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Does gravity have anything to do with Van Der Waals forces? Does gravity have anything to do with Van Der Waals forces? Just throwing this out there, I was wondering if they do because gravity is such a weak force and the VdW forces at the molecular level could seem to be a good intermediary force between gravity and the forces acting within atoms. Given that there are so many atoms and molecules within objects like the earth doesn't it seem possible that an extrapolation of the VdW forces could make a good candidate for a theory of gravity?
| An author called Zhang has indeed suggested they are related: https://arxiv.org/abs/1303.3579. As Zhang alludes to, "dark energy" is basically the new cosmological constant.
And I have seen other work by an author called Dmitriev - https://arxiv.org/abs/physics/0611173 - that suggests gravity is experimentally dependent on temperature - which Zhang relies on in theory.
I'm not endorsing these authors' work but I can't seem to find anything that criticises them.
| {
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Varying pressure in container with aperture The following diagram pictures a simple container filled with water. At the bottom of the container is a chamber with bottom B. Also directly above the chamber is an aperture that can completely open and close, sealing off the lower chamber or allowing unrestricted water flow.
My question is this: as pictured, is the pressure at B the same as if the aperture were completely open? If not, what is it?
I may be mistaken, but I think the pressure on B is proportional to the width of the aperture opening relative to the width of the chamber itself.
| What one could think is that what makes the pressure increase underwater is the weigh of the water column above it.
That is half true. In fact, it is the whole water above that pushes. So the water above the aperture also pushes.
So the pressure doesn't depend on the size of the aperture (at least when the equilibrium is reached). If it was not, the water inside would move until the pressure is homogeneous.
To prove it, you can use fluid statics :
$$\mathbf{\nabla}P = \mathbf{f_v} = -\rho \mathbf{g}$$
With $\mathbf{f_v}$ the force density, $\rho$ the density of water.
And as there is a little hole, one can find the pressure everywhere by following a track from a reference point.
| {
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What is the universal definition of the order parameter that is valid irrespective of the nature of the phase transition? Plausible definition Consider a phase transition from phase 1 to phase 2. The order parameter is zero in one of the phases 1 or 2 and nonzero in the other.
For example, in normal (phase 1) to superfluid (phase 2) transition, the order parameter is zero in the normal phase and nonzero in the disordered phase. So in this case, the above definition works good.
However, in the case of gas (phase 1) to liquid (phase 2) transition, the order parameter is taken to be $\mathcal{O}=\rho_{liq}-\rho_{gas}$. But $\mathcal{O}$ is nonzero in both the phases 1 and 2, and only vanishes above the critical temperature $T_c$. So, in this case, the above definition doesn't hold good.
Does it mean that the definition
Consider a phase transition from phase 1 to phase 2. The order parameter is zero in one of the phases 1 or 2 and nonzero in the other.
is wrong?
Is there a universal definition of order parameter such that it hold's good in both the cases?
| The order parameter is discontinuous for first order phase transition at the transition point. However, it need not be zero in any of the phases between which the first order transition takes place. For example, the liquid-to-gas transition or vice-versa below $T_c$, the order parameter is $\rho_{liq}-\rho_{gas}$, is non-zero in both liquid and gas but changes discontinuously. This also holds good for first order magnetic transition from up-aligned Ising ferromagnet to the down-aligned phase.
However, the same order parameter $\rho_{liq}-\rho_{gas}$ is zero in the phase above $T_c$ (the disordered phase) and nonzero below it for the second order phase transition at $T_c$. This also works for paramagnetic to ferromagnetic transition at $T_c$. This resolves the problem.
| {
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Why does acceleration due resulting force depends on mass while acceleration due gravity doesn't? Objects intrinsically resist to be accelerated due to their masses. A clear example would be kicking a soccer ball vs kicking a bowling ball. The latter ball will resist much more to be accelerated than the first one due to its greater mass (intrinsic property).
What if we position them in a inertial frame of reference in space? If we push both previous balls with the same force we will obtain different accelerations due to the balls' different masses, isn't it?
| Because measure of inertia (mass $m$) and the gravitational charge (mass $m_g$) happen to be the same.
That's a good question, actually, even if not that clearly formulated.
From Newton's 2nd law, $F=dp/dt$, with constant net force and mass, one gets
$$a=F/m$$
And Newton' gravitation law says that $F_g= GMm_g/R^2$, so for $F=F_g$, then
$$a=\frac{GM}{R^2}\frac{m_g}{m}.$$
The "gravity" you have in mind probably the one we feel in day to day, close to the surface of Earth, which means $R\approx \mathrm{const}$ and
$$a\propto\frac{m_g}{m},$$
from where you get that, if $m_g=m$, then $a$ is a constant: which we commonly denote by $g$.
| {
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Does increasing the resistance in a branch of a parallel circuit decrease the overall current?
In the above question, why does R3 increase? If R2 increases, wouldn't the parallel combination's resistance increase? If so, wouldn't the circuit have less current? Then why would the voltage across R3 increase?
| When "the circuit has less current", then there will be a smaller voltage drop across R1. It's (a little bit) harder to calculate the currents in the individual branches R2-4, but you don't have to. If the sum of the voltage across R1 + R2-4 is constant (that is not explicitly stated, but I will assume you have a constant voltage source across the network), then lower current through R1 means lower voltage drop there, and more voltage across the parallel network.
| {
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If someone were to put really small objects 10x spaced on a background, would they see the objects or the background? If someone were to put a sheet full of 1 micron x 1 micron black squares as a grid on a white piece of paper, and spaced them 10 microns apart (up-down, left-right, obviously diagonal doesn't count), would someone looking at it from a normal distance see black or white? If white, would they be able to see the black at all?
| American pop artist Roy Lichtenstein made some relevant art:
^Above, if you are close enough to the image, you can clearly see black dots on white background
^Same image, just resized... the face appears gray instead
So in general, the answer to "How do black dots on white paper look?" will depend on viewing distance. More specifically, the angular resolution of the human eye is around 0.03 degrees. If you pick what you mean by "normal distance", you can calculate what the resulting resolution in microns would be (assuming that the 0.03 degree rule of thumb holds for said "normal distance").
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Definition of symmetrically ordered operator for multi-mode case? As I know, Wigner function is useful for evaluating the expectation value of an operator. But first you have to write it in a symmetrically ordered form. For example:
$$a^\dagger a = \frac{a^\dagger a + a a^\dagger -1}{2}$$
For single mode case where there is only one pair of creation and destroy operator the symmetrically ordered operator is defined. But for multi-mode case,how is it defined? For example, how would we write
$$a_1^\dagger a_1 a_2^\dagger a_2$$
in a symmetrically ordered form (such that we could easily evaluate its expectation value using Wigner function)?
| Symmetrically order expansion of the ladder operator is written as follows;
a1b1=1/2(a1b1 +b1a1)= a1b1+1/2
a=creation operator b= anihilatinoperator, also
a1b1a2b2=1/2(a1b1 +b1a1)1/2(a2b2+b2a2)= (a1b1+1/2)(a2b2+1/2)=a1b1a2b2+1/2(a1b1)+1/2(a2b2)+1/4
| {
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Could a computer unblur the image from an out of focus microscope? Basically I'm wondering what is the nature of an out of focus image. Is it randomized information? Could the blur be undone by some algorithm?
| In this article the limits of the details that can be recovered using deconvolution are derived. It's explained that noise leads to limits on how effective deconvolution can be to recover details. In the ideal case there will only be Poisson noise due to the finite number of detected photons. The smallest recovarable details scale as $N^{-\dfrac{1}{8}}$. So, getting ten times more details requires 100 million times more exposure time. Clearly, a wide aperture without a lens and attempting to focus using deconvolution, is not going to work in practice.
| {
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Moment of a force about a given axis (Torque) - Scalar or vectorial? I am studying Statics and saw that:
The moment of a force about a given axis (or Torque) is defined by the equation:
$M_X = (\vec r \times \vec F) \cdot \vec x \ \ \ $ (or $\ \tau_x = (\vec r \times \vec F) \cdot \vec x \ $)
But in my Physics class I saw:
$\vec M = \vec r \times \vec F \ \ \ $ (or
$\ \vec \tau = \vec r \times \vec F \ $)
In the first formula, the torque is a triple product vector, that is, a scalar quantity. But in the second, it is a vector. So, torque (or moment of a force) is a scalar or a vector?
| Torque (Force Moment) is a vector that describes the location of the Force line of action.
*
*Lemma: If you give me a force vector ${\vec F}$ and a moment vector about the origin ${\vec M}$ then I can define a line whose points obey the relationship $\vec{M} = {\vec r} \times {\vec F}$. This line has direction parallel to the force ${\vec F}$ and passes through a point (closest to the origin) defined by $${\vec r} = \frac{ {\vec F} \times {\vec M} }{ \| {\vec F} \|^2 } $$
Proof:
Use $\vec{M} = {\vec r} \times {\vec F}$ into the equation for the point.
$$ \require{cancel} \frac{ {\vec F} \times {\vec M} }{ \| {\vec F} \|^2 } = \frac{ {\vec F} \times ({\vec r} \times {\vec F}) }{ \| {\vec F} \|^2 } = \frac{ \vec{r} ( \vec{F} \cdot \vec{F}) - \vec{F} (\cancel{\vec{F} \cdot \vec{r}} ) }{ \| {\vec F} \|^2 } = \vec{r} \frac{\| {\vec F} \|^2}{\| {\vec F} \|^2} = \vec{r} $$
This requires that $\vec{F} \cdot \vec{r}=0$ which is true for the point on the line closest to the origin.
It is true for both statics and dynamics that a moment is just a force at a distance. Only when the net force is zero (force couple) the moment is a pure moment and it does not convey any location information.
| {
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Why does an yellow object absorb blue instead of all colors other than yellow? I've always thought that an object appears to be of certain color X because it absorbs all other colors and reflects only X. But my current textbook(and some quick googling) tells me this is not the case.
As far as our eye is concerned, there is no difference between "yellow" and "red + green". This is because the cones in our eyes respond the same way to both of them. I guess I understand this.
But I still don't get why an yellow object would absorb only blue and reflect "red + green". Makes me wonder what's happening to all other frequencies... Appreciate any help. Thanks!
EDIT : This question is not a duplicate as I'm not asking about simple biology / physics of color. I think my question is specific : differences among "yellow", "red + green", and simple yellow reflection of an yellow object.
| @Mike already explained the color bit, but no one yet did quite answer "what's happening to all other frequencies".
Typically, for pigments, those photons get absorbed by the material of the object and converted to heat, as detailed in a very accessible manner in The Physics Classroom and here.
Another way photons can "disappear" is through destructive interference and that's indeed another possibility to generate color, the so-called structural color, generated by microscopic patterns in the material.
And the photons might also be not "destroyed", but rather be sent somewhere else (instead of your eye), as happens in the Rayleigh scattering, which is responsible by the most characteristic hues of the sky.
| {
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Special Relativity: Does non inertial frame of reference work in SR? I started on my own learning about GR and SR two months ago, and I still do not have clear if it is possible or not. The following example was explained to me by someone who affirmed: "SR applies only on inertial reference frames":
Let's imagine we have two different reference frames : A' and A. Reference frame (RF) A' is moving with constant velocity (v), meanwhile RF A has no velocity (A' moves relative to A with constant v).
RF A' has a wire underneath and RF A has an aerial above. When both interact, clocks start running in both RFs (clock A' and clock A) and a light ray emerges (from the wire-aerial interaction and with the same velocity vector direction RF A' has).
Then we agree distance can be determined from both RFs.
i.e. : x = x' + vt'
Then I asked myself: why would not be correct consider the case where A' is an accelerated RF and distance is determined from RF A (i.e.) as x = x' + at'?
My doubts about if "SR applies only on inertial reference frames" sentence was true increased when I checked out more sources and they affirmed accelerated reference frames were possible in SR.
| The claim that a certain physical theory "applies only in inertial reference frames" is not even logically possible. "Physical theories" describe physical quantities, which by definition are independent of one's reference frame or choice of coordinates. At most, one could claim that "many of the equations found in standard textbooks on SR only apply to inertial references frames," which is indeed the case.
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How does a man inside bubble ball accelerate without an external force? As Newton's Laws states an object should be in rest or in constant velocity if no external force is applied. A man inside a stopped car cannot push the car as he is not giving any external force.But a man inside a bubble ball can make it move. What is the theory behind this?
https://www.holleyweb.com/images/human_sized_hamster_ball_free_walking.jpg
| By moving his weight around inside the bubble ball, and utilizing friction between his body and the ball, coupled with friction between the ball and the ground, the man can cause gravitational force to topple the ball and move it over the ground.
If the center of mass of the ball moves to one side of the ball, gravity will topple the ball toward its center of mass. Because the ball is in a uniform gravitational field, the ball's center of gravity is also its center of mass. By changing the ball's center of mass, the man changes its center of gravity. When the center of gravity changes, gravity moves the ball.
| {
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Time dilation from different frames of reference I am trying to understand time dilation as a total newbie. This question is all about my trying to understand enough such that in a work of fiction an offhand comment (about FTL being a bit silly) makes sense.
If a ship were to travel for a period of say 25 years at a very modest speed such that time dilation is nominal (for this question, negligible too). Then, from the same destination, a ship was to travel at some speed a tiny fraction less than C (or even C, for simplicity) and catch up (in let's say a day) what would the experiences of the two crews be?
Obviously "ship B" would experience 25 years of waiting plus 1 day of travel before they meet. Would "ship A" experience 25 years plus 1 day, or much more time?
I found one answer which explained the complexities of the twin paradox well enough but that's not what I am looking for as I cannot use the answer to solve my own question.
|
Obviously "ship B" would experience 25 years of waiting plus 1 day of travel before they meet. Would "ship A" experience 25 years plus 1 day, or much more time?
Well that is quite simple:
A travels 25 years then waits N * 1 days,
where N is how many times faster A's clock runs, according to A. (How many days A's clock proceeds when B's clock proceeds 1 day)
N is the time dilation factor, also known as Lorentz factor.
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How could I measure the colour spectrum of a light bulb and investigate how closely it matches a black body radiation curve? Here is my research question:
What is the colour/spectrum produced by each globe type? What is the temperature equivalence? How closely does a globe match a black body radiation curve?
I will be testing this on halogen, compact fluorescent, and LED bulbs. How could I measure the colour specturm of the bulb? I have been suggested to take photos of the glowing bulb then use photoshop to analyse the colour. Is that a possible solution?
How could I match it to a black body radiation curve? Would I have to plot it out? From research it doesn't seem like a curve that could be hand-drawn, especially CFL and LED:
| If you need to do the experiment yourself, and you dso not have a spectrometer available, you can make a crude spectrometer using a diffeaction grating and a cylindrical lens. Make a thin slit in a piece of black paper, and let some of the light from your bulb pass through the slit. Put the diffraction grating against the slit. Downstream from the slit and grating, place the cylindrical lens. Adjust the position and orientation of the lens and grating until you get a focused rainbow "needle" on a piece of white paper further downstream. Use a photographic light meter to measure the intensity of the light at every position in the rainbow. This will give you a pretty good approximate spectrum of the light from the bulb. Compare the spectra from your various sources.
You can buy a grating and a lens from Edmund Optics, for example. You can use cello tape to hold the components in place on cardboard mounts. Have fun!
| {
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Speed of electrons in a wire What is the speed of electrons in a copper wire, used to charge a device? If there is a fixed speed, how is it determined?
| There is no point in speaking about the "speed of an electron in a copper wire". You may ask the drift velocity of the electron under an applied potential. The electrons are randomly scattered by phonons (lattice vibrations) as well as the metal ions. Under an applied field, in addition to the thermal random motion, the electron moves from region of negative potential to the region of positive potential with an average velocity, a motion which is known as drift. The drift velocity of an electron is very low: about $1 mm/s$. However, the Fermi velocity is as higher as several ten or hundred thousands of meters per second for a metal. The drift velocity of an electron in a metal is given by
$$v_{d}=\frac{eE\tau}{m}$$
where $e$ is the electronic charge, $m$ is the electron rest mass, $E$ is the applied electric field and $\tau$ is the relaxation time.
To know the drift velocity of electrons in copper, all you have to do is just measure the resistance of copper.
Then the conductivity of copper is given by
$$\sigma=\frac{l}{RA}$$
where $R$ is the resistance of the copper wire, $l$ and $A$ are the length and cross-sectional area of the wire. If the applied potential difference across the length of the wire is $V$, then the electric field can be approximated as
$$E=\frac{V}{l}$$
Now, the Fermi velocity of copper can be found out if you know the Fermi energy of copper ($7.00 eV$):
$$v_F=\sqrt{\frac{2E_F}{m}}$$
Next, you need the mean free path length, which is given by
$$\lambda=\frac{mv_F\sigma}{ne^2}$$
where, $n$ is the electron density in copper, and is given by
$$n=\frac{N_A\rho}{A}$$
where, $N_A$ is the Avogadro number, $\rho$ is the density of copper and $A$ is its atomic weight. Knowing $\lambda$, you can calculate the relaxation time as
$$\tau=\frac{\lambda}{v_F}$$
Substitute all these results in the first expression and you are done.
Note:
*
*This is an experimental way of doing the job. Standard values are available on textbooks and internet.
*Caution: Use a very long copper wire (10 m or above) or use mV range potential. The current density of copper is very large.
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Observing Two-Beam Interference at Home I want to know how difficult it would be for me to observe two-beam interference at home.
I have:
*
*A laser pointer.
*A non-polarizing beam-splitter.
*A mirror.
*Two concave lenses.
*An uneven shaky floor, some chairs, and tape.
*Patience that spans an entire day.
This is the sketch of the setup that I have in my mind:
where the laser, lenses, and so on are taped to the chairs.
How close do the two paths lengths have to be to each other? In my case, the two path lengths will differ by dozens of centimeters.
The laser spot from the laser pointer is not even uniform. Will that be an issue?
Are there any other ways this could go wrong? Is there any advice on how I could observe two-beam interference?
| I think you need less than that, though my setup may result in a very narrow viewing region.
A concave lens form a virtual image behind it, reflect the image with a mirror and you get two point sources.
| {
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Pure geometry proof of parabolic mirror property Is there a proof of the property that parallel rays of light incident upon a parabolic mirror converge to its focus that does not resort to Cartesian coordinates?
| Yes, there's plenty such proofs, but the choice between them depends on which of the multiple equivalent definitions of a parabola you take, and exactly what restrictions you place on the allowed proofs.
Physicists normally think of parabolas as the locus of equations of the form $y=x^2$, but you're explicitly looking for something with no formulas, so you probably need to change that as well. The next likeliest candidate is defining the parabola as the locus of points equidistant from a point and a line, which ties you in to all the proofs of classical geometry. Under that understanding, the proof of the reflection property is a staple of euclidean compass-and-straightedge geometry, and Wikipedia has a suitable proof, based on a construction of the form
Image source
In short, with $C$ on the directrix, you define $B$ as the midpoint of $\overline{FC}$, which means (since $\overline{FE}=\overline{EC}$ by definition of the parabola) that $\angle FEB=\angle BEC$, so you just need to show that the line joining $B$ and $E$ is tangent to the parabola. The Wikipedia proof relies on some facts from calculus, though if you want a calculus-free proof you can probably find one by rooting around in the toolbox for the euclidean geometry of conic sections.
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What does the spikes and curves in the spectral graph for compact fluorescent lights represent?
I have read from sources that the curves are generated by the phosphors in the bulb, and the spikes are caused by the mercury vapor.
However, if the mercury vapor's release of uv particles combine with the phosphors to produce visible light, then how can they give out light in different wavelengths separately?
| A fluorescent lamp is filled with mercury (Hg) gas. When you switch on the lamp the gas starts emitting light. However in contrast to the sodium lamps which are widely used in street lighting (the orange lamps), mercury emits UV light (mainly 254 nm) which is not only invisible for us, but could also harm us in the same way UV light from the sun can harm us.
To overcome this problem scientists use phosphors. These phosphors are insulating materials to which impurities are added. These impurities absorb UV or blue light (depending on the material and the application) and reemit visible light (e.g. green or red). If you take a closer look at the lamp you will be able to see these phosphors since they are applied as a coating on the inside of the lamp and are responsible for the white color of the glass.
The most widely used phosphors contain lanthanides as impurities. In the case of your spectrum the red lines come from trivalent Europium whereas the green emission comes from trivalent Terbium. However on the above wikipedia link you will find a huge list of used phosphors containing also transition metal ions like Manganese and various others.
A final note which is not really related to your question but it is closely related to my research and nice to know: the glow in the dark materials work in a similar way, they only have the possibility to store the absorbed energy before reemitting it again.
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How fast can you spin a proton or electron without breaking it? If you spin a single nucleus containing multiple nucleons fast enough it will fly apart.
Is there a speed limit to a spinning proton or electron assuming it's held at a fixed location with a strong magnetic field?
What speed would either have to attain before breaking up, assuming they can be broken up? What would they break up into?
| You can't spin up an electron. The only way that is possible is if as a string you can put angular momentum on it that way. The energy required to do that would be near Planck scale energy. So I will say this is not possible with the electron FAPP.
For the proton you can spin it up. This is the Regge trajectory that has the angular momentum $J$ as the abscissa and $M^2$as the ordinate. At the bottom here is the $940MeV$ particle, corresponding to the proton. The proton is made of three quarks and they have intrinsic spin $\frac{\hbar}{2}$ that adds up to the same. If I spin this up, we think of these three quarks as being like the masses on a bolos the Argentine gauchos use to rope cattle. The next higher state has a mass of $1680MeV$. This state is not stable and the additional $740MeV$ of rest mass can enter into the production of mesons. We may think of the proton with $udd$ quarks such that one of the $d$ quarks has its "gluon string" break so it becomes coupled to a $d$ anti-quark and the $d$ quark remains with the baryon. the $d,\bar d$ meson carries off this additional mass energy.
For a whole nucleus you have similar physics, though a bit more complicated. There are people who work with rotating nuclei.
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Can transverse sound waves be polarized? I know that polarization only occurs in transverse waves and polarization of light occurs as EM wave is a transverse wave. But sound waves are both transverse and longitudinal in solids. So can polarization occur for the transverse part? But we cannot stop the sound wave from propagating by any medium except vacuum. Because it will propagate through the stopping medium(like an analyzer but for sound).
Even if it gets polarized somehow(I don't think it can get polarized) then how can we observe it, since any sound reaching our eardrums will be longitudinal as the medium in front of our eardrums will be air, and so no polarization will occur in longitudinal waves.
See the 7th and 8th line in this image(source:- https://en.wikipedia.org/wiki/Polarization_(waves)).
I am a little confused now.
P.S. This may seem as a possible duplicate but all other answers didn't clarify my doubt.
EDIT:- Based on the answers, it seems that shear waves can be polarized. So my question is how to polarize these shear waves?
| "Sound" is a pressure phenomenon, and has no polarization.
It is possible to send acoustic shear waves through an elastic solid (and that transverse component can have a direction) - but not through a gas.
Just to confuse you more - in an anisotropic medium, different directions of shear may propagate at different velocities, resulting in an apparent rotation of the direction over time (and in fact it can go from linear to circular polarization, etc).
| {
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Traveling Wave Equation $\sin(kx-wt)$ vs $\sin (wt-kx)$ In my textbook most if the times it uses $A\sin(wt-kx)$, but occasionally there is a problem using $A\sin(kx-wt)$
So i just changed it from $A\sin(kx-wt) \to -A\sin(wt-kx)$
but does the amplitude change to $-A$?
Is the wave going downwards first?(as negative amplitude would imply)
| Yes that is correct.
If we look at the addition formulae for sine functions we see that $$\sin (A - B) \equiv \sin A\cos B - \cos A\sin B$$ and that $$\sin(B-A) \equiv \sin B\cos A - \cos B\sin A$$We can easily see that the second equation is just $-1$ times the first equation. By adding amplitude all we are doing to the equations is multiplying both sides by a scalar as such $$X(\sin (A - B)) = X(\sin A\cos B - \cos A\sin B)$$ So we can say that $$X(\sin (A - B)) = -X(\sin (B - A))$$
Hope this answers your question :)
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What is an order parameter? Thermal average or spatial integral of a thermal average? Sometimes the the order parameter is defined as the thermal average of a spatially varying field $\textbf{m}(\textbf{x})$ i.e., $\langle\textbf{m}(\textbf{x})\rangle$.
Sometimes the order parameter (density) is defined as the spatial integral $\frac{\textbf{M}}{V}=\frac{1}{V}\int\textbf{m}(\textbf{x}) d^3\textbf{x}$ where $\textbf{m}(\textbf{x})$ is supposed to be some coarse-grained microscopic variable such as spin.
Which one is the correct and most general definition of the order parameter?
A note: Chaikin and Lubensky's book says that when $\textbf{m}(\textbf{x})$ is independent of $\textbf{x}$, the order parameter is given by $$\langle\textbf{m}(\textbf{x})\rangle=\frac{\textbf{M}}{V}$$ where the left hand side is thermal average and right hand side is defined by the the spatial integral expression. The definitions will be consistent with each other if in the integral expression $\textbf{m}(\textbf{x})$ is actually $\langle\textbf{m}(\textbf{x})\rangle$.
| In the magnetic system, most of the time, people just calculate the thermal average of the magnetization at an arbitrary spatial point, say, $\langle m(\mathbf{x}) \rangle$. If you want to get the $M$ (the total magnetization), you just need to do a spatial integration over that.
In your problem, I think the authors are discussing the ferromagnetic phase, in that case, there is a translational invariance about the order parameter $\langle m(\mathbf{x}) \rangle$, which means at every spatial point $\mathbf{x}$, the value of the thermal average of $m(\mathbf{x})$ is the same, so you can integrate over the whole space (which is simply equivalent to time the volume of the system) and divide that by the volume of the system.
In some other cases, like spin density wave (or antiferromagnetic), when we calculate the $\langle m(\mathbf{x}) \rangle$, we would get some spatial dependence of the magnetization, e.g.
$$m(\mathbf{x}) \propto \cos(\mathbf{Q} \cdot \mathbf{x}+\theta )$$
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What does the refractive index for e.g. alpha mean? When I look for some materials like https://en.wikipedia.org/wiki/Barium_sulfate and want to extract the refractive index then there is written: (nD)=1.636 (alpha). And sometimes also for beta and gamma.
What does this mean? The refractive index is mostly dependent on the wavelength so why is alpha, beta and gamma the only value given?
| The Greek letters are not related to wavelength but to the directions in crystals. In anisotropic crystals the speed of light (and so the index of refraction) depends on the polarization of light and the direction of propagation relative to the crystalline axes. As your link shows, Barium sulfate has an orthorombic structure so probably is biaxial. You can learn about those by searching "Biaxial crystal optics". For example, this:
http://edafologia.ugr.es/optmine/intro/indibiaw.htm
Here you can find all three indices for this type of crystal - orthorombic (see section 4.3):
https://pubchem.ncbi.nlm.nih.gov/compound/barium_sulfate#section=pH
You will see that they are labeled with alpha, beta and gamma and they are all for this orthorombic crystal and not for three different crystal forms of the same compound.
| {
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Does a glass of water at room temperature emit (infrared?) radiation While reading the introduction to Feynman's lectures, it's mentioned how a glass of water cools down through evaporation, when some molecules get a bit extra energy and break free. If it's not a closed system, energy will be gradually taken away from the cup, hence blowing at the soup helps move those molecules away so that they don't reenter the surface.
But I thought that all bodies also radiate heat? Does a cup of water also emit low frequency radiation, or is my understanding incorrect?
| As @lemon explained, every body emits electromagnetic radiation if its temperature is above absolute zero
The peak wavelength of the radiation emitted by a blackbody is given by Wien's law: $$\lambda=2900/T$$ where T is the temperature in K, and the wavelength ($\lambda$) is in $\mu m$.
For a temperature of 300K (27C), approximately room temperature, the peak wavelength is $9.7\mu m$.
More details can be found at this site. The site has a calculator that shows the peak wavelength, given the temperature, as well as a graph that shows the spectrum at the selected temperature.
| {
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Two-dimensional collision conservation of energy According to wikipedia (https://en.wikipedia.org/wiki/Elastic_collision), and the sources cited in the article, the new velocity after a two-dimensional, elastic collision, can be calculated by rotating the reference frame so that components perpendicular to the line of contact do not change, and the use the formulas for a one-dimensional elastic collision for the remaining components along the line of contact.
But that does not seem to be right to me, because these equations are derived using both conservation of momentum and the conservation of energy. And while it seems perfectly fine to apply the conservation of momentum to this "reduced, one-dimensional problem", I don't see why it should be possible to "split" the conservation of energy into parts.
Can anyone explain to me why that's possible?
| We can write the kinetic energy of a particle moving in 2 dimensions as $$E=\frac{1}{2}mv^2=\frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2$$
Where $v_x$ and $v_y$ are the perpendicular components of velocity. If we take $v_x$ to be the velocity perpendicular to normal, the conservation of energy for $2$ colliding particles $1$ and $2$ becomes:
$$\frac{1}{2}m_1(u_{1x}^2 +u_{1y}^2) +\frac{1}{2}m_2(u_{2x}^2 +u_{2y}^2)=\frac{1}{2}m_1(u_{1x}^2 +v_{1y}^2)+\frac{1}{2}m_2(u_{2x}^2 +v_{2y}^2)$$
Where I used the fact $v_{1x}=u_{1x}$ and $v_{2x}=u_{2x}$. Thus, after cancellation, we get:
$$\frac{1}{2}m_1u_{1y}^2 +\frac{1}{2}m_2u_{2y}^2=\frac{1}{2}m_1v_{1y}^2+\frac{1}{2}m_2(u_{2y}^2)$$
Which is the same as that of the $1$ dimensional case.
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How to tell if spherical waves reach infinity? Say you consider spherical waves (momentum eigenstates) propagating outwards from some starting point $r = r_{0} $ (not defined for $r < r_{0}$) with $k \in \mathbb{R} $:
\begin{equation}
\psi\left(r, t\right) = \left( A {e^{ikr} \over r} + B {e^{-ikr } \over r } \right) e^{-ikt}
\end{equation}
We have boundary conditions at $r = r_{0} $:
\begin{equation}
\psi\left(r_{0}, 0\right) = {e^{ikr_{0}} \over r}
\end{equation}
This automatically sets $A = 1$ and $ B = 0 $.
Question: Does this wave reach infinity? At first glance it looks like the answer is definitely no:
\begin{equation}
\lim_{r \rightarrow \infty } \psi\left(r,t \right) = \lim_{r \rightarrow \infty} {e^{ikr} \over r } e^{-ikt} = {\lim_{r \rightarrow \infty} e^{ikr- ikt} \over \lim_{r \rightarrow \infty} r} = 0
\end{equation}
Since the the $e^{ikr-ikt}$ is just oscillating for any value of $r$ and $t$. However, if we consider the energy of the wave (let $I$ be the intensity):
\begin{equation}
I\left(r\right) = \left \vert \psi\left(r,t\right) \right \vert^2 = {1 \over r^2}
\end{equation}
And so the energy in a spherical shell at distance $r$:
\begin{equation}
4 \pi r^2 \cdot I\left(r\right) = 4\pi
\end{equation}
Notice the energy is constant independent of $r$, so in this sense the energy of the wave reaches infinity. What is happening, is the wave physically reaching infinity or not?
I am looking for an this question that might be generalizable to more complicated situations. Thanks!
| As stated, your question does not have a well-defined answer. If you had a definition for what you meant by "reaches infinity," then we would have something to work with.
Certainly the wave $\psi$ is not eventually $0$ as the term $e^{ikr}$ keeps it alive for infinitely many $r$. On the other hand, as you note, the limit is $0$, which does not provide any information about the wave itself being nonzero.
Physically, once $r$ is sufficiently large, the amplitude of the wave would be so small that no apparatus we invent could measure it. In that sense, it doesn't reach infinity.
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What are practical uses of over-damping? We have been given this task of preparing some small research on critical damping and comparing its behaviour and uses with over-damping. I am done with everything else but have been unable to find practical uses of over damping. It'd be great if someone could explain where it's desired.
| 'Critical Damping' is a descriptive term given to 2nd order linear dynamic systems where the damping factor is ~ 1.0. And for the 2nd order system critical damping provides a settling towards your equilibrium point as quickly as possible without overshoot or bouncing about the equilibrium state: a smooth however rapid transition. If you specify critical damping you might be trying to get your system to settle with an asymptotic trajectory as quickly as possible like targeting a data track on a disk drive for example.
But with overdamping you are further reducing speed for smoothness of settling to your equilibrium value. Any example of public transportation braking systems would be good examples where the desire is to provide the rider with comfort over the speed of coming to a stop. Like a train, elevator or automobile.
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Why are there more trapped protons during solar minima? Reading about trapped protons and eletrons in van Allen belts, I get the understanding that the number of trapped electrons increase during a solar maximum (which I find logical because we have more solar flares etc.), but that the number of trapped protons increase during a solar minimum. What is the explanation?
Also: what is "atmospheric cut-off"?
| Without the influence of an external force, the protons will happily gyrate around the magnetic field while bouncing from pole-to-pole as they drift around the Earth in the radiation belts. During a geomagnetic storm, the Earth's field changes rapidly causing protons in the radiation belts to either precipitate into the atmosphere or hit the magnetopause. In both cases, they are generally considered lost and no longer trapped.
Geomagnetic storms are caused by solar eruptions called coronal mass ejections or CMEs (and high speed streams). CMEs are produced by enhanced magnetic activity on the sun, which occurs more frequently during solar maximum.
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Does work done depend on the frame of reference? Suppose I am sitting on a bench and looking at a moving car. Force is applied on the car by its engine, and it makes it displace, hence some work is done on the car. But what if I am sitting in the car and looking at the bench? The bench covers some displacement, but who has applied force to it? Is any work done on it?
| Wikipedia says
In physics, work is the product of force and displacement.
Or to put it mathematically $$\overrightarrow W =\int_{\overrightarrow x_{i}} ^{\overrightarrow x_{f}}\overrightarrow F\cdot \overrightarrow {ds}$$ here ${\overrightarrow x_{i}}$ and ${\overrightarrow x_{f}}$ are initial and final position vectors and that dot is dot product.
As both $\overrightarrow F$ and $\overrightarrow {ds}$ are both frame dependent thus their product will change with frame transformation(unless there is some specific reason). It really depends on the context.
The bench covers some displacement but who has applied force to it. Is some work done on it?
Note that you are observing bench in a non inertial frame so you have to take into account pseudo forces which seem to do negative work on the bench on your system and decrease its energy(in that frame).
So, yes work is frame variant, it has to be, as follows from it's definition.
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How much power in a lightning strike? The Biblical story of 1 Kings 18:20-40 relates how a lightning strike ignited a bulls carcass, as well as a wood pyre, both of which had been thoroughly wetted with abundant water. Is there enough power in a lightning strike to do this? Does the power of a mountain top lightning strike differ from the power of a lightning strike elsewhere?
| A lightning strike can blow a tree or power pole into splinters. You can see it on Youtube. A "bolt from the blue" can strike from a cloud 10 miles away. Thunderheads reach up to 90,000 feet, so the height of a mountain makes little difference. Some of the deadliest strikes have been in the Tetons. Can what you read in Kings happen? Certainly.
Take a look at fulgurite fragments in the site below, or search "fulgurite and check images. If it can fuse meter long pieces of glass from sand, and as big as your arm, it can easily set a nice fat bull on fire despite being wet.
http://www.sciencealert.com/scientists-are-using-fossils-to-work-out-how-much-energy-is-in-a-lightning-strikes?perpetual=yes&limitstart=1
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What are the k-vectors in Ewald summation? Ewald summation is a common technique for computation of forces on charged particles in "infinitely periodic" (crystalline) systems.
I'm trying to understand Ewald force calculation from this website. The below image summarizes the two contributions to the force on a given atom $i$, from k-space and real-space.
What is the highlighted sum "summing over"? Is it all of the inverse positions of particles? Specifically how many instances of $\textbf{k} \ne 0$ should there be?
| The $k$-space contribution is intended to be the reciprocal space. The mighty power of Ewald summation is to split a slow converging summation into two rapidly converging series: this trick is made employing the Fourier transform and evaluating the summation on the phase space.
The definition of the vectors $k$ are the ones such that the following holds
$$e^{i\vec{k}\cdot\vec{r}}=1$$
that is nothing but
$$k_i\cdot r_j=2\pi\delta_{ij}$$
where $r_i$s (respectively $k_j$) are intended as the basis vectors of the Bravais lattice (respectively reciprocal lattice).
The summation should go in theory over all $\mathbb{R}^n$ but as you can see numerically, after a cutoff of few $k$s (tipically $10\times k$) you can truncate your summation.
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How can I calculate the sensitivity of a seismometer? I would like to know how to find the minimum ground motion a seismometer can measure, specifically if a certain seismometer can measure 1 micron/sec velocity. I have a few specs from the datasheet but I'm not a seismologist and am trying to figure out how to relate the specs to one another.
Velocity output band: 30s (0.03Hz) to 100 Hz
Output Sensitivity: 2400 V/m/s
Peak/Full scale output: Differential: +- 20V
Sensor dynamic range: 137 dB @ 5 Hz
What I initially tried was dividing Peak/Full Scale Output by Output Sensitivity but the number seems too large.
Thanks in advance!
| The dynamic range in system design describes the relationship between full scale output to the noise floor. Signal levels below the noise floor cannot be measured directly. If the signal path is digitized the digital resolution would normally be taken into account for the noise floor assessment.
The full scale range in voltage is +/- 20V which equates to 20/2400 = 0.00833(m/s) of full scale velocity.
The 137 dB dynamic range means the noise floor is 0.00833 (m/s) / 10^(137/20) = 1.17 e-9 (m/s)
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What happens to photons over distance? According to this diagram:
And a few articles I have read, the number of photons emitted from a point source is exactly the same, and the reason why we detect fewer photons as we move away from the source is because the photons spread over distance. If that's true, then why is there specific distance that light can travel in a medium like tap water (2 m) or distilled water (8 m) or ice beneath the south pole (about 100m - 200m)? Is it because photons are absorbed by that medium? And if I were to have a point source in "total" vacuum which emitted about 10,000 photons. If I went away from that source by 1 kilometer or two and surrounded the entire source by a sphere, will I detect all of those 10,000 photons?
| As you suggest the fall in intensity with distance in water, ice, etc is because the photons are absorbed. If you did your proposed experiment in space you would indeed detect all 10,000 photons.
There is a minor complication that we need to be aware of. If you shine a beam of light through some material then the light can be scattered as well as absorbed. So for example clouds are white not because they absorb the light but because the water droplets in the cloud reflect the light sideways and scatter it. If the water/ice/whatever just scattered the light then you would still detect all 10,000 photons but their original directions would have changed.
However water does absorb photons as well as scatter them. Water doesn't absorb light strongly, which is why the path length is quite long. However any dielectric, even apparently clear ones like water, will absorb light to some extent due to the interaction between the electric field of the light and the electrons in the dielectric.
| {
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Closed container buoyant force A closed container of water contains a Styrofoam block attached to the bottom of the container by a massless string. When the system is accelerated upwards, what happens to the tension in the string?
Background: I am an MCAT Physics teacher and this question was on an MCAT. Assuming the water is compressible, Newton's first law would indicate that a higher than normal pressure would result at the bottom of the container and a lower than normal pressure would result at the top of the container. The net result would be an increase in buoyant force (force due to a pressure gradient); thus, increasing the tension. The question I have is, "If the water were considered to be incompressible, would the tension still increase?"
| First, how is it that Styrofoam can float?
Because when partially immersed in water the weight of the displaced water (upthrust) is equal to the weight of the Styrofoam.
If the Styrofoam is fully immersed why must it be tethered by a string?
Because the upthrust (weight of displaced water = mass of displaced water $\times g$) is now greater than the weight of the Styrofoam (mass of Styrofoam $\times g$) and the difference is equal to the tension in the string.
Tension = (mass of displaced water - mass of Styrofoam) $\times g$
If the container and its contents has an upward acceleration of $a$ then the effective value of $g$ is now higher ($= g+a$).
So the tension is now higher = (mass of displaced water - mass of Styrofoam) $\times (g+a)$.
This increase in tension is to be expected because if the experiment was performed in space or with the container in free fall ($g_{\rm effective} =0$) the tension in the string would be zero.
The upthrust is due to the pressure change from the bottom to the top of the Styrofoam or pressure gradient within the water.
If the effective value of $g$ increases then so does the upthrust and the weight of the Styrofoam in the same proportion.
So the tension in the string will also increase.
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Is the fact that 100 kPa equals about 1 atmosphere accidental? Typical atmosphere near sea level, in ambient conditions is around 100,000 pascals.
But the pascal, as the unit, is not defined through Earth atmospheric pressure. It's defined as one newton per square meter. The newton is $\rm{kg \: m}\over s^2$. So, $\rm[Pa] = [ {kg \over {m \: s^2}} ]$.
Nowadays, definitions of units are often fixed to various natural phenomena, but it wasn't quite so when they were being created.
The Second is an ancient unit, derived from a fraction of day, 1/86400
of synodic day on Earth. The meter is derived from circumference of Earth, $10^{-7}$ the distance from north pole to equator. The kilogram came to be as mass of a cubic decimeter of water.
100,000 pascals, or 1 bar, though, is about the average atmospheric pressure at sea level. That's an awfully round number - while Earth atmosphere pressure doesn't seem to have anything in common with the rest of the "sources" of the other units.
Is this "round" value accidental, or am I missing some hidden relation?
| This is a coincidence. There's nothing about the atmosphere that would make it have a nice relationship with the Earth's rotation or diameter, or the fact that water is plentiful on the surface.
On the other hand, it's important to note that the coincidence isn't quite as remarkable as you note, because of a version of Benford's law. Given absolutely zero prior knowledge about how much air there is in the atmosphere, our guess about the value of the atmospheric pressure would have to be evenly distributed over many orders of magnitude. This is akin to throwing a dart at a piece of log-scale graph paper:
Note that the squares in which the coordinates start with $1.\:{{.}{.}{.}}$ are bigger than the others, so they're rather more likely to catch the dart. A similar (weaker) effect makes the probability of the second digit being 0 be 12% instead of the naive 10%.
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What is the meaning of the Hermann-Mauguin symbol $R$? In the International Tables for Crystallography A, table 4.3.2.1, the trigonal lattices are identified as having space group IT numbers 143 through 167. However, only space groups 148, 155, 160, 161, 166, and 167 have $R$ in their Hermann-Mauguin symbols, the others have $P$ in their HM symbols. All symbols have $3$ or $\overline{3}$ in their symbols, as is required for trigonal symmetry. Section 4.3.5.1 claims that the current 'P' HM symbols were a replacement for an early symbol 'C'-which indicates a "double rectangular C-centered cell", and that the replacement was done "for reasons of consistency", but this doesn't actually illuminate the situation very much.
If I were to attempt a definition of $R$ in the HM symbol, I'd say all groups with a $3$-fold or $\overline{3}$-fold rotational symmetry about a body diagonal, without a $4$-fold axis about a primitive vector. But this appears to contradict the usage in the International tables.
So what does the $R$ indicate?
| All those spacegroups labelled with $R$ have two settings, one with a hexagonal unit cell and one with a rhombohedral unit cell. The complete notation would be e.g. $R3 :\!\!R$ for the latter and $R3 :\!\!H$ for the former. The standard setting is the hexagonal one.
Let's take the example of $R32$. In rhombohedral axes (Hall symbol $P3^*2$), the space group is generated by
*
*the threefold rotation about $\renewcommand{\vec}[1]{\mathbf{#1}}\vec{a}+\vec{b}+\vec{c}$, which circularly permutes $\vec{a}$, $\vec{b}$, and $\vec{c}$;
*a twofold rotation about $\vec{a}-\vec{b}$, which sends $\vec{a}$ to $-\vec{b}$, and $\vec{b}$ to $-\vec{a}$.
In hexagonal axes (Hall symbol $R32''$), it is generated by
*
*the threefold rotation about $\vec{c}$, which "generates the hexagon";
*the twofold rotation about $\vec{a}+\vec{b}$, which permutes $\vec{a}$ and $\vec{b}$;
*the centring translations $\frac{1}{3}(1,2,2)$ and $\frac{1}{3}(2,1,1)$.
It is not too difficult to see how those transformations are geometrically related.
Now let's take the example of $P3_221$ (Hall $P3_22''$). This is very similar to the above $R32''$, as conveyed by the similarity of the Hall symbols, but with the threefold rotation replaced by a threefold screw about the same axis $\vec{c}$ and an intrinsic translation $\frac{2}{3}\vec{c}$. But contrary to the previous example, there is no way to use rhombohedral axes.
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Why is unit of pressure (psi) used to determine things like bite force of an animal? Whether in tv documentaries or journals, whenever they talk about an animal's bite force, it's measured in PSI anytime imperial units are used (ex: National Geographic, NIH Journal). Many even seem to highlight the fact that it's pounds per square inch.
But when it's used in SI unit context (ex: BBC Earth), it's written in Newtons, a measurement of force.
It seems to me that a force should be measured in a unit of force and not pressure. Additionally, pressure alone would seem to tell us nothing since area is never specified, thus unknown force.
So why is a unit of pressure so dominantly used to indicate an animal's force?
For example, this science daily shows the croc having a bite force of 3700lbs. But then other sources (along with every other web search) will say 3700psi. This National Geographic article even writes psi side by side with newtons as if it was pound-force.
| I absolutely agree with Jon Hilden. Bite strength is a torque. The limit of an animals bite must be a function of jaw muscle strength and jaw geometry, unless you're about to crush your teeth, which involves a whole different set of numbers than those given for bite strength (e.g., 30,000 psi vs. 150 psi). A couple of examples:
*
*If you take an animal and dull it's teeth (e.g. give it 100x more tooth surface area), it wouldn't have 100x as strong a bit.
*If you want to bite something really hard, you bite it with your back molars, because you can clamp down far harder (far more 'bite force') with the teeth closer to the hinge of your jaw.
*You an chip your tooth on a rock. Dentin has a compressive strength in the 10,000s of psi, which you are therefore achieving because the surface area between tooth and rock is very close to zero. If it were actually psi rather than torque, that small area would mean you'd be applying almost no force.
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Klebanov-Witten and Conifold equation I'm a bit confused by some that is understood in the following papers: 1 and 2. I understood that there is a GLSM with fields $A_i,B_i$ with $i=1,2$ whose moduli space is the conifold described with the constraint $z_1z_2=z_3z_4$ (page 5 of 1). But when I define the "mesons" $z_1=A_1B_1$ ecc, $A$s and $B$s are matrix valued fields (infact the superpotential $W$ has a trace), and so are the $z$s. So:
*
*how can I interpret the conifold equation from the matrix equation $z_1z_2=z_3z_4$?
*what tells me that the equation that I have to impose is morally $W=0$?
| *
*The snippet you're looking for here is at the top of page 10. The conifold coordinates are of course never matrix-valued. When the $A_i$,$B_i$ are matrix-valued, the conifold coordinates are $z_i = Tr A_i B_i$.
*I'm not sure exactly where you're referring to, but I would suppose that it's because $W$ is not renormalizable.
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How to show a sum of positive definite operators is still positive definite? Consider a Hamiltonian $H$ of the form
$$\begin{split}H&=\sum_{i,j}A_{ij}^\dagger A_{ij},\\
A_{ij}&=(1-\sigma_i^z\sigma_j^z)e^{-g (\sigma_i^x+\sigma_j^x)},\end{split}$$
where $g\in\mathbb{R}$ is a real parameter and $\sigma_i^x$ and $\sigma_i^z$ are the Pauli matrices $x$ and $z$ acting on the $i$th spin (qubit). Obviously, each term $A_{ij}^\dagger A_{ij}$ in the summation is positive definite. What I found in numerics is that the eigenvalues of $H$ are all non-negative. How should I prove (or disprove) that $H$ is positive definite?
I am aware that supersymmetric quantum mechanics Hamiltonians are positive definite. Can the above Hamiltonian be cast into a supersymmetric form explicitly?
| Three remarks. One, a sum of positive (semi)definite matrices is again positive (semi)definite. The proof is really easy. Let's work over the reals for simplicity. A matrix $H_I$ is positive definite iff for any vector $v$, we have
$$v^T \cdot H_I \cdot v > 0\,.$$
Now let $H = \sum_I H_I$ be a sum of a finite number of positive matrices $H_I$. Then for any vector $v$, we have
$$v^T \cdot H \cdot v = \sum_I v^T \cdot H_I \cdot v > 0$$
because a sum over positive numbers is again positive.
Two, the operator $A_{ij}$ is only semidefinite, because it annihilates certain states. Consider for instance the two-qubit state
$$ | \psi_{ij} \rangle = | + \rangle_i \otimes | + \rangle_j - | - \rangle_i \otimes | - \rangle_j\,.$$
Then I think that $A_{ij} | \psi_{ij} \rangle = 0$ (unless I made a mistake). To check this you first show that
$$\exp\left[-g( \sigma_i^x + \sigma_j^x)\right]$$
leaves $|\psi_{ij} \rangle$ invariant, and second that it's killed by $1 - \sigma_i^z \sigma_j^z$.
Third, you want to cast the Hamiltonian into a form $H = Q^\dagger Q$, right? In linear algebra this is known as a Cholesky decomposition. Since $H$ is positive semidefinite, it's certain that such a $Q$ exists (and that it's upper triangular), but it will not be unique in general. There are standard algorithms to compute (at least one realization of) $Q$ if $H$ is a concrete matrix.
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What do they mean when they say that it does not require any work to move a charge from one point to another in an equipotential surface? In the textbook it says that no work is required to move a charge from one point to another on an equipotential surface. Do they mean work by the electric field or work by anything? Because clearly the object cant just magically move sideways with nothing.
| It takes work to transfer kinetic energy into the charged object and get it moving, sure. But if the object was already moving, it wouldn't lose any energy by moving along the equipotential. Furthermore, you could decrease the work needed to stop and start by just moving slower, with no work needed in the limit of infinite time taken. The work of starting and of stopping can cancel each other out so that all your energy can be recovered in an ideal system. So if work is needed to move a charge along an equipotential, it isn't because of the electric field.
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Do gases mix faster when of unequal temperature? Or,
Is my room aired quicker during winter?
I always feel that in winter, after opening the window, my study room becomes breathable quicker than in summer. Now, this is of course highly subjective, but even though I see no rational explanation supporting such a phenomenon, I neither can intuitively outrule it.
Surprisingly, Google was no help here.
| There are a few things that would be going on here.
1. Convection
There's a common expression "heat rises". That is because things are heated, their density often lowers. In fluids (such as air and water), if there is a density difference in an open space, you will create a flow from high to low density (going from high to low pressure, and the diffusion will both contribute).
The cooler it is outside compared to inside; the greater this effect.
On the same note, once you get air moving, it not only cools the room quicker; but it feels cooler if it moves more.
2. Humidity
Warm air can pick up more moisture, and humid air feels warmer. It's harder to cool yourself off when it is humid.
I imagine this is a large part of why it becomes breathable quicker. The humidity in the air is a big part of comfort. When you open the window and it's humid outside,there is less diffusion of moisture (on really bad days I've had it bring moisture in). In the winter the cool air doesn't have as much moisture, so the humidity is quick to start mixing with the dry air outside.
| {
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Does Earth's precession rate change with the seasons? As Earth's precession is caused by the differential solar, etc., attraction to its 'spare tire', then does the rate go to ~zero at the equinoxes and ~twice the average at the solstices? (preferred answer: yes :-)
Or is there some 'carry-over' 'precessional momentum' that keeps the annual rate more constant? (preferred answer: no :-)
| The contribution to the precession from the sun does depend on the time of year (basically because tidal forces have a azimuthal symmetry around the line connecting the bodies but no polar symmetry), but there is also a contribution of similar magnitude from the moon, so the overall effect has a more complicated time dependence and only rarely drops to zero.
The inertia in the system is the angular momentum of the planet which will have a constant direction if there is no torque. Nor should you be imagining the tidal torque just disappearing, it rises and falls smoothly.
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Has the curvature of spacetime been measured at the human scale? The curvature of spacetime has been observed many times from the deflection of light around massive astronomical objects. But has it been observed around small objects in a lab?
In the Cavendish experiment, the gravitational attraction between two masses does sufficient effect for it to be measured on Earth. Thus, it raises the question whether light deflection from curved spacetime is also measurable in the lab.
If it has not been achieved, how far are we from it? How much precision would be needed?
| Yes it has, on 11th Feb last year - it did make many mainstream news reports. In one of the largest "labs" ever built, each arm is 2.5km long. I guess that fits the criteria "human scale" within certain limits.
http://en.wikipedia.org/wiki/LIGO
You may also read about various other methods here
https://en.wikipedia.org/wiki/Gravitational_wave#Detection
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Can quarks tunnel out of a proton? Can a virtual electron tunnel away from its antiparticle? I know that particles can tunnel through an impassable barrier. But what happens if a quark tunnels out of a proton?
Has that ever been observed?
And what about virtual particles?Can they tunnel away from each other?
| As you may know, it is only possible to tunnel if the barrier reaches a maximum and then starts decreasing again, so as to reach an area of potential < than the energy of the particle, where this can exist.
The quarks are bound by the strong force, which is quite weakly binding at very short distances ("asymptotic freedom") but becomes linearly stronger as you get out of the nucleus. The physical reason is that the strong force is mediated by gluons, that, as opposed to photons, can couple and radiate from/to themselves, they do not get sparse but on the contrary keep the 'density' of gluons constant.
So no, free quarks are never observed. This is also called "colour confinement". Strong force interacting particle are assigned a 'colour' charge in the same way as EM interacting particles have an 'electrical' charge - charge is particle physics and similar means something that is conserved. Colour + anti-colour makes colour neutral, same as blue+red+green. In nature you will always find colour neutral particles, never bare coloured particles. Because of the above discussion of the strong force getting stronger as quarks are moved far apart. As you are pulling quarks apart, at some point it would becomes energetically favourable just to convert some energy into the mass of other quark-antiquark pairs and form new bound colour neutral particles.
I should probably add that technically the top quark is found as a bare coloured change. But it's just because its mass is so high that it decays before it has time to form a strong force bound state.
Virtual particles do not exist, so no they cannot tunnel away from each other. I think you'd be better off asking that as a separate question.
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Bernoulli's theorem In Bernoulli's theoram as stream lines cannot intersect each other and even water is nearly incomprehensible then how is it possible to all the water which got in come out, eg 100 stream lines goes in from one side then 100 stream lines should come out the other end whereas the area is decreased and even water is incomprehensible
| Stream lines have no radial dimension (width).
You can keep squishing them infinitely close together, because they just describe the motion of the fluid.
What you describe does have implications for continuity though. Since mass flow has to stay the same (no mass being created/destroyed), if the area decreases, the fluid must move faster. That is to say that the spacing between the lines would have an effect on the velocity of the fluid.
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Why is translation energy a continuum and not quantized? When starting the study of quantum theory, usually the teacher separates the functions in electronic, vibrational, rotational and translational parts.
As far as I know, we should solve Schrodinger equation for each movement; and as far as I know, rotational and vibrational energies are quantized.
Is there any good explanation about why translational energy is not quantized? And how do we know $\psi$ is separable into vibrational, translational and so on?
| If you are in a finite system, translational energy is quantized as well.
The simple example is the infinite square well potential, which is just a finite one-dimensional free particle (also often called a particle in a box). The fact that the wave function must goes to zero on the boundary means that the energy levels are sinusoids with wavelengths that evenly divide into $2L$ where $L$ is the size of the system.
This gives a series of energy levels $E_n = \frac{1}{2m}\frac{h ^2}{\lambda_n^2} =\frac{h^2}{8mL^2}n^2$ where $n$ is a positive integer, so the energies are quantized. However, notice that the spacing between the energy levels will be proportional to $\frac{1}{L^2}$ so as the system size becomes large, the energy levels are approximately continuously spaced.
For a rotational system, the energy spectrum will also smooth out as the size gets large (so you don't see any quantum effects on bicycle wheels) but the molecule does not have a macroscopically large extent for its rotational motion (only the size of the molecule) whereas for its translational degree of freedom it has the whole container.
As for vibrational degrees of freedom, a tiny "block on a spring" isn't a lot different in principle from a particle stuck in a box. But here the particle stays confined by the spring microscopically close to home, so it's effectively a small box and so there is quantization.
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Derivation of Newtons second law of motion from the principle of conservation of energy Is newton's second law a consequence of the principle of conservation of energy? How can we arrive at
net force = rate of change of momentum
using only the law of conservation of energy?
| Conservation of energy isn't strong enough to reproduce Newton's second law. As a counterexample, consider a situation with no potential energy, $U(\mathbf{x}) = 0$. There is no force, so according to Newton's laws, the particle should move with constant velocity, $\mathbf{x}(t) = \mathbf{v}t$. But conservation of energy tells us that the kinetic energy $\frac{1}{2}m\|\mathbf{v}\|^2$ remains constant, thus only constraining the magnitude of the velocity to be constant. This allows for nonphysical motions like $\mathbf{x}(t) = (\cos t, \sin t)$, whose velocity magnitude is constant but velocity is not.
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Why is energy usually concentrated on low frequency modes in dynamics? In all structural dynamics applications I have seen, the motion is mostly governed by low frequency modes. For example, a pretty accurate approximation of buildings dynamics can be obtained with the two or three lower frequency modes, provided the motion is linear.
That might be a silly question, but why is it so? The answer might be in the concept of modal mass but the physical interpretation of this observation is not obvious to me.
| I believe there are two factors in play:
*
*Ease of excitation. If you have wind driving a tall building, there is a force along the entire length of the building. This is most likely to excite the lowest frequency mode - in order to drive a higher mode, the driving force must be out of phase between different parts along the height of the building in just the right way.
*Damping. Typically, higher frequencies will experience greater damping per unit time (same damping factor would result in same decrease in amplitude after a number of cycles - but that number of cycles is reached more quickly for the higher frequencies)
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Why do the clouds near the horizon appear two dimensional? Look at the photo. The clouds right above us seem real, but those near the horizon appear as if they are from a painting.
They lack depth and are visually flat. Why is that?
| I believe it's because of two reasons: relative distance and angular size. If we see cumulus clouds near overhead (stratus clouds would be a separate argument due to height and shape), they're relatively near and apparently 3D. But on the horizon in any direction, much further away and generally speaking smaller angular size (same exact cloud near overhead would have a much larger angular size). It's fairly easy to visualize in 3D with practice. The reason I say "near overhead" and not overhead is because we'd be looking at the bottom of a cloud and typically for cumulus, they're taller than wider. So that could also contribute to the perceived difference.
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Would mechanically moving electrons create a super-strong magnetic field? According to this Veritasium video, the magnetic field in a wire with a non-zero current is an artifact of special relativity. A moving charge sees a speed difference between the wire and the electrons in it (since the electrons are moving). Due to Lorentz contraction there appears to be a net positive charge and a repelling force arises. The physical speed of electrons supporting a current is very slow, a few micrometers per seconds, but due to the number of electrons and the strength of the electric field the effect is macroscopic.
What if you could mechanically increase this speed? Say you had a charged capacitor, and one of the plates was given a parallel velocity. E.g. two concentric cylinders with a charge imbalance, with one of them attached to a motor. Wouldn't that create a very strong apparent magnetic field?
| The magnetic field is created by a current as in the number of electrons per second. The speed of the electrons is thus irrelevant, only the current value in Amperes is important. Therefore the problem you are describing does not exist. However, if you rotate charged objects, such as by attaching them to a fan, they would indeed create a magnetic field. This would not be nearly as efficient as coiling a wire into multiple loops, because the magnetic field is proportional to the number of loops given the same current.
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When is the heat transfer between a solid and fluid conduction or convection? I know that the heat transfer between solids and liquids occur via both conduction and convection.
However, I am not sure about the fine line that separates them. For example, what is the mode of heat transfer when a hot piece of steel is put out in the air. Does the wind change the situation?
Or is, for instance, the heat transfer between a hot solid and a cold static liquid, which are in the same container conduction or convection and why?
| I will use the term fluid, that refers to both liquid and gas phase.
It is conduction if single fluid particles transport heat via Brownian motion. It is convection if a macro scale movement of fluid particles is involved. Usually both are playing a role, but for smaller differences of fluid and solid temperatures conduction is more important and conversely induced convection is more important in larger temperature differences.
Convection can be induced or forced. Forced convection occurs when for example a fan blows fluid to the solid and induced convection is when the solid for example is so hot compared to the surrounding fluid that the nearest fluid change their density and become more buoyant and start moving up. Then colder fluid converges the solid from around and heat transfer is enhanced due to larger local temperature differences.
Like the comments above mentioned radiation is also important, although it is more important in gases where the conductivity is smaller than in liquids.
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Why is the S-Matrix element essentially the residue of the Green function (LSZ formula)? On Wikipedia, quite similar to the script I am following the LSZ formula is given as
$$
_{out}\left<p_1,...,p_n| q_1,...,q_m \right>_{in} =\\
\int \prod_i^m \left(\textrm{d}x^4\, i e^{-q_ix_i}(\square_{x_i}-m^2)\right)\prod_j^n \left( \textrm{d}y^4\,i e^{-p_jy_j}(\square_{y_j}-m^2)\right) \left<0| T[\varphi(x_1)\ldots \varphi(x_m)\varphi(y_1)\ldots\varphi(y_n)]|0\right> \\ \equiv \,
\prod_i^m \left(-i (p_i^2-m^2)\right)\prod_j^n \left( -i (p_j^2-m^2)\right) \hat\tau(p_1...p_n,-q_1...-q_m)
$$
up to some renormalization Factors $Z$ and where $\hat \tau$ denotes the Fourier Transform of the time ordered correlation function $\left<0|T[...]|0\right>$.
It then says "..., this formula asserts that S-matrix elements are the residues of the poles that arise in the Fourier transform of the correlation functions as four-momenta are put on-shell."
How do I come to this realization? I know, that if I compute the time ordered two-point correlation function of a free field
$$
G_F(x) = \lim_{\epsilon \downarrow 0} \int \frac{d^4k}{(2\pi)^4} \frac{e^{ikx}}{m^2-k^2-i\epsilon}
$$
and perform the $x^0$-integration here I get essentially the residue at $k^0 = \pm (\omega - i \epsilon)$, but I wasn't able to generalize this result to the above case.
| The argument is completely general. The right hand side of your equation should really have a limit, where the squares of the four momenta are taken to $m^2$; i.e., they are put on shell. Now, the correlation function is multiplied by $p^2-m^2$, so the only way the limit is gonna be nonzero is if the correlation function has a pole at $p^2=m^2$; if it was regular, when multiplied by $p^2-m^2$ the result would just go to zero as $p^2 \to m^2$. The result of the limit is the residue of the pole, almost by definition.
Let's make it explicit: suppose that all the momenta except one are off shell; let's call the one that gets put on shell $p$, and suppose that for $p^2 \approx m^2$ the correlation function takes the form
$$\tau(p^2, \dots) \approx \frac{Z}{p^2-m^2} + \text{finite},$$
where $Z$ is some number and $\text{finite}$ represents things that don't diverge as $p^2 \to m^2$; this number is the residue of the pole since it is the coefficient of the $1/(p^2-m^2)$ term in the Laurent expansion. Then when put into the LSZ formula we get
$$\lim_\limits{p^2 \to m^2} (p^2-m^2) \tau(p^2, \dots) = Z,$$
so the residue $Z$ is what ends up in the RHS of the LSZ formula. Again, note that if the divergence of $\tau$ took any other form (for example a double pole, an essential singularity or a delta function), the result of the limit would be either zero or infinity. The only way to get a sensible result is to have a simple pole.
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Is there a "square root" version of the Einstein field equation? It is well known that the Klein-Gordon equation have a kind of "square root" version : the Dirac equation.
The Maxwell equations can also be formulated in a Dirac way.
It is also well known that the metric of general relativity have a kind of "square root" version : the tetrad field (or vierbein) of components $e_{\mu}^a(x)$ :
\begin{equation}\tag{1}
g_{\mu \nu}(x) = \eta_{ab} \, e_{\mu}^a(x) \, e_{\nu}^b(x).
\end{equation}
Now, a natural question to ask is if the full Einstein equations :
\begin{equation}\tag{2}
G_{\mu \nu} + \Lambda \, g_{\mu \nu} = -\, \kappa \, T_{\mu \nu},
\end{equation}
could be reformulated for the tetrad field only (or other variables ?), as a kind of a "Dirac version" of it ? In other words : is there a "square root" version of equation (2) ?
| *
*Since Nature has fermionic matter we are anyway ultimately forced to rewrite the metric in GR in terms of a vielbein (and introduce a spin connection). See e.g. my Phys answer here. The fermionic matter obeys a Dirac equation in curved spacetime. This however would not amount to a square root of EFE.
*There exist supersymmetric extensions of GR, such as, SUGRA.
*Another idea is to consider YM-type theories as a square root of GR, or GR as a double copy of YM. See e.g. the Ashtekar formulation or the KLT relations.
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Construct operator such that division of expectation values is equal to expectation value of the operator Is is possible to construct an operator $\hat{C}$ out of $\hat{A}$ and $\hat{B}$ such that:
$$\frac{\langle \psi|\hat{A}|\psi\rangle}{\langle\psi|\hat{B}|\psi\rangle} = \langle \psi|\hat{C}|\psi\rangle,$$
for any state $|\psi\rangle$?
| Let's say that this equation holds for particular $|\psi \rangle$. Putting $N |\psi \rangle$ would result in:
$$
\frac{\langle \psi| \hat{A} |\psi \rangle}{\langle \psi| \hat{B} |\psi \rangle} = |N|^2 \langle \psi| \hat{C} |\psi \rangle
$$
Which of course can hold only for $|N|^2 = 1$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/359046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What happens to the period of a pendulum if a spherical bob were to spin around the axis of the string? Consider a normal pendulum with a spherical bob oscillating back and forth. Would the period of the pendulum be longer, shorter or unchanged if the bob were to spin around the axis of the string that holds it?
| Rotation will create a gyroscopic tilting force perpendicular to the plane of oscillation, as the gyroscope "tries" to remain oriented the same way. This causes precession, giving the gyroscope oscillations of it's own.
Simplistically, It may be modelled as a compound pendulum, one pendulum on the end of another. There will be two periods, and these will combine to give a displacement waveform as two superimposed sine waves. The overall system will have a period that is the period of the beats of the two waveforms, which is
$\frac{1}{T_{sys}}=\frac{1}{T_{main}}-\frac{1}{T_{gyro}}$
There are other effects, however, making the above approximate; and the actual motion is probably chaotic - (meaning it is multi-stable with almsot random transitions).
As the forces the gyroscope experience depend on the position of the main pendulum as well as gravity, this will introduce a further harmonic.
Additionally if the motion of the gyroscope increases its height, this could shorten a swing. If its height is lowered it could lengthen a swing.
Finally of course practically the motion is damped by air resistance and friction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finite dimensional representations of Lorentz group I am trying to understand the topic in the title but I found some difficulties.
For example, I understand that $\left(\frac{1}{2},0\right)\otimes\left(\frac{1}{2},0\right)=\left(1,0\right)\oplus(0,0)$ which is a consequence of Clebsch-Gordan decomposition and the scalar
representation is given by the antisymmetric product. The representation
$(1, 0)$ can be represented by an antisymmetric, self-dual second rank tensor.
However, if I consider the following:$\left(\frac{1}{2},\frac{1}{2}\right)=\left(\frac{1}{2},0\right)\otimes\left(0,\frac{1}{2}\right)$, I don't understand how to decompose it.
I know that represents a four vector field, with a temporal scalar component (spin 0) and vector component (spin 1) but I don't really understand why.
|
[...]However, if i consider the following: $(\frac{1}{2},\frac{1}{2})=(\frac{1}{2},0)\otimes(0,\frac{1}{2})$, i don't understand how to decompose it. I know that represents a four vector field, with a temporal scalar component (spin 0) and vector component (spin 1) but i don't really understand why.
There is nothing to decompose with respect to the full $\mbox{SL}(2,\mathbb{C})$ because the spinor tensor with one undotted index and one dotted index is irreducible. Actually, you should write it backwards and apply the same rules as you already did for the 1-form representations:
$$\left(\frac{1}{2},0\right)\otimes \left(0,\frac{1}{2}\right) = \left(\frac{1}{2},\frac{1}{2}\right) $$, and this because $\frac{1}{2}+0 = \frac{1}{2}-0 = \frac{1}{2}$ for both terms.
Now here is something that addresses your question. $\mbox{SU}(2)$ is a proper subgroup of $\mbox{SL}(2,\mathbb{C})$. So one may ask if the representation $\left(\frac12,\frac12\right) $, which as I said is irreducible with respect to $\mbox{SL}(2,\mathbb{C})$, is perhaps reducible with respect to $\mbox{SU}(2)$. The answer is positive. The general formula below
$$D^{(u,v)}|SU(2) \cong D^{(u)}\otimes D^{(v)} \cong \sum_{w=|u-v|}^{u+v}\oplus D^{(w)}$$
becomes
$$\left(\frac12,\frac12\right)|_{\mbox{SU}(2)} \equiv D^0 \oplus D^1 $$
| {
"language": "en",
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Do fields describing different particles always commute? Is it true that field operators describing different particles (for example a scalar field operator $\phi (x) $ and a spinor field operator $\psi (x) $) always commute (i.e. $ [\phi (x), \psi (y) ]=0, \forall x,y $) in interacting theory?
Or is it true only at equal times? (i.e. $ [\phi (t,\vec x), \psi (t, \vec y) ]=0, \forall \vec x, \vec y $)
Or is it in general not true even at equal times?
Finally, if the fields in account are both fermionic must the commutator be replaced with an anticommutator?
| The Schrödinger-picture annihilators for different sorts of particle commute; so e.g. if you've got creation and annihilation operators for neutrinos as well as electrons in some sort of box, those operators commute across species but within one species do not commute with their adjoints.
Note that if you add time-dependence to them to put them into the Heisenberg picture, this is no longer true in general. Several interaction pictures do have this property, in particular when the baseline Hamiltonian that we "Heisenbergize" for lack of a better term has the quadratic form $\hbar\omega~\hat a^\dagger \hat a$ so that $\hat a(t)$ takes on the form $e^{-i\omega t}~\hat a_0.$
| {
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Variational Baeriswyl wavefunction for 2 dimensions I am reading this article.
The model hamiltonain of 2D square lattice for spinless fermions is written as:
$$H=H_{kin}+H_{int}=-\frac{J}{2}\sum_{<n,m>}c_n^\dagger c_m+\frac{V}{2}\sum_{<n,m>}n_nm_m$$
with J=hopping, V=interaction potential, $<n,m>$ nearest neighbor pairs and $n_n=c_n^\dagger c_n$=Number operator.
Baeriswyl wavefunction (BWF) can be written as:
$$|\psi_B>=N_B^{-1}\exp{(\tilde{\alpha}H_{kin})}|CDW>$$
and expression for CDW in k-space can be written as:
$$|CDW>=\Pi_{k\epsilon RBZ}\frac{1}{\sqrt{2}}(c_k^\dagger +c_{k-Q}^\dagger)|0>$$
If we convert H into k-space also we will get something like this:
$$H_{kin}=\sum_k\epsilon(k)c_k^\dagger c_k$$
$$H_{int}=-\frac{V}{N}\sum_{k,k',q}\epsilon(k)c_{k+q}^\dagger c_k c_{k'-q}^\dagger c_{k'}$$
To get final expression for $|\psi_B>$ (equ.3 in mentioned article) one have to apply $e^{\tilde{\alpha}H_{kin}}$ on $|CDW>$. I tried to solve this but couldn't succeed. Can anyone help me in this?
My Attempt
$$|\psi_B>=\Pi_{k\epsilon RBZ}\frac{N_B^{-1}}{\sqrt{2}} \exp{[\tilde{\alpha}{\sum_k\epsilon(k)c_k^\dagger c_k}]}(c_k^\dagger +c_{k-Q}^\dagger)|0>$$
$$|\psi_B>=\Pi_{k\epsilon RBZ}\frac{N_B^{-1}}{\sqrt{2}} [\exp{[\tilde{\alpha}{\sum_k\epsilon(k)c_k^\dagger c_k}]}c_k^\dagger |0> +\exp{[\tilde{\alpha}{\sum_k\epsilon(k)c_k^\dagger c_k}]}c_{k-Q}^\dagger)|0>]$$
In the article they are saying that this is equal to
$$|\psi_B>=\Pi_{k\epsilon RBZ}\frac{N_B^{-1}}{\sqrt{2}} [\exp{[\tilde{\alpha}{\sum_k\epsilon(k)}]}c_k^\dagger |0> +\exp{[\tilde{\alpha}{\sum_k\epsilon(k)}]}c_{k-Q}^\dagger)|0>]$$
But how?
| $H_{kin}$ can be written as:
$$\begin{bmatrix}c_k^\dagger & c_{k-\pi}^\dagger\end{bmatrix}\begin{bmatrix}\epsilon(k) & 0 \\0 & \epsilon(k-
\pi)\end{bmatrix}\begin{bmatrix}c_k \\ c_{k-\pi}\end{bmatrix}$$
and $\epsilon(k-\pi)=-2*t*\cos(k-\pi)=2*t*\cos k=-\epsilon(k)$ and at half-filling $Q=\pi$. As $c_k^\dagger$ and $c_{k-\pi}^\dagger$ are eigenstates of $H_{kin}$ so one can write $\exp(\alpha H_{kin})(c_k^\dagger+c_{k-\pi}^\dagger)=\exp(\alpha \epsilon(k))c_k^\dagger+\exp(-\alpha \epsilon(k))c_{k-\pi}^\dagger$
| {
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Why are work and energy considered different in physics when the units are the same? There is a question that explains work and energy on stack exchange but I did not see this aspect of my problem. Please just point me to my error and to the correct answer that I missed.
What I am asking is this: Why in physics when the units are the same that does not necessarily mean you have the same thing.? Let me explain.
Please let me use m for meter, sec= second , and kg = kilogram as the units for brevity sake.
The units for work are kg * m/sec^2 * m. The units for kinetic energy are kg* (m/sec)^2. They look that same to me. I need them to be the same so I can figure out the principle of least action. Comments are welcome.
| If your velocity changes from 5 m/s to 8 m/s, you say your velocity has changed by 3 m/s (assuming same vector direction) and your new velocity is 8 m/s. This seems like a very obvious statement; 3m/s represents change and 8 m/s the measure. In essence, a change in a vector or scalar quantity will have the same units as the quantity itself.
Work is nothing but change in energy and hence has the same units as energy itself.
It is a short answer but this is it!
| {
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Location of lens having effective focal length We know a/c to Gullstrand's equation that the effective focal length of two lenses separated by a distance $d$ is given as $$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1f_2},$$ but the equation doesn't clarify on the position of the lens having this effective focal length, how do i calculate that?
| Well, the distance will simply be the EFL - i.e. given the position of the final spot/image the lens will be at the EFL distance.
| {
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If I say time is the fourth dimension am I wrong? As far as I know the prevailing view is that time is the fourth dimension, but I've read there is also a spatial fourth dimension and even higher spatial dimensions after that so I hesitate to say that time is the fourth dimension. So, if I say time is the fourth dimension am I wrong?
| Whatever you say you are almost certainly wrong (this is nothing personal). If history teaches us anything, it is that (despite our tendancy to be enamoured with our more recent achievements) any given theory is sooner or later proven to some extent "wrong".
Probably including that one, (but neglect that for now :-) )
Anyway I suppose a less pedantic answer would be that if a theory is useful, then its paradigms are supportable, at least in its area of application. General Realtivity, in which time is treated as a 4th dimension is pretty useful, for example.
Clearly, though, there is more to time than just the dimensional aspect - a proper understanding requires consideration of "the flow of time" - the relative rates of interactions between particles in this hypothesised temporal dimension, and of entropy and its relevance to the irreversibility of time.
| {
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Relationship between displacement field and dislocation density tensor Let $u$ denote the displacement field in a solid body $\Omega \subset \mathbb{R}^3$ in the realm of continuum mechanics. Suppose we know that the restriction of $u$ to the boundary $\partial \Omega$ is discontinuous i.e. $u|_{\partial \Omega}$ is not continuous. Does it imply that the dislocation density tensor within the body is not zero? In other words, does the information about the displacement of boundary points (only) tell us anything about the dislocation density tensor?
| Consider a dislocation loop inside the body. There is no discontinuity of the displacement field on the boundary.
Consider now that a dislocation went through your body. The displacement has a jump even there is no dislocation in the body.
You may want to restrict $u$ to $u$ modulo $b$, where $b$ is the size of the Burger's vector, in which case, if a dislocation went through your body, and there are no other dislocations, $u$ is constant. But now you say $u$ is not constant and has a discontinuity.
If you integrate ${\mathbf{u}}$ over a loop you get the (signed sum of the dislocation's) Burger's vector.
$${\mathbf{b}} = - \oint {d{\mathbf{u}}} = \int_A {{\mathbf{\alpha }}d{\mathbf{A}}} ,$$
where ${\mathbf{\alpha }}$ is the dislocation tensor. Because dislocation lines must end and at the surface or in another dislocations fulfilling the conservation of Burger's vector, the dislocation must go inside the body, therefore, it is not $0$ inside.
| {
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Is energy $E$ in Schrödinger equation an observable/ Can $E$ be measured? Take this quantum approach to estimate mean energy of a molecule:
$$\langle\psi|H|\psi\rangle=\overline E$$
Question:
Is $E$ an observable? How we can compare it to an experimental value? i.e how to experimentally measure it and what are the states involved (as energy is all about differences there must be two states)
Edit
It is not a question about how is theoretically defined an observable.
Any help?
| That is the expectation value of the energy. $\hat{H}$ is the Hamiltonian operator which corresponds to the energy of the system. So by evaluating $\langle\psi|\hat{H}|\psi\rangle$, you get the expectation value of the energy. The expectation value is the average of measurements performed on particles that are all in the state $\psi$. So either you make sure that after you measure the particle, you return it in the original state, or else you prepare an ensemble of particles in the same state $\psi$, and measure all of them.
An observable is a special kind of operator. Its eigenvalues are always real and it describes a physical quantity. That is why all observables are described by Hermitian Operators, because Hermitian Operators are self-adjoint ($A=A^{\dagger}$).
| {
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Turbulence Model on Unsteady Navier Stokes I am asking you if the Unsteady (Time-Dependant) Navier-Stokes Equation is able to predict accurately the Flow Turbulence? I know that the RANS (with different Turbulence Models like Spalart–Allmaras, k–ε and k–ω models...) is the most used method for simulating the Turbulence.
I'd appreciate a constructive response.
Thanks
| Turbulence, being a random, chaotic, unsteady and 3D phenomena, is not straightforward to be computed. However, there are some approximations in the form of turbulence models that have made it possible to predict the flow.
The models you mentioned Spalart–Allmaras (one-equation model), k–ε and k–ω models (two-equation models) are most commonly used as you said but in the engineering and insustry application. It is the case due to their rather simple implementation, which do not require high computational power. This is also the reason why these models have been implemented in all commercial CFD solvers.
On the other hand, there are more advanced models for approximating turbulence which are mostly used in science and for research due to their complexity and hifgh computational demand. Examples are Large Eddy Simulations (LES) and Direct Numerical Simulations (DNS).
The title of your question is a bit off since for steady flow, tuebulence modelling is obsolete since turbulence is regarded for unsteady flows.
To get a more clear view you can read the first 2, 3 sections of Turbulence chapter of any fluid mechanics books. I also strongly recommend Turbulence for CFD book by Wilcox. It is a greeat route to approach this.
| {
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Interpretation of Hubble Diagram
According to my professor's notes, this is the Hubble Diagram. Unfortunately, I do not know what the y-axis is referring to. Is it the absolute luminosity?
| The quantity $m-M$ is the difference between the apparent magnitude and absolute magnitude, and is referred to as the distance modulus, $\mu$. The relationship between $\mu$ and distance, $d$, is logarithmic, i.e.
$$\mu=5\log d+5+\text{corrections}$$
where the correcting terms account for observational effects. In cosmological cases, we substitute $D_L$, the so-called luminosity distance, for $d$, where $D_L=d(1+z)$, for redshift $z$. Therefore,
$$
\begin{align}
\mu&=5\log D_L+\text{constants}\\
&=5\log(d(1+z))+\text{constants}\\
&=5\log d+5\log(1+z)+\text{constants}
\end{align}$$
We should therefore see a linear relationship between $\mu$ and $\log(1+z)$ - which we do.
| {
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Electric field charged disc and L'Hôpital's rule I have been looking at the electric field of a charged disk and have a question about the use of l'Hopital's rule for the limiting case of electric field at points along the axis $z\gg$ disc radius $R$.
$$E = \frac {q}{2\pi\epsilon R^2} \left(1 - \frac {z}{\sqrt{z^2+R^2}}\right)$$
I have applied l'Hôpital's rule in the limit of $R$ approaching zero, and see that the electric field approaches that of a point charge, as intuition suggests. HOWEVER, when I use l'Hôpital's rule in the limit that $z$ approaches infinity, I get a repeating loop of indeterminate forms that doesn't arrive at the point charge expression.
My question is does this difference in results using l'Hopital's rule have any physical or mathematical significance?
| To use L'Hôpital you either have to solve a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ kind of limit. I'll rewrite your expression to better show if this is the case:
$$E = \frac{q}{2 \pi \epsilon}\frac{\sqrt{z^2+R^2}-z}{R^2 \sqrt{z^2+R^2}}$$
(I just calculated the common denominator and separated the constants from the variables.)
As $R$ approaches 0, we can see that both numerator and denominator go toward 0, so we can use L'Hôpital.
Conversely, as z approaches infinity, we find $\frac{\infty - \infty}{\infty}$, so we cannot use L'Hôpital in this case. We first have to solve the $\infty - \infty$ indeterminate form. The easiest way (or the standard trick, if you prefer) is to multiply numerator and denominator by $\sqrt{z^2+R^2}+z$, so that the formula becomes:
$$E = \frac{q}{2 \pi \epsilon}\frac{R^2}{R^2 \sqrt{z^2+R^2}(\sqrt{z^2+R^2}+z)}$$
At this point $z$ disappears from the numerator and thus the indeterminate form is no more indeterminate and we can easily say that the limit goes to 0, as does the field of a point charge.
| {
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What is the total energy of the Universe? The Law of Conservation of Energy states that:
Energy can't be created nor can be destroyed. It only changes from one form to another.
According to this the total energy in a closed system never changes. I was wondering what this constant energy is when the closed system is the whole Universe.
Is there any estimate of what the total energy of the Universe is? If there is, kindly give a reference for it.
| In fact, for our Universe, the total mass-energy and angular momentum are undefined and undefinable. In addition, note that the total mass-energy of a system in general relativity cannot be generally defined. There are, however, a few tools one can employ to measure the total mass-energy of a system in the case of asymptotically flat spacetimes. (Which our universe being FLRW-type is not!) The first is the ADM mass, defined by:
\begin{equation}
M_{ADM} = \frac{1}{16 \pi} \int_{\partial \Sigma_{\infty}} \sqrt{\gamma} \gamma^{jn} \gamma^{im} \left(\gamma_{mn,j} - \gamma_{jn,m}\right) dS_{i},
\end{equation}
which requires the space-time to be asymptotically flat. Another tool is the Komar mass:
\begin{equation}
M_{K} = \frac{1}{4 \pi} \int_{\Sigma} d^3x \sqrt{\gamma} n_{a} J^{a}_{(t)},
\end{equation}
which also requires the space-time to have an asymptotically flat region.
The problem of course is that our Universe, or any spatially homogeneous and non-static one, that is, one that does not contain a global time-like Killing vector, is necessarily not asymptotically flat. So, in general, such definitions of mass and energy are ill-defined for our Universe.
| {
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Distribution of loss in a transmission line to minimize power dissipation This post will ask how to distribute loss in a transmission line so that the line has a known total loss, while dissipating the least amount of power.
We'll refer to "gain" of a transmission line, but we're thinking of the case where the line is lossy, so the gain is always less than one.
This post is in a sense a warmup for a somewhat more relevant and complex question that I will post after this one is resolved.
Discrete case
Consider a short section of transmission with a gain $G$, meaning that if a signal goes into that section with squared amplitude $A^2$, then it comes out with squared amplitude $G \, A^2$.
If this gain is really coming from losses in the line, then $G<1$.
If we cascade many section of transmission line with gains $\{G_1, G_2\ldots \}$, then the total gain is
$$\prod_{i=1}^n G_i = \exp \left(\sum_{i=1}^n \ln G_i \right) \, .$$
Each section of line dissipates power $P_i$ where
$$P_i = P_\text{in} - P_\text{out} = A^2 - G A^2 = A^2 ( 1 - G ) \, .$$
Continuous case
Now suppose we have a continuous transmission line of length $L$ where the gain per length at each point $x$ along the line is $g(x)$.
Extending the formula for the discrete case given above, it's clear that the total gain of the line is (remember that $g(x)<1$)$^{[a]}$
$$G = \exp \left( \int_0^L dx \, \ln g(x) \right) \, . \tag{$\star$}$$
The power dissipated in a bit of line of length $\varepsilon$ at position $x$ is
\begin{align}
P(x)
=& A(x)^2 \left[ 1 - \exp \left( \int_x^{x+\varepsilon} dx \, \ln g(x) \right) \right] \\
\approx & A(x)^2 \left[ 1 - \left( 1 + \varepsilon \ln g(x) \right) \right] \\
=& -A(x)^2 \varepsilon \ln g(x) \, .
\end{align}
The total power dissipation is of course
$$P \equiv \int_0^L dx \, P(x) = - \int_0^L dx A(x)^2 \ln g(x) \, . $$
The problem
Given a fixed value of $G$, calculate $g(x)$ that minimizes $P$.
This is a constrained optimization problem and I think some kind of variational calculus is needed.
Before we get to that, however, we should write the thing we're minimizing, $P$, in a better way by replacing $A(x)$ with an expression involving $g(x)$.
In particular, for an input amplitude $A_\text{in}$, the amplitude at a particular point $x$ along the line is
$$A(x)^2 = A_\text{in}^2 \exp \left( \int_0^x dx \, \ln g(x) \right) \, .$$
Therefore,
\begin{align}
P
=& - \int_0^L dx \, A(x)^2 \ln g(x) \\
=& -A_\text{in}^2 \int_0^L dx \, \ln g(x) \exp \left( \int_0^x dx' \ln g(x') \right)
\end{align}
How do we minimize $P$ subject to the constraint $(\star)$?
It's pretty obvious that if the total gain is fixed, then the power dissipation is also fixed because they're the same thing.
In other words, the form of $g(x)$ should not matter.
Therefore, I suppose a rewording of this question could be "how do we prove using variational calculus that the form of $g(x)$ doesn't matter?".
$[a]$: It's weird to have $\ln g(x)$ because $g$ has dimensions of length$^{-1}$. I suppose we can imagine multiplying $g$ by some length unit and dividing $dx$ by that same unit.
| Define
$$\Phi(x) \equiv \int_0^x dx\, \ln g(x) \, .$$
Note that $\Phi(0)=0$ and $\Phi(L)=\ln G$.
Then
\begin{align}
-\frac{P}{A_\text{in}^2}
&= \int_0^L dx \, \Phi(x)' \exp ( \Phi(x) ) \\
&= \int_0^L dx \, \frac{d \exp \Phi }{dx} \\
&= \exp (\Phi(L)) - \exp (\Phi(0)) \\
&= G - 1 \, .
\end{align}
Therefore, $P$ doesn't depend on $g(x)$, so the distribution of gain doesn't matter.
This answer was motived by a comment from Qmechanic.
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Does the temperature coefficient of a material depend on temperature? In my textbook, a relationship is plotted between the resistance of the material and the temperature, and the temperature coefficient is defined as the slope of that graph divided by an arbitrary resistance $R_1$ on the graph. Does that mean that the coefficient varies depending on the temperature I choose to divide the slope by?
If so, how does it stand as a valid reference to the material’s resistance growth rate with temperature?
If not, then where did I go wrong with this train of thought?
| Yes,temperature coefficient of resistance depends on temperature to some extent. At small temperature changes resistance versus temperature graph is linear which implies temperature coefficient is independent of temperature. At higher temperature the graph is nonlinear which implies temperature coefficient is now not independent of temperature.
| {
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Use of negative frequency for the sake of simplifying mathematics? How can we use the idea of negative frequency for the sake of simplifying mathematics if negative frequency does not exist (to my knowledge) in nature ? For example, when plotting the spectra of a Fourier series.
| As other answers, if it makes the mathematics smoother or more tractable than equivalent mathematics without the use of negative frequency, why wouldn't you embrace the technique?
However, this is a physics site, so a more satisfying answer is going to give a meaning for negative frequency. Already you have the beginnings of a good answer in user safesphere's comment:
Oscillation is equivalent to rotation where frequency is equivalent to rpm (revolutions per minute). A negative frequency is simply equivalent to rotation in tbe opposite direction.
and his/ her linked DSP SE answer here that fleshes this comment out.
An application of this idea in physics is the diagonalization of the Maxwell curl equations through the use of Riemann-Silberstein vector $\vec{F} = \vec{E} + i\,c\,\vec{B}$, which I discuss in more detail in my answer here. Both Maxwell curl equations are replaced by one:
$$i\, \partial_t \vec{F} = c\,\nabla \times \vec{F}$$
and recover electric and magnetic fields through the real and imaginary parts. The positive frequency parts of this solution represent the left-handed circularly polarized field, the negative frequency parts are the right-handed circularly polarized field.
In the equivalent notation of the exterior calculus, one can build self and anti-self dual parts of the electromagnetic field $\tilde{F} = F + i\,\star F$, but you may not have come across this yet. Its positive and negative frequency parts have the same interpretation in terms of oppositely handed circularly polarized parts of the field.
However, note that, as discussed in the other answer, the modern use of the Riemann-Silberstein / Self+Anti-Self Dual notation is to use two separate Riemann Silberstein vectors $\vec{F}_\pm = \vec{E} \pm i\,c\,\vec{B}$ fulfilling the separate equations $$i\, \partial_t \vec{F} = \pm c\,\nabla \times \vec{F}_\pm$$ and then to keep only the positive frequency parts. In this usage, the left and right hand circularly polarized field components are separated and given by $\vec{F}_\pm$, respectively.
| {
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Physical explanation of Joule heating The heat $Q$ generated in a wire, for a current $I$ flowing through a wire of a given resistance $R$, for a time $t$ is given by $Q=\mathscr{k}I^2Rt$ where $\mathscr{k}$ is the proportionality constant. For a given wire the resistance R is fixed. Is it possible to explain physically why $Q$ is proportional to the square of $I$?
| The pd, V, across the wire tells you how much energy is transferred to thermal in the wire per coulomb flowing. But the current, I tells you the rate of flow of coulombs. So VI tells you the energy transfer per second. Now here's the key thing: in a metal wire at constant temperature, I is proportional to V, that is $V=IR$ in which R is a constant. So we have:$$\text{energy transfer per second}=I\ \times IR =I^2R.$$ But you might object: the wire gets hot! In that case R isn't constant (it increases)! However if the wire is made of an alloy (as most heating 'elements' are), then R doesn't change much with temperature.
| {
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In physics sometimes we find energy that is negative. What does the negative sign indicate? Sometimes we see energy that is negative, for example, the energy of an electron in orbit. We know energy is something that can do something. In this view does negative energy mean something opposite someway?
| Take the simplest atom, the H atom. A hydrogen atom consists in a proton and an electron. When a proton and an electron bind to form a hydrogen atom, there is a release of energy. So in order to break the atom into its constituent parts, one has to provide an amount of energy. The amount of energy required to break the atom into its constituent parts is equal to the amount of energy released when the atom has formed. This is considered negative because you have to give energy to the system in order to break the bound. When it reaches zero, the particles become free.
| {
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Does providing more heat to a pan of boiling water actually make it hotter? Sometimes my wife has a pan of water 'boiling furiously'. Is the extra heat (wasted in my opinion) actually making any difference, apart from reducing the amount of water in the pan - which could be done by pouring some away?
|
Is the extra heat (wasted in my opinion) actually making any difference,
It is if you put a lid on the pot to retain the heat.
apart from reducing the amount of water in the pan
"all" the vigorous boiling"
*
*agitates the water (which can be important for cooking)
*indicate a hot burner, which will quickly bring the water back up to temperature if you add something cooler.
Bottom line: your Significant Other might be wasting energy, or she might be doing the right thing for that particular recipe.
| {
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Directions of static & kinetic friction? Static friction :
Static friction opposes relative motion between two surfaces. The car is moving at some tangential velocity relative to the road. Thus, static friction should act in the opposite direction. Yet it acts down the incline rather than parallel to the motion of the car. If it acts down the incline, it's essentially helping the car maintain the same relative velocity w.r.t the incline rather than opposing this relative motion.
Kinetic friction :
Consider a block placed on a long, rough plank which is moving to the right at some constant speed, and the block is moving to the right, but slower compared to the plank, so relative to the plank, the block is sliding towards the left. Will kinetic friction act to the left to reduce the kinetic energy of the block, or will it act towards the right to oppose the relative motion of the block and plank? If it acts to the right, doesn't that increase the kinetic energy of the block, which is something kinetic friction should not do?
| Why shouldn't kinetic friction increase the kinetic energy of the block?
It shouldn't increase the relative kinetic energy between the block and the plank; but because the friction isn't between the block and the static surroundings, there is no reason that it can't increase the kinetic energy of the block relative to the surroundings.
With the static friction on a care tire, it makes sense that the force points forward. The bottom of a tire is trying to slip backwards relative to the road. Static friction opposes this, and faces forward (also allowing the car to actually move). The way I think of it is that the car wants to pull the road backwards from underneath it.
| {
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Why steel can be blunted by ice? As an ice skater I cannot understand why I need to sharpen my blade very often since the hardness of steel is higher than that of ice for sure. Since steel is much harder than ice, how can ice change the shape of steel?
| Friction affects both surfaces. The material which is 'harder' is not immune to wear while the 'softer' material wears down. Both materials are worn down, to a different extent. Some of the ice breaks off the floor, some of the steel breaks off the skate.
The relative amount of wear depends (inversely) on the relative hardness of the surfaces. Two 'hard' surfaces made of the same steel wear down equally. The fact that they are equally 'hard' does not mean that they are both immune.
The amount of wear is greatest where the pressure between the surfaces is highest. This is at the sharpest edges or points. Skates need a sharp edge to control sideways motion. But a sharp edge wears down quicker than a blunt edge.
See also Why is it easier to glide on sharp ice skates than on dull skates? and Ice skating, how does it really work?
| {
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Validation of Work-Energy Theorem Is the work-energy theorem valid when there's an impulsive force during motion of considered body?
For eg: Consider a man jumping from some height into a swimming pool of certain depth, if we apply work energy theorem from his initial position to final position (Change in Kinetic Energy would be 0), Should we consider the impulsive force that water would provide when man hits the surface of water?
Even if for an instant this impulsive force should act in opposite direction of man's motion there is some displacement and once he enters water buoyancy force would take over, Am I Getting It Correct?
| Work requires both force and displacement. You could say that the displacement is a result of the force, which depends on the duration that the force is applied.
So work done might be small, even though a huge force has acted. If that huge force just applied at an instant, as an impulse, then the work is not as big as we would expect. But some work is still done.
How much this work, however small or big, influences a free object, is described by the work-energy theorem. This theorem combines the work $W_{total}$ done by external forces $\vec F$ on an object (the energy added to it) with the object itself (it's mass $m$) and it's mechanical state (it's speed $v$ and position $\vec s$). And that combination is done through energy conservation:
$$\begin{align}
W_{total}&=\Delta K\\
\int \vec F \cdot d\vec s&=\frac12 mv_2^2-\frac12 mv_1^2
\end{align}$$
Work is in general an integral. Meaning, it is a sum of all force-times-displacement products at every instant, however small. So at impact with the water surface in your example, the large force that acts for a very short time (it acts over a tiny displacement) is indeed included and added to the total amount of work. While the object sinks further, the fluid resistance, buoyancy force etc. add additional work.
Which that is more significant depends on many things, but in the end they must all be added together into the total work done that causes the change in kinetic energy as the theorem tells.
| {
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How to show barometer height is independent of area I have seen this question asked before (https://forums.studentdoctor.net/threads/please-help-explain-this-concept.724066/) however, I don’t fully understand the explanation given. I am trying to understand why a mercury barometer will always rise to the same height irregardless of the surface area of the tube or container. I have seen it explained as P = F/A
so F = ρhgA and P = ρhg. But I don’t understand why the last step is allowed. It seems that on the left side of the equation you are dividing by the area of the whole barometer dish but on the right hand side you are only dividing by the area of the tube.
| Pascal's theorem states pressure is the same in all directions. This means that the pressure at the surface of the barometer dish is the same in all directions.
Bernoulli's theorem states that the sum of the inertial, potential and pressure components in a steadily flowing incompressible fluid is constant. Well, the fluid in a barometer isn't moving, so we just say that pressure + potential components are constant
So, now consider two barometers, one with a large area and one with a small area.
the pressure on the surface of both is P_atm. And using 1 & 2, we see that pressure (not force!) can be traced to the top of the barometer tube (no matter how narrow/wide the tube is!)
| {
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Correction factor for my formula obtained by Rutherford scattering results I did an experiment on Rutherford scattering , finding the number of counts at an angle $\theta$. The problem is that I had a single sensor i.e. I measured the counts at a single position , and the scattering occurs over a cone. I was thinking of integrating my obtained function, $f(\theta)$ over the area of a hoop and so obtaining :
$N(\theta) = \int_{0}^{2\pi}f(\theta)d\phi$
Where $d\phi$ is the angle subtended by part of the hoop. Therefore I get just
$N(\theta) = 2\pi f(\theta)$
Because surely
$f(\theta) = \frac{N(\theta)}{d\phi}$
| The angle $d\phi$ that your detector subtends (out of a possible $2\pi$) changes with $r$; if the circumference of the "hoop" is $2\pi r \sin\theta$, and the dimension of the detector perpendicular to the $\theta$ direction is $d$, then
$$d\phi = \frac{d}{2\pi R \sin\theta}$$
where $R$ is the distance to the detector.
After that, your argument works - so the fraction $F$ of counts you observe is
$$F = \frac{d}{2\pi R\sin\theta}$$
and the conversion from $f(\theta)$ to $N(\theta)$ is
$$N(\theta) = \frac{f(\theta)}{F}=\frac{2\pi R\sin\theta}{d}f(\theta)$$
The term $\frac{2\pi R}{d}$ is something that will be constant for the apparatus; you are left with a $\sin\theta$ scaling term. The approximation will obviously break down when the detector dimension gets large compared to the size of the cone... i.e. for very small values of $\theta$ you have to be careful.
| {
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Why is electromagnetic wave (light) a sine wave? If I remember correctly solutions to the wave equation are any periodic function. My question is related to light.
Is there a reason why light waves are sine waves?
Theoretically where do sine waves come from as representation of a photon?
| In your comment to eranreches answer you ask
If we are able to measure the E and B fields in time of light would we measure sine wave?
Yes, if you accept a indirect measurement. Generating radio waves one accelerate a lot of electrons in the antenna rod and by this a huge number of photons in phase get emitted. Although each photon is in the range from infrared to X-rays and for this frequencies it’s not easy to measure the electric or magnetic field component, in radio waves in-phase photons follow each other in the duration of e.g. kHz to MHz and this is measurable. In the near field the radio wave looks like this:
The near field of a radio wave
The sine (cosine) wave is a harmonic oscillation and the sine for the electric field component together with the cosines for the magnetic field component perfectly conserves the energy content of the photon. Unfortunately scientists find out that both field components of the photon oscillate in phase.
| {
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Co and contravariant: tensors or components? I am learning Special Relativity and have a question: given a four vector $\vec{x}$ whose contravariant components are $x^\mu$, do the covariant components $x_\mu = g_{\mu\nu}x^\nu$ make reference to a different physical/geometrical object other than $\vec{x}$?
I mean, for the physical/geometrical object $\vec{x}$ we can say
$$ \vec{x} \underset{\text{has components}}{\longrightarrow} x^\mu $$
Then, who is $\vec{?}$ in the following expression?, is it $\vec{x}$ to?
$$ \vec{?} \underset{\text{has components}}{\longrightarrow} x_\mu $$
| For orthogonal coordinate systems, the covariant and contravariant components are the same. The difference shows up when you have oblique coordinate systems. Here's a pretty good explanation of the difference with some illustrations:
http://www.farmingdale.edu/faculty/peter-nolan/pdf/relativity/Ch04Rel.pdf
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Why is it "bad taste" to have a dimensional quantity in the argument of a logarithm or exponential function? I've been told it is never seen in physics, and "bad taste" to have it in cases of being the argument of a logarithmic function or the function raised to $e$. I can't seem to understand why, although I suppose it would be weird to raise a dimensionless number to the power of something with a dimension.
| The reason your instructor called it 'bad taste' rather than just outright wrong is because people will do this all the time with the logarithm. The logarithm is unique because it lets you split out multiplicative factors into additive terms, so people will write something like
$$\log(r/r_0) = \log(r) - \log(r_0) = \log(r) + C.$$
The most common way to do this accidentally is through an integral,
$$\int \frac{\mathrm dr}{r} "=" \log r + C.$$
This is technically wrong but almost everybody writes it this way. At the end of the day, you can always combine the constants back into the logarithm so the arguments have the right dimensions.
| {
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Commutator of intrinsic derivatives in NP formalism when timelike and null congruence are both given Suppose we have a congruence of affinely parametrized null geodesics (light rays), with tangent vector $\ell^a$, and a congruence of timelike curves (observers), with tangent vector $u^a$, such that the observers measure a circular frequency of $\omega$. This, of course, means that $u^a\ell_a = \omega$, and we can construct a null vector dual to $\ell^a$ by letting $n^a = \omega^{-1}u^a - \frac{1}{2}\omega^{-2}\ell^a$. Letting $r$ be the null parameter and $t$ be the timelike parameter, we thus have
\begin{align}
D \equiv \ell^a\partial_a &= \partial_r, \\
\Delta \equiv n^a\partial_a &= \omega^{-1}\partial_t - \frac{1}{2}\omega^{-2}\partial_r,
\end{align}
where $D$ and $\Delta$ are the intrinsic derivatives of the NP formalism. Thus, when acting on a scalar field we find that
$$
[\Delta,D] = \frac{D\omega}{\omega}\Delta - \frac{D\omega}{2\omega^3}D,
$$
where $[\Delta,D]$ is the commutator. However, it is common knowledge that in terms of the spin coefficients
$$
[\Delta,D] = (\gamma + \bar{\gamma})D + (\epsilon + \bar{\epsilon})\Delta - (\bar{\tau} + \pi)\delta - (\tau + \bar{\pi})\bar{\delta},
$$
when acting on a scalar field. The crux lies in that $(\epsilon + \bar{\epsilon}) = 0$ whenever the null geodesics are affinely parametrized, seemingly implying that $D\omega = 0$, in effect claiming that $\omega$ is constant along the lightrays, regardless of observer velocity and curvature.
Obviously, I am making some fundamental error above, but where is it?
| As usual when it comes to apparent contradictions like this, the error lies in an assumption. Specifically, when I say that $r$ is the null parameter of a congruence of null curves $c_1$ and thus $D = \partial_r$, what I am actually doing is taking a chart such that $c_1^{-1}$ is one of the coordinate maps. So far so good (as long as there are no self-intersections), but when we wish to simultaneously do this with two different congruences of curves, $c_1$ and $c_2$, we run into a problem. Because this requires that we have $(c_1^{-1} \circ c_2)(t) = \text{constant}$, and conversely $(c_2^{-1} \circ c_1)(r) = \text{constant}$, which obviously requires us to fix the parametrization.
If we wish to retain the affine and proper time parametrizations, or indeed either of them, we can use either $r$ or $t$ as a coordinate, but not both.
| {
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Rationale behind the linearised Navier-Stokes equation Some applications of fluid dynamics consider the linearised Navier-stokes equation where the advection term $(\vec{u}\cdot\vec{\nabla})\vec{u}$ is dropped.
I am trying to build a convincing argument for this based on scale analysis. I tried to find something about this out there but haven't been very lucky so far, hence this question.
Here is what I have come up with so far.
If I set the scale of the velocity as $U_0$, that of length as $L_0$, of time as $T_0$ and pressure as $P_0$. For simplicity I take $P_0=\rho U_0/T_0$ with $\rho$ being the (constant) density. I can then write the dimensionless equation:
$$
\partial_t\vec{u}+\frac{U_0T_0}{L_0}(\vec{u}\cdot\vec{\nabla})\vec{u}=-\vec{\nabla}p+\frac{\nu T_0}{L_0^2}\nabla^2\vec{u}
$$
where $\nu$ stands for the kinematic viscosity. The relative scale of each term in the above is contained in their respective pre-factor.
It would then make sense to neglect the second term in the regime $U_0T_0/L_0\ll1$.
Is this argument valid ? If not, any guidance and/or reference would be greatly appreciated.
Thanks.
| This is exactly what we do to for the Stokes flow.
Instead of $P_0$, we use the dimensionless quantity, Reynolds number (Re), for linearising the Navier-Stokes in the limit $Re \rightarrow 0$.
Refer to the wiki page on Stokes flow and Reynolds number for more details.
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Why does entropy explain why a colder object won't spontaneously give off energy to a hotter one? I'm very new to my studies of entropy, but in my view entropy is the amount of disorder in a system, which I know is unimaginatively true and not really demonstrative of any true understanding I might have on the subject. It has to do with the number of ways a system has of arranging itself, which is why a cup of cracked, scattered ice has less entropy than a neat glass of water as the molecules in the water have more ways of arranging themselves in the glass while still being water while the ice, with its solid structure has pretty fixed positions and arrangements of its molecules.
However, I still can't tie in my understanding to explain why a cold object won't spontaneously give off energy to a hotter one. I know the answer is tl;dr "because entropy" but my understanding of it doesn't tie this in. I know the hotter object has more disorder, as the system have a higher multiplicity i.e. more range of properties to exhibit (like more arrangements of varying speeds as the cap for speed for each particle has raised) but I don't see how that means it can't spontaneously gain energy from the object with less.
| I guess the focus needs to be put on the cold object. If we place a cold object in a hot environment, intuitively, we would say the cold object will not give heat out to its hot environment. Using the concept of entropy and $ds = \frac{\delta Q}T$, we find, if $\delta Q$ is negative, entropy will decrease, which violate the law. In terms of disorder, by giving heat out, the cold object will get even colder and thus more ordered, which is not right as well.
| {
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Tensorpart of NN potential The potential of the nucleon nucleon interaction includes a tensor part which is given by:
$S_{12}(\hat{\pmb{r}}) = \hat{\pmb{r}} \cdot \pmb{\sigma_1} \hat{\pmb{r}} \cdot \pmb{\sigma_2} - \frac{1}{3} \pmb{\sigma_1}\cdot\pmb{\sigma_2}$
Where $\hat{\pmb{r}}$ ist a unit coordinate space operator and $\pmb{\sigma_k}$ are vectors of the Pauli matrices acting on particle 1 and 2.
Why is the second term substracted?
| Consider the general symmetric rank 2 tensor:
$S_{ij} = \frac{1}{2}(T_{ij}+T_{ji})$
It has 6 independent terms. However, one of them is a scalar times the unit tensor:
$\frac{1}{3}{\rm Tr}({T})\delta_{ij}$.
so it doesn't really count as a rank 2 tensor part. The so-called natural form for a rank 2 tensor is:
$ T^{(2)}_{ij} = S_{ij} - \frac{1}{3}{\rm Tr}({T})\delta_{ij}$
which has 5 terms.
The antisymmetric (vector) part has 3 terms:
$ T^{(1)} = A_{ij} = \frac{1}{2}(T_{ij}-T_{ji}) $,
and the scalar part--the trace:
$ T^{(0)} = \frac{1}{3}T_{kk}\delta_{ij} $
has 1, for a total of 9, as required. Note how they match-up with the spherical harmonics: $Y^{m}_{l=2}$, $Y^{m}_{l=1}$, and $Y^0_0$.
@SAKhan is saying that the 1st term in your equation counts the scalar part, so that it must be subtracted off to make pure rank-2 tensor interaction.
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Can ferromagnetism be described by classical physics? It may sound as a trivial question, but I am very confused about the origin of ferromagnetism. According to Bohr–van Leeuwen theorem, ferromagnetism cannot be predicted by classical physics. Therefore quantum physics is required. Actually, people say that the quantum mechanical phenomenon known as exchange interaction is responsible for ferromagnetism (note that, according to Wikipedia, exchange interaction has no classical analogue).
However, as far as I understand, the Ising model of ferromagnetism is usually solved in the classical limit, and Onsager's solution to the 2D case predicts the formation of ferromagnetism through a phase transition. Therefore it seems that a classical model can predict spontaneous magnetization, which contradicts the Bohr–van Leeuwen theorem.
Interestingly, according to this post, there shouldn't be any difference between the classical and quantum Onsager solution (since Onsager considered the case $h = 0$, i.e a model without external magnetic field).
Could you explain how all these results can be reconciled? Can ferromagnetism occur in classical-physics models?
| I have found the following interesting answer to the question, from "Introduction to the Theory of Ferromagnetism, Amikam Aharoni".
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The electric field derivation from Liénard-Wiechert versus Griffiths are subtly incompatible I have a follow-up question from my answer to a previous question.
The electric field of a point charge moving at a constant velocity, as derived from the Liénard-Wiechert equation, reads
$$ E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|^2},$$
where $ n =\overline{r-r_s}$ and $ \beta = v/c $.
There is nothing controversial about this, it is repeated in various textbooks and the derivation gets taught. For example, these notes or these ones.
In contrast, here is the electric field for the same point charge moving at constant velocity derived from the relativistic field tensors,
$$ E = -\frac{e}{4\pi\epsilon_0} \frac{\gamma}{(1+u_r^2\gamma^2/c^2)^{3/2}}\mathbf r,$$
from this link.
Here is the E field for constant velocity derived by Griffiths,
$$ E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|},$$
as reviewed in Wikipedia.
Griffith's version is symmetric about $y$, Liénard-Wiechert is not.
If $n \cdot \beta = |\beta| \sin\theta$, then
*
*$ \sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-|\beta|\sin\theta)(1+|\beta|\sin\theta)} $ in one case, and
*$ (1-n \cdot \beta) = (1-|\beta|\sin\theta) = \sqrt{(1-|\beta|\sin\theta)^2} $ in the other case.
Presumably someone has made a sign error. One or the other of these formulas is wrong.
Have I misunderstood this? Are they both right?
| I bought Griffifths's book and looked at his derivation.
I thought he had something wrong because I didn't read it carefully enough. His angle theta is not the angle I thought it was. We have the direction from the source at retarded time to the target, $n$. And we have the velocity vector, $v$. I thought theta was the angle between $n$ and $v$.
But instead his theta is the angle between $n-v$ and $v$.
Also, he does not consider the distance between the source at retarded time and the target, $R$. He instead considers the distance between the source at present time and the target at present time, $R'$.
So when he accounts for distance, he uses not $\frac{n-v}{R^2}$ but instead $\frac{\overline{n-v}}{R'^2}$
When I make those changes, the result is exactly the same as L-W.
It's possible to make a physical interpretation for both graphs. In the first case, we have a bunch of source charges which are equidistant from a target at retarded time, and they all travel at velocity V. Since they are equidistant, their force arrives at the target at the same time, now, while the charges themselves each move. The direction each force pushes is tilted from its direction at retarded time, and that direction happens to coincide with the charge's location at present time. The magnitude of the force is also changed.
With constant velocity, we can instead consider the following physical interpretation: Consider a bunch of source charges which are equidistant from a target at present time, and they all travel at velocity V. Imagine that the speed of light is infinite. Then the forces will all arrive at the same time, now. None of the forces are tilted relative to their sources, but their magnitudes are reduced inline with velocity, and increased normal to it.
This fanciful interpretation works only for constant velocity. If the sources do not reach their proper place at present time, the illusion is destroyed because in fact the forces depend on the location and velocity at retarded time.
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Is torque in an electric motor generated from repelling magnetic dipoles and the Lorentz force on a solenoid? So in an electric motor the torque is generated from the Lorentz force on the current carrying wire by the interaction with the outer magnetic field. There is also another interaction between the magnetic dipole created by the solenoid (or wire) and the external magnetic field which would drive the motor as the B-fields repel and attract each other periodically.
Is an electric motor driven by both these forces or are they the same force (if so, how?)
| The torque which you get via the Lorentz force comes out to be $BIA$ where $A$ is the area of the coil ie length $L$ times the width of the coil.
The magnetic dipole moment of the coil is $IA$ and the torque is the cross product of the dipole moment and the magnetic field which works out to be $IA\,B$ with the orientation of the coil and magnetic field as shown in your diagram, which is exactly the same magnitude for the torque as found using the Lorentz force.
It is two different ways of looking at the situation and getting the same result not two different ways of a torque being applied to the coil.
| {
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Where does the energy for electron degeneracy pressure come from? I just watched a web video about white dwarf stars. It mentions that the star’s electrons flow among the degenerate matter. Since most of the electrons can’t be in the lowest state, due to Pauli’s exclusion principle, they get bumped to higher states and replace the radiation pressure that disappeared when fusion stopped. But where did the energy to boost the electrons to distinct states (modulo spin) come from? What lost energy to make up for it?
| The answer is that you have to do work on the gas to compress it enough for it to become degenerate. The work done is $\int P\ dV$ and this work is used to increase the internal (kinetic) energy of the electrons.
In a white dwarf star, the work is done by the gravitational force. As the core of a proto-white dwarf shrinks then about half the gravitational potential energy that is released is lost as radiation or convection to its surroundings and about half goes into raising the internal energy of the gas.
| {
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Light bulb longevity I have a summer home in North Carolina where there are more people in the summer than in the winter. Light bulbs seem to last longer in the winter when there is less demand on the system. I suspect the line voltage drops with the higher use. Would this affect the life of the bulb?
| The bulb life highly depends on the voltage. The lifetime is limited due to tungsten evaporating from the filament. This process is more intense at a higher voltage, especially because the heat power is proportional to the square of the voltage.
Having said this, the voltage usually drops with higher use, which is in summer. So your observation is reverse to what would be normally expected. Is it possible that the bulbs last longer in winter, because you are not there in winter as often and don't use them as much as in summer?
| {
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How to find the direction of velocity of a reference frame where two events are simultaneous in case of a space-like interval Suppose in a inertial reference frame $S$, an event $A$ occurs at $(ct_A, x_A, y_A, z_A)$ and event $B$ occurs at $(ct_B, x_B, y_B, z_B)$.
Now the invariant interval of these two events is,
$$I = -c^2 (t_A - t_B)^2 + (x_A - x_B)^2 + (y_A - y_B)^2 + (z_A - z_B)^2 = -c^2 \Delta t^2 + \Delta \bar x^2,$$
where I'm using the $(-, +, +, +)$ metric.
Now there can be $3$ particular cases of interest corresponding to time-like, space-like and light-like events.
For $I = 0 \implies c^2 \Delta t^2 = \Delta \bar x^2$, events are light-like.
For $I < 0 \implies c^2 \Delta t^2 > \Delta \bar x ^2$, events are time-like and a reference-frame $\bar S$ exists(accessible by appropriate Lorentz Transformation) for which these two events occur at the same location. The velocity(magnitude and direction) can be computed.
For $I > 0 \implies c^2 \Delta t^2 < \Delta \bar x^2$, events are space-like and a a reference frame $\bar S$ exists(again accessible by appropriate Lorentz Transformation) for which these two events are simultaneous.
I know how to calculate the velocity(direction and magnitude) of the $\bar S$ frame relative to the $S$ frame in case of a time-like event. I also know how to calculate the magnitude of velocity of the $\bar S$ frame relative to the $S$ frame for a space-like event.
How to find the direction of the $\bar S$ frame relative to $S$ for a space-like event?
| This problem would have given me fits when I first learned relativity, but since I've started using the geometric point of view as the tool I reach for first it is almost trivial.
The direction is easy: boost in the direction from the earlier to the later event as measured in your current frame. Why? Because you want the space-like axis to tilt upward in that direction.
Getting the speed is also surprisingly easy. You need the new space-like axis to have a slope (in your current coordinate system) equal to $(\Delta x)/(c \,\Delta t)$, which is exactly the $\beta$ of the boost you need.
| {
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Geocentric frame of reference and superluminal speeds I have a couple of questions on frames of reference.
From my understanding, we can do math in an accelerating frame of reference as long as "fictitious" force terms are correctly added. From this point of view, is there anything wrong with viewing the Earth as stationary, and the rest of the universe rotating around it, at least kinematically? And, if so, wouldn't several cosmological objects move faster than light in this frame of reference? How can this be?
I'm sorry if this is a dumb question, I'm not asking this from a religious perspective I just want to understand how frames of reference work.
| You're right - but so too are the photons. They're moving "faster than light", too, in this frame!
You see, talking of "light speed limits" is really just pseudo-Newtonian mechanical heuristics on top of the actual gist of relativity theory, which is that information can transmit between some places in space-time and not others. These relations, or predicates on events, are what must not be violated.
However the predicates that tell you "yes or no" to "Can you send a message from A to B?" take different forms depending on the coordinate system. For "flat" spacetime, in a certain special class of coordinate systems, they show up as a "speed limit", but are more subtle in others. This is why the theory here is called "special" relativity (SR).
All the "weird" effects - time dilation, length contraction, and so forth can be thought of as what a Universe with minimum imposed communication latencies looks like from the viewpoint of someone inside it subject to those constraints.
| {
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Information contained in a quantum simulation
The system in the picture has a box filled with $N$ classical particles which can either be on the left or the right side. Let this system have a Hamiltonian $$H= \frac{\mathbf p^2}{\sum 2m_i} + V(x, t) $$ where V becomes infinite at $\pm \frac{L}{2} $. Let it be represented by a point in $2N$-dimensional phase space $$ (x_1, p_1, ... , x_N, p_N) $$ If we say that a positive x coordinate corresponds to 1, and negative to 0, then this system will encode a string of bits $N$ digits long.
Suppose I want a quantum computer to track the information in the string of bits above. The computer will contain $M$ qubits where $M = \log N$. Assuming this is a perfect quantum computer with no errors, the general state of the computer will be a density matrix in $2^M$-dimensional Hilbert space. $$ \rho = \begin{bmatrix} c_{1,1} & c_{1,2} & \dots & c_{1,N} \\ c_{2,1} & c_{2,2} \\ \vdots & & \ddots \\ c_{N,1} & & & c_{N,N} \end{bmatrix}$$
Now we will quantify the amount of information in both systems.
In order to quantify how much information the box of particles contains, we can use the Shannon Entropy. For a random variable $P_i$ representing the probability of finding a one of a zero in the string at position $i$, the information entropy is$$ S = - \sum_i P_i \log P_i $$ If we assume the particles are sufficiently mixed up in the box, then every $P_i$ will have a probability $\frac{1}{2}$, and then total entropy will be $\frac{N}{2} \log 2$
Next we turn the quantum computer, where the measure of the amount of information is Von Neumann Entropy. Given a density matrix $\rho$ for the quantum computer, the amount of information is given by $$ S = - Tr(\rho \log \rho) $$
Question 1: Does there exist a density matrix that contains the same amount of information as the classical system?
Question 2: Does there exist a Hamiltonian to control the evolution of the quantum computer so that the information contained in both systems are always equal?
| In general, the entropy of a system is written
$$
S_\text{tot}=S_{EE}+S_\text{thermal}
$$
where EE denotes the entanglement entropy. In a thermal classical system, you have no ignorance associated with the possibility of states being entangled. However, a quantum computer simulates the classical system using some quantum circuit that has non-zero entanglement entropy so clearly the information content of the two systems are different.
Ignoring the quantum computer itself, the information pertaining to the classical simulation the quantum computer performs contains the same amount of information as the classical system being simulated.
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When must an alien civilization send off a signal that can reach us today? Given the rate of expansion of the universe and the speed with which galaxies separate: Can a rough estimate be given, at which time T (in its proper time) a civilization that lives on a planet which is X light years away from Earth today would have had to send off a light signal or spaceship that can reach Earth in our days?
For which distances X would T be much smaller than the 13.80 - 4.28 = 9.52 billion years that it took life to appear on earth (only 120 million years after the first appearance of liquid water). Especially so small, that not even liquid water could have appeared on that planet? This might restrict the distance from which we could receive signals significantly below the event horizon, resp. radius of the observable universe.
But maybe I'm completely off the track.
| What you are looking for is the size of the past lightcone of our present moment, trying to find the time $T$ when it is $X$ lightyears in radius. Measured in co-moving coordinates the distance is $$\chi(T)=c\int_T^{t_{now}} \frac{dt}{a(t)}$$ where $a(t)$ is the scale factor of the universe. Note that by now the expansion has turned the distance into a (proper) distance $X=\chi(T)/a(T)$ (where we have used that $a(t_{now})\equiv 1$).
Annoyingly, the actual scale factor does not have a closed form expression. During the matter dominated era it grew as $t^{2/3}$ which might be enough for a rough approximation in this case. But you can for example use Edward Wright's cosmological calculator to find the co-moving distance for a given light travel time. That calculator gives $\chi(9.52\cdot 10^9)=14.758$ Gly. Unfortunately it does not give the scale factor, but using for example this calculator and a bit of search I get $a(T)=0.723$, so the current distance to where the civilisation that sent the signal/spacecraft was is 20.41 Gly.
(Note that this distance is somewhat bigger than our current event horizon, $\approx 5$ Gpc $\approx 16.3$ Gly, so we can never reach that spot ourselves in the future.)
Generally, the further back in time a civilisation could have been possible, the tougher the Fermi paradox becomes since the volume of sites where civilisations could have originated grows a lot. It does not get infinite, though. It approaches a finite co-moving volume with radius $\chi\approx 45$ Gly as $T$ approaches 0.
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Eigenvalues of the exchange operator determined by the particle type (boson or fermion) in a two particle system While dealing with a two particle system in QM (the particles are identical),
the net wave function of the system would be simply the product of the wavefunctions of the individual particles in the given potential(assuming no interactions between the particles) if they were distinguishable. When the particles aren't distinguishable (as in this case), the eigenfunctions of the Hamiltonian must also be eigenfunctions of the exchange operator P (Griffiths section 5.1) The eigenvalues of P are 1 and -1 and based on whether the particles in question are bosons or fermions, we choose the corresponding eigenvalue of P(1 for bosons and -1 for fermions) and thus construct the correct 2 particle wavefunction.
I don't understand why it is the type of the particle(boson or fermion) that determines the eigenvalue. In other words, why must all fermions necessarily have an eigenvalue of -1 in all possible scenarios ( and bosons 1)? We group particles into bosons and fermions based on the eigenvalue they take, but why must a given kind of particle be forced to take a fixed eigenvalue in the first place?
| The spin-statistics theorem relates the bosonic/fermionic nature of a particle to the type of exchange symmetry of indistinguishable many-particle wavefunctions. Possibly it is easiest to consider the consequences of this exchange symmetry in 2-particle systems.
For fermions, states must be antisymmetric so that
$$
\vert \Psi_-\rangle = \frac{1}{\sqrt{2}}\left(\vert \psi\rangle_1\vert\phi\rangle_2 - \vert \psi\rangle_2\vert\phi\rangle_1\right)\, .
$$
In particular, if $\vert\phi\rangle=\vert\psi\rangle$, $\vert\Psi\rangle=0$, thus enforcing the Pauli condition that two fermions cannot occupy the same state.
Contrariwise, for bosons
$$
\vert\Psi_+\rangle=\frac{1}{\sqrt{2}}\left(\vert \psi\rangle_1\vert\phi\rangle_2 +\vert \psi\rangle_2\vert\phi\rangle_1\right)\,
$$
permutation symmetry of identical bosonic states leads to a bunching effect demonstrated experimentally in Hanbury-Brown-Twiss-type of experiments.
Thus, permutation symmetry and the related eigenvalue of the exchange operator is compatible with the observed experimental behaviour of fermions and bosons under exchange.
| {
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What is the Bandgap energy of Rubidium? Could anyone please tell me the bandgap energy for alkali metals like rubidium and cesium?
| An early paper is "Energy Bands of the BCC Metals rubidium and cesium", A. M. Radwan, Cryst. Res. Technol. 23(6) 785-791 (1988). The results shown are focused on the bands near the Fermi Energy. It shows the parts of the next higher band(s) that are overlapping with the conduction bands.
There is also "Method for Performing LCAO Band Structure Calculations in Crystalline Solids: Application to Rubidium", Werner Obermayr, Int. J. Quantum Che. 78 212-225 (2000). It also shows the bands close to the Fermi Energy.
Google pulls up a 2001 masters thesis by A.J. Zukaitis that shows the deeper valence bands.
| {
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What is the equation of relative motion for two objects moving in straight lines? If two objects, A and B, are moving in the same direction along straight lines in a plane, they might be diverging, converging or moving in parallel.
If we wish to describe B's motion with respect to A, what is the equation of motion?
For example, imagine that A is moving at 10 knots along the line described by the parametric equation:
x = 30t
y = 20t
and B is moving at 9 knots along the line described by the parametric equation:
x = 35t
y = 10 - 15t
what is the motion of B with respect to A? In other words, if we hold A to always be at the origin, what would be the parametric equation (and/or non-parametric equation) for B's motion?
I guess the shape will be a parabola or hyperbola, but am not sure how to compute it.
| As pointed out by other writers, the relative distance of B to A is given by $\vec{r'}=\vec{r_B} - \vec{r_A}$. So it is necessary to calculate the positions of both objects for a given time. The unit vector for the direction of the first object is $\vec{e_1} = \frac{3}{\sqrt{13}}\hat{i} + \frac{2}{\sqrt{13}}\hat{j}$ .
We know the direction, to describe the movement completely, we need an initial position as well, which we can pick as any point on the trajectory. Let's pick $(0,0)$ arbitrarily. So the position by time becomes $$\vec{r_A} = 10t\vec{e_1}=10t(\frac{3}{\sqrt{13}}\hat{i} + \frac{2}{\sqrt{13}}\hat{j})$$
The units are in knots. The time is typically given in second, however this t is different from your trajectory parameter t.
We can do the same for the second trajectory. The unit vector along the path is $\vec{e_2} = \frac{7}{\sqrt{58}}\hat{i} - \frac{3}{\sqrt{58}}\hat{j}$. In other words, this is the direction you need to go if you start on any point on the trajectory you described and you want to stay on the trajectory. Now we can pick a starting point, which we pick as $(0,10)$. So the position by time will be
$$\vec{r_B} = 10\hat{j} + 9t\vec{e_2}=10\hat{j} + 9t(\frac{7}{\sqrt{58}}\hat{i} - \frac{3}{\sqrt{58}}\hat{j})$$
Now all you need to do is subtract one from another.
$$\vec{r'}=\vec{r_B} - \vec{r_A}\\
=10\hat{j} + 9t(\frac{7}{\sqrt{58}}\hat{i} - \frac{3}{\sqrt{58}}\hat{j})-10t(\frac{3}{\sqrt{13}}\hat{i} + \frac{2}{\sqrt{13}}\hat{j})$$
Sorting out the vector components, we get
$$\vec{r'}=(\frac{63}{\sqrt{58}}-\frac{30}{\sqrt{13}})t\hat{i} + (10-(\frac{27}{\sqrt{58}}+\frac{20}{\sqrt{13}})t)\hat{j}$$ Bear in mind that this is the A's point of view, which happens to have the same unit vectors ($\hat{i}$ and $\hat{j}$). From the last equation I wrote you can see that B, according to A, moves in a line characterized by the vector $(\frac{63}{\sqrt{58}}-\frac{30}{\sqrt{13}})\hat{i} -(\frac{27}{\sqrt{58}}+\frac{20}{\sqrt{13}})\hat{j}$ which can be turned into a unit vector too if desired. The exact equation of the relative position may change as we picked the initial positions arbitrarily. However, the direction of motion doesn't change.
| {
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Why is work done equal to $-pdV$ only applicable for a reversible process? In thermodynamics, when we're interested at gases, I know that the work done can be written to be $-pdV$ for a reversible process ($p$ is the pressure of the system, and $V$ is the volume of the system).
This is because $$dW=Fdx=-pAdx=-pd(Ax)=-pdV$$
However, why is it not true also that the work done is $-pdV$ for non-reversible processes as well?
| Please note that work is always $-pdV$ irrespective of any method. It is valid for reversible as well as irreversible. But then how to obtain full expression for work? The expression $-pdV$ represents the infinitesimal quantity of work. So we need to get finite quantity of work. So, we need to look one step forward: whether the external pressure is constant or variable. If it is constant, we get full work as $-p(V_{\text{final}} - V_{\text{initial}})$; meant for irreversible process.
If the external pressure is variable, we obtain the expression from integration. This is meant for reversible process. In the reversible case, the external pressure is considered slightly smaller than internal pressure. Mathematically, them are considered same and used any one in the expression.
| {
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What is the universal definition of 'inertia' in fields? In my physics class we recently did the Gravitational Field unit, and the idea of gravitational field strength was introduced:
$$g=\frac{F}{m}=\frac{GMm}{r^2}\frac{1}{m}=\frac{GM}{r^2}$$
After doing that unit, now we're on to Electric Fields, and the idea of electric field strength was introduced:
$$E=\frac{F}{q}=\frac{kQq}{r^2}\frac{1}{q}=\frac{kQ}{r^2}$$
I saw the main similarity; mass was replaced with charge (and a different proportionality constant). My teacher explained that phenomenon in general for all fields as:
$$Field \,\,Strength=\frac{F}{Intertia}$$
He said in the gravitational field, the inertia is mass. In electric fields, the inertia is charge. My understanding of inertia was the classical mechanics one; the property of matter that resists change in motion. But my teacher explained that each field has it's own 'inertia'. I've been searching the web about specifically this, the formal definition of inertia across all fields, and have not come across anything. Does such a definition even exist?
| This is an odd use of the term inertia. I would usually define the field strength as the force per unit charge, where the word charge refers to the property generating the field. Note that electrical charge is one form of the general term charge - the terminology can be a bit confusing. The term charge has a precise definition, though this is probably a bit involved for many readers.
For gravity the charge is mass, so the gravitational field strength is the force per unit mass i.e. Newton's per kilogram (which is just the gravitational acceleration).
For electric fields the charge is electrical charge, so the field strength is the force per unit charge i.e. Newtons per Coulomb.
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Mass and Newton's Second Law While trying to understand the second law of Newton from "An Introduction to Mechanics" by Kleppner and Kolenkow, I came across the following lines that I don't understand:
"It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector. Although this turns out to be the case, it is not obviously true. For instance, if mass were different in different directions, acceleration would not be parallel to force and force and acceleration could not be related by a simple vector equation. Although the concept of mass having different values in different directions might sound absurd, it is not impossible. In fact, physicists have carried out very sensitive tests on this hypothesis, without finding any variation. So, we can treat mass as a scalar, i.e. a simple number, and write $\vec{F} = m\vec{a}$."
The lines above lead me to question:
*
*Why is it not" obviously true" that force behaves like a vector?
*Why is it not impossible for mass values to be different in different directions?
| Consider a solid sphere with mass $m_1$. Now lets say there is a hole through the center that runs horizontally from end to end. Inside this hole is another mass $m_2$ that slides without friction.
If you were to push the sphere horizontally you will have $F_x = m_1 a_x$. But if you push vertically you would get $F_y = (m_1 + m_2) a_y$ since both masses would need to be accelerated equally.
$$ \mathbf{F} = \mathcal{M}\, \mathbf{a} $$
$$ \pmatrix{F_x \\ F_y \\ F_z} = \begin{bmatrix} m_1 & & \\ & m_1+m_2 & \\ & & m_1 + m_2 \end{bmatrix} \pmatrix{a_x \\ a_y \\ a_z} $$
So the relationship between forces and accelerations is not always a simple vector scaling operation, but more complex.
The only physical stipulation for $\mathcal{M}$ is that it is symmetric matrix.
In robotics, whenever you have multiple connected rigid bodies you have the concept of Articulated Inertia which is exactly as shown above. This matrix in general is not a scalar multiple of the identity matrix, but has varying diagonal elements and cross terms in the off-diagonal.
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