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How to find the direction of an eddy current? Suppose there is a magnetic field going from left to right. Suppose a thin sheet of metal conductor (e.g. a 1m*1m square) is dropped through the magnetic field such that the plane of the conductor is PERPENDICULAR to the magnetic field.
Now I know that by Faraday's Law, there will be an induced emf that will induce eddy currents in the conductor which oppose the motion of it by Lenz's Law. However, I have no idea which way the eddy currents are flowing, i.e. clockwise or anticlockwise.
Does there exist a simple hand rule which can predict the direction of eddy current?
| as a thought
$$\nabla \times j = M$$
$$\nabla \times M = j$$
this current density has three manifestations shown in amperes material derviation:
$$ \nabla \times B = \mu_0(j_f + j_p + j_m) + \epsilon_0 \frac{\partial E}{\partial t}$$
Current due to change in polarisation
$$j_p = \frac{\partial P}{\partial t}$$
Current due to rotation of magnetisation
$$j_m = \nabla \times M$$
Current due to p.d
$$j_f = I$$
$$ \therefore \nabla \times \frac{B}{\mu_0} - M = j_f + \frac{\partial( \epsilon_0 E + P)}{\partial t}$$
as $$\epsilon_0E + P = D$$
and
$$\frac{B}{\mu_0} - M = H$$
then
$$ \nabla \times H = j_f + \frac{\partial D}{\partial t}$$
By going in and out of the free space and material representations of Ampere's Law, you can see that a magnetic field next to a material would produce such a current. i.e the the curl of H creates a change in the displacement current or creates a magnetic current density.
| {
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What is the curve that describes the Daytime line in a Day and Night World Map? A Day and Night World Map shows which parts of the Earth are in daylight and which are in night at a given instant.
At one side of the Daytime line they are in daylight and at the other side they are in twilight.
The shape of the day and night areas are different at the different days of the year.
For example, on UTC time = Wednesday, 23 September 2015, 08:20:00; during the September Equinox it was like:
While on UTC time = Sunday, 21 June 2015, 16:38:00; during June Solstice it was like:
And on UTC time = Friday, 20 March 2015, 22:46:00; during March Equinox it was like:
I believe that this has to do with the Earth's axial tilt with respect to the Ecliptic plane.
What I would like to know is:
How is the calculation of the shape of the areas in a Day and Night World Map done?
What is the curve that describes the Daytime line in a Day and Night World Map?
I suspect it has to do with the projection of a great circle on an Equirectangular projection of the world map... but I would like to know if this is correct or not.
| Absent a small amount of distortion from the oblateness of the Earth, yes: it is a great circle. To generate the corresponding curve on a flat map, you have to use the map's projection rules to project the coordinates of the circle's circumference onto the map. That completely depends on the projection rules for the specific map you're using.
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Why does a ray passing through optical centre remain undeviated? How can it be explained using the laws of refraction that a light through optical centre of a lens passes undeviated?
If we assume the portion of the lens in the middle to be made of even number of alternately place up and down prisms, then it's clear, but why can the number of prisms not be odd?
| The middle part of the lens will just act as a rectangular glass slab. We may verify it by cutting the lens horizontally. Now we know that rectangular glass slab refract light in such a way that emergent is parallel to incident. Actually the same happens when the ray passes through optical centre. This can be observed in a thick lens. In thin lenses the perpendicular distance between extended incident ray and extended emergent ray is negligible. So we can say that light ray passes through optical centre without deviation.
| {
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Are chemical bonds matter? So it recently blew my mind that chemical bonds have mass. And that a spring that's wound up similarly weights a little more.
But there is a distinction between mass and matter.
I believe that a chemical bond, even though it has mass, is not considered matter and is instead a form of energy.
If I'm getting any of that wrong, I'd love to hear the rational.
| Chemical bonds form when atomic orbitals of the nuclei to be bonded interact and form a molecular orbital.
The simplest case of bond formation is the formation of dihydrogen ($\mathrm{H_2}$) from 2 hydrogen atoms. The latter have (in the ground state) each one electron in a $1s$ atomic orbital and these orbitals then combine into a $\sigma$ molecular orbital, see schematic below:
What's most noteworthy is that the energy level of the $\sigma$ orbital is lower than that of the $1s$ orbital. Of course that means that energy is being released when hydrogen bonds (dihydrogen) forms.
To summarise, chemical bonds aren't really made of matter but small amounts of matter do convert to energy when they form.
And due to the mass-energy equivalence this also means that a very, very small amount of matter is converted to energy. The bond Enthalpy of $\mathrm{H_2}$ is about $-435\:\mathrm{kJ/mol}$ (a $\mathrm{mol}$ of hydrogen is about $2\:\mathrm{g}$), so you can work out just how little matter disappears though!
To summarise, chemical bonds aren't made of matter but small amounts of matter do convert to energy when chemical bonds form.
| {
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What it sounds like when I'm travelling at the speed of sound totally hypothetical here:
lets say a man is playing a song on a guitar and I begin travelling quickly away from the guitar, if I were to reach the speed of sound, what will I hear? (my assumption is that I will hear a single note humming in a constant state...like pressing a key on a synth).
assume im not in a vehicle and the sound of air wizzing past me isn't involved...not a practical situation, just hypothetical.
total noob here, my apologies.
and to take it a step further...if i can speed up or slow down (move forward or backward) ever so slightly from the current note "im in", then back to the speed of sound at another note, would this be possible?...to move from one note of the song to another?
|
my assumption is that I will hear a single note humming in a constant state.
A sound wave is not a thing that you can hear. Assume for a moment that you are just standing in the coffee shop, enjoying the music. What you are hearing is not the waves. What you are hearing is the guitar.
The waves carry acoustic energy from the guitar to your ear. The guitar causes fluctuations in the pressure of the air that immediately surrounds it. "Wave" is our word for how those fluctuations propagate through the air. Your eardrum experiences the same fluctuations as the wave passes by, and you hear the sound.
If you could somehow magically keep pace with the waves and not feel the supersonic blast of wind in your face then you would hear nothing because the wave is not passing you by. You would experience only the steady-state pressure of one peak of the wave or one trough. As far as your ears are concerned, a steady-state pressure equals silence.
| {
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What does the discovery of a pentaquark signify? at a particle collider a pentaquark was discovered. My question in short is what does the discovery of a pentaquark signify? Is there a theory that it supports or something like that?
| You're asking about the significance of the discovery of the Pentaquark. It would shed light on some mysteries surrounding the nuclear force, to begin with. It could perhaps lead to a change in what we understand about Neutron stars, too.
In terms of "some theory it supports or something like that", generally the stuff we already think still makes sense. Good!
| {
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Is it (practically) possible for a large building to be a Faraday cage? In my sophomore year of high school, my P.E. teachers kept on complaining about how phones didn't have a network connection in our gym, regardless of model, service provider, etc. A couple of feet outside the gym, cellular reception was crystal clear, however, as you moved your phone towards the gym wall, it rapidly diminished in strength. A couple of months after taking AP Physics C, my mind randomly drifted back to this event, and it came to me that this is what one would expect to happen if the gym were a giant Faraday cage. Is this even possible or likely considering usage of fairly standard building materials and structural design? Here's an image of the building (it's the one in the foreground) if that helps.
| Just to add to what Floris has said. It is frequent (in the UK) that institutional settings would have toughened glass in windows, particularly in bathrooms, gyms etc. that would have the form of a wire mesh (of order 1cm grid) embedded in the glass. That would do a particularly good job of blocking phone signals that would otherwise penetrate the glass.
| {
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What is the smallest item for which gravity has been recorded or observed? What is the smallest item for which gravity has been recorded or observed? By this, I mean the smallest object whose gravitational effect upon another object has been detected. (Many thanks to Daniel Griscom for that excellent verbiage.)
In other words, we have plenty of evidence that the planet Earth exhibits gravitational force due to its mass. We also have theories that state that all mass, regardless of size, results in gravitational force.
What is the smallest mass for which its gravity has been recorded or observed?
(By the way, I hoped this Physics SE question would contain the answer, but it wound up being about gravity at the center of planet Earth.)
| The classic gravitational measurement is the Cavendish Experiment, and the masses involved were a pair of 0.73 kg lead weights. So that forms an accessible reference. Other versions of the experiment may have used smaller weights, though.
| {
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Why can't I push myself in a chair? If I am sitting in a chair with wheels and someone pushes on the back of my chair with sufficient force it will role along the ground. However, if I push on the back of the chair with the same force it will not move the chair. Why?
| If you sit on the chair and push with same force, it is not an external force. Newton's law of motion: if external force is not acted on a body it will remain same in state. So the chair will not move. Your acted force is an internal force.
| {
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Why do tall buildings have low resonant frequencies? I know that tall buildings have low natural frequencies, hence they're more vulnerable to earthquakes, but why do they have low natural frequencies?
| Real pendulums are pendulums with non-point distribution of mass. For these, to calculate time period as a function of length, we can apply the regular formula for harmonic motion after considering the length of the oscillator to be the the distance between the pivot and the center of mass. A tall building would obviously have pretty high center of mass, hence it'll have a very low frequency. It's like considering a really long simple pendulum, though the equations are somewhat different.
More specifically, we can model the building as a standing wave with one open end. The equation for position in a standing wave is $$y=2A\cos(\omega t)\sin(\frac{2\pi x}{L})$$where $L$ is the total length of the string, $x$ is the distance of the particular point from the end of the spring, and $$\omega = 2\pi f$$ Also, $${f=\frac{1}{2L}\times k'}$$ $k'$ is dependent upon mass per unit length of the body and tension. $L$ is the length of the whole body.
Clearly, there's an inverse proportion between frequency and length, so a tall building will have a low frequency.
| {
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Does quantum tunnelling drill holes in solid state drives? The solid state drive (SSD) consists of numerous data storage elements. Each element (NAND flash memory element) reminds me of a microscopic battery. An electron is supposed to enter the element through the oxide layer and remain inside. A transistor beneath the element is used when reading the memory element. However, the oxide layer is prone to deterioration.
After I attended a series of lectures on this technology, I was struck with a question in mind. The electron moves through the oxide layer by means of quantum tunneling. Does that mean that the tunneling in principle is a process that deteriorates the barrier?
| The electron does not deposit energy in the oxide layer. The oxide layer provides for a potential barrier, the electron moves though it despite the fact that in a classical picture the electron should have a negative kinetic energy there. You can consider a process where the electron is going to do some damage to the oxide layer, but such a process is only possible if in the final state the electron can end up with a positive kinetic energy.
| {
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What is the definition of linear momentum? Every where and book I search I get that the definition of linear momentum is the amount of speed (quantity of motion) contained in it or simply it is mass $\times$velocity? So, what is an appropriate definition of linear momentum? What did Newton think when he discovered it? He certainly did not think it as the amount of speed in a body.
| Newton (if I recall correctly) typically referred to the concept of inertia, which was an objects resistance to changes in velocity when subjected to external forces. You are right about him not thinking about it as just the speed of the object, because this is where the mass term comes in. Many people think of Newton's second law as being written as $F = ma$, and while this is true, I think that Newton liked to write it in terms of momentum as
$$F = \frac{dp}{dt}$$
Obviously, not with this notation since that came later. So, this can be thought of in the following way. The change in an objects momentum is equal to the force applied and the time interval over which the force is applied. Or,
$$\Delta p = F\Delta t$$
Starting from rest, this will give you the total momentum of the object, $p$. Considering inertia, objects with a higher mass will exhibit more resistance to velocity change. If the object doesn't change mass, then $\Delta p = m\Delta v$. Starting from rest, where $v_0 = 0$ and $p_0 = 0$, you end up with $p=mv$ for the total linear momentum of an object.
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What prevents me to accelerate an object to near light speed in space? As far my limited knowledge go, things in space aren't slow down unless something interferes with them, so what prevents me to build a spaceship powered by nuclear power that will keep accelerating until we get to the limits of physics?
Like the voyager ship that is now outside our solar system, it had by know plenty time to accelerate to be much more faster than it's right now (about 17030 m/s)?
Wouldn't that greatly reduces the time you need to get to a distant star, since you can increase your speed exponentially ?
|
so what prevents me to build a spaceship powered by nuclear power that
will keep accelerating until we get to the limits of physics?
Now matter how long the spaceship is able to accelerate (as measured by an accelerometer attached to the spaceship), there is always an inertial reference frame in which the spaceship is instantaneously at rest.
Indeed, at any instant, there are an infinity of inertial reference frames in which the spaceship has a speed arbitrarily close to $c$.
So, it isn't clear to me what you mean by "until we get to the limit physics".
Certainly, if the spaceship accelerated (linearly) long enough, the CMB would become an intense gamma ray beam in front of the spacecraft so that is at least one type of limit.
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Derivation of $E=pc$ for a massless particle? In classical mechanics, massless particles don't exist because for $m=0$, $p=0$.
The relativistic relation between energy, mass and spatial momentum is: $E^2= (pc)^2 + (mc^2)^2$ . So it is said that setting $m=0$ in the first equation you get $E=pc$.
How could setting $m=0$ in that equation give you $E=pc$ whilst $p$ appears in the equation and we know $p=γmu$? If you set $m=0$ you will have indeterminacy due to "$γm$". It seems to me like we are doing a "trick" in order to get the $E=pc$. Perhaps there is another proof for this relation?
| If one considers that the deBroglie relationship holds for photons we have $$p=\frac{h}{\lambda} = \frac{hf}{c} = \frac{E}{c}$$
which immediately gives us $$E=pc.$$
This is consistent with the Lorentz invariant energy four-vector magnitude which yields the mass of a particle: $$ mc^2=\sqrt{E^2-(pc)^2}=0.$$
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Why do helium balloons rise and fall? I understand why a regular party balloon filled with helium falls over time due to leakage of the helium. However I've also noticed that recently filled helium balloons put outside rise and fall. At one point in the afternoon they were dropping but later in the evening it was fully upright again.
Why is this? Is it because of a change in atmospheric pressure? The weather did go from a little rainy and overcast to dry and slightly clearer skies later. It must be something to do with P = VxT.
| The upthrust on the balloon is equal to the weight of air displaced, so we get:
$$ F = V_b \rho g \tag{1} $$
where $V_b$ is the volume of the balloon and $\rho$ is the density of the air. Assuming air is approximately an ideal gas it obeys the equation of state:
$$ PV = nRT $$
so the molar density is:
$$ \rho_M = \frac{n}{V} = \frac{P}{RT} $$
where $n$ is the number of moles of air. The density in kg/m$^3$ is given by multiplying the molar density by the (average) molar mass of the air $M_{\text{air}}$, and substituting this in equation (1) we get:
$$ F = V_b M_{\text{air}} g \frac{P}{RT} \tag{2} $$
Now let's consider what happens to the volume of the balloon. We'll take the two extreme cases where the rubber skin is infinitely rigid and where it's infinitely compliant.
First consider the case where the rubber skin is infinitely compliant i.e. it doesn't exert any force on the helium inside it. In that case the volume of the helium is (approximately) given by the ideal gas equation:
$$ V_b = \frac{n_{\text{He}}RT}{P} $$
where $n_{\text{He}}$ is the number of moles of helium. Substituting this into equation (2) we get:
$$ F = n_{\text{He}} M_{\text{air}} g $$
which is constant. So in this case we find that the bouyancy is unaffected as the pressure and temperature change.
Now consider what happens if the rubber skin is infinitely rigid, in which case the volume $V_b$ is constant. We end up with:
$$ F \propto \frac{P}{T} $$
In this case the bouyancy is affected by the pressure and temperature. Assuming the pressure is approximately constant the bouyancy is inversely proportional to temperature so the balloon will rise when it gets cold and fall when it gets hot, which matches your observation.
I've taken the two extreme cases because I don't know the equation for the force produced by the rubber skin of the balloon. However it is presumably somewhere in between the two extremes I've discussed, so we expect the behaviour of the balloon to fall between those two extremes. That means we expect the bouyancy will increase as the temperature decreases and vice versa.
| {
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Hyperfine lifetime calculation: what is the spin eigenfunctions?
I'm trying to calculate the lifetime of the 21 cm line in hydrogen and have the following expression:
$$\frac{1}{\tau} = \frac{4\alpha}{3}\omega_{if}^3|\langle a_f|\vec{x}|a_i\rangle|^2.$$
The initial state is $a_i = a|{F=1, F_z=1}\rangle + b|F=1, F_z=0\rangle + c|F=1, F_z=-1\rangle$ and the final state is $a_f= |F=0, F_z=0\rangle$.
My problem arises when calculating the matrix element $|\langle a_f|\vec{x}|a_i\rangle|^2$, what are the spin wave functions?
So far I've thought to use
$a_i = |1s\rangle(a|\uparrow\uparrow\rangle + \frac{b}{\sqrt{2}}(|\uparrow\downarrow\rangle +|\downarrow\uparrow\rangle) + c|\downarrow\downarrow\rangle)$ and
$a_i = |1s\rangle\frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle -|\downarrow\uparrow\rangle)$ but then the matrix element becomes zero.
So I must be missing something.. Are there any other ways to represent $|1, \pm 1\rangle, |1,0\rangle$ and $|0,0\rangle$?
| So the matrix element I tried to calculate is indeed zero for the dipole moment. In order to find the hyperfine splitting, one must calculate $|\langle a_f|\mu|a_i \rangle|^2$.
| {
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What experimental evidence shows that sound velocity is the same for all wavelengths? I'm studying sound waves with Halliday's book, and after reading the whole chapter, one of the questions suggested was:
What is the experimental evidence that allows the assumption that the sound wave velocity in air is the same for all wavelengths?
I was surprised that such an apparently simple question really caught me. I can't seem to think of a better answer than
"The fact that we make experiments and see that the velocity is the same for all wavelengths".
But of course there must be a better answer, this one is just Why?? Because yes.
What would be a better answer to that question?
| The spectrum of various resonant tube arrangements (half-open, fully-open fully closed) is something that can be measured in a very basic laboratory and gives solid evidence that the claim is true over the kinds of frequencies that are accessible in such a lab. Say a few hundred to a few thousand hertz.
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Does turning a spoon in water raise the temperature? I read about Joule's experiment proving the transformation of mechanical work into heat. But say I have a bowl with some water, and I start turning a spoon in it very fast, thus doing work — the water won't get hotter! What am I missing?
I think maybe the work I put is simply kinetic, and won't turn into heat. But then how do you explain Joule's experiment?
| One of the reasons that makes you not to believe that one cannot heat up water by stirring it, might be that we usually experience the opposite effect. Namely, one usually stirs a hot tea or soup to cool it down. Why a cup of hot tea or a bowl of warm soup cools down when one stirs it? The reason is that the liquid/air interface where the heat exchange occurs is increased by stirring and therefore the warm liquid cools faster. The other contribution comes from the spoon which is usually a metal and sucks the heat fast from the liquid and dissipates it to air or to one's fingers.
The other reason that makes you not to believe might be that you have never experienced a glass of water warmed up because of stirring. There are couple of reasons why we normally do not experience that. The most obvious one is that the increase in temperature that one can induce is way too lower than our senses can detect. The other reason might be that if one does such an experiment in an ordinary glass the added heat dissipates so fast that one never observes the increase in temperature.
I do not know where do you live but never you rubbed your hands to each other to heat them up in a cold winter day?
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How does mass change with speed? While reading a textbook on Physics, I came across this :
Mass is a universal constant. It does not depend upon the position of the body on the Universe but it changes with speed of the body.
It's just two contrasting statements with no further explanation. Maybe the author left it to our curiosity but I can't even find out how mass changes with time.
Thanks
P.S : This is not a homework question.
| The textbook writer is referring to the concept of relativistic mass, which is the idea that accelerating a body tends to become harder and harder as its speed approaches the speed of light. This is sometimes thought of in terms of an increase in the object's mass as the speed increases.
However, you should think of this as a deprecated concept that most modern physicists consider unnecessary and misleading. Nowadays, we prefer to think that the force-acceleration momentum needs to be rephrased to something along the lines of
$$
\mathbf F=\frac{\mathrm d}{\mathrm dt}\left[\frac{m_0\mathbf v}{\sqrt{1-v^2/c^2}}\right],
$$
with an invariant mass $m_0$, than to introduce a variable "relativistic mass" $m_R=m_0/\sqrt{1-v^2/c^2}$ in an attempt to clean up that relationship and make it look more classical.
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Why can we replace a cavity inside a sphere by a negative density? I have a sphere with radius $R$ and inside this sphere there is a smaller sphere with radius $\frac{R}{3}$. This small cavity has its center at $\frac{R}{2}$, it doesn't matter in which direction.
If I want to find the electric field at the point $(0.0.0)$, why can I not enclose it with a surface and apply Gauss' law, for example with a small little sphere around $(0,0,0)$ but not touching the cavity. Is it because the sphere is not symmetric outside the Gaussian surface?
I find lots of solutions on the internet that say you can replace the cavity with a negative density, why?
http://jkwiens.com/2007/10/24/answer-electric-field-of-a-nonconducting-sphere-with-a-spherical-cavity/
| "I find lots of solutions on the internet that say you can replace the cavity with a negative density, why?"
Because they use a trick to calculate the potential easier. They assume that the empty hole is neutral, but composed of a positive charge density equal to that of the sphere plus a negative charge density of the same amount. In this way you can compute the potential of a sphere with the hollow sphere inside as the potential of a uniform large sphere and that of a negatively charged sphere located at the hole.
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Could gravity accelerate light? Gravity causes anything with energy to accelerate toward the source. Black holes, for example, have such strong gravity that they pull in light and don't let any escape. But can acceleration still apply to light? The speed of light is constant, of course, but why are photons affected by gravity yet aren't accelerated by it?
Edit: My main question is why photons aren't affected in the same way as most other particles. I'm perfectly aware that it cannot surpass lightspeed, but I want to know what makes it unaffected by acceleration while other particles are affected.
| I would suppose the short answer is no, but photons are affected by gravity.
What is happening is naturally quite relativ to the observer. Suppose you are sitting on the photon, travelling past the gravity source with the speed of light. As has been argued above you would experience the force of the gravitational pull as an acceleration toward the source. Also, if you were to observe the passage of time during the journey, you would find that your clock is running slower than the clock of an observer travelling further away from the gravity source. To this "further away traveller" you would therefore appear to be travelling faster, even though your both travelling at the speed of light.
Ultimately, what this means is, while the gravity-source affects the path of the photon, the speed isn't affected due to the relative change in time.
At least thats how I understand it anyway. :)
| {
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Distortion of body in Schwarzschild black hole Suppose I toss a cloud of matter into a Schwarzschild black hole; for the sake of argument, have it be timelike dust. As we know, the dust is "spaghettified" by tidal forces: simultaneously compressed in the tangential and elongated in the radial directions relative to the singularity.
Consider any particular volume in the dust cloud. Is its fate thus to become infinitely sparse (due to the radial elongation) or infinitely dense (due to the tangential compression)?
EDIT: It has been correctly pointed out that the tidal forces remain always finite, so the cloud is not infinitely anything. The modified question: is the net effect on the cloud a rarefaction or a compression?
| The volume of your dust cloud will not change. The shape changes but the volume does not, so the density of the cloud remains constant.
The Schwarzschild solution is a vacuum solution and the Ricci tensor is everywhere zero. If you look at section 5.2 of this paper it shows that the change in volume of an infinitesimal volume element is proportional to the Ricci tensor, so when the Ricci tensor is zero the volume is constant.
| {
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Visualising gas temperature and gas pressure Gas pressure is created when gas molecules collide with the wall of the container creating a force.
Gas temperature is a measure of how fast the molecules are moving / vibrating.
However, they both seem to be concerned by "kinetic energy" of the molecules, or in other words, the "collision" they impose on the target.
How do we visualize the difference between pressure and temperature of gas?
Is there any obvious difference between the two?
The same question in another form:
*
*A gas is hot when the molecules collided with your measuring device.
*A gas have high pressure when the molecules collided with your measuring device.
So, what is the difference between the two "collisions" in the physical sense and how do we visualize the difference?
For Simplicity,
How can a Hot gas be Low Pressured? ( They are supposed to have High Kinetic Energy since it is Hot. Therefore should be High Pressured at all times! But no. )
How can a High Pressured gas be Cold? ( They are supposed to collide extremely frequently with the walls of the container. Therefore should be Hot at all times! But no. )
| An example of a difference where the pressure of a reasonably dilute gas depends on something else other than the kinetic energy of the particles is actually just the air on Earth. A classic exercise in statistical mechanics is to consider an ideal gas subject to gravity and find how the pressure varies with altitude.
Of course, in reality the temperature of the air on Earth varies with altitude, but doing this problem by assuming that the gas has a constant temperature provides a pretty reasonable result, that the pressure goes as $P(z) \sim \exp\{-mgz/kT\}$ (don't quote me on this) where $m$ is mean molecular mass. In this case, to a decent approximation, the pressure of the gas varies with height, but the temperature does not, because one now takes into account the gravitational potential and not just the kinetic energy.
| {
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The dual of a surface element in 4-space In reading the classic text, "The Classical Theory of Fields", Third Edition, by Landau and Lifschitz, I found an "obvious" statement not so obvious to me. It is on p.19, the statement of the normality of the dual of a surface element, $df^{*ik}=\frac{1}{2}e^{iklm}df_{lm}$ to the element $df^{ik}$. Yes, the contraction is zero, as one can see if he lists the 24 terms of the sum and takes account of the alternations of the sign of the completely antisymmetric tensor coefficient and the sign changes of the surface elements. That is a bit of tedium that I found necessary, because I did not find it obvious. Maybe that is because I was not clever about the way I listed the terms.
Question: Is there some way of listing the terms that would have quickly made clear that for every positive term there would be a negative one? One thought that suggested itself to me, after I did the work (!) was that if the terms were not all of the same sign, there would have to be an equal number of positive and negative terms because of the symmetry of the form and, thus, normality of the two surface elements. Is that the obvious quality that I first missed?
| It is OK to use an explicit form in a local orthonormal coordinate system (in the Minkowski sense). The dual for each component would then be just a possible flip of signs (see, e.g., https://en.wikipedia.org/wiki/Hodge_star_operator#Four_dimensions). Raising the indices would amount to doing nothing in that coordinate system. This works even if the space-time curvature is nonzero.
| {
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FWHM increase with energy (gamma spectra) Below I have two plots from a gamma spectrum which I've been analyzing. The first plot is between a low energy range, the second between a significantly higher energy range. It is clear that the FWHMs (Full Width Half Maxima) of the peaks in the spectrum increase with energy. This is apparently on each of the spectra I've seen and seems to be an integral property of all of them.
Generally speaking, the peaks/Gamma lines in a Gamma spectra are wider at higher energies.
What is the exact reason for this? I do assume intuitively that there would be more inherent variance in the energy of peaks at higher energies. Whether this is a correct assumption is part of my question. Is the increase in width to do with error in the measurement of a particular Gamma? Do the Gammas at higher energies genuinely have a larger fluctuation in their energy? Is it a combination of both?
| This is something I never really understood, but Glen Knoll offers the following in pp. 116 of his book "Radiation Detection and Measurement":
The energy resolution of the detector is conventionally defined as the FWHM divided by the location of the peak centroid $H_0$. The energy resolution $R$ is thus a dimensionless fraction conventionally expressed as a percentage. Semiconductor diode detectors used in alpha spectroscopy can have an energy resolution less than 1%, whereas scintillation detectors used in gamma-ray spectroscopy normally show an energy resolution in the range of 3-10%.
$R={\rm FWHM\over H_0}$
So, from a purely definitional point of view, if we hold the resolution constant, and increase the energy, the FWHM must also increase.
I think really, the direction of inference in this question is the wrong way. You have an observation (FWHM increases as energy does). Now, the task is to come up with a theory as to why this is true (and test of course).
| {
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Why can't I see the blue color scattered by the lower atmosphere of the earth? I understand that the blue colour of the sky is because of the scattering of blue light by molecules in earth's atmosphere. The scattering appears to be happening from molecules that are far above in the earth's atmosphere. What about the scattering that happens because of molecules near the surface of the earth? Why can I not see the blue light scattered by molecules closer to the earth?
| Two reasons:
The scattering separates red/orange and blue in different directions. At sunset you'll see the red parts that are missing from the blue skies by day. This isn't noticeable for objects close by, because those objects surround you. The blue from some objects mixes with the red from others.
Secondly, there is a lot of air between you and the sun. The atmosphere is kilometers thick. In comparison, there's only a few meters of air near you. As each air molecule has a small chance to scatter the light, you need a lot of molecules before you notice significant scattering.
| {
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A false proof of drag force being conservative Consider a particle moving along some trajectory in the $x$-$y$ plane, in a viscous medium.
Then its equation of motion is given by:
$$\mathbf{F}_d = - b \mathbf{v} .$$
it's well-known from the Gradient theorem(fundamental theorem of line integral) that if there exists a scalar-valued function $\varphi$ that satisfy:
$\mathbf{F}_d=\nabla\varphi$,then this implies $\mathbf{F}_d$ is conservative.
I wanna show through a proof by contradiction that $\varphi$ does not exist for $\mathbf{F}_d$.
Let(for the sake of Reductio ad absurdum) $\mathbf{F}_d=\nabla\varphi$.
Consider an arbitrary curve which is parameterized by the position vector $\mathbf{r}(t)=<x(t),y(t)>$.
consider that our particle is moving on that curve.
Therefore $\mathbf{F}_d$ by definition is given by:
$$\mathbf{F}_d=<-b\dfrac{dx(t)}{dt},-b\dfrac{dy(t)}{dt}>.$$
And let $\nabla\varphi$ be given by : $<\dfrac{\partial \varphi }{\partial x},\dfrac{\partial \varphi }{\partial y}>$
From our hypothesis $\mathbf{F}_d=\nabla\varphi$ We have that:
$$\varphi(x,y)=\int -b\dfrac{dx(t)}{dt} dx = \int -b\dfrac{dy(t)}{dt} dy. $$
Assuming that our functions are well-behaved then we get:
$$\varphi=-b( x\dfrac{dx}{dt} + y\dfrac{dy}{dt} ).$$
So Although I expected that I'd arrive at some contradiction I did not, In a sense I proved that $\mathbf{F}_d$ is conservative (Although it's not!)
So what possibly I did wrong?
| The issue is that the formula that connects force and potential gets an extra term when the force depends on velocity ${\bf v}$. The formula reads (see e.g. Ref. 1)
$$\tag{1} {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}},$$
rather than just $$\tag{2}
{\bf F}~=~ - \frac{\partial U}{\partial {\bf r}}. \qquad\qquad(\leftarrow\text{Wrong}!)$$
Velocity dependence of conservative forces and potentials is also discussed in my Phys.SE answer here. To see why the friction force
$$\tag{3} {\bf F}~=~-b {\bf v}$$
has no velocity-dependent potential $U$, see e.g. this Phys.SE post and this MO.SE post.
References:
*
*H. Goldstein, Classical Mechanics; eq. (1.58).
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Equivariant cohomology formula I'm studying equivariant cohomology on three references:
*
*Szabo's review about equivariant localization (S);
*Libine's note on equivariant cohomology (L);
*Berline, Getzler, Vigne's book "Heat Kernels and Dirac Operators", Springer (BGV).
I'm deriving all formulas by myself. I face two problems:
*
*In (S) when he derive (2.83) he uses a commutator while (L), in the longest formula below Definition 34 writes
$$
i_X\nabla + \nabla i_X.
$$
I agree with (L) because it's just the square:
$$
(\nabla - i_X)^2,
$$
so why (S) put the commutator? (BGV) seems to agree with (S).
*Then I compute
$$
\begin{align}
(i_X\nabla + \nabla i_X)\alpha
& = i_X(d\alpha + \omega\wedge\alpha) + (d + \omega)i_X\alpha\\
& = i_Xd\alpha + i_X\omega\wedge\alpha - \omega\wedge i_X\alpha + di_X\alpha + \omega\wedge i_X\alpha\\
& = (i_Xd\alpha + \omega(X)\alpha) +di_X\alpha\\
& = \nabla_X\alpha + di_X\alpha
\end{align}.
$$
So I get an extra $di_X\alpha$. This must be correct because also (BGV) write (page 211)
$$
[\nabla,i_X]=\nabla_X
$$
which is, apart for the mysterious commutator instead anticommutator, the same formula.
| The main issue seems to be that BGV uses a supercommutator notation
$$\tag{1} [a,b]~:=~ab-(-1)^{|a||b|}ba,$$
where $|a|$ and $|b|$ denote the $\mathbb{Z}_2$-grading of operators $a$ and $b$,
respectively. So if $a$ and $b$ have odd gradings, then $[a,b]$ is actually the anticommutator.
References:
[BGV] N. Berline, E. Getzler & M. Vergne, Heat Kernels and Dirac Operators, 1991; p.39.
| {
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Is Newton's second law tautologous? Newton's second law
$$\mathbf{F} = m\mathbf{a}$$
where $\mathbf{F}$ is the force, $m$ the mass, and $\mathbf{a}$ the acceleration, seems at first blush to be a simple tautology, since $\mathbf{F}$ and $m$ are not defined anywhere else in the formalism. Of course we can use the more general formulation
$$\frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt}$$
with $\mathbf{p}$ the linear momentum and $\mathbf{v}$ the velocity, but given the usual elementary definition of linear momentum $$\mathbf{p} \equiv m\mathbf{v}$$
this seems hardly an improvement.
Have I missed something? Perhaps in the Lagrangian or Hamiltonian formulation one can use Noether's theorem etc. to give a less trivial definition of $\mathbf{p}$? Or is there something even more elementary than that which I've missed?
Now, the law is clearly not tautologous in concert with some kind of force law, such as Coulomb's law or the law of universal gravitation. In that case we know what force means independently of mechanics. I'm just wondering if mechanics can be viewed as logically self-contained without recourse to such an addition.
|
F and m are not defined anywhere else in the formalism
The force F is generic. In each particular case it is replaced by a force that is defined. For example $F = kx$ or $F = KMm/r^2$.
$m = Weight/g$ and the weight is measurable with a balance.
So everything is defined.
What I know is that the Aristotelian view, before Newton, was that:
$F=kv$
where $k$ was an universal constant and $v$ the speed.
The second law of dynamics $\mathbf{F} = m\mathbf{a}$ is not evident at all.
| {
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What depth should water be filled to in a cylinder flask to make it the most stable As said in title, a cylinder flask with a mass of 100g and radius of 3 has a center of balence 10cm above the base. Assuming negligible wall thickness, to what depth should water be added to this flask to make it the most stable.
The question doesn't give any other information and I'm at a loss of how to approach this question. Any help or tips in the right direction would be great
| The flask become most stable when its centre of gravity is at the smallest height. If you start pouring water, you will notice that the effective centre of gravity gets down to a lower postion. As you keep on filling, it would be at the lowest height for some level of water and rises again, afterwards. You will have to find that point of minimum height. Just frame an equation relating height of water and height of resultant centre of gravity, find the derivative and equate to zero
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How do we know that the rate at which a body loses heat is proportional to the difference between its temperature and that of its environment? Did someone do an experiment, or was that fact derived from other ideas we had about how the world works?
|
How do we know that the rate at which a body loses heat is proportional to the difference between its temperature and that of its environment?
In classical physics this is a law.
"Fourier's law
The law of heat conduction, also known as Fourier's law, states that the time rate of heat transfer through a material is proportional to the negative gradient in the temperature and to the area, at right angles to that gradient, through which the heat flows. We can state this law in two equivalent forms: the integral form, in which we look at the amount of energy flowing into or out of a body as a whole, and the differential form, in which we look at the flow rates or fluxes of energy locally."$^1$
$^1$Source
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Does Newtonian physics work on a galactic scale? I'm currently working on a simulation that aims to use Newton's Law of Gravitation to simulate how a galaxy behaves gravitationally. While I haven't gotten the simulation finished yet, I have had a few people tell me that Newtonian Physics don't work on a galactic scale, and that I need a different model to accurately simulate a galaxy gravitationally.
Is this true?
| Yes, Newtonian Physics works on a galactic scale.
Still, for long distance interactions on fast objects you might want to take into account the finite speed of gravity, but I don't think it is necessary for ordinary galaxies simulations.
Conversly a lot of phenomena occur that impact the galactic material: writting a decent simulation is not easy.
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How does order of scalar $\phi$ interaction impact feynman diagrams? On page 60 of srednicki (72 for online version) for the $\phi^{3}$ interaction for scalar fields he defines
$Z_{1}(J) \propto exp\left[\frac{i}{6}Z_{g}g\int d^{4}x(\frac{1}{i}\frac{\delta}{\delta J})^{3}\right]Z_0(J)$
Where does this come from? I.e for the quartic interaction does this just become
$Z_{1}(J) \propto exp\left[\frac{i}{6}Z_{g}g\int d^{4}x(\frac{1}{i}\frac{\delta}{\delta J})^{4}\right]Z_0(J)$
and for the feynman diagrams the $\phi ^{3}$ theory has 3-line vertices whereas the $\phi^{4}$ has 4-line vertices? Then how do the feynman diagrams change as we change the order of g?
| Yes for the $\phi^{3}$ theory the vertex has 3-lines, whereas for the $\phi^{4}$ theory this becomes 4 lines meeting at the vertex. g just refers to the number of vertices in the diagrams, so for $g^{1}$, you're summing all diagrams with one vertex, for $g^{2}$, you're summing all diagrams with two vertices, and so on.
| {
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Special Relativity in water If there are 2 observers in water moving relative to each other as well as to water, they measure different speeds of light. So does time dilation occur for them? (since Time dilation is based on constancy of speed of light)
| The opening statement "If there are 2 observers in water moving relative to each other as well as to water, they measure different speeds of light" is false. Both observers will measure the same speed for light, and therefore, relativistic effects will occur.
Unlike sound, light does not require a medium through which to travel, so the speed of the medium (in this case water) is irrelevant.
Search the web for "submarine paradox" for proof.
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The maximum distance for which Coulomb's law has been verified? We know that Coulomb's law, $F_{12} = \frac{kq_1q_2}{r^2}$, was experimentally verified for small distances by Coulomb himself at the and of the XVIII century.
The question is what is the maximum distance, experimentally confirmed, between two charges for which Coulomb's law still holds?
| I might be erring something basic here, so downvotes are welcomed, but I would love if they include comments to correct this answer, or just erase it.
I do not believe the Coulomb law has been tested beyond the order of a few meters. Arguing that light remains unchanged across the universe should be irrelevant. The reason is that the electrostatic and electrodynamic parts of Maxwell's equations can be decoupled.
The way I see this is that you can argue that Maxwell's equations are a result of imposing Lorentz invariance on Coulomb law. The same was attempted with gravitation, and there are several ways to do it. Einstein's equation proved to be the most successful, both experimentally and aesthetically, but there are other potential generalizations that result in alternative gravitomagnetic effects (i.e., Nordstrom theory).
The reverse is also true; you should be able to keep electromagnetic waves in vacuum unchanged by modifying Gauss law only. The reason being that electromagnetic propagation in vacuum only requires $\nabla . E \neq 0$. Using $\nabla . E = 0$ still describes electromagnetic waves in the absence of a static force. A modification of the Gauss law (similar to those used in Lorentz invariant MOND's
theories) could be in principle made (I am not aware if this has been shown impossible) that results in a modified Coulomb force, is Lorentz invariant, and leaves the propagation of light in vacuum for infinite distances unchanged.
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Light and Gravity
The gravitational force does not affect the speed of light rather affects the frequency of light (hence light changes colors, red to blue and vice-versa).
I know this has been verified but I have 2 questions:
1.Why the speed of light is not accelerated due to gravity?
2.Why the frequency is affected? (if the energy has to be altered, then instead of frequency, wavelength could have been changed)
| The laws of special relativity, of which the constant speed of zero mass particles in vacuum is a basic tenet , have been tested innumerable times with many experiments, particularly in particle physics. The Michelson Morley experiment has shown that there exists no luminiferous ether, i.e. there is no medium on which light propagates with this velocity c. If there were a medium it should obey Lorenz transformations , keeping the velocity as c.
The quantized nature of electromagnetic radiation is equally well established, giving an energy to photons h*nu where h is the Planck constant and nu the frequency.
General relativity in its theoretical framework, incorporates special relativity. This means that the velocity of light in vacuum is a constant in General Relativity too, impsed in the mathematics.
Thus it is the mathematics that predicts that photons, since they carry energy, have to be affected by the gravitational fields, and that this effect will manifest in the energy of the photons affected. This prediction has been verified by observations, from star and galaxy systems to solar radiation.
All the observations have resulted in a coherent cosmological model, consistent with the observations, i.e. validated.
Please keep in mind that gravitation from Newtonian physics will not "see" a zero mass particle at all. The fact that redshifts to the known spectra appear when the photons pass a strong gravitational field is expected in General Relativity.
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When does normal force equal to $mg$? Can someone once and for all explain when does normal force equal to mg?
I know for sure that when there is no friction, normal force will be equal to mg.
But, i encountered some questions when there is some mass on an incline with friction, and then the normal force was the y component of mg.
It does not make sense to me, because as i understood when there is friction, we cannot assume that mg will be equal to normal force.
| Normal Force arises due to the Newton's Third law.
Normal Force will be always acting opposite to the force falling on the surface.
Normal Force is a reaction force. Remember
Normal force is equal to mg only when the object is placed horizontally, and the force is acting in the direction of the gravitational field.
Now your second question
Here you will see that the weight of the body is passing through the Centre of gravity and acting in direction of the centre of the earth.
But the component of weight on the incline is not mg it is cos component.
In order to satisfy the Newton's third law Normal reaction to the object is the cos component
$$N=Wg\cos \theta$$
even if friction is there or not there this will be the same
| {
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Why aren't pictures of Pluto darker than those of other planets? Just as the intensity of the light from a candle decreases with distance, I would expect the light from the sun to illuminate the distant planets less than the closer ones. However, the pictures of Pluto from New Horizons don't seem any "darker" than those of the other planets. Have the photos been edited in some way to correct for the lower light? Or is there something else going on? My first thought was that this problem could be solved with longer exposure for each image, but I imagine that would cause blur problems given the speed of the probe.
| This question has been asked and answered (by me) on Astronomy Stack Exchange:
It's brighter on Pluto than you think.
NASA developed a tool called Pluto time, which tells you when at your place the ambient light conditions are similar to the ones on Pluto. This occurs when the Sun is only 2° below the horizon! That's quite shortly after sunset, and considerably before the end of civil twilight, which is when it's 6° below.
All of these photos were taken at local "Pluto time":
Pluto time, according to NASA. Source: NASA
To answer your question: all it takes is a slightly longer exposure time / larger aperture than taking photos closer to the Sun. It's easily bright enough for outdoor activities, so have fun glacier hiking on Pluto!
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Do a receiving antenna interfers with an emitting antenna? As I understand:
*
*Accelerating electrons generate electromagnetic waves.
*An emitting antenna have an alternating current (electrons are moving) which generates an electromagnetic waves.
*The electromagnetic waves reach the receiving antenna and makes the electron inside move.
*Greate, the communication is done.
However, since the electron in the receiving antenna are moving too, the receiving antenna is generating an electromagnetic wave too.
How is this not affecting the incoming wave? Does it have a different wavelength?
| Yes they both operate to different frequencies.
Two-way radios can be designed as full-duplex systems, transmitting on one frequency and receiving on another, this is also called frequency-division duplex.
How is this not affecting the incoming wave?
Source : Wikipedia
A radio transmitter is an electronic circuit which transforms electric power from a battery or electrical mains into a radio frequency alternating current, which reverses direction millions to billions of times per second. The energy in such a rapidly reversing current can radiate off a conductor (the antenna) as electromagnetic waves (radio waves). The transmitter also impresses information such as an audio or video signal onto the radio frequency current to be carried by the radio waves. When they strike the antenna of a radio receiver, the waves excite similar (but less powerful) radio frequency currents in it. The radio receiver extracts the information from the received waves.
As it says the incoming wave excite but with less power, so it do not effect other incoming waves..
The best example you can find is Transceiver using RF Module
This article can explain different frequency used difference Why do you need two antennas in a radio?
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Inconsistency with electrostatic energy formulas The energy of point charge configuration can be written as:
$$W = \frac{1}{2}\sum_{i=1}^{n}q_{i}V(r_{i}) \, ,$$
which can take both positive and negative values.
However, when we integrate the equation to get the energy of a continuous charge dustribution:
$$W = \frac{1}{2}\int\rho Vd\tau \Rightarrow W = \frac{\epsilon_{0}}{2}\left [ \int E^{2} d\tau + \oint VE\cdot da\right ] \, .$$
Take the volume to integrate to be all space, then the second term vanishes:
$$W = \frac{\epsilon_{0}}{2}\int E^{2}d\tau \, .$$
This formula can take only positive values.
So there is a discrepancy between the two formulas. What caused the discrepancy?
According to Griffith's Introduction to Electrodynamics, it says in the former equation $V(r_{i})$ represent the potential due to all charges but $q_{i}$, whereas later $V(r_{i})$ is the full potential. But why would the original charge has any potential when there is no other charge already present?
| Your formula for the first energy is incorrect. Instead use:
$$W = \frac{1}{2}\sum_{i\neq j}q_{i}V_j(\vec r_{i}) \, .$$
Or even:
$$W = \frac{1}{2}\sum_{i\neq j}\frac{q_iq_j}{4\pi\epsilon_0|\vec r_i-\vec r_j|} \, .$$
And now you see right away that you are avoiding the energy of the point charges themselves. Because naively it would be zero.
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Can light have zero wavelength? As you increase the energy of a photon it's wavelength shortens. Is it theoretically posible for light to not have a wavelength? Like a still pond?
| A still pond would be equivalent of infinite wavelength and the frequency of 0. The equivalent of 0 wavelength would be an infinitely tall tsunami wave in the pond. It is not possible to have wavelength at exactly 0, but you can get arbitrarily close to 0. Proof: any positive non-zero wavelength, no matter how small, can be made smaller via blueshift (approaching towards the source). At some point, though, you will be limited by the amount of energy needed to gather in order to create such radiation.
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What is the correct way of writing the antisymmetrized state of two identical fermions? I am just confused:
If I have 2 identical fermions, where one of them is in state A and the other one is in state b, and they are normalised and orthogonal, which statement is right:
1) $|\Psi\rangle=\frac{1}{\sqrt{2}}\left(|a_1\rangle|b_2\rangle-|a_2\rangle|b_1\rangle\right)$
or
2) $|\Psi\rangle=\frac{1}{\sqrt{2}}\left(|a_1\rangle|b_2\rangle-|b_1\rangle|a_2\rangle\right)$
when '1' and '2' denote the coordinates.
I think it should be 2, but in my script its different...
edit: Okay, I now think 1) should be right, but it arises another question:
If I calculate $\langle\Psi|\Psi\rangle$ I get
$\frac{1}{2}(\langle a_1|a_1\rangle\langle b_2|b_2\rangle+\langle a_2|a_2\rangle\langle b_1|b_1\rangle-\langle a_1|a_2\rangle\langle b_2|b_1\rangle-\langle a_2|a_1\rangle\langle b_1|b_2\rangle)=\frac{1}{2}(1+1-1-1)=0$
| Your second option is correct.
Writing
$$ \lvert a_1\rangle\lvert b_2 \rangle - \lvert a_2 \rangle \lvert b_1\rangle$$
does not make sense. The notation $\lvert \psi_1\rangle\lvert\phi_2\rangle$ is shorthand for the state $\lvert\psi_1\rangle\otimes\lvert\phi_2\rangle$ in the tensor product $\mathcal{H}_1\otimes\mathcal{H}_2 $ of the Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ of the individual partices. But $\lvert a_2\rangle\lvert b_1\rangle$ would live in $\mathcal{H}_2\otimes\mathcal{H}_1 $, which is a different (although isomorphic) space, and you cannot add/subtract elements that lie in different vector spaces.
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Nature of Pressure Pressure is defined as force per unit area. Therefore, pressure = force divided by area. Since force and area are both vectors, how can we perform this division without violating the rule of vectors?
If pressure is a scalar quantity, how did this scalar nature came about from the vector nature of force?
| The pressure is defined by:
$$ \text{d}\mathbf{F}_n = -P\text{d}\mathbf{A} $$
where $\text{d}\mathbf{F}_n$ is the normal force on a surface element $\text{d}\mathbf{A}$. The pressure $P$ is the (scalar) constant of proportionality linking the two. Since the two vectors are in the same direction we can rewrite the equation using the unit vector normal to the surface $\mathbf{n}$:
$$ \text{d}F_n\mathbf{n} = -P\text{d}A\mathbf{n} $$
where $F_n$ and $A$ are now just scalars, and this gives us:
$$ P = \frac{F_n}{A} $$
This last equation only involves scalars.
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Work done in moving a body out of Gravitational influence By "out of Gravitational influence ", I suppose it's meant that Gravity of a planet has no effect whatsoever on the body.
Am I correct?
Could we do it by making the object reach Escape Velocity by doing Work = GMm/2D? {assuming circular orbit}
(m, mass of object;M, mass of planet; D= distance from center)
It could also be done by taking the object to infinity (hypothetically)
This would require a work of
GMm/D
In first case total mechanical energy is zero but in second it is equal to Kinetic Energy.
In which of these is the body actually out of Gravitational influence of the Planet?
Or is there some other way?
What should the total work done be in putting it out of Gravitational influence of a planet?
|
By "out of Gravitational influence ", I suppose it's meant that Gravity of a planet has no effect whatsoever on the body. Am I correct?
Yes, that's correct.
An object that is subjected to a central gravitational field has potential energy $U$, given by:
$$U(r)=-\frac{GMm}{r},$$
where $G$ is the universal constant of gravity, $M$ the mass of the central object, $m$ the mass of the object and $r$ the distance between the centres of gravity of the objects.
For $r \to \infty$, $U \to 0$ and the object is no longer under the influence of the gravitational field. Due to conservation of energy the work $W$ needed to move the object from $r$ to $\infty$ is:
$$W=U(\infty)-U(r),$$
$$W=0-(-\frac{GMm}{r}),$$
Or:
$$W=\frac{GMm}{r}.$$
If the object was launched from position $r$ with an initial velocity $v$ (and with no means of propulsion/thrust) then if:
$$\frac{mv^2}{2}=\frac{GMm}{r},$$
$$v_e=\sqrt{\frac{2GM}{r}},$$
would be the escape velocity.
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Combination of Simple Harmonic Motions Will the combination of 2 Simple Harmonic motions will be an SHM in itself?
For example for simple functions such as
$$\ f(t)=\sin\omega t-\cos\omega t$$ I can use trigonometry to show that it can be expressed as $$\ f(t)=\sqrt 2\sin(\omega t-\pi/4) $$.
But what about functions given in the questions given below?
[Ref: “NCERT Class 11th (XI) Physics, Part 2”, Digital Designs; notes on p. 357 and Problem 14.4, p. 359 <link> ]
In (b) I can express the function as a combination of
$\sin\omega t$ and $\sin3\omega t$.
Each of these 2 terms can independently express an SHM but will their combination do the same?
As an answer to part (b) and (d) ,the book says that the superposition of two SHM is always periodic but never an SHM. (I believe that this is incorrect.Maybe a typo)
Moreover, at the end of the chapter there is a note:
I am getting pretty confused.
Can anybody tell me when the combination of 2 SHM's be an SHM/periodic/not periodic?
| Consider the superposition of two simple harmonic motions
\begin{equation}
x(t) = x_1(t) + x_2(t) = A_1 \cos \left( \omega_1 t + \phi_1 \right) + A_2 \cos \left( \omega_2 t + \phi_2 \right).
\end{equation}
The first motion $x_1(t)$ is periodic with period $T_1 = \frac{2\pi}{\omega_1}$ and the second motion $x_2(t)$ is periodic with period $T_2 = \frac{2\pi}{\omega_2}$. Clearly the sum of both is only periodic if $n T_1 = m T_2$ where $n$ and $m$ are positive integers (thanks to user fibonatic for pointing out the most general case). To see this, simply write
\begin{equation}
x\left(t+nT_1\right) = x_1\left(t+nT_1\right) + x_2\left(t+mT_2\right) = x_1(t) + x_2(t) = x(t).
\end{equation}
Moreover if the period of both harmonic motions is the same $\omega_1 = \omega_2 = \omega$, we can write
\begin{align}
x(t) & = A_1 \left[ \cos(\omega t) \cos \phi_1 - \sin(\omega t) \sin \phi_1 \right] + A_2 \left[ \cos(\omega t) \cos \phi_2 - \sin(\omega t) \sin \phi_2 \right] \\
& = \left[ A_1 \cos \phi_1 + A_2 \cos \phi_2 \right] \cos(\omega t) - \left[ A_1 \sin(\phi_1) + A_2 \sin \phi_2 \right] \sin(\omega t) \\
& = A \cos \left( \omega t + \phi \right),
\end{align}
where used the sum rule $\cos\left( \alpha + \beta \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$, and we defined
\begin{align}
A \cos \phi & = A_1 \cos \phi_1 + A_2 \cos \phi_2 \\
A \sin \phi & = A_1 \sin \phi_1 + A_2 \sin \phi_2 .
\end{align}
This can be generalized to an arbitrary sum of harmonic motions with the same period:
\begin{equation}
\sum_i A_i \cos \left( \omega t + \phi_i \right) = A \cos \left( \omega t + \phi \right).
\end{equation}
Another way to understand this, is to note that the harmonic equation is a linear differential equation; any linear combination of solutions is also a solution.
Conclusion
The sum of two harmonic motions with frequencies $\omega_1$ and $\omega_2$ is periodic if the ratio $\frac{\omega_1}{\omega_2}$ is a positive rational number. If the ratio is irrational, the resulting motion is not periodic.
If, moreover, the frequencies of the two harmonic motions are equal, the resulting motion is also a harmonic motion with the same frequency.
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Relativistic acceleration in sinusoidal electric field Consider a relativistic charge $q$ moving with an oscillating electric field $E_z$ with phase velocity $v_p=c$ in direction $\hat{z}$ (e.g. radially polarized laser coprogating with electron). What is the energy gain of this charge as a function of time?
I set this up from the relativistic force
$$F=\frac{dp}{dt}=\frac{d}{dt}(\gamma m \dot{z})= qE_0 \sin{((c-\dot{z})t k)}$$
where $t$ is time, and $k$ is the usual wave number $k=2\pi/\lambda$.
My confusion arises from the $\dot{z}$ on the RHS. I don't have much experience with differential equations, and so I wonder if it is necessary to actually write it as $\dot{z}=\int_0^t dt' \ddot{z}$ or if the LHS' attribution of $\ddot{z}$ automatically leads to an appropriate $\dot{z}$?
Thank you for any help.
After some thinking the equation written above is actually wrong. It should indeed be written like so:
$$
F=\frac{dp}{dt}=\frac{d}{dt}(\gamma m \dot{z})= qE_0 \sin{(k(ct-\int_0^t\dot{z}dt' +z(t=0))}
$$
I would appreciate help solving this equation if someone has experience. TY!
| The very first thing you should do is to Lorentz transform into the initial rest frame of the electron. This leads to a doppler shift in the electric field. If you really want to omit the magnetic field your equation of motions become 1 dimensional because the electron will only oscillate in the direction of the polarisation of the light field. The equations of motion become (I use a cosine field here):
$$\frac{d}{dt} p = q E_0 \cos(\tilde{\omega} t) $$
$$\Rightarrow p = \frac{E_0}{\omega} \sin(\tilde{\omega}t)$$
$$\frac{d}{dt} z = \frac{p}{m_e \sqrt{1+\frac{p^2}{m_e^2 c^2}}} $$
This can be solved after inserting the solution for $p$ by simple integration and yields
$$z(t) = \frac{c}{\tilde{\omega}}\left(\arctan{\alpha} - \arctan{\left( \frac{\alpha \cos(\tilde{\omega}t)}{\sqrt{1+\alpha^2 \sin^2(\tilde{\omega}t)}} \right)} \right)$$
with $\alpha = \frac{E_0}{m_e c \omega}$
Now in the last step we have to transform back into the lab frame. This can be done with the replacement
$$t = \gamma_0 (t_\text{lab}+\frac{v_0 x_\text{lab}}{c^2}) $$
and trivially
$$z_\text{lab} = z$$
$$x_e = v_0 t_\text{lab} $$
finally
$$\tilde{\omega} = \sqrt{\frac{1-\frac{v_0}{c}}{1+\frac{v_0}{c}} }\omega $$
Below is a plot of the solution for different $\alpha$. You can clearly see that it goes from simple harmonic, which is the newtonian limit to a triangular function, which is the relativistic limit.
[1
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Do liquids creep from cold to hot like gases? Consider the U-tube water experiment below with the left red block
at 10C and the right red block at 90C. I think the right level
will become about 3% higher than the left level simply because
hotter water is 3% less dense than colder water, but I'm
assuming negligible pressure change along the bottom portion of the
tube. I'm guessing this effect is small since I can't find papers
for liquids like I can for gases (gas moves from cold to hot via
thermal transpiration).
So, what is the pressure change along the bottom?
| If there was a horizontal pressure gradient, this unbalanced force would begin to redistribute the fluid. So by assuming the fluid settles in a static equilibrium configuration, you've implied that there is no pressure change along the bottom.
Of course, in a realistic situation (with a tube that is not infinitesimal in cross-section), you'd expect to see convection occurring inside the tube between the two blocks, transferring heat via dynamic and chaotic turbulence of the fluid (i.e., across the junction the pressure would continually fluctuate slightly).
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Velocity of a leak in a closed water tank Bernoulli's equation states
$P_1+{1\over2}\rho v_1^2+\rho g h_1 = P_2+{1\over2}\rho v_2^2+\rho g h_2$
In a classic "water tank with an open top and a leak" scenario, "point 1" is the surface water in the tank, and "point 2" is the leak. The equation could be rewritten for $v_2$ as
$v_2= \sqrt{2g\Delta h}$
This is simple, but suppose it involves a tank where the top is closed off. The above simplification will no longer yield the correct velocity.
How do I apply Bernoulli's equation for "water tank" scenarios in which the water tank is closed?
| For a tank that is either open at the top or has a (sufficiently large) venting hole in it above the fluid line, the outflow speed is indeed approximately:
$$v= \sqrt{2g\Delta h}$$
But with a closed tank pressure $p_0$ above the water and thus also hydrostatic pressure:
$$p=p_0+\rho g\Delta h,$$
drops and flow rate drops. This is because the air in the closed tank gets to occupy a larger volume and according to the Ideal Gas Law that means pressure must drop.
What really happens you can observe by overturning a half filled bottle of pop or such like. Very quickly the flow of liquid starts slowing down, then it becomes erratic but without actually stopping. The reason is that atmospheric pressure now pushes air through the bottom opening because atmospheric pressure has become higher than the pressure inside the bottle.
For that reason the flow never stops, rather it proceeds with stops and starts as liquid is trying to get out and air trying to get in, at the same time.
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Rolling Friction Problem The handbrake of a vehicle of mass $1.5\ \mathrm{tonnes}$ completely fails while it is parked on a $30^\circ$ slope. It rolls $20\ \mathrm{m}$ down the slope before colliding with, and locking on to, a parked vehicle of mass $0.9\ \mathrm{tonnes}$.
(a) estimate the velocity of the two vehicles immediately after the collision.
(we are given that the coefficient of static friction of rubber on asphalt $= 0.9$ and the coefficient of kinetic friction is $0.7$)
For part (a):
I realize I need to find the net force down the slope, however I am unsure regarding how rolling friction works. Do I take into account static friction to be in the opposite direction as motion? Or do I ignore it? Why? (Also if I ignore friction I realize it would be easiest to just use conservation of energy.)
Once I understand this (-> how rolling friction works) I can find the velocity $20\ \mathrm{m}$ down the slope using $v = u + at$, and then using conservation of momentum find the initial velocity of the two vehicles when they collide.
Thank you so much!
(I just need help understand how rolling friction applies in this problem.)
| Rolling friction is neither static nor kinetic (sliding) friction. It's caused by inelastic forces between the wheels and the road. For example, compressing the tire may take a lot of force, which is pointed up and toward the back of the vehicle. But when the tire expands (as that part of the tire is leaving the road), it may expand with less force, which is directed up and toward the front of the vehicle. Since the rearward component of the compressing regions is less than the forward component of the expanding regions, there is a net rearward force.
A rolling friction coefficient would be much less than either static or kinetic friction. Since you're not given a rolling friction coefficient or a different model for rolling friction, it looks like you can ignore it.
This page (http://www.phy.davidson.edu/fachome/dmb/PY430/Friction/rolling.html) has some decent diagrams.
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What does "the fabric of space and time" actually mean? I've heard the term "the fabric of space and time" in both physics and science fiction, and although I know it has something to do with general relativity, I don't understand what, specifically, they're referring to.
| In Einstein's theory space and time is single entity called spacetime. We treat spacetime as a smooth fabric which is distorted by presence of energy.Thus the term the fabric of space and time.The term fabric is used to help us visualize how spacetime works in GR.
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How does one get the first few terms of the $S$-matrix expansion? According to a set of notes I'm reading
$$\langle p_f | S | p_i \rangle = \delta(p_f-p_i) + 2 \pi \delta(E_f-E_i) \bigg[\langle p_f | V | p_i \rangle + \cdots\bigg]. \tag{1.29}$$
I don't understand where the $2 \pi \delta$ factor comes from in the second term. Could someone please help me see this?
(I'm trying to understand Eq. 1.29 here: http://www.physics.umd.edu/courses/Phys851/Luty/notes/renorm.pdf)
| The Hamiltonian is: $H=H_0+V$, where V is the interaction part. The scattering matrix (expansion form to the first order) is:
$$S=1-i\int_{-\infty}^{\infty}V(\tau)d\tau+...$$
where $V(\tau)=e^{iH_0\tau}Ve^{-iH_0\tau}$ (interacting formalism). So:
$$S^{(1)}_{fi}=-i\int_{-\infty}^{\infty}\langle\psi_f|e^{iH_0\tau}Ve^{-iH_0\tau}|\psi_i\rangle d\tau=2\pi\int_{-\infty}^{\infty}e^{i(E_f-E_i)\tau}\langle\psi_f|V|\psi_i\rangle d\tau=2\pi\delta(E_f-E_i)\langle\psi_f|V|\psi_i\rangle$$
(we can do for other terms similarly).
In physical sense, this Dirac delta function guarantees the energy conservation of the system (so in Feynman diagrams, there is always a Dirac delta fucntion in each vertex function ).
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Can we speed up the evaporation of black holes manually by accelerating it? If we throw an object to pass near a black hole, to bypass it, it will change the speed of the black hole, just like gravitational assist for a space probe. Does an accelerating black hole evaporate faster because:
*
*When object accelerates, mass increases
*When mass increases, gravity increases
*When gravity increases, the black hole collects virtual particles more rapidly
Is above true?
| Isn't this just like the twin paradox? The twin that undergoes acceleration is the one that ages the slowest. So if you want to speed up the evaporation of black holes (from your viewpoint), you should accelerate yourself.
Maybe black holes work differently, but offhand I don't see why they should.
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Question regarding gravity and time I and my friend (Age 15) were discussing about light and speed of light when we thought of a question. Imagine you are travelling in a spaceship at the speed of 2.9*10^8 m/s circling the earth. According to our calculations, 1 second on the spaceship is 30 seconds on earth. Lets take the gravity on earth to be 10m/s^2. Now, what will be the force of gravity acting on a person in the spaceship? We tried to find concepts regarding this but could not find anything.
| Start in the Earth frame i.e. the frame of an observer standing on the Earth watching the orbiting spaceship. In this frame the acceleration is given by the usual expression for circular motion:
$$ a = \frac{v^2}{r} $$
where $v$ is the speed of the spaceship and $r$ is its distance from the centre of the Earth.
This is less than the acceleration measured on board the spaceship because on the spaceship time is dilated. Suppose someone on the spaceship drops an object and we on Earth watch it for a time $dt$, then distance we see the object move is given by the standard Newtonian expression:
$$ dr = \tfrac{1}{2}a\,dt^2 \tag{1} $$
The observers on the spaceship also see the object fall a distance $dr$ because radial distance is the same in both frames. However time is dilated on the spaceship because of the motion, so the time the object takes to fall is $dt'$ where:
$$ dt' = \frac{dt}{\gamma} $$
So on the spaceship the distance the object moves is given by:
$$ dr = \tfrac{1}{2}a'\,dt'^2 = \tfrac{1}{2}a'\,\frac{dt^2}{\gamma^2} \tag{2} $$
whre $a'$ is the accelkeration on the spaceship. Since it's the same $dr$ in both equations (1) and (2) we can equate the right hand sides of the two equations to get:
$$ \tfrac{1}{2}a\,dt^2 = \tfrac{1}{2}a'\,\frac{dt^2}{\gamma^2} $$
And this gives us:
$$ a' = \gamma^2 a = \gamma^2 \frac{v^2}{r} $$
To calculate a value for $a'$ you need to specify the distance $r$ at which the spaceship is circling, and you haven't specified this in your question.
One last note, the Lorentz factor $\gamma$ is given by:
$$ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $$
At $v = 2.9 \times 10^8$ m/sec I get $\gamma \approx 3.91$ so one second on the ship is 3.91 seconds on Earth. I don't understand where you get the factor of 30 from.
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Does 'focal length' mean something different with lenses and pinhole cameras? Sometimes different but related things have the same name by some tradition or accident, causing a lot of headache to newcomers to a field.
I would like to come to clear terms with this: does the expression 'focal length' mean something distinct when applied to pinhole cameras vs. lens cameras?
With pinhole cameras it's the distance from the pinhole to the image plane. With lenses, it's the distance from the lens to the point where parallel incoming rays meet. Why are these things called the same? Am I right that this is just clumsy nomenclature or are these related at the limit of some infinities?
EDIT: Apparently some other diagrams label the lens-to-image-plane as "focal distance". Is "focal distance" something else than "focal length"? Or are people using inconsistent definitions? Something's fishy here.
| Have a look to the beginning of this video https://www.youtube.com/watch?v=ecuDGXZjyl0 The presenter is trying to make links between the people comming from computer vision and optics and explicitely discuss this.
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Does the Entropy of the Visible Universe Decrease? My understanding is that the entropy in a closed system increases. However, I would also assume that the number of particles in the open system of a visible universe decreases because of accelerated expansion.
(1) What would be the entropy of the visible universe when only one particle is left? Is the entropy higher than the entropy in the visible universe today?
(2) How does the entropy of the visible universe change in time?
| *
*With one particle, it would be nearly impossible to convert energy into work. So the entropy would be significantly higher than the entropy of our current universe, in which energy to work conversion is possible.
*Overall, the entropy of our universe increases in accordance to the second law of thermodynamics. The reason being that a percentage of energy released in reactions is produced as heat, which cannot be regained.
| {
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What is the energy source if a tall water tank is used to transfer floating objects upwards instead of cables with motors? Suppose I want to transport some logs from the ground to the roof of a tower. Originally I can use a lift, or some cables, or even move the logs upwards manually; then the energy is converted to the potential energy of logs.
Now, if I build a tall water tank from ground to the roof of the tower, and fill it with water, and suppose that I can push the logs from the bottom of the tank with a well shaped door, then let the logs float to the roof of the tower. What is the energy source that used to convert to the potential energy of logs?
| When you put the log inside (from the side) you need to place the water from the log position somewhere else. The relevant "somewhere" is at the surface of the tank, so you have to lift the water there (it takes a lot of energy to get a log-equal volume of water to the roof level). The surface is now higher than before (the log-equal volume divided by the area of your tank top).
When the log floats up, the water goes under it (a lot of friction is involved here, some energy get converted to warmer water, but you would probably not notice it).
When the log reaches the top (and everything is quiet some time after), the level of water surface will be a little higher than before you inserted the log, as the log still occupies some space in the tank (enough that a volume of water with weight equal to the weight of the log is displaced - ask Archimedes for more details).
Finally, when you remove the log on the roof, you return the surface level to its starting position (minus water spilled and soaked into the log).
Should you use some pipe for inserting the log and avoid lifting the water up, but just remove it from other side, the insertion could be easier, but afterwards there would be less water in the tank and lifting the water up to return it would you cost the energy which you borrowed to lift the log (with the "trick" with pipe).
| {
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Can you compress pure carbon into diamonds? I'm doing a science project, and we're wondering if it is possible to compress pure carbon (C) to the point where it becomes diamonds? What would the process have to be and how much energy would this take? Has this been done and is this feasible?
| Theoretically yes but those diamonds would be industrial, not worth much and wouldn't have the exterior of natural diamonds. Like what was said in a previous answer, these diamonds can be made in an explosion too! I think there's a myth busters episode on it if I'm not wrong!
| {
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Simple quadrupole field not yet in Lorenz gauge? I'm having trouble reproducing some of the results regarding gravitational waves in the Wald's General Relativity.
In section 4.4 of gravitational radiation, eq.4.4.49 shows the far-field generated by a variable mass quadrupole:
$$ \gamma_{i j}(t,r)=\frac{2}{3R} \frac{d^2 q_{i j}}{dt^2} \bigg|_{t'=t-R/c} $$
I want to verify that this simple field satisfies the Lorenz gauge (eq.4.4.25)
$$\partial_{i} \gamma_{i j}=0 $$
I wrote the $q_{i j}$ for a simple rotating binary
$
\ddot{q}_{i j} =
\begin{bmatrix}
2 \omega^2 \cos{2\omega(t-R/c)} & - 2 \omega^2 \sin{2\omega(t-R/c)} & 0 \\
- 2 \omega^2 \sin{2\omega(t-R/c)} & - 2 \omega^2 \cos{2\omega(t-R/c)} & 0 \\
0 & 0 & 0
\end{bmatrix}
$
then, I wrote $R=\big|(x,y,z)\big|$
$\partial_{i} \gamma_{i 3}$ trivially cancels, but when I compute the other components of the divergence I get
$$ \partial_{i} \gamma_{i 1} = \frac{2 \omega^2([-c x+2 y R \omega]\cos{2\omega(t-R/c)}+[c y+2 x R \omega]\sin{2\omega(t-R/c)})}{c R^3} $$
$$ \partial_{i} \gamma_{i 2} = \frac{2 \omega^2([c y+2 x R \omega]\cos{2\omega(t-R/c)}+[c x-2 y R \omega]\sin{2\omega(t-R/c)})}{c R^3} $$
Which as you might have noticed, are not zero in general
Given that the Lorenz gauge is used everywhere one wants to study gravitational wave propagation, it seems unexpected that the far-field of a simple binary quadrupole is not automatically in the Lorenz gauge
Question: I want to understand what is wrong here, if there is anything wrong. Am I wrong/naive in expecting this simple field to be
in the Lorenz gauge? Is there a simple transformation that can be
applied to this field in order to be manifestly in harmonic
coordinates?
| I think the issue here is that you are simply forgetting the $\gamma_{0\mu}$ components of the metric. As mentioned in Wald, these are to be obtained using the (Lorenz) gauge condition, and are given explicitly in terms of the spatial components of the metric in eq. (4.4.45).
When those components are included in your metric, Lorenz gauge should be manifestly satisfied (by construction).
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Electrostatic induction, induced charges Is it true that if a conducting object is not grounded, the nearby charge will induce equal and opposite charges in the conducting object?
It is mentioned on Wikipedia (electrostatic induction) but it is also mentioned that charges will appear such that the total electric field inside the conductor becomes zero.
My doubt is that which statement is true whether the charges induce such that the electric field inside the conductor becomes zero or the induced charge is equal in magnitude to the inducing charge (the charge which causes induction).
| What happens is that for a conductor limited in real space and not grounded the net charge can't change (if is 0 it has to remain 0), so on the face towards the external charge it accumulates a charge to balance the electric field inside the conductor, but on the other face it accumulates a charge opposite in sign, to balance the net charge on the conductor.
If the conductor is unlimited in space (that happens only in physics exercise) the "other face" is moved to infinity, so also the charge that should balance that on conductor, and so if you integrate the charge density in space ($\mathbb{R}^3$ doesn't include the "infinity") you find a net charge on the conductor, but space-unlimited conductor don't really exist.
| {
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Is it possible that a person with myopia will see a blurry picture as normal? I am trying to process an image in good quality to appear blurred to a normal person and good to a person suffering from myopia
as seen in this source.
Is it possible that a picture that is blurry will appear normal to a person suffering from myopia (farsightedness)?
| A quick footnote to Nathaniel's answer:
If an image looks blurred to you it's because you are viewing it in a plane that isn't the focal plane.
If you put a screen where I've drawn the red dotted line then the image on the screen will look blurred.
If you measure the light in the red dotted plane then at every point in that plane the light wave will have an intensity and a relative phase. If you know the intensity and phase then you can reconstruct the in focus image using the Huygens construction, and indeed the process is known as Huygen's deconvolution. The trouble is that when you take a photograph the photographic process only records the light intensity and it loses the phase.
So if you're starting from a photograph you've lost half the information originally present, i.e. the relative phase, and that means it's impossible to reconstruct a perfectly focussed image. A blurred photograph won't look normal to anyone - myopic or otherwise. However it is usually possible to improve the blurred picture to some extent, which is why Huygens deconvolution software is so widely available.
| {
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How do I figure out the totally airborne height for a given machine? Technically "airborne" can just mean to move through the air, but I would like to know how high you have to be before you are entirely supported by air in a helicopter-like machine, as opposed to benefiting from the reaction from the earth (or whatever platform you are taking flight from). I am effectively asking for an equation that will tell me how high I have to be before the effect demonstrated in the following image is effectively zero.
I assume that this has something to do with mass, but am unclear on how to proceed beyond that.
| The effect is never zero, unless you happen to stop producing lift.
Lift is produced by air being continuously pushed downwards. This downward motion continues for several minutes, but is dissipated eventually. But in all cases the motion produces a pressure increase on the ground underneath. This is clearly audible for aircraft flying at supersonic speeds, because then the pressure change will not spread ahead of the aircraft. For subsonic aircraft, the pressure change is much more gradual and widely spread, but still there nonetheless.
To answer your question: Eventually, the aircraft is always supported by the earth.
Maybe you know the question about the closed truck full of birds. Someone bangs at the outside, so all birds fly up. Will the truck be lighter? Same story: The truck will not change weight, because now the air transmits the weight of the birds into its structure.
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Symmetry factor for Feynman diagrams in $\phi^4$-theory for $n$-points Green function I'm working with two theories.
Theory A: $H_{int} =\int d^3x \frac{Mg}{2}\phi\varphi^2$
Theory B: $\phi^4$-interaction: $H_{int} = \int d^3 x \frac{\lambda}{4!}\phi(x)^4$
Where $M$ is the mass associated to the field $\phi$.
I have to calculate the $n$-points Green-function at tree-level for these theories, $G^A(p_1,…,p_n)$ , $G^B(p_1,…,p_n)$, considering the limit $M\rightarrow + \infty$ for the scattering of scalar particles of $\varphi$ field
In every following diagrams, $v$ is the total number of vertices and the external propagators (indicated by the arrows) are $n$.
Feynman diagram for theory A
Let's work with the theory A, trying to calculate the symmetry factor of this diagram.
If we have $n$ external propagators, then $n = v+2$. The number of propagators of field $\phi$ is $\#_{\Delta_\phi} = v/2$, while the number of $\Delta_\varphi$ is $\#_{\Delta_\varphi} = v/2-1$ and so the total number of propagator is $\#_\Delta = v-1 = n-3$.
I calculate the symmetry factor as
$$
\frac{1}{S!} = \frac{1}{2^v}2^2 3^{\#_{\Delta_\varphi}}2^{\#_{\Delta_\varphi}} = \frac{1}{2^v}2^2 6^{\frac{v}{2}-1} = \left( \frac{3}{2}\right)^{\frac{v}{2}-1}
$$
because the initial and final vertices have 2 possibles equals configurations (point $p_1$ can be contracted in two different way with the initial vertex, same thing for $p_n$) and then they contribute with a $2^2$ factor. Then, for each internal $\varphi$-propagator we have $3\cdot 2$ ways to contract external momenta and vertex legs to obtain such diagram.
Is it true or am I missing something?
Feynman diagram for theory B
In this case $n = 2(v+1)$, $\#_{\Delta_\varphi} = v-1 = (n-4)/2$.
I would say $1/S! = 1$ because for each vertex there are $4!$ possible ways to contract legs and external momenta and so,
$$
\frac{1}{S!} = \frac{1}{(4!)^v}(4!)^v = 1
$$
But I'm not sure this is correct...
Which is the symmetry factor of such diagram? How to derive it?
| The symmetry factor should be 1 in both theories.
First a few remarks. If you are computing amputated diagrams the external propagators are removed from the evaluation of the amplitude, however it is unnatural (to me) to remove them from the count when considering symmetry factors. In Theory A I would say that $\#\Delta_{\varphi}=\frac{3v}{2}+1$ and in Theory B that $\#\Delta_{\varphi}=3v+1$.
BTW in Theory A $n$ is $v+2$ and not $v-2$.
Wick contraction calculation in Theory A:
The symmetry factor is $\frac{1}{v!}\frac{1}{2^v}C$ where $C$ is the number of Wick contractions which produce the desired shape. You have to choose which internal vertex attaches to $p_1$, $p_2$ (gives a factor $v$) then which vertex attaches to the previous one etc. So you get a factor of $v!$. Then you contract the $\phi$'s for which there are no choices. Finally, you contract the $\varphi$'s. The extreme vertices give $2^2$ as you rightly noted, but so do the vertices internal to the chain. This is because for each of the two $\varphi$ legs you need to choose who connects to the exernal leg and who connects to another chain vertex. So altogether you have $C=v!\times 2^v$.
Wick contraction calculation in Theory B:
Same reasoning.
You can learn more about symmetry factors in my answer Problem understanding the symmetry factor in a feynman diagram
or for the systematic theory using Joyal's theory of combinatorial species in my article "Feynman diagrams in algebraic combinatorics" but that's not an easy read.
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Origin of radio waves In the same way as the origin of X rays is the excitation of electrons, what is the origin of radio and infrared radiations in this respect?
| There are a number of ways of generating radio waves.
The simple acceleration of electrons in a transmitting dipole will generate radio waves. I guess as you might have been using wi-fi to write your question, you knew this and are more concerned with "natural" sources of radio waves?
Basically, electromagnetic radiation of all wavelengths can be produced by bound-bound, free-bound or free-free transitions of electrons that may be bound in atoms or molecules.
Where bound electrons are involved, then it is typically the spacings between molecular rotational and vibrational levels that are responsible for infrared emission. Longer wavelength bound-bound transitions can occur between hyperfine transitions with small energy differences. e.g. The famous 21cm radiation comes from a hyperfine transition in hydrogen atoms.
The are also a number of mechanisms involving the acceleration of unbound electrons that produce continuum emission at long wavelengths. The most commonly occurring are bremsstrahlung -the acceleration of electrons in the electric fields of ions - which can produce all wavelengths down to a short wavelength roll-off that depends on the gas temperature; and synchrotron continua that are produced by electrons spiralling at relativistic velocities in magnetic fields.
| {
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Hot Object In a Cooler Space How would you calculate the heat given by an object that is hotter than its surroundings? I know there's Newton's Cooling Law, but what about any heat given off by radiation? Are these additive quantities?
| Just use Newton's cooling Law to calculate heat transfer:
$$\frac{dQ}{dt}=hA(T-T_0)$$
where:
*
*$\frac{dQ}{dt}$ is the heat transferred from the glass to the room.
*$A$ is the total surface area of the glass.
*$T$ and $T_0$ are the temperatures of the glass and the room respectively.
$h$ is the overall heat transfer coefficient and can be found on various engineering sites, for various materials and temperature ranges.
Make sure your calculation is dimensionally consistent (use compatible measuring units).
| {
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How can I convert Right Ascension and declination to distances? I am calculating galaxy rotation curves for various galaxies in the Ursa Major cluster and I want the distance of those galaxies from the center of the Cluster. The values referred to as coordinated are RA and dec and I don't know anything about these coordinates. How/Where can I get the distances of galaxies?
| If the centre of your cluster has coordinates $\alpha, \delta$, the right ascension and declination in radians$^{1}$, and you have a galaxy's coordinates $\alpha_g, \delta_g$ in radians, then the following formula gives the angular distance $\theta$ in radians.
$$ \cos \theta = \sin \delta \sin \delta_g + \cos \delta \cos \delta_g \cos (\alpha - \alpha_g)$$
[A previous version had a sine instead of a cosine in the last term, but that cannot be right because if you set $(\alpha, \delta) = (\alpha_g, \delta_g)$ you did not get $ \cos \theta = 1$ and thus $\theta=0$. You can check that the current expression now does satisfy this requirement: if the angles are equal, then $ \cos \theta = \sin^2 \delta + \cos^2 \delta \times \cos (0) = 1$.]
From there, to get a physical distance you need to to know the distance to the cluster $D$. The projected separation$^2$ of the galaxy from the centre of the cluster $r$ is then
$$r = D\tan(\theta) \simeq D\theta.$$
$^{1}$ To convert right ascension given in hours, minutes and seconds and declination, given in degrees, minutes and seconds, to radians you do
$$\alpha = (RAh*15 + RAm/4 + RAs/240) \times \pi/180$$
$$\delta = (DEd \pm DEm/60 \pm DEs/3600) \times \pi/180,$$
where in the latter formula you use plus signs for objects in the northern hemisphere and minus signs for objects in the southern hemisphere.
$^2$ You see the galaxies on the plane of the sky; there is no way at present to get the deprojected 3-dimensional distance from the cluster centre.
| {
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We say Light is Red-Shifted or Blue-shifted from faraway stars and galaxies We say Light is Red-Shifted or Blue-shifted from faraway stars and galaxies. Can we find out the distance at which it changed its frequency.
So in another solar system, it might seem to be Green shifted or something similar. (Assuming that those beings can see only in our visible region)
And why do stars still appear white in the sky? If the rest of the universe is going away from us.
Is it only a misconception or misunderstanding. Please let me know.
| It is my understanding that virtually all galaxies are moving away from us, and so red shifted, except for a few galaxies (including dwarfs) in the local group, like Andromeda, which would be blue shifted. So, most galaxies are red shifted, and the further away they are, the more red shifted they are. So the the change in frequency is relative to the distance the galaxy is from us.
The current theory(?) is that no matter what solor system you are in, you would see the same effect, that most of the galaxies are moving away from you, and the further away they are, the faster they are moving away, and thus red shifted.
Stars emit all colors of light, so they appear to be white. Small stars emit much more red light, the sun emits more yellow light, and large stars emit more blue light. I could be wrong here, but I'm pretty sure it's spectroscopy that helps us determine a star's 'color,' not what we see with the naked eye.
https://en.wikipedia.org/wiki/Spectroscopy
https://www.youtube.com/watch?v=hI8gifMbeVU
https://www.youtube.com/watch?v=Dq7fmftneaY
the last two videos should answer most of your questions
| {
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What is difference between polarization and polarizability and how do we define it? The book of physics that I have, uses the word "polarization" sometimes and sometimes uses the term "polarizability" and I am getting confused. And I even checked the dictionary for the term "polarizability" and it is not even a word in the English Dictionary.
What might be the case here, spelling mistake or conceptual difference?
| In short, polarization (the noun) is the displacement of positive charges relative to negative charges in a system (i.e. an atom's nucleus vs its electrons). Polarizability refers to the difficulty with which such a displacement can be achieved.
As discussed in Griffiths Introduction to Electrodynamics 4E, we can imagine a hydrogen atom with an electron cloud surrounding the nucleus. If an electric field $\vec{E}$ is applied across the atom, the electron will be pulled against the field while the nucleus will be pushed along it in the opposite direction. The charges are thus separated and a vector $\vec{p}$ can be drawn from the positive to negative charges.
$$\vec{p}=q\vec{r},$$
where q is the electric charge and $\vec{r}$ runs from the positive to negative charges. $\vec{p}$ is known as the dipole moment and could be thought of as analogous to a moment of inertia-- a measure of the difficulty of torquing something.
$\vec{p}$ for the example of the H atom in an electric field, would be given by $\vec{p}=\alpha\vec{E}$, where $\alpha$ is polarizability-- the relationship between the strength of the field and the induced separation of the charges.
With the dipole moment defined, we can then speak of the polarization of the atom. The polarization is described by $\vec{p}$. For a dielectric material with many atoms, each with a $\vec{p}$ of their own, we could define the polarization of that material as the dipole moment per unit volume.
| {
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Why didn't the glass break? In the figure below, a needle has been placed in each end of a broomstick, the tips of the needles resting on the edges of filled wine glasses. The experimenter strikes the broomstick a swift and sturdy blow with a stout rod. The broomstick breaks and falls to the floor but the wine glasses remain in place and no wine is spilled. This impressive parlor stunt was popular at the end of the nineteenth century.
What is the physics behind it?
| Well, the fact is that the needle lies atop the edge of the glass, and aas a rod stikes at the middle of the broom, a vertical component of force, acts on the needle, mind you, there is no horizontal vector, the needle produces a normal force, so in the end, no force is actually transmitted to the glasses. However a the strike has to be quick and powerful lest shockwaves are produced, and the needles don't produce a sufficient normal force to break the rod.
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What is the trajectory of a photon moving through a vacuum? Since electromagnetic energy is carried by photons and moves in forms of waves, does it mean that a single photon when propagating through space doesn't follow the straight path but instead always moves up and down, up and down like a wave. If so another question arises the speed of propagation of light in vacuum is fixed meaning that it will always take the same amount of time for it to travel from point A to point B, but if a photon always moves up and down it will also mean that it travels longer distance than the distance between A and B and so it ill travel faster than light propagates, is it even possible, could you please clarify these concepts for me?
| Your confusion comes from combining two different concepts (although they are related). Photon is a discrete particle. A wave is a continuous. You can look at light as a discrete particle or a wave, but if you think of them the way you are thinking of them, things get confusing.
*
*A photon does not travel among the wave's amplitude function. It
travels in all possible paths and we observe one path at a time (And
usually if the path from A to B is simple, the many possible paths
cancel out).
*A photon always travels at speed c.
*The up and downs
of a electromagnetic wave are the consequences of a photon moving.
They are self inducing oscillations in the electromagnetic field.
*You can also think that the ups and downs of an electromagnetic wave can be represented as a photon, which is carrying information about a change in the electromagnetic field at speed c. (A change caused perhaps by moving a charged particle like an electron.)
By the way, if I didn't answer your question, try to think about why we started thinking of light as particles. (The problems that lead to the concepts you are studying). You can start at the photo-electric effect.
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How can I prove that for a Killing vector $\nabla^a \nabla_a \xi^\mu = -R^b_a \xi^a$? I'm taking a course on General Relativity and I'm trying to prove that for a Killing vector field $\xi^\mu$ the following equation holds:
$$\nabla^a \nabla_a \xi^\mu = -R^\mu_a \xi^a$$
Where $R_ab$ is the Ricci tensor. To prove this I thought of applying the operator $\nabla^a$ to the equation that $\xi$ satisfies due to being a Killing vector field. Then I get:
$$\nabla^a \nabla_a \xi_b = -\nabla^a \nabla_b \xi_a$$
And then I wanted to prove somehow that the RHS is very closely related to the expression that I want to obtain. A little problem is that in the LHS appears $\xi_a$ instead of $\xi^a$ so I would have to raise the index using the metric tensor.
However, I haven't been able to prove this result because I end up with a lot of terms involving different covariant derivatives. I've also tried using Ricci's identity and even the definition of the Riemann tensor but to no avail thus far.
How should I proceed to prove this result?
P.S. I wasn't sure if I should post this question here or in Math SE if it doesn't fit here please flag it to move it to Math SE.
| Let us look at $-g^{ac} \nabla_a \nabla_b \xi_c$. Because the part symmetric in $(bc)$ vanishes, we have $$-g^{ac} \nabla_a \nabla_b \xi_c = -\frac{1}{2} g^{ac} (\nabla_a \nabla_b \xi_c - \nabla_a \nabla_c \xi_b).$$
Now by the definition of the Riemann tensor, $\nabla_a\nabla_b \xi_c = \nabla_b\nabla_a \xi_c + R_{abc}{}^\mu \xi_\mu$, so $$-g^{ac} \nabla_a \nabla_b \xi_c = -\frac{1}{2} g^{ac} (\nabla_b \nabla_a \xi_c - \nabla_c \nabla_a \xi_b + R_{abc}{}^\mu\xi_\mu) .$$ In this contraction, the first term vanishes by Killing's equation, the second is $\frac{1}{2}\nabla^a \nabla_a \xi_b$ and the third is $-\frac{1}{2}R_b^\mu \xi_\mu.$ The result follows.
| {
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Is there any physical interpretation for $\nabla\cdot(\nabla \times F)=0$? It is well known that the divergence of the curl is always 0. Mathematically I understand why this happens ($d^2=0$ where $d$ is the exterior derivative) but today I was wondering what is the physical meaning of this.
The divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point (from Wikipedia) and the curl describes the infinitesimal rotation of a 3-dimensional vector field (also from Wikipedia).
Does this mean that the rotation of a vector field is always stable and doesn't go inwards or outwards? What is the physical meaning of the divergence of the curl equals 0?
| First I should say that it is a mathematical definition and does not carry any physical meaning necessarily. But as an example it relates the Maxwell equations:
$$\nabla\times E=-\frac{\partial B}{\partial t}$$
and
$$\nabla \cdot B=0$$
$$\Longrightarrow \nabla\cdot(\nabla\times E)=-\frac{\partial \nabla\cdot B}{\partial t}=0$$
and the continuity equation can be deduced from
$$\nabla\times H=J+\frac{\partial D}{\partial t}$$
and
$$\nabla \cdot D=\rho$$$$\Longrightarrow \nabla\cdot(\nabla\times H)=\nabla\cdot J+\frac{\partial \nabla\cdot D}{\partial t}=0$$
which is the continuity equation:
$$\nabla\cdot J=-\frac{\partial \rho}{\partial t}$$
If one day in future a magnetic monopole is discovered in reality, as some people are interested in it for some reasons, then $\nabla \cdot B$ is not zero anymore, but, mathematically $\nabla\cdot(\nabla\times E)$ has to be zero.
This means that a new entity has to be defined instead of $B$ in order to hold the $\nabla\cdot(\nabla\times E)=0$. This is how math can help physics.
| {
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How exactly can a speaker produce the huge number of frequencies in musi at he same time? I know the basics of how a speaker works, but when I think about the fact that just one sound from an instrument has many frequencies happening at once and there are other sounds such as the guitar pick, echo in he chambers, etc. AND there are many instruments and there many be delay and Echo effects added as well. So it seems impossible rocks that just a basic thing moving back and forth can be at a given moment moving back and forth at all those frequencies at the same time.
| Adding on to the above answer, let's think of how the speaker's motion would then be generated. If a waveform has a time varying amplitude x(t), a voltage $V$ is applied so that at any time t, the integral of that voltage's action on the speaker membrane yields an amplitude of x(t). Thus, the desired waveform is realized.
A waveform may correspond to the linear combination of various frequencies, but it will always be a smooth curve. This means that we can have "complicated" sounds being played by a simple mechanism.
| {
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I do not understand this comparison? It is frequently discussed that to find solutions having some sort of supersymmetry is easier than solving Einstein’s equations of motion.
I do not understand this discussion though. Specifically, why are Einstein's equations difficult to solve? Is it because they are second order equations and the supersymmetric ones are first order? Another thing that makes me not understand this is the fact that Einstein's equations are not supersymmetric while supersymmetric ones obviously are supersymmetric, so how can we compare two equations from two different theories?
| The problem with the Einstein equations is that they are non-linear partial differential equations and, as you know, there are no general algorithms to solve them. Sometimes you can impose some symmetries - make an ansatz (guess) such that the Einstein equations becomes ODE and then solve them (that is how the Schwarzschild solution was found).
On the other hand, if you have a supersymmetric background, then, besides the Einstein equations, it also satisfies the Killing spinor equations (KSE), which are much easier to solve. Different theories have different KSE, but they are always simpler than Einstein's. So usually people start with Killing spinor equations, solve them to obtain some restrictions on the metric/fluxes and then check the Einstein equations.
| {
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Why do we only feel the centrifugal force? After spending some time researching about the centrifugal force, I now understand that it is needed in a non-inertial reference frame for Newton's Laws to hold true. However, I don't understand why we only feel the centrifugal force when moving in a circular path. For example, if you imagine yourself being spun around in a circle, then in your frame of reference you would feel yourself being pushed outwards. But since in your frame of reference you are stationary, the outwards force must be balanced by an inward pull (for you to remain stationary). So why would you only feel the outwards force but not the inwards force?
| Suppose you have your feet pointing outwards and your head pointing towards the centre of rotation:
Now ask your self what direction the force you feel is pointing. Well, the force is pushing upwards on your feet, so the force points from your feet towards your head. In other words the force you feel is pointing inwards not outwards i.e. it is the centripetal force.
| {
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Liouville's theorem for systems with dissipation described by a single hamiltonian Following this link, one can treat dissipation by using a factor $e^{\frac{t \beta}{ m}}$ in addition to the Lagrangian $L_0$ of a system without disspation:
$$
L[q, \dot{q}, t] = e^{\frac{t \beta}{ m}} L_0[q, \dot{q}] \, ,
\quad \text{where}\quad
L_0[q, \dot{q}] = \frac{m}{2}\dot{q}^2 - V[x]\, .
$$
The equation of motion associated to $L[q, \dot{q}, t]$ is:
$$
e^{\frac{t \beta }{ m}} \frac{d}{d t} \frac{ \partial L_0}{\partial \dot{q}} - e^{\frac{t \beta}{ m}} \frac{ \partial L_0}{\partial q} = e^{\frac{t \beta}{ m}} \frac{\beta}{m} \frac{ \partial L_0}{\partial \dot{q}}
\qquad \Rightarrow \qquad
\frac{d}{d t} \frac{ \partial L_0}{\partial \dot{q}} - \frac{ \partial L_0}{\partial q} = \frac{\beta}{2} \dot{q}^2
$$
This equation describes the motion of a particle in a potential under the effect of dissipation forces proportional to its velocity.
Question: If I use Hamiltonian mechanics, this system is described by a single Hamiltonian and the canonical equations of motion. That means that Liouville's theorem holds, and therefore the state density would be constant along the physical trajectories of the system in the phase-space. I can't imagine that this is possible for a system with dissipation, where I would expect the phase-space volume to shrink and the density to grow.
Consider for example a damped harmonic oscillator: The trajectories would be circles with exponentially decreasing radius in the phase space, so the phase-space-volume decreases.
Where is my mistake? I thought the only requirement for Liouville's theorem is that the system is described only by the Hamiltonian. So why does it hold where it clearly should not?
| In my opinion, we should define the Lagrangian only when the force is conservative. Eventhough the force is not conservative, we can write down the Lagrangian which is mathmatically correct. However, the Lagrangian and the Hamiltonian do not contain correct nature inside it. Only mathmatically, those work. So, the divergence is still zero, but this does not means that the density in the phase is constant.
| {
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LCD's working principle
In the image's information it says ''Liquid crystal device windows are in the group of electrically activated smart windows in which normal "off" condition of the glazing is a translucent milky white. When an electric current is applied, it turns clear.''
Shouldn't it be the reverse?
| The diagram you reproduce shows that the liquid crystal elements align in the presence of a potential difference - and when they do, they no longer scatter the light. This makes the window "clear".
Note that this is different than the situation in a monochrome LCD, where the liquid crystal elements are placed between crossed polarizers which have a surface texture that tends to align the crystals (weakly). This construction means that light, after passing through the first polarizer, is slowly rotated until its polarization is aligned with the second polarizer. This means the default state is "transmission"; and adding an electric field disturbs the crystal-to-crystal alignment and makes the element turn dark.
Whether the default state is "aligned" or "not aligned" depends on the structure of the molecules of the liquid crystal - but one could argue that a material whose default structure is "not aligned" doesn't really deserve the moniker "liquid crystal"...
I would argue that what you have here is a suspended particle device or a polymer dispersed liquid crystal device - this is different than a conventional LCD device. But for such devices the "on" state is the transparent one, as correctly stated in the image.
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Motion of an object in rotating frame Yesterday I was looking at an old sloan video that describes motion in inertial and non-inertial frame. An experiment was actually like this. Two persons are sitting on the opposite side of a table fixed to a turning platform. The platform is rotating in uniform circular motion. Now Guy1 pushes a ball over the frictionless surface of the table in a straight line towards Guy2. The question what will be the motion of the ball from a viewer inside the rotating frame and to someone outside in fixed frame of reference. I got little confused as to how to conceive the fictitious force. What will the motion be? In general, how to derive equation of motion in non-inertial frame. Please also add some good references on intuitive understanding of these type of problems.
| The motion for viewer inside rotating frame will be a straight line while for inertial frame observer or outside person it will be circular as motion is relative to the frame.
| {
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Why does moving air have low pressure? According to Wikipedia lift in an aircraft is due to an area of low pressure formed above the wings of an aircraft due to the fast moving air there. So why exactly is an area of low pressure created due to fast moving air?
| Moving air can have any pressure. However if air is moving away from a surface without enough new air flowing in to replace it, then the pressure at that surface drops. For a wing with lift the cause is inertia / momentum conservation. Since the air collides at higher rate with other parts of the wing there is a net force, hopefully up.
| {
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Is spin-orbit coupling really necessary for topological insulators I have heard that for an insulator to be non-trivial, large spin-orbit coupling is necessary.
However, I have read the definition of $Z_2$ topological invariant and chern number. In no way can I recognize what role spin-orbit coupling plays in topological transition. The only relationship is that spin-orbit coupling can drive a band inversion, which I suppose some other factors may also lead to.
So, is spin-orbit coupling really necessary for topological insulators?
| I guess no. Su-Schrieffer-Heeger (SSH) chain is one of the simplest topological insulators and has no spin-orbit coupling.
| {
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Is causality a total order? I've read that it is physically not possible to violate causality defined as a total order on the spacetime graph. So I was wondering if at least causality can be broked down to a partial order and if phenomenon in the universe exist that induce such an order.
Meaning if we have a set of events $E=\{A,B\}$ and an existing total order $A<B$, is it possible to create $A\leq B$, such that our successor function $f:\mathbb{R}^2 \rightarrow\mathbb{R}^2$ that maps an event in time to another event in the future $\{e_0,t_0\}\mapsto\{e_1,t_1\}$ could map an event to another event without changing the time variable $f(e_0,t_0)\mapsto\{e_1,t_0\}$. So that the event $A$ that triggers $B$ either happend before or simultaneous with $B$. I don't expect that $f(B,t_0)\mapsto\{A,t_1\}$ is possible.
I hope my question is understandable.
| Partial orders are permitted in Causal Set Theory (Raphael Sorkin) and it does make predictions, which relate to an information universe at least, and these might be regarded as physical. That said, at present such predictions are rather limited. Separately, violation of Bell's inequality, in which the result of an experiment in one part of the universe immediately affects (in a highly confined circumstance) the results in other areas of the universe, predicted by quantum mechanics but not permitted by special relativity, suggests that a partial ordering might relate. There is also some consideration of closed timelike curves (see for example 't Hooft's discussion in "A Locally Finite Model for Gravity." of Gott, J.R.: Phys. Rev. Lett.66, 1126 (1991) and Ori, A.: Phys. Rev. D 44, R2214 (1991) which suggest the possibility of backwards causation (though this might have been sorted since he wrote that paper in 2008). So, if I am understanding your question correctly, partial order is a possibility as far as competing theories now stand.
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How the LHC bump can be a mere coincidence? Speaking of http://www.nature.com/news/lhc-sees-hint-of-boson-heavier-than-higgs-1.19036.
I understand that such a bump can be a statistical fluctuation.
What troubles me is that the bump has been seen in two completely disconnected and independent experiments. How can this happen by chance? Is there some systematic error that has been overlooked by both of them? Is there any common mistake in background subtraction?
| I rolled two sixes in a row when I sat down once and rolled a six sided die five times. Same thing happened to my friend once! Instead of asking how that can be, you can just ask about the chance of that happening.
When the chance gets too small you are starting to get good evidence that something is going on. When the chance is too high, you have to accept that these things literally happen every so often.
| {
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Orbifold actions and twist operators A twist operator $\sigma$ is the operator that acts on the untwisted vacuum $|0\rangle$ to create a twisted vacuum $\sigma|0\rangle$. States belonging to the twisted sector of an orbifold are built on such twisted vacua (which are also in 1-1 correspondence with the fixed points of the orbifold action).
My question is given a specific twist operator $\sigma$, how can I possibly deduce the orbifold action that corresponds to it?
More specifically, let
$$\sigma=\psi^1+i\psi^2$$
the operator that arises after bosonizing two real fermions. If the boundary conditions for the fermions are known, let's say $\psi^1\rightarrow \psi^1$ and $\psi^2\rightarrow -\psi^2$ then we have some information on $\sigma$. We should be able to interpret this $\sigma$ as the twist operator that generates a twisted sector of some orbifold, but what would the orbifold action in this case be?
| For a toroidal orbifold, suppose you know the boundary condition in the k-twisted sector and the spin structure $\alpha, \beta \in \lbrace 0,1/2\rbrace $. In light cone gauge you have:
$\psi^{j}(\sigma^0, \sigma^1+2\pi)=- e^{2\pi i \alpha }e^{2\pi i k v_j }\psi^{j}(\sigma^0, \sigma^1)$
where:
$v_j=a_j/N$
in which $a_j$ are some integer, fixed by the possible crystallographic actions. From this boundary condition you can get N of the $Z_N$ orbifold.
Check "Basics concepts of String Theory", (Lust, Theisen, Bluemenhagen), pag.306
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Is polarization a stable state? Do a polarized light beam stays polarized over large distance or does it kind of relax and eventually become unpolarized?
| It depends on the material the light beam is traveling through.
If the light is traveling through some media - vacuum, glass, etc. - then its polarization state will remain the same. However, if it is traveling through certain birefringent materials, then its polarization state may change.
Mathematically, this difference arises from the description of the permittivity of the medium. Generally - in isotropic1 materials - this is a scalar quantity. However, in materials exhibiting birefringence - anisotropic materials - it is a tensor quantity. Thus, there is a more complex relationship between the electric field and the electric displacement field, leading to a change in polarization of the electromagnetic field.
1 Isotropy refers to uniformity in all directions; anisotropy refers to some lack of uniformity in a given direction.
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Physics after a Theory of Everything There is a lot of controversy over the existence of a Theory of Everything (ToE), and as far as I know, we are a long way from having a possible candidate. But what interests me is, what after that? If a Theory of Everything is truly found, will there be anything left for physics to do? Is a ToE actually the end of Physics? Or are there things that will remain unexplained even after a ToE is found out?
Edit (7/1/2016):
Just to make it fit for this site, my question is What are the areas of physics that would require working on, even if a ToE is discovered?
I hope I have been able to "isolate an issue that can be answered in a few paragraphs".
| It's a question of definition what really is a physics, where the engineering and applied sciences begin etc. In this point of view:
$A)$ There is really lot of uncertainties and blank spaces when it comes to human senses and cognitive sciences and their connections to "objective" physics.
$B)$ There is much of cases that is not "staring at the face of creation" but are still unsolved (or not fully solved). The most significant cases (in my humble opinion) are solutions of Navier-Stokes equations. And there are plenty more of this topics waiting for new formulations (as e.g. Lighthill did for vortex sound generation in case of Navier-Stokes equations).
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How does calblock (water softener) work? This "calblock" device claims to prevent limescale accumulation on water heaters in a washing machine. It sits between the water faucet and the washing machine; the water passes through it.
I cannot imagine how it could work. This video, made by a supplier, describes its function this way:
Thanks to 2 magnets, at a low temperature, molecules of calcium bicarbonate are splitted in aragonite, smaller molecules, that remains sospended, is not deposited and is discharged with the waste water
which is not satisfying at all.
How do the magnets convert calcium bicarbonate to aragonite? I cannot imagine how can a chemical transformation be caused by magnets. Also, why 2 magnets? How are they arranged?
|
How do the magnets convert calcium bicarbonate to aragonite?
They can't.
I cannot imagine how can a chemical transformation be caused by magnets. Also, why 2 magnets?
If one magnet would do it, then wouldn't two be even better?
How are they arranged?
The magnets are arranged in such a way to separate your money from your wallet. They won't do anything to the water.
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Effect of Particle Mass on Thermal Conductivity Fouriers law of thermal conductivity is
$$ \vec{q} = -k\nabla T $$
where $q$ is the heat flux, $k$ is the thermal conductivity.
Mass does not seem to appear in the equation. So I'm wondering what if keeping all else constant, but if we change the particle mass (i.e. increase it) in a solid, how would $k$ change?
It would be nice if someone could refer me to a book/article too.
| The thermal conductivity contains the cross-section for particle collisions.
The cross-section itself contains the mass of the particles that are carrying the thermal energy (as well as the mass of the "thing" the particle is colliding with). Sometimes one of these masses can be ignored.
The exact dependence of the cross-section on the mass of the particles depends on the regime you are considering (density, temperature, type of particles involved). But it's in there.
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Conservation of angular momentum while sitting on a spinning chair Today my friend was sitting on a spinning char. By moving his top part of the body left to right and his bottom part of the body the opposite he managed to spin. As I understand Conservation of angular momentum if he was just sitting still he should not be able to spin. He has not touched anything (wall, table). The question is how has he changed his angular momentum (and the angular momentum of the chair) just by moving his body while sitting on the chair.
| You do not say, but I assume his feet were in the air and not on the floor which is why you are wondering about conservation of angular momentum.
I suspect the spin axis of the chair was not exactly vertical and the plane of all possible positions was tilted from the horizontal, one side of this plane is actually lower and the other is actually higher. So, when he shifted his body rearranging it side wise and changed the position of his body he was actually moving the heaviest portion of his body from the lower side to the higher side. In doing so he raised his center of gravity and did work. As the heavier side of his body was now on the higher side of the plane of spin, the chair rotated a half turn returning the heavier side of his body to the lower position on the plane of possible positions. He was sliding down hill on a tilted circular track. The chair/human system was returning to its lower energy state.
Depending on friction etc his position may have over shot and oscillated to some extent, but after time it will settle with the center of gravity in the lowest permitted position. Once the CG is at the lowest permitted position, the system is in the lowest possible energy state and will remain there.
If he continues to shift his position with each half turn of rotation, he can continue to add energy to the system and repeatedly move it from its lowest energy position causing a restoring force/torque to be generated that will continue to rotate the chair/person system.
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Biot-Savart law from Ampère's with multivariate calculus Let us assume the validity of Ampère's circuital law $$\oint_{\gamma}\mathbf{B}\cdot d\mathbf{x}=\mu_0 I_{\text{linked}}$$where $\mathbf{B}$ is the magnetic field, $\gamma$ a closed path linking the current of intensity $I_{\text{linked}}$.
Can the Biot-Savart law $$\mathbf{B}=\frac{\mu_0}{4\pi}\oint\frac{Id\boldsymbol{\ell}\times\hat{\mathbf{r}}}{r^2}=\frac{\mu_0}{4\pi}\int_a^b I\boldsymbol{\ell}'(t)\times\frac{\mathbf{x}-\boldsymbol{\ell}(t)}{\|\mathbf{x}-\boldsymbol{\ell}(t)\|^3}dt$$where $\boldsymbol{\ell}:[a,b]\to\mathbb{R}^3$ is a parametrisation of a closed (or infinite) wire carrying the current $I$, be inferred without using Dirac's $\delta$, by using the tools of multivariate calculus and elementary differential geometry only, at least if we assume the validity of the Gauss law for magnetism or other of the Maxwell equations? All the proofs I have found (such as this, where, as far as I understand, $$\nabla^2\left[\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d^3r'\right]=-\mu_0\mathbf{J}(\mathbf{r}) $$ is derived by using the $\delta$) use the expression
$$\mathbf{B}=\frac{\mu_0}{4\pi}\int_V\frac{\mathbf{J}\times\hat{\mathbf{r}}}{r^2}d^3x$$ and Dirac's $\delta$, but I wonder whether, both assuming a linear current distribution as when we use the expression of the magnetic field as$$
\mathbf{B}=\frac{\mu_0}{4\pi}\oint\frac{Id\boldsymbol{\ell}\times\hat{\mathbf{r}}}{r^2}$$ and assuming a tridimensional spatial current distribution, it is possible to prove the Biot-Savart law from Ampère's without the use of the $\delta$. I heartily thank any answerer.
| I really like the proof contained in the paper Derivation of the Biot-Savart Law from Ampere's Law Using the Displacement Current from Robert Buschauer (2013)
It's simple and it fulfills the role of convincing the reader.
Basically the author works with one point charge $q$ situated in origin of Z azis $(0,0,0)$. He supposes a particle moving in Z axis to positive Z values with velocity $v$. He creates a magnetic field line in a arbitrarious circle with $c$ radius, by symmetry, with center in $(0,0,a)$. The angle between any point in the circle and the center of circle starting from origin $(0,0,0)$ is $\alpha$.
Starting point is a part of 4th Maxwell's Equation of electromagnetism, the Ampere-Maxwell Law that consider changing electric flow with time in a area produces magnetic field circulation. This law generates a magnetic force that can be verified using special relativity that in another reference frame it's just a plain electric force.
$$\oint B\, dl = \mu_0\epsilon_0 \; d/dt(\int_A E.dA)$$
In the left side, the solution consists of integrating the $\oint B dl$ in this circle (butterfly net ring). As $B$ is constant by symmetry, we have
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \oint B\, dl = 2\pi c B \qquad\qquad$ (1)
In the right side $\;[\;\mu_0 \epsilon_0 d/dt (\int_A E\; dA \,)\;],\;$ as the surface (butterfly net) we choose a sphere of radius $r$, to ensure that all points have the same value of electric field:
$$ E = q / 4\pi\epsilon_0r^2$$
Let's first calculate the right-hand integral in the right side. We adopted here a slightly different standard in spherical coordinate. Just to remember,the element for integration into spherical coordinates is $\; r^2 \sin \phi \, dr \, d\phi \, dq $
Let $\theta$ (XY axis) vary from $0$ to $2\pi$ and by consider the angle $\phi$ with the vertical (Z axis) from 0 to $\alpha$.
$$\Phi_E = \int_A E\; dA = q/4\pi \epsilon_0 r^2 \int_A dA = q/(4\pi \epsilon_0 r^2) r^2 \int_{0,2\pi} d\theta \int_{0,\alpha} \sin \Phi\; d\Phi = $$
$$q/4\pi \epsilon_0 2\pi ( -\cos \alpha + 1) = q/2\epsilon_0 (1 - cos\alpha)$$
Thus
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Phi_E = \mu_0 q /2 (1 - cos\alpha)$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad d \Phi_E / dt = - q/2\epsilon_0 d \cos \alpha/dt\qquad$(2)
Putting $\alpha$ as a function of $z$, we have, by the chain rule:
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad d \cos \alpha/dt = (d \cos\alpha/dz) \; (dz/dt)\qquad$(3)
However as $z$ is decreasing with the motion at velocity $v$, we have
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad dz / dt = -v \qquad $(4)
On the other hand:
$$ \cos \alpha = z / r = z / \sqrt{c^2 + z^2}$$
Using this online tool for derivation:
$d \cos \alpha/dz = c^2/r^3$ where $r = \sqrt{c^2 + z^2}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad 2\pi c B = q \mu_0 /2 v (c^2/r^3)\qquad$ By (1),(2),(3),(4)
$$B = \mu_0 q v c / 4\pi c r^3$$
but $\quad\sin \alpha = c / r\quad$ so we can add $\quad \sin \alpha\; r / c$:
$$B = \mu_0 q v \sin \alpha /4\pi r^2 $$
Vectorizing we have a cross product:
$$B = \mu_0 q \; v\uparrow \times r\uparrow /4\pi r^3$$
In some infinitesimal point we can consider a element of electric current as a point charge, so we can add other charge points by integration (any force is addictive!) for using in real applications. Thus we have in scalar notation:
$$dB = \mu_0 dq \; v \; r \sin \alpha /4\pi r^2$$
Considering $\quad dq = i\;dt\quad$ and $\quad v = ds/dt\quad $, we finally have reached to Biot-Savart law:
$$dB = \mu_0 i \; ds \; r \sin \alpha /4\pi r^2$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
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} |
Missile-like trajectory calculation First of all I want to let you know that I'm not a Physicist, I am a Video Game Developer. I can simulate physical and mathematical equations and can also use some built in physics.
For example I can move an object through parametric equation of projectile
$$x = vt\cos\theta$$
$$y = vt\sin{\theta} - \frac{1}{2}gt^2$$
I can get values by increasing time, $t = 0$,$ t = 1$, . . . So these values make a good projectile.
On the other hand if I use built in Physics of Game Engine then I have to only Apply some force with a 2d or 3d vector. For example I can apply force of $\vec f = 24,15,6$ with other things in hand and I can change it like mass, gravity.
So now my question is that how can I calculate $\vec f (second method)$ or $data$ to populate in first method that the body will land on a specific predefined point.
*
*I'm repeating the objective is to launch a projectile such that it will land on a specific predefined point?
| One of the pieces that you need is called the range equation (assuming that you are firing from the ground):
$$
R=\frac{v^2_0\sin(2\alpha)}{g}\tag{1}
$$
So if you start with knowing $v_0,\,\alpha$, you can figure out how far it will go. But you are starting with $R$ and trying to see what pairs of $v_0,\,\alpha$ would match this, which would be quite difficult as many pairs could lead to the same $R$.
However, you say that you are applying a force for some duration, so you can use the impulse to determine the (final) velocity:
$$
\int\mathbf F\,dt=\Delta\mathbf p=m\mathbf v_{fin}-m\mathbf v_{init}
$$
Since the projectile is initially at rest, the launch velocity would be
$$
\mathbf v_{launch}=\frac{1}{m}\int\mathbf F\,dt
$$
which, if the force is constant throughout, can be simplified to
$$
\mathbf v_{launch}=\frac{1}{m}\mathbf F\Delta t\tag{2}
$$
where $\Delta t$ is the length of time the force is applied.
So now you know the initial velocity, $v_0=\lVert\mathbf v\rVert$, and how far you want the projectile to go, $R$, so then you can solve Equation (1) to get $\alpha$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Why do the conserved charges in the case of SSB of a global symmetry not exist? Reading "From Linear SUSY to Constrained Superfields" by Komargodski and Seiberg, I got a bit confused regarding the existence of the conserved charges in a theory with spontaneous symmetry breaking (SSB) of a global symmetry:
More precisely, in the second-to-last paragraph on page 1 we have
"When a global symmetry is spontaneously broken, the corresponding conserved charge does not exist because its correlation functions are IR divergent. However the conserved current and even the commutators with the conserved charge do exist."
I know that in the case of global SSB we have $Q|0\rangle\neq0$ for the conserved charge $Q$. However, I don't have any insight about the correlation functions. Could somehow $Q|0\rangle\neq0$ imply something like $||Q|0\rangle||=\infty$ or $\langle Q\rangle\rightarrow\infty$? And how could one see that?
| This is called the Fabri-Picasso theorem. Their argument requires both the vacuum and the charge $Q$ to be translationally invariant: $P |0\rangle = 0$, and $[P, Q] = 0$.
The argument goes as follows: Since the charge originates from a symmetry, then according to Noether's theorem:
$$Q = \int d^3x j_0(x)$$
Consider the correlation function of the charge with itself:
\begin{align}
\langle 0| QQ |0\rangle& = \int d^3x \langle0|j_0(x) Q|0\rangle \\
& =\int d^3x \langle0|e^{iPx} j_0(0) e^{-iPx} Q |0\rangle \\
& =\int d^3x\langle0| e^{iPx} j_0(0) e^{-iPx} Q e^{iPx} e^{-iPx}|0\rangle\\
& =\int d^3x \langle0| j_0(0) Q |0\rangle
\end{align}
The integrand in the r.h.s. does not depend on the position, therefore its value is proportional to the total space volume, thus
$$||Q|0\rangle||^2 = \infty$$
Thus the operator $Q$ does not exist in the Hilbert space unless $Q|0\rangle = 0$.
However, the commutators of $Q$ with the fields for example exist because by Noether's theorem, they generate the symmetry, in other words the right hand sides of:
$$[Q, \phi] = \delta \phi$$
exist, since they are the symmetry transformed fields.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What does singularity mean in the context of black holes? Non-rotating big stars can be subject to a gravitational collapse increasing their density. When the density is so high that the mass volume shrinks below the event horizon, a black hole is formed.
Is such an "initial" black hole already considered as a singularity, or are there further shrinking processes required in order to consider the black hole to be a singularity? Do the characteristic dynamics of black holes include further shrinking processes until the mass is reduced to a point (or nearly)?
| A singularity is defined as a point of infinite density and they are believed to reside at the center of a black hole .
The "when does the singularity form" question, is answered concisely here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228586",
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"question_score": "1",
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What does the Dirac delta function physically do while deriving Gauss Law form Coulomb's law? While doing this derivation, the the source coordinates are mentioned as "$s$" and the coordinate of the point at which field is to be calculated is mentioned as "$r$". Kindly follow this Wikipedia link and click on the "Outline proof" under "Derivation of Gauss Law from Coulomb's law".
Finally it comes out that $$\nabla\cdot E(r)= \frac{\rho(r)}{\epsilon_0}. $$But $\rho$ is actually defined for the "$s$" coordinates and $\rho(r)$, where $r$ is the point at which electric field is calculated is 0. Here I can not understand how the $\nabla\cdot E(r)$ is equal to $\frac{\rho(r)}{\epsilon_0}$.The information about $\rho(s)$ is totally lost in the final equation. What does the Dirac delta function actually do?
| I think I have found the answer.
Let us consider the case of a uniformly charged sphere of radius $R$ and charge density $\rho$. The field inside this sphere is
$E_{in}=\frac{\rho\times r}{3\epsilon_0}$.
Here $r$ is the distance from the centre and $r < R$.
If we calculate the divergence of $E_{in}$ then
$$\nabla.E(r)=\frac{\rho}{\epsilon_0} $$
Kindly note that $\rho$ is actually $\rho(r)$ which is constant for r < R.
The electric field $E_{out}=\frac{\rho\times R^3}{3\times\epsilon_0\times r^2}$ for $r>R$.
Calculating the divergence of this field we get
$\nabla.E(r)=0$, Kindly note that for this point $(r > R)$ $\rho(r)=0$.
This is the reason why I disagree with the interpretation $\rho(r)$ by Prof Shonku
It proves that $\nabla.E(r)=\frac{\rho(r)}{\epsilon_0}$ where $\rho(r)$ is the charge density exactly at the point where the field $E(r)$ is measured. If $r$ is such that the point is inside an extended charge distribution then $\nabla.E(r)$ is non zero. If $r$ is such that the point is outside an extended charge distribution, then $\nabla.E(r)$ is zero. Thanks
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Isn't LIGO basically measuring the luminiferous aether? I am bit confused about this one. I am not very acknowledgeable about gravitational waves and LIGO. But if it is basically a Michelson interferometer and can detect shifts in vacuum, doesn't this means that we detected the luminiferous aether and if not, why? Is there an analogy or direct correlation between gravitational waves and aether?
| You are right, spacetime is the luminiferous aether (that kind of aether that Einstein abhorred) and gravitational waves are ripples in that aether. But, the Michelson-Morley experiment showed us in 1887 that there is no luminiferous aether, therefore gravitational waves do not exist or they cannot be detected by means of a LIGO interferometer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
} |
What size particle would be attracted to this 340 ton sculpture's gravitational field? I know that gravity is a weak force, but this rock has mass, so it has gravity. The Levitated Mass sculpture is big enough to attract something. What is the largest particle that would "stick" to the underbelly of this sculpture because of its gravitational attraction?
I am interested in this sculpture because you can walk right under it! If the particle is big enough to see in a microscope, seems like this would be a great experiment for the youngins.
| This experiment is bound to fail, because a particle underneath the large mass will still be feeling the force of mg of the earth. The attraction of the large mass will have an upward force of (m_1xm_2/r^2)xG .
From the value of the gravitational constant one sees that the equivalent "g" for the attraction of the levitating mass is very small for 340 tons to overcome the force of earth gravity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
visible light spectrum Why do we see black objects? Colors of objects are formed when the spectrum of that color is reflected. Example Green objects are green because they reflect the green spectrum of light, red objects are red because they reflect red spectrum of the visible light and white objects because they reflect all the visible spectrum of light. When an object absorbs all the visible spectrum of light, it becomes black. But if nothing is reflected from the object, shouldn't it be invisible instead of black?
| I will address the "see" part.
Why do we see black objects? Colors of objects are formed when the spectrum of that color is reflected. Example Green objects are green because they reflect the green spectrum of light, red objects are red because they reflect red spectrum of the visible light and white objects because they reflect all the visible spectrum of light.
This is only partially true. We "see" the spectrum colors because the retina of the eye responds to the frequency. But at the same time the retina and the brain have a system that builds up colors from combinations of frequencies, even two frequencies are enough to produce color perception in the brain.
Thus one can have the black of the other answers, no reflected frequencies, but one can create a black paint that gives the same perception from the three primary colors. The eye perceives "black" from the confluence of the reflected frequencies.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Can the Heisenberg Uncertainty Principle be explained intuitively? I have heard several pseudoscientific explanations about the Heisenberg Uncertainty Principle and find them hard to believe.
As a mathematician mainly focusing on functional analysis, I have a considerable interest in this matter. Although I still have no idea about it, I aim to truly understand it one day. For now, the question is:
Can the Heisenberg Uncertainty Principle be explained so that a non-scientist can at least get a correct idea of what it says and what not?
One of the most heard explanations is the one that says naïvely that while trying to determine the position of a particle you send energy (light) so you modify its speed, therefore making it impossible to determine the speed. If I understand well this should only be a some sort of metaphor to explain it, right?
| Imagine that the information that describes position and momentum is digital and of limited precision. There is a constant total precision for both of them, but you can slice it up differently. If you dedicate more bits to momentum, you get fewer bits for position and vice versa.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "115",
"answer_count": 18,
"answer_id": 11
} |
How are orbits made stable? I understand the concept of object staying in some orbit due to centrifugal force and gravity.
However I do not understand how is orbit of a body like satellite or planet has perfect balance between gravitational pull and centrifugal force of revolution?
because if the angular velocity is even a little bit more than required then the object will move away and gravitational pull will reduce with square of distance and so centripetal force will decrease leading the object to stray further.
so is everything in orbit is either slowly moving away or falling in?
or is it that only those bodies whose orbital velocity is just perfect stays in orbit with everything else spiraling in or out of it?
so are satellites put with very accurately calculated velocity to prevent them from falling in or moving away or somehow it balances itself?
| The reason why planets in our solar system have stable orbits is because, during the formation of the solar system, debris disk which consisted mostly of gas was orbiting the sun, During this period when protoplanets started forming they were interacting with this debris disk, due to this interactions(frictional forces) on the planets from the debris disk, Planets achieved a more or less a circular orbit. Afterwards our solar system continued to evolve and This debris disk disappeared(Asteroids, Comets were formed), From this point planets were fixed at an orbit they achieved.
The image bellow represents a star system with one planet orbiting also with debris disk.
how is orbit of a body like satellite or planet has perfect balance between gravitational pull and centrifugal force of revolution?
No it doesn't have a perfect balance. If that were the case all the orbits would have been perfectly circular which is not he case. at a given distance $r$ from the earths there is a specific orbital velocity at which an object has a circular orbit, If an object at that orbit accelerated to a larger velocity gravitational pull wouldn't increase just to keep that object in a circular path, gravitational pull stays the same Therefore that object will have an elliptical orbit. To explain this mathematically, For an object to be in a circular orbit it must have centripetal force equal to gravitational force:
$$\frac{mv^2}{r} = G\frac{Mm}{r^2} \Rightarrow v^2=G\frac{GM}{r}$$
$$v_{orbital}=\sqrt{\frac{GM}{r}}$$
From the equation we can see that if we increase the orbital speed the above equality doesn't hold between centripetal acceleration and gravitational attraction Thus it no longer has a circular orbit.
From the picture bellow we can see in red that if orbital velocity equals the above equation we calculated then it has a circular orbit, If it is less than or more it achieves an elliptical orbit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Kinetic energy of a rotating object in an exercise, a linear molecule is being subject to a force applied on the edge in its axis. Then $K_1=\frac{1}{2}mv^2$, all is well.
Then in the second point of the exercise, the force is applied on the same edge but in an orthogonal direction to its axis. Then the molecule begins to rotate. So its kinetic energy is composed of two terms: $K_2=\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$,$\omega$ being the angular velocity of the rotation. The thing is that the linear velocity is the same as before and the correction says that $K_2=K_1+\frac{1}{2}I\omega^2$. But how come the same force can give two different energies to the molecule? I thought that $v$ would decrease in the second case, because of the apparition of the angular velocity $\omega$ so that the energy would be conserved. So in the second case the molecule goes as fast as before but in addition it rotates on itself?
| The mistake in your reasoning is to assume that the same force does the same work. That is simply not true - as the force operates over a different distance.
To analyze the problem, you have to think in terms of impulse ($F\cdot \Delta t$) or work done ($F\cdot \Delta x$).
Let's assume that the same impulse is applied. Then indeed the linear momentum of the molecule will be the same in both cases, but we also cause rotational angular momentum $L=F\cdot r\cdot \Delta t$ where $r$ is the distance between the line of action of the force and the center of mass of the molecule.
Now the work done by the impulse depends on the distance traveled. Because the molecule starts rotating as we hit it, the force is applied over a greater distance (the center of mass moves less than the side we hit). We can compute the distance moved (and thus the work done) in different ways - but the easiest way is to use conservation of energy...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Thermodynamics and Newton's second law Is it correct to say, that the Newton's laws (or a Newtonian system) is reversible if the friction isn't considered (the fact that the time is of second order $\frac{d^2x}{dt^2}$) and an isolated thermodynamic system is irreversible due to the second law ($\frac{dS}{dt}\ge 0$)?
Can somebody elaborate that, because my teacher seems to disagree.
|
What I try to say, is that it makes equally sense if a ball rolls down
of tray or up (without friction) in a newtonian system. The "proces"
is then reversible. In thermodynamics a proces will only go one way in
a given situation therefore irreversible. – Hamid Mohammad 18 hours
ago
When you think of a ball rolling up or down a hill it is easy to imagine that you do not have any friction.
"In thermodinamics a process" is often irreversible due to the fact that some part of the energy in dissipated in heat due to what microscopically are frictions phenomena between molecules.
So your question is not strickly fair
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Where does the factor of half appear from in the Klein-Gordon Lagrangian for a real scalar field? The lagangian density of a scalar field or a Klein-Gordon field has the form of
$$\begin{align}
\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2.
\end{align}$$
Why is there a factor of half appearing in the lagrangian? Being a constant, when entered into the Euler-lagrange equation, should yield the same equation when compared to a lagrangian without the factor of half, so what is the reason for having that factor in the equation?
| According to these Cambridge lecture notes it's "conventional". I'm guessing that choice of constant gives you the right T-V energy when you integrate the Lagrangian density over a certain volume.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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