length, because in the next instant they quickly calculate the length. Sloppy words, that's all.
*
*$I$ is defined as the amount of charge passing through a cross section every second. Just like speed is the distance passed per second. Regardless of direction. That's just a definition.
This definition of $I$ doesn't care about direction, typically because wires are one-dimensional. When direction is needed, people usually use $I/\vec A$, so that the current is seen in comparison to the area it passes through. Since $I$ is defined without direction, the area must be given a direction instead (defined as the normal pointing perpendicularly out from it). This quantity is usually called current density.
This is just definition. When you see current as a vector $\vec I$ anywhere, they are re-defining (making their own slightly tweeked definition) this quantity, which they must/should state clearly. As a matter of fact I asked that same question some time ago, since I agree with your doubt, and you'll find answers in on that post.
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Current constraints on Dark Matter self-interaction from galactic profiles The self-interaction of dark matter may be small but it cannot be negligible if it is able to dissipate energy to relax into galactic clumps (necessary to explain galaxy rotation curves).
According to some answers in this old question: How Does Dark Matter Form Lumps?, the gravitational self-interaction alone is enough to allow dark matter clumping (via n-body interactions). Although two answers suggest something other than gravity is needed (one states considering the weak force is necessary, while another answer argues for why gravity alone doesn't explain how in cosmology dark matter could clump first).
I am curious about:
*
*Have the measurements of dark matter profiles of galaxies become good enough to provide indirect measurements of dark matter self-interactions?
*Can this self-interaction be used to say anything about the mass of the dark matter particles? At the very least, can we say with certainty they have mass above some threshold (ruling out very light particles such as axions or neutrinos, and ruling out some kind of unseen massless particles)?
*Since the strength and radial distribution of the gravitational force vs the weak force differ so strongly, is it possible to determine from the self interaction whether dark matter interacts via the weak force?
| Dark matter is still hypothetical - name given to excess gravity that can not be explained in terms of known bayionic matter.
Your question assumes proven existence of dark matter, which is not necessarily true.
Properties attributed to dark matter are - transparent, cold, and non-interactive (except via gravity). These properties resemble those of empty space more than anything else.
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The direction of the induced electric field Recently I got stuck witht the following problem.
Imagine we have uniform a magnetic field which induction points upwards. The fields strength is steadily decreasing. If we put an iron coil perpendicular to the magnetic induction vector, then, obviously, there will be electric current induced in the coil.
However, as I understand, the coil itself is only a 'marker' that displays the electric field lines that actually make the electrons move. It means that the elcetric field is there even when there is no coil.
Now the problem:
I can imagine some coils being close to each other. It will essentially mean, that it in one of them the current will go one way and in the other - the opposite. How can this possibly be?
I looked at this answer as it is phrased very close to what I want and still I couldn't get the idea. Could the answer be presented in more layman terms .
| I think what is missing in the analysis is the charge carriers that carry the current (i.e. the electrons). Electric field would be indeed there, if there is no coil, but without coil there are no charges that could be affected by this field and consequently there is no current. If the two coils in the image are joined together, there will be indeed no current through the central part.
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Is light deflected by external electric and magnetic field? I recently read about the Maxwell's laws of Electromagnetic Waves and I found that Light is made up of both Electric and magnetic fields.
So now if i pass the light through a capacitor such that the plates are parallel to the light will the light be deflected?
If it is deflected then what about the particle nature of light in which the photons are neutral without any charge(as far as i know charge do not exist without mass).
If not why are the electric and magnetic fields not affecting the light in the wave nature if it is solely due to the wave nature why are electrons being deflected in the external fields.
I hope someone give me a clear idea of what is wrong with my idea.
| Light is not deflected when it passes through a capacitor.
Light is classicly an electromagnetic wave, and visualizing it as the water waves in a pond might help you solve this cognitive dissonance.
In the finite region in which there is a electric field generated by the capacitor, the electric field of the wave and the previous one interfere through superposition (because, as you might have studied, Maxwell equations are linear), but after that, since the wave hasn't got charge anywhere in it, there is nothing the Lorentz force can interact with, and thus nothing is accelerated, nothing is deflected, and the wave exists the capacitor as it entered it.
This is of course consisten with the quantum mechanical notion that light is made of photons which have no charge, and therefore cant be deflected by electric fields.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why mass & energy bend spacetime? I understand how light / matter bend spacetime but I'd like to understand WHY. Is there some kind of interaction?
| Given that you understand the how part, I guess you also realize that the how part is actually "how much". I mean you realize that what you understand, is the quantitative analysis, not the root cause analysis.
And I agree with you that quantitative analysis does not always reflect on the root cause. But if someone tells you the root cause, then you will ask what is the "root of that root" and it will be never ending.
Therefore, while it is good to think about the why part, you have to figure it out yourself. When you do that, then your mind will settle on it, otherwise, you will always have the next "why".
Hint to you is that the answer will be something that will explain its own root along with itself. Meaning the answer will be rooted in itself. Alternatively, you can master the concept mathematically, and you will not bother about the why part. Just like, you never ask why 2+2 is 4?
| {
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How to calculate error using logic? The title may seem a bit off topic. I will explain my doubts with an example.
Let there be a situation where we are measuring gravity using the formula
Now, if the least count or error in the measurement of $l$ and $T$ is given, I can easily find the net error or relative error by taking a natural log on both the sides of the equation and differentiation.
But how to include the number of trials? For example, I learned somewhere that if the least count of a stop watch is 1 second, and the number of trials, say to measure the oscillation period of a pendulum is 20, then should the error in measuring the time period be cut down to 1/20 seconds?
| There's two possible things going on in such a measurement:
a) measure the time taken for $n$ oscillations and then your systematic error will indeed be reduced, e.g. minimum stopwatch interval $ / n$; and
b) do the $n$-oscillation measurement $N$ times to estimate the statistical uncertainty.
As described by the answers at How to combine measurement error with statistic error (thanks to Emilio for the link), these error sources should be added in quadrature. The statistical error will converge to zero as $N \to \infty$, but the systematic limitation on $T$ remains fixed... unless you make $n$ bigger, assuming that $T$ remains constant through a long "run".
| {
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What is the relation between image velocity, object velocity and mirror velocity? Suppositions used:
Velocity of image = VI
Velocity of object = Vo
Velocity of mirror = VM
I Know the fact that VI=-Vo supposing mirror at rest
and VI=2VM supposing object at rest
Now considering both mirror and object in motion, VI=2VM - Vo I ended up with this equation but my reference book suggests VI=2VM + Vo I am stuck on this for last 4 hours. I searched over internet and found the same expression like that of mine in a youtube video, I did not find much reference on this topic though. Tried many ways but all ended up on this simple argument, which equation to follow?
Help
| I think its for two cases
case I; when object and mirror moves along same direction Vi=2Vm -Vo
case II: when object and mirror are in motion in opposite direction Vi=2Vm + Vo
| {
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How to deduce the charge of an electron from the charge of an oil drop in the Millikan experiment? In the Millikan experiment we measure the charge of one oil drop. But how can I measure charge of one electron when I’m not sure how many electrons are contained within one oil drop?
| I think the key was that when you measure enough oil drops, you will notice that the charge is always a discrete multiple of some number. So it shows that charge is quantized. With enough data points, you'd see that the smallest charge difference possible is about $1.6\times 10^{-19}$ coulombs.
| {
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In cosmology, what is meant by a 'scale', e.g. 'scales smaller than the horizon'? I'm currently reading a cosmology textbook, looking at inflation in particular. I have been seeing "scales smaller than the horizon [Hubble length]" or "horizon crossing" in the context of inflation. I'm just wondering what those phrases mean. What are they using as a scale? What does it mean to 'cross' the horizon? Is this all related to the causality between two events (say, two different regions of the CMB - or in this case, the 'future' CMB) within the observable universe at that time? Or am I getting all this completely wrong?
| You're on the right track. Inflation solves, amongst other things, the homogeneity/horizon problem: the problem that distinct patches of the sky are
*
*causally disconnected; but
*the same temperature.
To solve this, the Universe underwent a period of inflation. 'Scales', that is, distances between patches of the Universe, inflated during inflation. Before inflation, the patches were causally connected and equilibrated. After inflation, those scales expanded, crossed the horizon and thus the patches now appear to have been causally disconnected. At late times, those scales may re-enter the horizon.
Note that smaller scales are the last to exit and first to re-enter the horizon.
In this figure, taken from these notes, 'scale' relative to the Hubble radius is shown on the y-axis (which decreases up the page, as the Hubble radius grows with time) and time is shown on the x-axis (which increases to the right of the page). You can see both horizon crossing (first exiting, then re-entering once inflation has finished).
| {
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Final velocity when falling to the Earth's center Could someone tell me what the final velocity of an object will be if it fell from the Earth's surface to its center (assuming no air resistance). Since it involves a constant changing acceleration due to the object's distance to the Earth's center decreasing, I'm sure it involves calculus (which I don't know how to do).
I wrote a program that recalculates the acceleration at every 0.3m, and it gave me a final velocity of 9.6 x 10^7 m/s. Is this close to the exact answer?
| I make it $7900\ \text{m}\ \text{s}^{-1}.$ This is using Simple Harmonic Motion theory.
| {
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When the angle of incidence is less than the critical angle, why is some light still reflected? When the angle of incidence is less than the critical angle, why is some light still reflected? When the angle of incidence is greater than the critical all light is reflected, why is not all light transmitted when it's less?
| In the general case, there are 3 components, namely, the incident light, refracted light and reflected light. However, when the light is plane polarised along the plane of incidence,one can show using wave theory of light that if $E_I,E_R,E_T$ denotes the amplitudes of incident,reflected and refracted electric fields, then the following holds:
$E_R = \frac{(\alpha - \beta)}{(\alpha + \beta)}.E_I$, $E_T = \frac{2}{(\alpha + \beta)}E_I$
where $\beta = \frac{\mu _1 . n_2}{\mu _2.n_1}$ ($\mu _i$ denote the permeability of the 2 media and $n_i$ denote the refractive indices.)
and $\alpha = \frac{cos (T)}{cos(I)}$ ($T$ denotes the angle of refraction/transmission and $I$ denotes the incident angle).
Neglect the mathematics if you don't want them, but just notice that for the case $\alpha = \beta$, $E_R$ becomes zero. So there is no reflected ray. This happens for a particular angle of incidence given by:
$tan(I) = n_2/n_1$. This is known as Brewster's Angle. At this angle, plane polarised light(w.r.t the plane of incidence) is transmitted 100% into the second medium. This can be shown using rigorous maths, and you can read that up from Griffith's Electrodynamics as mentioned in the other answer by Renan.
| {
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How is it possible to define work for friction in several dimension? I have been taught that, given a force field F, the work done by the force over a certain curve $\gamma$ is defined as the line integral of said field along $\gamma$.
But this makes sense only if force can be written as a function of position, as it is the case with gravity, or a spring. Unlike these, friction does not depend on position only: the same body might go through a point in space at two different times and experience a different friction (the magnitude wouldn't change, but direction and sense might).
So how can it make sense to talk about work done by friction if you can't define a force-field for it in the first place?
| For friction with some medium:
From Rayleigh dissipation function you can take the friction as function of velocity, and define it as a gradient (special one) of some scalar field.
$$\vec{f}=\vec{f}(\vec{v})=\vec\nabla_v \:(\mathcal{F})$$
where $\vec\nabla_v$ is defined as follow
$$ \vec\nabla_v =\dfrac{\partial}{\partial v_x}\hat x + \dfrac{\partial}{\partial v_y}\hat y +\dfrac{\partial}{\partial v_z}\hat z $$
In this case you have to know the object's velocity instead the position to know the force at one moment.
For "normal" friction:
In fact you can define the friction due to movement over a surface (over the xy plane for example)
$$\vec f =
\begin{cases}
f_x\hat x +f_y \hat y& \text{over the surface} \\
0 & \text{otherwise}
\end{cases}
$$
| {
"language": "en",
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Is really the electric field felt in every thing in space? I was wondering if the action/force of the electric field is really felt everywhere. I know it does reduce as you get further, but my thoughts concerned more about materials.
So, I know that the electric field is strictly related to Coulomb's Law, and Coulomb's Law is different between materials. So my questions are,
*
*Does a net elctric field (generated by two oppositely charged object)
is felt in every material in space? (In the picture, bottles,wood,copper,air,etc..)
*If it does, does it change its intensity based on materials? How?
I made a picture to try to help you understand my strange question, I drew the electric field reversed, because I like to imagine electrons moving.
If there are problem with the question, please let me know in a comment. I will try immediately to fix them, editing the question
| If I understood the question correctly, yes.
Since every material, at least at the scale you seem to be interested, is made up of atoms, which is a collection of (balanced) positive and negative charges, it will "feel" the effects of an electric field, since the charges that constitute the material will "feel" it.
Of course different materials (e.g. different types of atoms and arrangements) respond differently to the same electric fields. A major distinction is to be made between conductors and insulators (or dielectrics). You can find information between the behaviour of these categories of materials in virtually any resource covering electromagnetism.
| {
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What is the difference between $\psi$ and $|\psi\rangle$? My understanding is that $\psi(\vec{r}, t)$ and $|\psi(\vec r,t)\rangle$ are the same thing yet one expressed as a wave function and the other expressed as a vector in the Hilbert space. Is this true? Or is there a deeper difference between the two notations?
| It is convenient to think of $\vert\psi\rangle$ as a vector with components $\langle x\vert\psi\rangle=\psi(x)$ for various values of $x$. If you imagine discrete rather than continuous values of $x$, then the vector $\vert\psi\rangle$ would be the infinite column vector
$$
\left(\begin{array}{c}
\vdots \\
\psi(x_{n-2})\\
\psi(x_{n-1})\\
\psi(x_{n})\\
\psi(x_{n+1})\\
\psi(x_{n+2})\\
\vdots
\end{array}\right) =
\left(\begin{array}{c}
\vdots \\
\langle x_{n-2}\vert \psi\rangle \\
\langle x_{n-1}\vert \psi\rangle \\
\langle x_{n}\vert \psi\rangle \\
\langle x_{n+1}\vert \psi\rangle \\
\langle x_{n+2}\vert \psi\rangle \\
\vdots
\end{array}\right)
$$
obtained by decomposing the vector $\vert\psi\rangle$ on the basis of states $\{\ldots, \vert x_{n-2}\rangle,\vert x_{n-1}\rangle,\vert x_{n}\rangle,\vert x_{n+1}\rangle,\vert x_{n+2}\rangle\ldots\}$
| {
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Potential energy for a crystal lattice I would like some help understanding the following passage:
Consider a crystal lattice such that its unit cell has 27 ions arranged such that there are alternative positive and negative ions for same magnitude.
Then the electrical potential of the crystal lattice is just the potential of arrangement of any one of the ion taken with all the ions in the crystal times times the total number of ions halved.
i.e,
$$U = {1\over2}nN_0\sum^{nN_0}_{k = 2} {q_1 q_k \over r_{1k}}, $$
where $n$ is the number of moles of the solid.
I understand what author is saying but not why he is saying so. I don't understand why taking total potential of one ion multiplied by total number of ions gives the total potential. For instance take a ion at the bottom right corner of the crystal, clearly the potential of this ion wrt to a ion in middle and an ion at top left corner of lattice will differ. What is the author's reasoning here?
|
For instance take a ion at the bottom right corner of the crystal, clearly the potential of this ion wrt to a ion in middle and an ion at top left corner of lattice will differ.
The author here is treating the problem as a symmetric one and ignoring the edge effects (i.e. crystal is extremely big). Say you have N atoms in the crystal. So imagine you have an atom somewhere in the lattice. The total potential energy due to that individual atom is: $$ \frac{1}{4\pi\epsilon_0}\sum_k^N{q_1q_k/r_{1k}}$$
I.e sum of potential energies of interaction of this atom with every other atom in the crystal. Now the author is assuming, that wherever in the crystal you look, it all looks the same (i.e. crystal is infinite). Although, you are right, on the edge of the crystal this does not work, this is a common assumption. Now, because you have N atoms in the crystal, multiply energy of each atom by N to get the total potential energy. But you need to divide by 2 because if do not do that, you will be double counting.
| {
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Why can gluino (superpartner of gluon) have a Majorana mass? I read in a paper by Scott Willenbrock that gluinos can have a Majorana mass although they have SU(3) color symmetry. The explanation was that gluinos transform under the adjoint representation which is real.
Here is my understanding: gluinos are fermions in the adjoint representation of SU(3) and we can write a Lorenz invariant mass term for it because in the group notation we have
$8 \times 8 = 1 + \cdots$ So, because there is a singlet we can have a Majorana mass term for gluinos.
Is my understanding right?
| The gauge bosons in QCD are gluons with the
$$
(c_i\bar c_j + c_j\bar c_i)/\sqrt{2},~i(c_j\bar c_i - c_i\bar c_j)/\sqrt{2}
$$
$$
(r\bar r - b\bar b)/\sqrt{2},~(r\bar r + b\bar b - 2g\bar g)/\sqrt{6},
$$
for $c_ i = (r, b, g)$. These gluons are 3 plus 3 as the root space vectors plus 1 plus 1 as the weights, or the diagonal Gel-Mann matrices. This defines the 8 of SU(3).
The gluons are combinations of color and anti-color charges. This means the gluon is its own antiparticle. In the supersymmetric setting the fermion similarly has a combination of color and anticolor charges. This requires the gluino be Majorana.
| {
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Michelson and Morley experiment time troubles When discussing the Michelson and Morley experiment (as though the aether existed) we say that the different beams of light would take different times to travel the two distances due to the aether wind and so would arrive out of phase.
I'm having a lot of trouble intuitively seeing how they take different times. I can't seem to grasp that the beam moving against/with the wind will be this take a longer time. I think this is as no equation made from it directly shows this conclusion. When I think about it I always get confused by the fact that the beam would be sped up and then a lower down, so in my mind it would still arrive at the same time.
If anybody could try and possibly explain why they arrive at different times in a simple way that would be lovely!
| Suppose the aether is flowing past the Earth at a speed $v$, then when the light is travelling with the flow its net speed is $c+v$ and when it's travelling against the flow the net speed is $c-v$.
We'll call the length of the arm $\ell$, so for the trip with the flow the time taken is:
$$ t_1 = \frac{\ell}{c+v} $$
and for the trip against the flow the time taken is:
$$ t_2 = \frac{\ell}{c-v} $$
The average speed is then just the distance travelled, $2\ell$, divided by the total time taken, $t_1 + t_2$:
$$\begin{align}
v_{av} &= \frac{2\ell}{t_1 + t_2} \\
&= \frac{2\ell}{\frac{\ell}{c+v} + \frac{\ell}{c-v}} \\
&= \frac{c^2 - v^2}{c} \\
&= c\,\left(1 - \frac{v^2}{c^2}\right)
\end{align}$$
So even though you're quite correct that one leg of the trip is speeded up and the other leg is slowed down, the average velocity doesn't stay constant.
| {
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Heisenberg uncertainty principle: If a particle is totally localized, does it go faster? In Feynman's lectures on physics, volume I, section 6-5, Feynman states:
$$\Delta x\cdot \Delta v \ge h/m $$
($\Delta x$ is the width of the probability distribution of the location, $\Delta v$ is the width of the probability distribution of the velocity, $h$ is the Planck constant and $m$ the mass) and goes on to say the following:
Since the right-hand side of this equation is a constant, it says that if we try to "pin down" a particle by forcing it to be in a particular place, it ends up by having a high speed.
My question: Why should the particle have a high speed if it is very localized? To my understanding the Heisenberg uncertainty principle only states that the velocity probability distribution will be very spread.
clarification: I'm not asking how can the particle be moving if it is localized, as asked before here, so this is not a duplicate to my understanding. My question is why should it go faster if it is more localized.
| The argument says nothing about the average velocity of the particle, which will remain at $\langle v\rangle = 0$ for bound states; moreover, $|v|=0$ will normally still be the most likely speed in that state. Nevertheless, if you want a distribution with a very large width $\Delta v$, then it will need to have support from very large values, so the probability that the particle has some high velocity will increase.
| {
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Why does gravity act at the center of mass? Sorry if this is a trivial question.
Why does gravity act at the center of mass?
If we have a solid $E$, shouldn't gravity act on all the points $(x,y,z)$ in $E$? Why then when we do problems we only only consider the weight force from the center of mass?
| The only time we need to know where a force acts is when we are calculating a torque. For contact forces, it is clear that the force acts at the point of contact. But for a force like gravity, that acts at a distance, it is less clear.
In reality, a rigid object is made up of many particles, and there is a small gravitational force and torque on each of them. When we only care about acceleration we only need the sum of all these forces, which is $\vec{F}_{tot} = \sum_i m_i \vec{g}= M\vec{g}$. But what about the torques?
We would like to pretend that this total gravitational force acts at a single point for the purpose of calculating torque. Is there a point $\vec{x}_{cg}$ such that $\vec{x}_{cg}\times \vec{F}_{tot}$ gives the same total torque as summing up all the small torques?
If we do sum up all the torques we find $\vec{\tau}_{tot} = \sum_i \vec{x}_i\times (m_i\vec{g}) = \left(\frac{1}{M}\sum_i m_i \vec{x}_i\right) \times (M\vec{g})$. This tells us to call $\vec{x}_{cg} = \frac{1}{M}\sum_i m_i \vec{x}_i$ the center of gravity, and if we pretend that the total force of gravity acts at this point, it will always give us the right answer for the gravitational torque. Finally, we notice that it happens to have the same form as the definition of the center of mass!
However! If you do the calculation yourself you might notice that if $\vec{g}$ varies from particle to particle then this derivation does not work. In this case the center of gravity is not actually well defined. There may be no $\vec{x}_{cg}$ that does what we want, and even if there is it is not unique, except in a few special cases.
| {
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Queries about rotational groups $\mathrm{SO}(3)$ and $\mathrm{SU}(2)$ in QM In a QM text I am using (Sakurai 2nd edition 'Modern Quantum Mechanics'), he describes two rotation groups, namely the $\mathrm{SO}(3)$ rotation group and $\mathrm{SU}(2)$ rotation group (unitary unimodular group).
He defines $\mathrm{SO}(3)$ as a group with matrix multiplication on a set of orthogonal matrices (which are matrices which satisfy $R^TR = 1 = RR^T$), he then states that this group only includes rotational operators (and not also inverse operators which would be the group $\mathrm{O}(3)$). He does not ever rigorously define 'rotational operation'.
*
*How you would distinguish between rotational operators and inverse
operators, would a sufficient definition be that rotational
operators is a transformation with one fixed point?
He also defines the the group $\mathrm{SU}(2)$ which consists of unitary unimodular matrices, and states that the most general unitary matrix in two dimensions has four independent parameters and it is defined as $$U = e^{i \gamma}
\left( {\begin{array}{cc}
a & b \\
-b^* & a^* \\
\end{array} } \right)
$$
where $|a|^2 + |b|^2 = 1,~~~\gamma^* = \gamma.$
*Am I right to assume that the $\mathrm{SO}(3)$ rotation group does
not have much of application in quantum mechanics but is rather used
more in classical mechanics whereas $\mathrm{SU}(2)$ is used more in
quantum mechanics, particularly for $s =\frac{1}{2}$ spin systems
where we work in a two dimensional Hilbert space?
*How does it follow that there are four independent parameters for the
general unitary matrix, the way I see it there are three independent
parameters, namely, $a$, $b$ and $\gamma$?
| When classifying representations of a group in QM, it is necessary to allow for projective representations, because states are actually rays (equivalence classes) in the Hilbert space. This means that in order to study the rotational symmetry of a system, you want the projective representations of $\mathrm{SO}(3)$, which are standard representations of $\mathrm{SU}(2)$, because the latter is the universal cover of the former. This is the reason $\mathrm{SU}(2)$ is important in QM.
| {
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Crash simulation on Mythbusters I remember an episode of mythbusters where they were busting myths to do with a head on collision between two cars.
They said that instead of crashing two cars into each other at 50mph they would crash a car into a stationary object at 100mph because the energy involved in the crash would be the same.
Later on they corrected themselves to say that the energy is not the same. But I can't figure out why this would be the case?
Can someone explain if these two scenarios are the same or not. And why?
| In first case: Total Energy $$E_1= \frac{1}{2}mv^2 + \frac{1}{2}mv^2=mv^2$$
In second case: Total Energy $$E_2=\frac{1}{2}m(2v)^2=2mv^2$$
Thus, total energy doubles i.e $$E_2=2E_1$$
| {
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Is Frobscottle from the movie 'The BFG' less dense than air? For those who have either read the book, or watched the movie "The BFG", you would know Frobscottle as a green drink the giant uses, and has bubbles fizzing "in the wrong way", which is downwards. Assuming the bubbles to be filled with air, and that gravitational force on the bubble is greater than buoyant force, does this imply Frobscottle is less dense than air? Furthermore, is a liquid possible that is less dense than air?
| Why assume that the bubbles are air? A mixture of xenon and oxygen in a bottle of water pressurized lower than the critical point of xenon might result in bubbles that sink. The BFG could maintain low pressure by sucking the Frobscottle out via a straw with a one-way valve.
| {
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Bell Inequalities - Expectation Values I'm currently reading Loopholes in Bell Inequality Tests of Local Realism by Jan-Ake Larsson
https://arxiv.org/abs/1407.0363
On Page 6, Equation 7, he has a short proof, where I am having a hard time seeing through the math. I'll re-state here for convenience:
$ \lvert E(A_{2}B_{1}) - E(A_{2}B_{2}) \rvert = \lvert E(A_{2}B_{1} + A_{2}B_{1}A_{1}B_{2}) \rvert \leq E(\lvert A_{2}B_{1}(1 + A_{1}B_{2}) \rvert ) = 1 + E(A_{1}B_{2}) $
where
$ A_{1}B_{1} = -1 $
and
$E(A) = \int A( \lambda ) \rho ( \lambda) d \lambda $
I'm not seeing how he goes from the 2nd expression on the LHS of the less-than-or-equal sign to the third expression on the RHS of the less-than-or-equal sign. I suspect I am missing something on the properties of expectation values. Mainly, I think the absolute signs (|) moving from outside of the E()'s to the inside of the E()'s have me the most confused. Further, I don't see how he goes from the 3rd expression to the 4th, either.
Can anyone offer any clarification here?
| Let's write $X$ for $A_2B_1$ and $Y$ for $A_2B_1A_1B_2$. And keep in mind that $X$ and $Y$ are each always equal to either $1$ or $-1$.
Now think of the expectation as an average. You've got a bunch of possible outcomes, and $E(X+Y)$, for example, is the average value of $X+Y$ over all those outcomes. To understand how these averages behave, it's enough to understand how the totals of the outcomes behave, since the average is just the total divided by the number of outcomes.
So write $T(X+Y)$ for the total of all the possible outcomes of $X+Y$. Those outcomes come in three types: Either $X=Y=1$, which contributes $2$ to the total, or $X=Y=-1$, which contributes $-2$, or $X=-Y$, which contributes $0$. If there are $m$ observations of the first kind, $n$ of the second, and $k$ of the third, then $T(X+Y)=2m-2n$, and $|T(X+Y)|=|2m-2n|$ (which is either equal to $2m-2n$ or $2n-2m$, whichever is positive).
Now $T(|X+Y|)$ is the same total, except that all of the $-2$'s have been converted to $2$'s. $So |T(X,Y)|=2m+2n$.
This makes it clear that $|T(X+Y)|\le T(|X+Y)|)$. If that's not already crystal clear, think of it this way: The left hand side involves some positive and negative terms that cancel, while on the right hand side all the negatives have been converted to positives, so no cancellation occurs.
To pass from $T$ to $E$, just divide by the number of possible outcomes. That tells you that $|E(X+Y)|\le E(|X+Y|)$, which is the inequality between the second and third expressions.
For the final equality, note that $A_1B_2$ is always either $1$ or $-1$, so multiplying by it doesn't change the absolute value of anything.
(This is all a special case of the more general theorem that Candyman quotes in his answer, but the special case is all you need and might be easier to grasp. On the other hand, it's well worth mastering the general theorem at some point.)
Now for the final step:
First, $A_2B_1$ is always either $1$ or $-1$. Either way, if $x$ is anything at all, $A_2B_1x$ is going to have the same absolute value as $x$. If we take $x=1+A_1B_2$, we get $E(|A_2B_1(1+A_1B_2)|)=E(|1+A_1B_2|)$.
Second, $A_1B_2$ is always either $1$ or $-1$. Either way, $1+A_1B_2$ is non-negative, and hence equal to its own absolute value. This lets you drop the absolute value signs.
| {
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Friction on cars It is known that friction is given as :
$F_{friction}=\mu F_n$ , where $F_n$ is the normal force, and $\mu$ is coefficient of friction.
For a car travelling down a hill with constant velocity, the component of the gravitational force which is parallel to the cars velocity must be equal and opposite to the frictional force, whereby the frictional force opposes the motion of the car.
However, when the car is going up the hill, for a constant velocity to be obtained, the frictional force must be going up the hill, in the same direction as the motion of the car, and equal and opposite to the gravitational force which is antiparallel to the cars velocity.
I thought friction always opposes motion?
How can a car accelerate with the same force (i.e. friction) which also causes it to slow down. If there is no friction, a car cannot accelerate?
| I think the misconception that arises is frictional force doesn't opposes the motion of a body but in deeper sense it opposes relative motion between two surfaces in contact which are different things. Hope it solves your problem.
| {
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How is physical information created? I am not a physist, so please forgive my ignorance. I am an avid reader of popular science though. So I read about the problem of information seemingly lost at a black hole horizon.
But my question is about the creation of the information. My limited understanding is that our Univers had a very low entropy at creation and the entropy of the entire Univese is increasing (is it a closed system where second law is applicable?). At the same time the information - presumably present at the creation - is preserved.
The question is: can/does new information arise as a result of local organization - as in dynamic equlibrium for example? Or is it neseccary that the overall increasing entropy results in a zero sum? In later case - can the increase of entropy be equated to the information contained in the newly created order. I suspect that the observer has something to do with that. But the observer itself contains the information. Is it a paradox?
| I am not a physicist either, but lately I've been entertaining new ideas relating to information, in particular. One idea is that all instruments we use to make measurements with are conspiring to give us the impression that all phenomena beyond what our senses can directly experience are really there. Like when you put a voltmeter to a battery. We can only trust that the instrument is not just pulling a value on the readout from something other than the measured value.
To take this a step further, how can we prove, without any shadow of a doubt, that our eyes are not just being fooled into seeing the world as we think it should be. This is where quantum weirdness comes in. If there is no observer, there is no measurement.
You could call this the "conspiracy to convince materialists that the universe is real."
| {
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Pulley System - How the blocks will move?
I have been given the above question, with the solution,
One correction I observed from the answer was, in question c part would be acceleration of m1 and m2 are same and equal to g
Now the problem I have with the solution is that it is not in line with the string constraints. I know that T=0 so the string is not taut but just as soon as the motion will begin, it would be a problem right? They are saying in the solution that m1 comes down with g acceleration and m2 goes up with g. But if we apply string constraint then for every x distance that m1 comes down, m2 would move x/2 distance up. (in which I assumed that pulley A is fixed) and if it is movable then I am not able to apply it properly.
Also I don't understand how the motion of the pulleys would be. If m1 is coming down then the motion of pulley A should be anticlockwise by my intuition. (I might be wrong here)
Kindly explain me what is right and why
| If the acceleration of the two masses is $g$ downwards then the acceleration of pulley $A$ is $3g$ downwards, so pullet $A$ does move.
Let the centre of pulley $A$, $a$ move down a distance $x$.
On pulley $B$ the string on side $b1$ moves down $x$ and the string on side $b2$ moves up a distance $x$.
If pulley $C$ did not move then on side $c1$ the string would have moved up a distance $x$ and the string on side $c2$ the string would have moved down a distance $x$.
However the pulley $C$ moved down a distance $y$ so the string at $c2$ must have moved down a distance $x+2y$.
If pulley $A$ did not move then on side $a1$ the string would have moved down a distance $x+2y$ and the string on side $a2$ the string would have moved up a distance $x+2y$.
However the pulley $A$ moved down a distance $y$ so the string at $a2$ must have moved down a distance $2x-(x+2y) = x-2y$.
This must equal $y$ as the distance moved down by the two masses is the same.
$x-2y = y \Rightarrow x = 3y$.
You will note from the right hand diagram that the acceleration of a pulley is half the vector sum of the strings.
| {
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Talking in Space Of course, it's impossible to speak in space due to the lack of a high enough density of atoms to allow for vibrations, however, is it possible for a person's vocal chords to vibrate in outer space without any air passing through?
Furthermore, if vocal chords can still vibrate in space - would touch another person's hand in space (assuming they are somehow able to survive without oxygen in the harshness of space) allow for communication through bone conduction?
I imagine it would work in a similar way of having two astronauts touch each other's suits and be able to communicate through vibrations in the suit.
| Air passing through is what makes your vocal cords vibrate.
Also, the words that a person speaks are not formed by his/her vibrating vocal cords: It's formed by changes in the complex-shaped resonant cavity formed by the person's throat and mouth. Note how, you can understand the words that a person whispers to you, but in whispering, the vocal chords do not vibrate at all.
With no air in the person's throat or mouth, there's nothing there to resonate, so no speech sounds would be possible.
| {
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Quantizing one real fermion It is well-known how to canonically quantize the Lagrangian
$$L = i \bar{\psi} \dot{\psi} - \omega \bar\psi \psi$$
I now wonder how one quantizes the Lagrangian with one real fermion
$$L = i \psi \dot\psi$$
Obviously there can be no mass term since it is anticommuting so $\psi \psi = 0$.
I find no obstacles when going through Dirac's procedure. First I get the conjugate momentum as
$$\pi = i \psi$$
which I view as a constraint since there is no time derivative in this relation,
$$\Phi = \pi - i \psi = 0$$
Next I construct the Dirac bracket (DB) from the Poisson brackets (PB)
$$\{\psi,\pi\}_{PB} = 1$$
$$\{\psi,\psi\}_{PB} = 0$$
$$\{\pi,\pi\}_{PB} = 0$$
by following the standard procedure. First I define
$$C = \{\Phi,\Phi\}_{PB} = - 2 i$$
and then
$$\{\psi,\psi\}_{DB} = \{\psi,\pi\}_{PB} C^{-1} \{\pi,\psi\}_{PB} = 1\cdot(-2i)^{-1}\cdot 1 = i/2$$
Quantizing amounts to replacing DB by anticommutator as
$$[\psi,\chi]_+ = i\hbar\{\psi,\chi\}_{DB}$$
In this case this gives
$$[\psi,\psi]_+ = i\hbar(i/2) = - \hbar/2$$
This amounts to
$$\psi \psi = - \hbar/4$$
contradicting the fact that $\psi \psi = 0$.
Is there no way out of this? Is it impossible to canonically quantize this theory? The path integral seems to exist and make sense.
| Comments to the post (v2):
*
*Concretely, a Grassmann-odd operator $\hat{\psi}$ does not have to square to zero, cf. e.g. this Phys.SE post, even though it is true that a Grassmann-odd number $\psi$ always squares to zero: $\psi^2=0$.
*It seems OP is interested in Grassmann-odd point mechanics rather than field theory.
*The quantization of fermions are discussed in various Phys.SE posts, e.g. here, here and links therein.
| {
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Is the event horizon of a black hole wavelength-specific? The event horizon is the boundary surrounding a black hole from which not even light can escape. There's a certain negative potential energy level associated with the event horizon. Now, some shorter-wavelength photons may have enough energy to escape the black hole's gravity well starting off at this level, whereas some longer-wavelength ones might not – they would have to get infinitely redshifted and more.
Does this mean that the event horizon is actually not the same for different wavelengths of light?
| No, it's all the same.
GR does not care about the mass of something, nor does it care about its energy, when it calculates its path. It's always a geodesic and for massless particles it's always a lightlike geodesic. Makes not difference what its freq or energy is.
If it did when light is deflected by gravity, such as the many light deflection experiments done over the years, they would split up into its colors, I.e., it would show dispersion. It does not.
It would be possible if light or gravity is dispersive, i.e., if the speed of light c was different for each freq., or at least in non-flat spacetimes. There is an alternative to GR that says it is dispersive in the large scales of the universe, but alas, it's never been founded in measurements. Do not remember the error bounds, but they've been pretty good. See Experimental bounds on Lorentz-violating dispersion relation
See also https://en.m.wikipedia.org/wiki/Variable_speed_of_light for a bunch of variable speed of light theories, none of which have had any support from measurements.
| {
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Field due to internal Induced charge on a conductor to an external point? A charge q is located at a distance r from the center of a conducting sphere with inner radius 2r. The charge induces charges on the inner surface of the sphere according to Gauss' law .
The electric field at point p is to be approximated.
Inside the material of the conducting sphere, the electric field due to induced charge will cancel out the electric field due to the charge inside the sphere. Accordingly the electric field lines will begin at induced charge and terminate at the inner charge.
Therefore the field due to internal induced charge on the point p must be zero , (note it may be nonzero due to external induced charge but the problem specifies internal)
The solution however says it to be $kq/17r^2$ and not zero
Isn't the electrostatic system shielded from the conductor?
| Due to q charge on the sphere there will be no charge(or total charge is only at R distance from center of sphere) but at point P:
1)distance of point P from charge q is $\sqrt{17R^2}$
just use the formula of electric field for point charge then we get
$$\frac{Kq}{17R^2}$$
| {
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Measurement location on a rigid body I am just wondering if it is possible to calculate/estimate the location of measurement point on a rigid body?
For example, lets say we have a rigid body that is in motion. We attach a sensor, say an accelerometer on the surface of the rigid body. Now can we estimate the location of the accelerometer by using motion equations or just mechanics? The location can be given as a vector from the center of the rigid body for instance. Things we know about the rigid boy are it mass, moment of inertia and rotation rate. Also, the axis of rotation is not at the center of the body
Thanks
| Of course this can be done: a rigid body moves, by definition, by Euclidean isomteries, i.e. its motion can be defined by a composition of a translation and a rotation of its orientation. If you know the translation of its center of mass and the rotation of its orientation, then you can work out the position of any point on it at any time by simple geometry.
The translational motion of the center of mass is defined by Newton's second law given the nett force on the body, and the body's rotation about the center of mass is defined by the Euler's second law - that the rate of change of angular momentum equals the nett torque on the system calculated about the center of mass.
| {
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Would a handspinner spin indefinitly in space? I'm having a argument with a colleague, I don't know how to explain to him that if you spin a handspinner in space it will spin indefinitly (if you don't hold it). I agree that if you hold it, it will slow down because of the friction with the center part.
Would it theoreticaly spin forever?
| Assuming a perfect vacuum, you're both partially right. The spinner would slow down due to friction with the bearing. However, this also speeds up the bearing so at some point, the entire spinner is spinning at the same rate, at which rate it would spin forever.
Furthermore, even if you do hold it, the entire spinner will eventually stop spinning with respect to your hand, but will transfer angular momentum to you, so that you and the spinner are both spinning in space. If you then let go of the spinner it would continue spinning at the new lower rate, again indefinitely.
| {
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Can we theoretically "derive" the mass of a particle? I read a pop sci book on the Higgs which said that particles get their mass due to interacting with the Higgs field. If that is true, could we use first principles to derive the mass of, say, an electron? After all, QED is built on the interactions of particles and fields, right?
| We could if we'd know the coupling between the Higgs field and other particles. Instead, we use the measured mass of particles to get the value of this coupling.
| {
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Master Equation under a classical fluctuating noise I have a system as a qubit with Hamiltonian
$H_S = \frac{\Delta}{2}\sigma_z$
The interaction Hamiltonian is
$H_I = \frac{V(t)}{2}\sigma_z$ where $V(t)$ is a stochastic fluctuating variable. One can for example assume it as a random telegraph noise(RTN). In this case, what is the general prescription to write down the master equation for the qubit?
| There is a procedure outlined here: A. A. Budini, "Non-Markovian Gaussian dissipative stochastic wave vector", Phys. Rev. A 63, 012106 (2000).
You basically take an ensemble average over the noise realizations, and use some techniques from functional calculus. It is based on Gaussian noises, but you can extend it to non-Gaussian noises too, however you will not reach a Lindblad-like master equation.
You can also see my paper, which is based on the above reference: https://arxiv.org/abs/1612.02628
The derivation is in section III.B (it is severely shortened though), and the key point is Eq. 49. If you want to use non-Gaussian noises, that equation has a generalization in
F. Moss and P. McClintock, Noise in Nonlinear Dynamical Systems:
Volume 1, Theory of Continuous Fokker-Planck SystemsRef. (Look for Eq. 9.4.1).
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Correct definition of an 'acoustic mode'? I am reading 'The Oxford Solid State Basics' by S.H.Simon in which on page 92 defines an acoustic mode as:
... any mode that has linear dispersion as $k\rightarrow 0$.
Whilst on page 94 he defines it as:
... one mode will be acoustic (goes to zero energy at $k=0$).
Unless all modes that tend to zero do so linearly and vice versa then these two definitions don't overlap. Thus my question is as follows: does one of these conditions imply the other and if not what is the correct definition for an acoustic mode?
| The vibrational modes that have linear dispersion close to $k=0$ are acoustic modes where the slope of the dispersion curve is the speed of sound in the material (different for different directions of $k$).
The frequencies of optical modes do not go to zero at $k=0$. I would guess the dispersion to be quadratic.
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Question regarding thermal expansion of a bi-metallic strip I was reading about bi-metallic strips and came to know that on heating it forms an arc like shape. I also read a sentence that said the radius of such an arc can also be calculated which will be taught in the future grades. So I just wanted to know if there is any direct formula to find the radius of curvature?
Also, I think it can be found out using $A=l/r$ where $A$ is the angle subtended. But then again, how can the angle be found? So what is the formula for radius?
| The angle subtended is given by the arc length divided by the radius:
$$\phi = \frac{L_2}{R+t/2}=\frac{L_1}{R-t/2} $$
where $L_2$ is the length of the longer strip (at $R+t/2$) and $L_1$ the length of the shorter strip (at $R-t/2$), $t$ is the thickness of the strips. $R$ is the radius to the middle of the strips. Assumptions here are basically small bending and thin strips: $R\gg L_{1,2}\gg t$
Solving this equation for $R$ gives:
$$R=\frac{t(L_1+L_2)}{2(L_2-L_1)}$$
The change in length is related to the thermal expansion coefficients $\alpha_{1,2}$ and the change in temperature $\Delta T$ of the materials:
$$L_{1,2}=L(1+\alpha_{1,2}\Delta T)$$
Plugging this into the equation for the radius gives:
$$R=\frac{t}{\Delta\alpha\Delta T}$$
where $\Delta\alpha = \alpha_2-\alpha_1$ and a small term $\propto \Delta T$ has been neglected, basically assuming that the change in length due to temperature is small: $\Delta L \ll L$
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How to calculate altitude from current temperature and pressure? In a certain project, I need to calculate the altitude of the current location given the current location temperature and current location pressure. Temperature, pressure and altitude of a 'reference-level' could be provided if necessary (using a fixed sea-level pressure constant is also acceptable). This project is done between altitudes of-100 meters to 2000 meters above sea level.
This website uses the 'hypsometric formula':
$$h=\frac{((\frac{P_0}{P})^\frac{1}{5.257}-1)\times(T+273.15)}{0.0065}$$
given current location pressure, $P$, pressure at sea level, $P_0$, and current location temperature in Celsius, $T$.
However, I was also told by my friend that finding the altitude could also be calculated by the 'barometric formula':
$$h=44330\times\left(1-\left(\frac{P}{P_0}\right)^\frac{1}{5.255}\right)$$
which is obviously not equivalent to the first equation. Furthermore, this formula doesn't allow changes in temperature.
The 'barometric formula' given in Wikipedia is also different;
$$P=P_b\times\left[ \frac{T_b}{T_b+L_b\times(h-h_b)} \right] ^ \frac{g_0M}{R^*L_b}$$
This formula uses more constant values including the universal gas constant, $R^*$, the gravitational acceleration, $g_0$ and the molar mass of Earth's air, $M$. However, this formula isn't what I was looking for because it appears that the temperature at current location isn't taken into account.
My question is what equation is used to calculate current location altitude given current location temperature and current location pressure (or, if no such equation exists, best suited in range from sea level to 2 km altitude).
| The barometric formula is the same as the hypsometric formula if you set T=15. The reason for T+273.15 is just to put the temperature in Kelvin. This formula works to an altitude of about 9000m where the change in pressure with altitude becomes less linear.
Source: BMP180 Datasheet
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Field between charged cylinders The question is the following:
"For an air-filled cylindrical capacitor, with inner radius a and outer radius b, show
that the electric field between the cylinders is
$$E = E_{in} \frac{a}{r}$$
where $E_{in}$ is the field strength at the inner cylinder and r is the distance from its axis.
A breakdown occurs in air if the electric field strength is larger than a particular
value $E_0$. This determines the maximum value of $E_{in}$. For a fixed outer radius b, find the
inner radius a for which the energy per unit length in the electric field is a maximum."
I managed to do the first part, but then I am stuck. I introduced $\sigma$ as a variable for the charge density on the inner cylinder, but I seem unable to find an expression which cancels it and only leaves me with an expression of b in terms of a. I am also unsure about what "energy per unit length" is referring to. The answer is supposed to b: $ a = b/ \sqrt(e) $.
Many thanks!
| Cylindrical capacitors have an inner radius, outer radius and also a length. So energy per unit length actually refers to the energy in a cylindrical capacitor with a unit length.
Now, the field at distance $r$ from the axis of the capacitor is $E = E_{in} \frac{a}{r}$. Thus the energy density per volume $\rho$ is given by:
$$\rho = \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2}\epsilon_0 E_{in}^2\frac{a^2}{r^2}$$
Thus, the total energy $\phi$ id given by integrating $\rho$ over the whole volume enclosed by the capacitor:
$$\phi = \int_{Volume} \rho dv = \int_{0}^{l}\int_{a}^{b} \frac{1}{2}\epsilon_0 E^2 * 2\pi rdr* dl $$
$$= \pi\epsilon_0 E_{in}^2a^2 \int_{a}^{b}\frac{dr}{r} \int_{0}^{l} dl$$
$$= \pi\epsilon_0 E_{in}^2a^2 \ln(\frac{b}{a})l$$
Thus, the energy per unit length is given by $\frac{\phi}{l}$
$$=\pi\epsilon_0 E_{in}^2a^2 \ln(\frac{b}{a})$$
$$=ka^2 \ln(\frac{b}{a})$$
Where $k=\pi\epsilon_0 E_{in}^2$.
To maximise the energy per unit length, we need to differentiate it with respect to $a$ and equate to zero to get:
$$\frac{d(ka^2 \ln(\frac{b}{a}))}{da} =0$$
$$\Longrightarrow 2a\ln(\frac{b}{a})-a=0$$
$$\Longrightarrow \ln(\frac{b}{a}) = \frac{1}{2} $$
$$\Longrightarrow a=\frac{b}{\sqrt{e}}$$
As required by the question.
| {
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Why is my intuition failing me in these cases? In cases where the string is rotating:
In both cases, the accelerations of both the masses along the direction of the string are different because one of them is undergoing rotation. I understand that.
But intuitively, the acceleration of two ends of a taut string along the string should be the same. What am I missing?
| The thing which is constant about an inextensible string is its length, not its shape or orientation. There is nothing which requires both ends to have the same vector acceleration, even if it is kept taut.
With the origin at the hole or pulley, the radial acceleration of each end is $a_r=\ddot r -r\dot \theta^2$. The total length of string $r_1+r_2=$ constant, so $\ddot r_1=-\ddot r_2$. But there is nothing to prevent the ends from having different angular velocities $\dot \theta$, and therefore different radial accelerations $a_r$.
| {
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Easily approximate center of mass of a person when sitting What is the simplest way to get the center of mass of a human body when sitting? I'm especially interested in getting this when sitting in a chair, so this center of mass would include the chair. I'm trying to make an exercise device for myself that attaches to a lift that I installed on my ceiling. I use a wheelchair and don't get much movement other than pushing my wheelchair, so I'm hoping this will improve my health.
My intention is to have a bar stabilizing the chair, but I don't want a lot of torque/stress (axle) on the chair. You can think of the axle as the rod that might connect the inner gimbal of a gyroscope. But this will only rotate on one axis. I will simply be able to change my pitch with this chair hanging.
| The chair can be hanging from a self leveling short (say 6 inches vertically by 2 inches horizontally) metal strap on each side. these brackets are attached to your bar through a bolt and nut on top passing through a flange welded on the bar, providin approximately 5 inch hanging distance between the floor of the chair and its support hinge!
Anybody can adjust the balance when they sit on the chair and let the gravity swing and suspend the chair and its occupant to CG of the system while the bar is locked.Then they tighten the screws on the bracket and unlock the bar!
| {
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Integral representation of Dirac distribution The Fourier transform of the Dirac distribution is given by
$$\tilde \delta(\vec{k}) = \frac{1}{(\sqrt{2 \pi})^3} \int_{\Bbb R^3} \delta(\vec{r})e^{-i \vec{k} \vec{r}} d^3r = \frac{1}{(\sqrt{2 \pi})^3}\tag{1}.$$
By transforming back again, we get an integral representation of the Dirac distribution, which is given by
$$\delta(\vec{r}) = \frac{1}{(2 \pi)^3} \int_{\Bbb R^3} e^{i \vec{k} \vec{r}} d^3k.\tag{2}$$
My question:
In some books (for example in the context of quantized modes $\vec{k}$ in a volume $V$), it reads
$$\delta(\vec{k}) = \frac{1}{V} \int e^{i \vec{k} \vec{r}} d^3r,\tag{3}$$
with a finite volume $V$.
How can one justify this formula - why can one use a finite volume $V$ instead of $\Bbb R^3$?
| Note that the Kronecker delta is dimensionless, while the Dirac delta has the dimensions of one-over its argument:
$$
\begin{aligned}
{}[\delta_{a,b}]&=1\\
[\delta(a-b)]&=[a]^{-1}
\end{aligned}
$$
With this, you should be able to tell if $(3)$ is a Dirac delta or a Kronecker delta. Hint: it's a Kronecker delta in disguise - the integral denoting a sum:
$$
\frac 1V\int\mathrm d^3r\ f(r)=\sum_i f(r_i)
$$
| {
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Does a capacitor have a resistance? Does a capacitor have a resistance? And why? When I asked my physics teacher, he said certainly not, but I didn't figure out why. Can anyone please clarify? Thanks in advance.
| If you are dealing with a real capacitor, it has for sure parasite resistances, you may model it as follow
Where $Rs$ is the equivalent serie resistance, $Rp$ the parallel one, and the capacitor in the circuit, is intended to be an ideal capacitor which of course has not parassite resistances.
| {
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Are all atomic collisions elastic? If this is the case, why? In kinetic theory of gases it is considered all atomic collisions to be elastic. But if collisions are non-elastic the molecules must lose energy.
| The kinetic energy of a moving atom $K~=~\frac{1}{2}mv^2$ has to be comparable to the atomic levels for the collision to be inelastic. The Rydberg levels of a hydrogen atom are $E_n~=~-13.6eV/n^2$, for $n$ the atomic level. For a transition from the $N~=~2$ level to the $n~=~1$ level the energy released is $\Delta E~=~E_2~-~E_1$ $=~10.2eV$ This energy is $16.3\times 10^{-19}j$. The temperature by the equipartition theorem $E~=~\frac{3}{2}kT$ is then $7.9\times 10^4K$ That is fairly hot. To excite a hydrogen atom requires a fairly high temperature.
The assumption of the elastic collision then means the kinetic energy of the atoms are small enough so the inner electronic structure of the atom is not perturbed.
| {
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What causes this triangle effect? (waterfall) I was in a friends garden and saw this:
My question is: what causes the water to flow towards the center? My first thought was that maybe the water in the center falls faster and thus creates a sort of force inwards, but because gravity doesn't care about weight I don't think thats correct...
I also noticed that the width of the water is everywhere the same, exept for the edges. The water which flows towards the center forms a small tube there.
What causes this effect? Or is it simply because of the design of this fountain? (I don't think so, I've seen this before on other designs)
| In regard to the effects on the edges, by the looks of it, it is quite reasonable to think of Kelvin-Helmholtz instabilities, since there will be a shear layer between the water falling at a specific speed and the air being pulled down at a different (probably slower) speed.
| {
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How to prove the constant speed of light using Lorentz transform? I read the light-clock example in my book which proved the time dilation formula by assuming that the speed of light is constant for all observers. But I've trouble in understanding it the other way around. Lorentz transformation is just a correction to Newtonian mechanics to account for the constant speed of light for all observers, right? I have trouble understanding how does applying this correction preserve the speed of light for all observers.
Can we start by assuming that the Lorentz transformation formulas are true and then prove that two observers $A$ and $B$ will see a light pulse moving at the same speed $c$ regardless of their relative velocity with respect to each other?
| We could use the relativistic velocity addition equation, which would show the speed of the light pulse to be independent of the relative motion between the two observers.
EDIT: Attached is a brief proof of the problem.
Let an observer in frame S see an object in a reference frame moving at velocity V w.r.t. S emit a photon which travels at c. Then photon relative velocity w.r.t.S, U':
$U'=\frac{c+V}{1+\frac{cV}{c^2}}\\=\frac{U+c}{1+\frac{U}{c}}\\=c(\frac{U+c}{U+c})\\=c$
| {
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Diverging Sound Horizon Cosmology So from my understanding the sound horizon equals
$$
r_s(z) = \frac{c}{3^{1/2}H_0}\int_z^{+\infty}{\rm d}z' \frac{1}{\sqrt{1+(3\omega_b)/(4\omega_r))(1+z')^{-1}}}
$$
However for all finite $z$ this integral diverges. Am I reading the formula in this paper wrong? (Equation 29). I omitted $E^{-1}(z)$ because according to that paper it is not a function of the integration variable. Can anyone help me understand how this formula can be useful if it diverges for all finite $z$.
| There is a problem in the paper you're citing. The sound horizon is (roughly) defined as the distance that waves have propagated prior to redshift $z$
$$
r_s(z) = \int_z^{+\infty}{\rm d}z' \frac{c_s(z')}{H(z')} \tag{1}
$$
Where $c_s(z)$ is the time-dependent sound speed
$$
c_s(z) = \frac{c}{\sqrt{3(1 + R(z))}} \tag{2}
$$
and $R$ is the ratio of baryon to photon momentum density
$$
R(z) = \frac{3\rho_b(z)}{4\rho_\gamma(z)} = \frac{3\Omega_{b,0}(1 + z)^{3}}{4\Omega_{\gamma,0}(1+z)^{4}} = \frac{3\Omega_{b,0}}{4\Omega_{\gamma,0}}(1 + z)^{-1} \tag{3}
$$
Finally, the Hubble factor is usually written as
$$
\frac{H(z)}{H_0} = E(z) \tag{4}
$$
Putting everything together you get
$$
r_s(z) = \frac{c}{\sqrt{3}H_0}\int_{z}^{+\infty}{\rm d}z' \frac{E^{-1}(z')}{[1 + (3\Omega_{b,0}/4\Omega_{\gamma,0})(1 + z')^{-1}]^{1/2}}
$$
| {
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Intuitive reason for the $T^4$ term in Stefan Boltzmann law The Stefan Boltzmann Law gives a relation between the total energy radiated per unit area and the temperature of a blackbody. Specifically it states that, $$ j= \sigma {T}^4$$ Now using the thermodynamic derivation of the energy radiated we can derive the above relation, which leads to $T^4$. But is there any intuitive reason for the $T^4$ term?
| There's roughly $kT$ energy in each active mode. The active modes are characterized by momenta which live inside a sphere of radius proportional to $kT$, which has volume proportional to $T^3$. Multiplying these factors gives $T^4$, and the result clearly generalizes to $T^{d+1}$ in general dimension.
| {
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What is an operator times one of its eigenstates? I am trying to get a hold of caluclating with matrix elements. I have a Hamiltionian $\hat{H}$ in a two-dimensional Hilbert space, having eigenstates $\psi_1$ and $\psi_2$. My professor wrote down these equations:
$$\hat{H} \psi_1 = H_{11} \psi_1 + H_{12}\psi_2 \\
\hat{H} \psi_2 = H_{21} \psi_1 + H_{22}\psi_2 $$
He said these are alternative way to write the matrix elements of $\hat{H}$. However, I fail to see why is this. I tries to look it up in a linear algebra textbook, maybe this is a special property of matrix multiplications.
I have tried the following:
$$H_{11} = \psi_1^* \hat{H} \psi_1 \\
H_{11} \psi_1 = \psi_1^* \hat{H} \psi_1 \psi_1 $$
I did the same thing with $H_{12}$, added the equations together, but still can not see anything that resembles the original system.
Where do those two equations come from?
| The idea is that the two states $\{\psi_1,\psi_2 \}$ form a basis for the Hilbert space, this means that any other vector can be written as a linear combination of these two states, in particular the result of operating $H$ on, say, $\psi_{1}$
$$
H\psi_1 = H_{11}\psi_1 + H_{12}\psi_2 \\
H\psi_2 = H_{21}\psi_1 + H_{22}\psi_2
$$
Note that up to this point, the coefficients $H_{ij}$ are just complex number in the expansion. Now, if $\{\psi_1,\psi_2\}$ form an orthonormal basis we can write
$$
\langle\psi_1 | H \psi_1\rangle = \langle\psi_1 |H_{11}\psi_1 + H_{12}\psi_2 \rangle = H_{11}\underbrace{\langle \psi_1|\psi_1\rangle}_{=1} + H_{12}\underbrace{\langle \psi_1|\psi_2\rangle}_{=0} = H_{11}
$$
You can test the rest, but in general
$$
H_{ij} = \langle \psi_i|H|\psi_j\rangle
$$
| {
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Will a Cylinder placed on a frictionless inclined plane keep on slipping at its place or skid and slide down the plane? I've been wondering after learning about rolling without slipping and how it needs static friction for an object to start rolling but my question is that if theoretically the surface is frictionless then due to the torque of the weight will the Cylinder keep rolling at a fixed place or slide down etc.
Would appreciate a lot if I got to know what really happened as it's I don't seem to find much on the web.
|
Taking torque about COM of the disc.
Torque due to mgsinα and mgcosα is zero.
Only friction causes torque i.e. fR.
Torque due to friction causes the rolling, hence in absence of friction there is no rotation which implies that disc won't be able to attain rolling condition.
| {
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Evaluating Potential Energy Integral in Quantum Chemical Calculations My question is what are the steps for taking an integral of the following form?
$$\int e^{-\alpha|\mathbf r- \mathbf R_a|^2} {1\over|\mathbf r- \mathbf R_b|} e^{-\beta|\mathbf r- \mathbf R_b|^2} dV$$
This integral is commonly seen when attempting to do Quantum Chemistry calculations with a Gaussian type basis set. I have tried to use wolfram alpha to solve this problem but it fails to give a solution.
| Here is a more formal complete solution going off the approach suggested from the comments.
$I=\int e^{-\alpha|\mathbf r-\mathbf R_a|^2} {1 \over |\mathbf r-\mathbf R_b|} e^{-\beta|\mathbf r-\mathbf R_b|^2}dV $
First changing the arrangement of the equation we can define $\mathbf \Delta \mathbf R$ and $\mathbf r'$as:
$\mathbf \Delta \mathbf R = \mathbf R_a - \mathbf R_b$
$\mathbf r'=\mathbf r - \mathbf R_b$
Substitution leads to:
$I=\int e^{-\alpha|\mathbf r'-\mathbf \Delta \mathbf R|^2} {1 \over |\mathbf r'|} e^{-\beta|\mathbf r'|^2}dV $
Aligning the $z'$ axis along the direction of $\mathbf \Delta \mathbf R$
$\mathbf r'=r'sin(\phi')cos(\theta')\hat x'+r'sin(\phi')sin(\theta')\hat y'+r'cos(\phi')\hat z'$
$\mathbf \Delta \mathbf R=0\hat x' + 0\hat y' + \Delta R\hat z'$
$|\mathbf r' -\mathbf \Delta \mathbf R|=\sqrt {(r'sin(\phi')cos(\theta')\hat x')^2+(r'sin(\phi')sin(\theta')\hat y')^2+(r'cos(\phi')\hat z'-\Delta R\hat z')^2}$
Returning to the integral in spherical coordinates and plugging in:
$I=\int^\infty_0 r' \int^{\pi}_0 sin(\phi')\int^{2\pi}_0 e^{-\alpha((r'sin(\phi')cos(\theta')\hat x')^2+(r'sin(\phi')sin(\theta')\hat y')^2+(r'cos(\phi')\hat z'-\Delta R\hat z')^2)} e^{-\beta r'^2} d\theta' d\phi' dr' $
This leads to
$I=2\pi\int^\infty_0 r' \int^{\pi}_0 sin(\phi') e^{-\alpha(r'^2-2r'\Delta R cos(\phi')+\Delta R^2)} e^{-\beta r'^2} d\phi' dr' $
$I=2\pi\int^\infty_0 e^{-\alpha(\Delta R+r')^2} {{e^{4\alpha \Delta R r'} -1} \over 2\alpha\Delta R} e^{-\beta r'^2} dr' $'
Finally, if I did everything correctly
$I = {\pi^{3 \over 2} e^{{-\alpha \beta\Delta R^2 \over \alpha+\beta}}Erf({\Delta R \alpha \over \sqrt{\alpha+\beta}}) \over {\Delta R \alpha \sqrt{\alpha+\beta}} }$'
| {
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What is the accepted meaning of atomic orbitals and elementary particles in Quantum Field Theory? If elementary particles are represented as oscillations of their respective fields, why are atomic orbitals said to represent the probability of finding an electron at a specific location in the electron cloud or orbitals like it is a solid particle? Isn’t it more plausible or intuitive to think of the atomic orbitals (wave functions, oscillations) as the actual particles (electrons) themselves? Like for example the way a single hydrogen atom is experimentally imaged here (specifically page 13, fig.3, if you don’t want to look through the whole article):
https://link.aps.org/accepted/10.1103/PhysRevLett.110.213001
Why are the results described as probability distributions in the above article?
Or the way electrons are imaged here as something that looks like typical normal modes of oscillation (specifically last four pages of the first article and last page fig.4 in the second one):
https://arxiv.org/ftp/arxiv/papers/0708/0708.1060.pdf
http://portal.research.lu.se/ws/files/2746286/3224376.pdf
I understand that for example in the last two articles the images show the momentum distribution of the electrons, but wouldn't that distribution correlate to actual fluctuations of the substance of whatever the electron field would consist of?
Or if I have a totally wrong understanding please correct me.
| At energy scales sufficient for the creation or destruction of particle-antiparticle pairs, we have to talk in terms of fields. At low energies, we can talk in terms of $N$-particle states because $N$ cannot change. Then wavefunctions can describe these states, e.g. Slater determinants can describe systems of identical fermions. Orbitals describe individual electrons' probability distributions in such fixed-$N$ systems. This is what we expect in atomic physics.
| {
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Can the 7-10 rule of thumb for radiation be understood theoretically? Is there a way to understand where the 7-10 Rule of Thumb for nuclear radiation comes from? A seven fold increase in time after explosion results in a 10 fold reduction in exposure rate.
From a FEMA page on responding to "nuclear threats":
From the exposure rate determined by a survey instrument, future exposure rates may be predicted from a basic rule known as the "7:10 Rule of Thumb."
The 7:10 Rule of Thumb states that for every 7-fold increase in time after detonation, there is a 10-fold decrease in the exposure rate. In other words, when the amount of time is multiplied by 7, the exposure rate is divided by 10. For example, let's say that 2 hours after detonation the exposure rate is 400 R/hr. After 14 hours, the exposure rate will be 1/10 as much, or 40 R/hr.
The exposure rate must be expressed in the same unit as the time increase. For example, if the time increase is expressed in hours, the exposure rate must be expressed as the radiation exposure per hour.
| As long as isotopes half-lifes are log-uniformly spread, decay of their mix will follow hyperbolic law. Actually we do not even need really good uniformity - even big random error will not break hyperbola. See my code
I suppose it is connected to the distribution of half-lives, but I have no idea how to explain in a good way why fallout mix is relatively log-uniform.
| {
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Marginal and relevant operators that a $\phi^4$ theory should contain as an effective field theory Consider the Lagrangian of $\phi^4$ theory in 4-dimensions $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4\tag{1}$$ For a term in the Lagrangian of the form $C_{m,n}\phi^n(\partial_\mu\phi)^m$, the corresponding coefficient in 4-dimensions scales as $$C^\prime_{n,m}=b^{n+m-4}C_{n,m}\tag{2}.$$ For a reference, see Eq. 12.27, page 402 of Peskin and Schroeder.
The Lagrangian $\mathcal{L}$ in Eqn.(1), contains all relevant and marginal operators except $$\phi^2(\partial_\mu\phi)^2, \phi(\partial_\mu\phi)^3,\phi^3(\partial_\mu\phi),(\partial_\mu\phi)^4,\phi^3.\tag{3}$$ Even if we exclude the $\phi^3$ terms by demanding a symmetry under $\phi\to-\phi$, or by demanding the Hamiltonian to be bounded from below, one is still left with the other 4 possibilities.
If $\phi^4$ theory is regarded as a low-energy effective theory, I do not understand why the first 4 terms in (3) are they not considered? Is it just for simplicity?
| your power-counting is not correct. In $d=4$ a gradient $\partial_{\mu}$ counts like a field $\phi$. So for instance you should have $C_{m,n}'=b^{n+2m-4}C_{m,n}$
| {
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What is the quantum structure of the interstellar matter? Consider an interstellar medium of one hydrogen atom/cm$^3$ with $500$ photons/cm$^3$ coming from CMB. This density of particles is very weak according to quantum decoherence. So, according to Schrödinger equation, the wave function of each atom could be quite flat/spread.
Is there an equilibrium coherence/decoherence due to Schrödinger equation and CMB?
In other words, is the wave function of each atom stable (on average)?
Can (the squared modulus of) this wave function be almost equidistributed in one cm$^3$?
Can then the interstellar matter be correlated?
If relevant, we can ask the same questions for an intergalactic medium of one hydrogen atom/m$^3$.
| CMB photons interact extremely weakly with interstellar matter (and with each other), so it is indeed the case that if the photons were coherent, it would take a long time for them to decohere. However, the CMB spectrum shows no coherence at all, the spectrum is a perfect black body (the most perfect black body ever observed). This is in agreement with our best theories of the evolution of the universe, which predict that CMB photons were emitted by an $ep\gamma$ plasma in thermal equilibrium (which recombined into $H\gamma$, with the photons falling out of equilibrium with matter, but retaining their thermal spectrum).
The interstellar medium in galaxies is very dilute, but it is interacting with the light produced by stars, and frequently re-ionized. It varies in temperature quite a bit, sometimes reaching thousands of K. Even at temperatures of a few K the thermal de Broglie wave length is very short (much shorter than the inter-particle spacing), so this is a classical gas.
P.S.: If you are interested in coherent matter waves in cosmology, the place to look is (speculative) ultra-light dark matter candidates (like axions).
| {
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Why are there two layers of clouds? Clouds form because the warm air has risen above the cool air, and in between those two entities the water vapor condenses.
So why do there happen to be three entities - that is, one layer of air, a cloud, another layer of air, then another cloud?
For example, cirrus clouds over low-level clouds.
| Clouds form because the air cools enough to condense the water out of them into droplets. So if the clouds end at a certain altitude, it means that enough moisture in the air has condensed out such that the amount remaining is low enough to stay gaseous.
Sometimes this happens at low altitudes, sometimes (such as thunderstorms), it happens at really high altitudes. In the case of thunderstorms, the speed of the rising air helps carry the moisture higher before it condenses, but even then you get the flat tops of the anvil clouds when it gets high enough to condense out the moisture.
Now, for why two layers may sometimes form. If you look at the International Standard Atmosphere:
and you look at the red line indicating temperature, you can see that it decreases up to about 11km. As that happens, the water condenses and may form the first layer of clouds. The remaining air may still have some water vapor in it, but it will be below 100% relative humidity.
Above this, for the next 10km, the temperature is actually constant. This means it isn't cooling enough to condense any more water. The pressure is decreasing, so it is possible the amount of water that can be supported will drop. Above 20km, the air actually gets warmer again, although not many clouds will be that high up. That second, higher layer of thin ice crystals you sometimes see in the sky is separated from the lower layer because the pressure has to decrease enough to condense it out, as opposed to cooling.
| {
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Why is $\gamma^5$ used to define the projection operator? The properties of the projection operators are defined as:
$$P_+ = \frac{1}{2}(1+\gamma^5)$$
$$P_- = \frac{1}{2}(1-\gamma^5)$$
where $\gamma^5 = -i\gamma^0\gamma^1\gamma^2\gamma^3$
and their key properties are that $P_+^2 = P_+, P_+P_- = 1, P_-^2 = P_-$.
But since $\gamma^0$ has the same property as $\gamma^5$ that ${\gamma^0}^2=1$, if we replaced $\gamma^5$ with $\gamma^0$ in the definition of the projection operators, the newly defined projection operators would also satisfy all the key properties. And it's not like ${\gamma^0}^2=1$ is basis dependent; to show that ${\gamma^5}^2=1$ we actually need to make use of that fact. Why do we then bother to define $\gamma^5$ at all if we could use $\gamma^0$ to define the projection operators?
| To understand the importance of $\gamma^5$ you need to understand chirality first. Let me describe briefly. Chirality is a property of asymmetry, an object/system is called chiral if it is distinguishable from its mirror image. In other words the object can not be superposed on its mirror image just by rotations. Chirality defines the handedness (right/left handed) of the object.
Now come to $\gamma^5$, this matrix is the generator of chiral transformations of spinors$e^{-i\gamma^5\alpha}$. So this matrix can be used to check the handedness of spinors. This kind of matrix is possible in Clifford algebra in only even space time dimensions. So in 3+1 D we defince $\gamma^5$ as you did.
Why not $\gamma^0$? Just because it is not the generator of chiral transformations. (I think) It has nothing to do with chirality.
Projection operators are the operators which separate left/right chiral parts of Dirac spinor. These operators would be constructed with $\gamma^5$. Hope it helps. You can discuss it further.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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