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Is Pascal's law applicable in a very large enclosed ocean? Pascal's law states that pressure is transmitted undiminished in an enclosed static fluid. Is this law applicable, say, We apply it to a small scale so molecules push each other to transmit the pressure to the whole fluid say water so what about in a very large enclosed ocean could molecules push all the way? and what is the percentage of the lost pressure?
| Yes, it works fine.
Pressure loss is generally associated with restrictions to flow. But when there is no flow, no such loss occurs.
Assume you are in the (again, static) ocean at a particular depth. If there were more pressure from one side than the other, then the pressure difference would accelerate some of the fluid and cause flow. The redistribution would decrease the pressure difference. Once the fluid stopped moving, the pressure would once again be identical.
In a huge fluid, this redistribution will take much longer than in a small tank. But given sufficient time and stability, it would happen.
| {
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Killing Vectors in Schwarzschild Metric Given the Schwarzschild metric with $(-,+,+,+)$ signature,
$$\text ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$
the lack of dependence of the metric on $t$ and $\phi$ allow us to read off the Killing vectors $K_1=\partial_t$ and $K_2=\partial_{\phi}$. These vectors, in their coordinate representations, are given by
$$K_1=\left(-\left(1-\frac{2M}{r}\right),0,0,0\right)$$
$$K_2=\left(0,0,0,r^2\sin^2\theta\right)$$
How does one immediately read off those vector components for $K_1$ and $K_2$? What is the logic behind reading them off? How would I "read off the Killing vectors" if I, while maintaining no explicit dependence on $t$ or $\phi$, added some off-diagonal terms to the metric? Please help me intuitively understand what's going on here.
| Conceptually:
If we leave the mathematical definition aside for a while, we can define the killing vector:
Killing vector $K^\mu(x)$ leaves metric unchanged under infinitesimal coordinate changes
Time coordinate
Change in $t$ does nothing to the metric:
Change in $t\rightarrow t+dt$:
$g_{\mu \nu}=g_{\mu \nu}(t)=g_{\mu \nu}(t+dt)=g_{\mu \nu}$
So one of the killing vectors should be along $t$:
$K^\mu=(1,0,0,0)$
That's it!
Note: The form you have has the indice lowered: $K_\mu=g_{\mu \nu}K^\nu=(g_{\mu0}K^0,0,0,0)=(-(1-2M/r),0,0,0)$
| {
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Effects of inner painting in the temperature of room which room will stay warmer for long time in winter night. The room with inner painting black or white?
| The scientific fact is that white reflects the radiant energy rays of the sun and black absorbs them.
Most major paint manufacturers can tell you the Light Reflectance Value (LRV) of any color paint chip. White reflects 80% of the light, black 5%.
So if you want to have a warmer room paint it black since more of the radiant energy of the light will be absorbed to keep the room warmer.
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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?
| The first thing to note is that if the loop is moving through a uniform magnetic field the induced emf and hence the induced current is zero as there is change in the magnetic flux through the loop.
The first thing you need to decide is whether or not the magnetic flux through the loop is increasing or decreasing.
The use Lenz's law which states that the induced current will be in such a direction as to oppose the motion producing it.
So the direction of the current will try and either decrease the magnetic flux through the loop.
You can use the right hand grip rule to decide on the direction of the induced current.
I am not so good at 3D diagrams so here is one with the external magnetic field into the screen and the loop moving down the screen.
I also have drawn a loop rather than a plate so the induced currents will flow in the plate in the direction shown in the diagrams probably more in the region where the magnetic flux is changing.
So the eddy currents in the conducting plate will flow anti-clockwise in the left hand diagram and clockwise in the right hand diagram.
| {
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Single Slit Diffraction I am trying to derive the intensity variation function for a single slit diffraction.
Sorry for the poor diagram...
So I decided to take the amplitudes of the waves originating from the slit on the left (wherein the variable that denotes distance within the slit is $l$) and integrate the amplitudes over the entire slit width, taking some point at a distance $x$ on the screen to achieve the resultant amplitude of the waves that strike the screen. With this function, I decided I would use the standard expression for intensity (i.e. $I=\kappa A^2)$
The amplitude for a wave originating from a point on the slit should be:
$$ y=a\sin{kr}$$
where $r$ is the distance between the point of origin on the slit and point of contact on the screen (and $k$ is the angular wave-number).
So:
$$ r^2=D^2+(x+l)^2$$
and on approximating:
$$ r\approx D+\frac{1}{2D}(x+l)^2$$
So I took the amplitude function (for the screen) as $A(x)$ and:
$$ A(x)=a\int_{-l/2}^{l/2}\sin{kD+\frac{k}{2D}(x+l)^2} dl$$
substituting $k(x+l)/2D=u$ (ignoring limits for now):
$$ A(x)=a\sqrt{\frac{2D}{k}}(\sin{kD}\int_{l_1}^{l_2}\cos{u^2}du+\cos{kD}\int_{l_1}^{l_2}\sin{u^2}du)$$
I looked these integrals up so I know that they are Fresnel Integrals, but more importantly that they are transcendental functions.
So my questions are:
*
*Are my assumptions flawed?
*Is there a flaw somewhere in the procedure?
*If what I've done is correct, how shall I proceed?
| Try using this method.
To study diffraction of light, laser light is passed through a narrow single slit and the
diffraction pattern is formed on a distant screen. An imaginary reference line is drawn
perpendicularly from the center of the slit out to the screen (see Figure 3), which is a
distance L away. The intensity variation of the diffraction pattern can then be measured
accurately as a function of the distance y from the reference line. In the theoretical
description of the diffraction pattern, however, it is more convenient to quantify the light intensity as a function of the sine of the angle θ defined accordingly by
$sin θ = y/\sqrt{y2+L2}$
The theory of diffraction predicts that the spatial pattern of light intensity on the
viewing screen by a light wave passing through a single rectangular-shaped slit is given
by
(4)
where I0 is the light intensity at θ = 0◦ and the quantities in parentheses are in radians.
http://www.physics.nus.edu.sg/~ephysics/documents/PC2232-Diffraction-revised.pdf
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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?
| Actually the same happens when a ray of light passes through optical centre the perpendicular distance between extended incident ray and extended emergent ray is negligible so...we can say that the ray which passes through the optical centre is undeviate.
| {
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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.
| John Custer's comment is the best answer so far, so I'll wait for him to make it into an answer for now rather than copying it here.
Owing to the unified description of the "amount of content" of non quantum ground state systems through the system's total energy, the distinction between what a 1910s scientist would call matter and energy is no longer useful. Look up the article Matter on Wikipedia: the word is not used in science now, aside from in in niches applications and then it often clashes with everyday usage. For example General Relativity theorists call anything that contributes to the $0\,0$ stress-energy tensor "matter" but this includes all kinds of stuff that most lay people would not call "matter" in the everyday usage of the word. By this definition, the bond energy is most certainly "matter". But in general nowadays when we speak of anything aside from the quantum vacuum state (all "matter" and energy in, say, the 1910s usage) one needs to specify exactly what we have in a system: photons, electrons, muons, atoms arranged in chemical elements and so forth.
| {
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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?
| I'm pretty sure you will hear nothing from the guitar, because you will be traveling at the same speed as the sound so the sound wave will not be able to make vibrations on your eardrum thus you hear nothing.
| {
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Liquid electromagnet Would it be possible to make a liquid electromagnet using ferrofluid and Ferris sulfate liquid. Having the core be 3/4s of it be ferris sulfate and 1/4 be ferrofluid. Then just having a normal coil. Would that work and if it does would it be stronger then normal electromagnets?
| If you start with a coil with nothing inside it then introduce a core of relative permeability $\mu$ the effect is to multiply the strength of the generated magnetic field by $\mu$.
So in your case the strength of the field would depend on the relative permeability of your mixture of 25% ferrofluid and 75% ferric sulphate. The relative permeability of a ferric sulphate solution is going to be pretty close to water (i.e. $\approx$ 1) so it depends on the ferrofluid. A quick Google failed to give me any clear answers for the permeability of ferrofluids - no doubt it depends on what type of ferrofluid you choose.
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GR: Pseudo Riemannian or Riemannian? Is General Relativityy described by Pseudo-Riemannian manifold or Riemannian manifold? I cannot understand the vast difference between the two manifolds. In books, General Relativity is looked as a pseudo-Riemannian manifold, though I am not sure after reading some threads on the web which confused me.
Now checking wikipedia, it says here:
After Riemannian manifolds, Lorentzian manifolds form the most important subclass of pseudo-Riemannian manifolds. They are important in applications of general relativity.
A principal basis of general relativity is that spacetime can be modeled as a 4-dimensional Lorentzian manifold of signature (3, 1) or, equivalently, (1, 3). Unlike Riemannian manifolds with positive-definite metrics, a signature of (p, 1) or (1, q) allows tangent vectors to be classified into timelike, null or spacelike.
This is the best I've gotten searching for an answer and it is still confusing and not clear.
| A pseudo Riemannian manifold is a manifold equiped with a metric of signature $(p,q)$, $p$ indicating the number of positive eigenvalues and $q$ the negative eigenvalues. For a Riemannian manifold, $q = 0$.
A spacetime of dimension $n$ is defined by a pseudo-Riemannian manifold of signature $(1, n-1)$, or alternatively $(n-1,1)$, also called a Lorentzian manifold.
| {
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Raw data acquirable from amateur astrophotography What raw data can I possibly acquire from an 8" Classical Dobsonian Telescope, and a DSLR? Could anything eye-opening to amateur astronomers be computed or calculated first-hand with such equipment? I'm sure scientists must've considered this equipment "advanced technology" at some point in history not too far back...Could I rediscover or calculate some Laws (like Kepler's laws) or some other things amateur astronomers would be amazed to calculate themselves (like the distance to a planet) using this equipment?
| In addition to CuriousOne's fine answer, I'll point out that Dobsonians, while easy and cheap to build, are rather difficult to combine with a tracking mount. For really good photos (or faint objects), you'll be much happier with a slightly more expensive 'scope with an equatorial mount and a USB interface to your PC.
In general, don't bother with an adaptive optics add-on (not cheap) unless you're going for greater than 8-inch or so primary.
In any case, even with a Dobsonian, people often discover emerging comets or interesting star clusters. The photography part will lead you to select a camera with extremely low noise (or cooled, as CO said). Consumer-grade cameras probably won't cut it.
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How do space probes identify molecules? How does a space probe identify molecules without actually obtaining the molecules? The common identification techniques I can think of are spectroscopy and magnetic resonance, but for both of them, the apparatus needs to interact with the molecule and get readings from them (first the absorption and then the transmission/emission), which, from thousands of kilometers away, sounds quite difficult.
Is there a completely different technique developed for space? Or does space spectroscopy or magnetic resonance work a bit differently than in earth?
| There are several broadband light sources in outer space such as quasars and blazars which basically can act as a light bulb. Earth bound telescopes as well as satellite telescopes can see absorption features in the light when the light passes through some cloud in outer space that contains molecules before the light reaches earth. In addition molecules in outer space can be excited by this radiation or by collisions with other molecules and emit light. In fact there are many masers in space such as water masers and methanol masers.
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Is Earth really a magnet? I am a student of class 9. When I was going through magnetism and read that an earth is a magnet I got some doubts. My question is: is earth really a magnet? Doesn anyone have any proof that earth is a magnet? Is there a magnetic core at the center of the earth? Has anyone reached the core of the earth?
| No there is no magnet inside earth.
I mean there is not any magnet like we see in our surroundings.Earth itself is a magnet or not is upto you, because there is no direct answer to it.
As explained by some scientists
there are convectional currents of molten iron and nickel in core
which generates magnetic field around earth.
Nobody truly knows the real reason behind Earth magnetism.
But still we know earth is magnetic.
the best proof of it is magnetic compass.
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Why do 3d movies have a red and blue "double image"? My question is why do 3d movies have a red and blue"double image" that is basically just a few inches to the right and left of the real image. And how does this help us see the image as "3d". Does it have anything to do with polarized light. And do you need a special computer screen to produce these images because when I use 3d glasses to loot at 3d images on google i still see the see image?
| red/blue double image is the (very) old technology to see 3D on paints, drawings and ordinary screens. See 'Anaglyph' on wikipedia. It requires red/blue glasses (very cheap on Ebay :-) ).
Polarized display is expensive, and is the one requiring polarized glasses.
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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.)
| I found a 1988 paper by Mitrofanov et al which describes a Cavendish style experiment where the "big" mass was 706 mg - where Cavendish used balls of over 150 kg. The "small" mass ( the one on the torsion pendulum) was only 59 mg.
This experiment was done to examine possible deviations of Newton's law at extremely short distances, and established a lower limit on the size of such deviations at a distance below 1 mm. The write-up is quite interesting.
This may not be the lowest value - so I encourage others to find credible publications that show the effect of smaller masses (note that Anna's answer sets a pretty high bar for "lowest mass that was subject to gravity" - but this is an attempt to find the "lowest mass that is seen to attract another mass gravitationally")
| {
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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?
| Actually, if you don't take into account the friction between you and your friend pushing the chair and the ground, the chair could have any velocity, but it would stay constant. The thing is that the force you exert on the chair is equal and opposite to the force applyied by your friend so the total force (resultant force) is zero. Knowing then that force is mass times acceleration, we find that in this case the chair has an acceleration of zero, which means that its velocity stays constant.
Now, if actually your chair stops it is because your feet don't perfectly slip on the ground, there is friction, which adds an additionnal force on your side and on your friend's side, until the chair stops completely in which case those additionnal force are equal.
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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?
| Excellent answers have been given above, so there may be considerable overlap with mine.
Consider a mass $m$ at the end of a spring with spring constant $k$. Admittedly, this is a very crude model but physicists have a way of making simple models. Let's call it a lumped 1D model of a building. The resonance frequency is $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$. For a taller building the spring constant $k$ goes down like $1/h$. At the same time the mass $m$ increases with $h$. So in this very simple model the frequency scales with $1/h$.
Using a tuned mass damper is one way to deal with this.
| {
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What is Pressure Energy? While deriving Bernoulli's Theorem, our teacher said that the sum of KE, PE and Pressure Energy per unit volume remains constant at any two points.
$$P + \rho g h + \frac{\rho v^2}{2} = \text{Constant}$$
In this, he stated that the first term is Pressure Energy per unit Volume. What exactly is meant by Pressure Energy?
I know we can write:
$$P = \dfrac{F}{A} = \dfrac{F\cdot d}{A\cdot d} = \dfrac{W}{V} = \dfrac{\text{Energy}}{V} $$
What is physical significance and expression of pressure energy?
| When a fluid is squeezed, as in a cylinder by a piston, work is done on the fluid.
This work 1) elevates the pressure (pressure energy), and 2) the temperature (heat energy).
(If the cylinder is insulated, this is called "adiabatic".)
In an ideal gas, these are all related by the ideal gas law,
which says roughly that volume times pressure equals heat.
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Splitting molecule The photon reacts with the binding electrons orbiting the two atoms.
The photons have the 'correct' wavelength for Bond Dissociation Energy (BDE).
'Splitting' the molecule involves applying the photon wavelength to separate the electron from the molecule.
With the photons being applied between the binding electrons in between the two oxygen atoms, does 'splitting' occur when there is one photon reacting with one binding electron, even when there are two binding electrons?
| This isn't really how it works. A photon doesn't interact with a single electron, it interacts with the entire molecule.
Suppose you take the example of ozone photolysis to $O_2$ and an oxygen atom. We can do a calculation for ozone and come up with a series of molecular orbitals, then put two electrons in each orbital. So far so good. But if you remove an electron, or even just excite it to a higher energy orbital, then all the molecular orbitals change and you have to recalculate them all. You can't do anything to one electron without affecting all the others and changing the properties of the molecule as a whole.
In the case of ozone it can absorb a photon and the whole ozone molecule rearranges into a higher energy state. From this higher energy state it can relax back into the ground state and re-emit a photon, or it can split into $O_2$ and an oxygen atom. Like most things in quantum mechanics this is a probabilistic process. We can calculate the probabilities of relaxing and spliiting, but it's impossible to predict what any individual excited ozone molecule will do.
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Difficulty understanding the energy transformations during resistance in a circuit I am currently having problems understanding the energy transformations that occur when resistance in a solid conducting wire restricts current flow.
From my understanding, resistance in a solid copper conducting wire, for example, is caused by:
*
*Charge carrier electrons colliding with other charge carrier electron while in random motion
*Charge carrier electrons colliding with stationary lattice copper cations that are in the way.
I have always thought these collisions were 'inelastic' and the kinetic energy of the electrons is converted into heat energy in a heating element, or light energy in a light globe.
However my internet sources always state that it is the potential energy of the electrons which is converted into heat energy in a heating element, etc.
So I am currently confused to whether it is the electric potential energy or kinetic energy of the electrons which is converted to other energy forms
| The way I imagine it is that the conduction electrons in a wire move under the influence of the applied electric field and then periodically transfer their kinetic energy to the surrounding atomic lattice. But this kinetic energy comes from the applied electric field (i.e, the electric potential energy). So I think that although you may be right in saying that kinetic energies do play a role in the process, that it should be recognized that all that electronic kinetic energy that's being transferred to the lattice ultimately comes from the interaction of the electrons with the applied electric field.
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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.
| What you're looking for is an intuitive explanation or how you could visualize momentum.
You can think of momentum as the quantity/amount of motion or "how much would I not want be in the path of this body."
I'm going to try and provide some intuition through a few examples:
A car of mass 1000 kg moving at 5 m/s would have the same "quantity/amount of motion" as a truck of mass 5000 kg moving at 1 m/s or a bicycle of mass 20 kg moving at 250 m/s.
If a rocket of mass 500 kg moving at at 250 m/s, uses up 100 kg of stored fuel to accelerate the rocket upto 312.5 m/s, although the rocket is moving faster, the mass has decreased by 100 kg and hence the "quantity of motion" has remained the same or "how much would I not want be in the path of this rocket" has remained the same.
Therefore, Netwon defined the force as something that changes the "amount of motion" that a body has in a certain amount of time, or:
$F = \frac{dp}{dt}$
Where p represents momentum and t represents time
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Why light moves sideways? Greatings!
I'm trying to understand special relativity and have one question bugging me.
In almost every book or video about the subject there is a thought experiment with moving light clock. I hope I need not elaborate on the sutup and the outcome of the experiment.
So the question is this: When a light source on the MOVING light clock emits a photon of light the path of the photon is triangular.
How this can be?
I thought since light speed is constant in every reference frame the movment of the "emmiter" can't affect the movment of the light so light should shoot right up (from the point of stationary observer) and thus move backward and up from the point of moving observer.
So you see my logic is: light speed is invariant => you can't make it move UP & FORWARD only UP. i.e. you can't change it's direction once it's emited in UP direction. Why the hell light moves sideways then?
And I'm puzzeled. There is flaw in the logic but I don't see it.
Thanks for help.
| The components of the velocity vector can change, it's the magnitude c to be invariant.
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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?
| For all particles, $p^{\mu}=(E,\vec{p})$ and $p^{\mu}p_{\mu}\equiv m^2$ (using the mostly-minus metric). Thus $E=\pm \sqrt{m^2+|\vec p|^2}$. If you set $m^2=0$, you get $E=\pm |\vec p|$. The non-trivial aspect of these definitions is that $E$ is to be literally identified as the energy, and $\vec p$ as the spatial momentum (so in the classical limit $E=p^2/2m + \textrm{const.}$ ).
For massive particles with positive-energy ($m^2>0$, $E>0$), the 4-momentum and 4-velocity are related by the equation
$$p^{\mu}=m\,u^{\mu}$$
whereas for massless particles with positive-energy ($m^2=0$, $E>0$), the relationship between the 4-momentum and 4-velocity is given by:
$$p^{\mu}=E\,u^{\mu}$$
where $u^{\mu}$ is a light-cone vector.
| {
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Calculating the color temperature and intensity of skylight with an algorithm I think it will be possible to calculate the natural sky light intensity and color balance (CCT) based on time of day/year, GPS location. Has anyone seen a formula/algorithm for this? I would like to work this out for an app that will present the values in 15 minute increments. Eg: 12 noon, November , tropical north coast Australia CCT in deg Kelvin = 5600 deg, Light intensity 32000 LUX. If anyone knows how to calculate this please chime in. Thank you.
| There are plenty of algorithmic sky models. see for instance paper (and previous work section) https://hal.inria.fr/inria-00288758
But this is not related at all to "color temperature", i.e. Plank's law (or black body). (well, there is the Sun's one at the begining, but after it's about selective absorption, Rayleigh and Mie (multiple) scattering, etc).
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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.
|
Why do helium balloons rise and fall?
At one point in the afternoon they were dropping but later in the evening it was fully upright again. ... The weather did go from a little rainy and overcast to dry and slightly clearer skies later.
Likely many droplets of water adhered to the helium balloons, when the weather was rainy, and made them heavier. When the water evaporated the balloons became lighter and rose again.
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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 $|0,0\rangle$?
| You misunderstood the eigenfunction of angular momentum or spin.
The state vector $|l,m\rangle$ is an abstract notion of the eigenstate of the commutated operators $\hat L^2, \hat L_z$ . They are always orthogonal: $\langle l,m|l',m'\rangle \propto \delta_{l,l'}\delta_{m,m'}$ .
Usually, we do not need to express them with other vectors. But you can always define some vectors as a set of new basis then express them, and the result should be the same as you do with basis $|l,m\rangle$.
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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?
| Do the maths and calculate how much energy is needed to raise the water's temperature by 1K. If you have a fast moving stirrer, you should be able to measure the increase in temperature of a liquid in an isolated pot.
By the way: the microwaves in your microwave oven turn around the water molecules very fast and heat up your food this way.
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What is the distance between two objects in space as a function of time, considering only the force of gravity? What is the distance between two objects in space as a function of time, considering only the force of gravity? To be specific, there are no other objects to be considered and the objects in question are not rotating.
For instance, say you have two objects that are 6 million miles apart. One is 50,000 kg and the other is 200 kg. Say I want to know how much time has passed when they are 3 million miles apart. How would I go about doing that?
EDIT: Looking at the other question I am having trouble following David Z's steps in his answer. Intermediate steps would be helpful. In particular I don't see how the integration step works. I also don't understand why the initial r value, ri remains as a variable after it's derivative has been set to 0, wouldn't the integral of that derivative (i.e. the function ri) be 0 + C? I also don't see how you wind up with a term that includes 2 under a square root sign.
I can not ask for the intermediate steps on the question itself because I do not have the reputation points.
I think it probably answers my question or will once I understand it, but I am not sure.
EDIT: I can sort of understand the integration step. But it seems like he is integrating with respect to two different variables on both sides, the variables being r on the left and the derivative of r on the right. There must be something I'm missing here.
| The two equations of motion reduces down to one equation of motion by considering the separation $x=x_2-x_1$ and the separating acceleration $\ddot{x} = \ddot{x}_2 -\ddot{x}_1$
$$ \ddot{x} = -\frac{G (m_1+m_2)}{x^2} $$
or $ \ddot{x} = -K/x^2 $ with $K=G (m_1+m_2)$
This can be re-written as $\frac{{\rm d} \dot{x}}{{\rm d} t} =\frac{{\rm d} \dot{x}}{{\rm d} x} \frac{{\rm d} x}{{\rm d} t} =\frac{{\rm d} \dot{x}}{{\rm d} x} \dot x = -\frac{K}{x^2}$
$$ \int \dot{x} {\rm d} \dot{x} = -\int \tfrac{K}{x^2}\,{\rm d}x + C_1$$
$$ \frac{1}{2} \dot{x}^2 = C_1 + \frac{K}{x} $$
If initially the bodies are at rest, separated by $d$ then
$$ \frac{1}{2} \dot{x}^2 = -\frac{K}{d} + \frac{K}{x} $$ or $$ \dot{x} = \sqrt{\frac{2 K (d-x)}{d\, x}} $$
This has a solution for time $t$ as a function of separation $x$ of
$$ t= \sqrt{ \frac{d^3}{2 K}} \cos^{-1} \left( \sqrt{\frac{x}{d}}\right) - \sqrt{ \frac{d^2 x-d x^2}{2 K}} $$
This means the time to reach collision $x=0$ is
$$ t_C = \frac{\pi}{2}\sqrt{ \frac{d^3}{2 K}} = \frac{\pi}{2}\sqrt{ \frac{d^3}{2 G (m_1+m_2)}}$$
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Difference between convexo-concave and concavo-convex lenses? What is the difference between concavo-convex and convexo-concave lenses? We dont say convexo-plane for plano-convex. Does that mean concavo-convex and convexo-concave are essentially the same?
| In both types, (convexo-concave or concavo-convex) the lens has one convex and one concave side.
Convexo-concave : The concave face has a greater degree of curvature than the convex face.
Concavo-convex : The convex face has a greater degree of curvature than the concave face.
The images are as follows:
Now a days these lenses are referred as convex-concave or meniscus. It is this type of lens that is most commonly used in corrective lenses.
Convex-concave (meniscus) lenses can be either positive or negative, depending on the relative curvatures of the two surfaces. A negative meniscus lens has a steeper concave surface and will be thinner at the centre than at the periphery. Conversely, a positive meniscus lens has a steeper convex surface and will be thicker at the centre than at the periphery.
| {
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Why is it easier to pull a lawn roller rather than pushing it? Does it have something to do with friction or weight distribution in pulling is favorable or anything else ?
| When you push a lawn mower, you are directing force down as well as forward because of the angle of the handle. This means the wheels of the mower are being pushed into the grass, which increases the rolling friction (when the wheel makes a hole in the grass, you have to "climb out of the hole" all the time; the deeper the hole, the steeper the slope you are climbing). When you pull, the force is "up and out" of the (smaller) hole - so it's easier.
A little cartoon to help:
This is ignoring the fact that most lawn mowers have a free wheel mechanism so that the blades only move when you push it forward. I don't think that's what you were asking about.
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Why don't galaxies orbit each other? Planets orbit around stars, satellites orbit around planets, even stars orbit each other. So the question is: Why don't galaxies orbit each other in general, as it's rarely observed? Is it considered that 'dark energy' is responsible for this phenomenon?
| They do! There's an entire class of galaxy, called a 'satellite galaxy' which is defined entirely based on them orbiting a larger galaxy (which would be called a 'central galaxy'). Our own milky-way is known to have many orbiting satellite galaxies, or at least 'dwarf-galaxies'. If dwarf-galaxies aren't enough, the milky-way itself is gravitationally bound to the andromeda galaxy, and they are effectively orbitting eachother. Because of the tremendous size-scales, however, the orbital period is billions of years --- in many cases, far longer than the age of the universe, so that a pair like the milky-way---andromeda 'local group' actually hasn't completed a single complete-orbit in the history of the universe. That's why we can definitely never (even hope to) see galaxies orbit in real-time.
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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?
| The answer to the question is due to the fact that the number of charge carriers increased.
Glen Knoll offers the following in pp. 117 of his book "Radiation Detection and Measurement":
The width parameter σ determines the FWHM of any Gaussian through the relation FWHM = 2.35σ. The standard deviation σ of the peak in the pulse height spectrum is then σ = K*sqrt(N). Here the K is a proportionality constant, and N is the charge carriers, particularly in a scintillator detector.
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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?
| It's because you're not looking far enough. From personal experience, it takes at least 10 km of atmosphere to build up a really obvious blue (see, for example, this picture), and if you're not in hilly country, the horizon is only 5km away. In contrast, most of the sky has distances to space on the order of hundreds of kilometers.
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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?
| To see that your integral expression does not make any sense, imagine that $\vec{r}(t)=( x(t),y(t))$ describes a circle. Then the line integral of the force around the loop gives the change in potential energy, which should of course be zero,
$$\oint \vec{\nabla} \phi \cdot \vec{dl} = \Delta \phi =0. $$
But if you insert the actual values from your calculations,
$$\oint \vec{\nabla} \phi \cdot \vec{dl} = -b \int \vec{r}'(t) \cdot \vec{r}'(t) dt = -b\int\left| \vec{r}'(t) \right|^2 dt \neq 0. $$
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Why is the blue line in the Balmer series sometimes not visible? So I've conducted an experiment to find the four visible hydrogen emission spectrum lines in the Balmer series in a laboratory. I don't have any background in quantum physics.
When I looked through the eyepiece, I saw the red light, the pale blue light, and the purple light as shown in the picture below:
I've asked my lab instructor why couldn't I see the blue light and he said it is well-known that sometimes not all the spectrum is shown. He told me to look it up on the internet, since I didn't take the quantum physics course yet.
Can someone explain to me this phenomenon? Or at least refer me to an article which discusses this issue? The experiment revolved around Balmer series only. (It was my first spectroscopy experiment)
The setup looked like this:
It looked like this:
Edit:
These are the wavelengths I've found:
So I think the missing wavelength is actually the $410 nm$.
What can be the reason for this?
|
Why is the blue line in the Balmer series sometimes not visible?
The human eye has difficulties in distinguishing dark blue lines on a black background.
You can use "Microsoft Word" to draw a black rectangle and a few dark blue lines of different thicknesses on the rectangle. The thinner the dark blue line the less visible it is. Even the thick dark blue lines are not quite visible.
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How does air travel after leaving air nozzle? I am working on a project which involves air nozzles. I am interested in estimating the time it takes for air particles after they leave the nozzle to reach a particular distance away. I have the following information--- a converging nozzle, the speed with which it is coming out of the nozzle (I know when the air is choked it comes out at Mach number). I am aware that this problem is complex and would be very difficult to solve it precisely without intensive Mathematical Models. I just want a rough idea/guideline. Any help will be appreciated.
http://postimg.org/image/cw53jnibx/
Above is a profile of the nozzle
| Without more details, we can't find a specific case that will match what you are doing. However, I did find a decent example to show you what to look for to answer your question.
Data for a nearly-sonic round jet can be found in this paper. If you look at Figure 5(a), you'll see how the normalized centerline velocity from several experiments collapses together and has essentially a hyperbolic tangent shape. You could find experimental data for conditions near your operating point (speed, Reynolds number) and create a tanh function that approximates the data and go from there.
You can find similar figures and/or correlations for the width of the jet as a function of downstream coordinate also. Standard textbooks on turbulence should have the subsonic jet correlations, but for a sonic jet I can't think of any off the top of my head that will definitely have it. But you can search through the literature using keywords learned in the paper I linked to and you should be able to find what you need.
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What constitutes a force? There are a few questions on here about why Gravity is not a force, but I'm having trouble grasping why exactly.
It sounds to me that Gravity is not a force because it is simply a by-product of mass on spacetime. How does this differ from the strong nuclear force? Are the other forces not an effect of the presence of a particle?
I apologize if this is an overly simple question, I'm not well-versed in physics.
|
It sounds to me that Gravity is not a force because it is simply a by-product of mass on spacetime.
The force in general, including its particular case the Gravity, is a byproduct of mass, space and time because it is measured in $kg*m/s^2$.
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Smallest possible size of nuclear reactor I recently began playing Fallout 4 and asked myself a question: Would it be possible to build a power armor with todays technology?
I did some investigation on servomotors and stuff and came to conclusion that it would be theoretically possible to build such a suit. The main problem, in my humble opinion as a non-engineer, would be the power supply, as strong servomotors need much more of it than for example solor cells could supply.
So here comes my question: What would be the smallest possible size of a functional nuclear reactor in theory (1) and in practice regarding todays technology (2).
Well, put on hold, okay... somehow restrictive. Can a moderator maybe move this Question to an appropriate StackExchange site please?
| Considering that the fuel element is called "fusion core", I guess they use fusion reactors, not fission. But in reality fusion reactors are not cost-effective and require extreme temperatures to initiate a fusion reaction (see "tokamak" to get an idea and see the size).
There were some news about a team of scientists who made some progress on that kind of reactors, but we'll see if it's true.
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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?
| Ultimately, Newton's law of cooling is a simplification that can be obtained from the full heat equation, i.e. $$\rho c\frac{\partial T}{\partial t} = - \kappa \nabla \cdot T.$$
The heat equation itself can be derived from first principles, assuming Fourier's law for heat flow, namely that it is proportional microscopically to the difference in temperature between two arbitrary regions. Fourier's law itself can be derived from modern statistical mechanics, although it's a little bit involved to do so: have a look at this.
Historically, Fourier's law was "just" an experimentally derived truth.
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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?
| No, Newton's Second Law of motion is only an approximation and doesn't work on anything larger than a solar system. When you get into the domain where the acceleration is on the order of $10^{-14}\space km\space s^{-2}$, then you can see limits of the approximation. Stars at the edges of spiral galaxies travel much too fast to be governed by Newton's Second Law of Motion. Even backfilling a galaxy with some imaginary matter won't fix the problem. Take a look at the data for Andromeda and you'll see a rising velocity curve beyond 30 kpc that clearly conflicts with the predicted Newtonian decay.
The second law can be easily fixed, however:
$$\sum_{i} F_i = m(a_F + a_0)$$
$$a_0 = 3.74\times 10^{-14}\space km\space s^{-2}$$
Applying this to a circular orbit gives:
$$v = \sqrt{\frac{M G}{r} + a_0r} $$
Where $a_F$ is the acceleration of the unbalancing force (gravity in the case of an orbiting star) and $a_0$ is the constant acceleration of the expansion of the universe. Go ahead and try it! It will fix any velocity curve problem you have without resorting to science fiction.
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Is there any limit to distance between two waves when used in Huygens principle? Apologising for not being clear.
When I used Huygens principle to determine the spherical wavuefront of radius $R/2$ created by reflection from a spherical concave lens of radius $R$, I got amazingly different results for different size of wave front created.
I used a hemispherical lens. When I used a wavefront of width $2R$ I really got a spherical wavefront of radius $\frac{R}{2}$. But in this case all rays were meeting at different position and not at $\frac{R}{2}$. Also laws of reflection was violated.
When I used a wavefront of much smaller width I got most of rays meeting at focus and obeying law of reflection. But the wavefront was not spherical.
Can anyone tell me what I did wrong and what I should do to get correct results?
Here a ray falls at pt. A. Consider a big and vertical wavefront here touching other side of mirror. After central ray of this wavefront, reach at pole of mirror, a new secondary wavefront is created at A of radius AB.
Now I drew a perpendicular from O on arc of this wavefront, intersecting at B. This created a ray AB. So, light should pass from A to B.
1)Law of reflection state , It should reflect along OD.
This is a contradiction.
2) Then, I used a new wavefront(look at point near D).
Made required drawings and found ray reflecting near focus.
All this helped me to draw a circular wavefront created due to all such secondary wavelets.
This circle tells, rays should focus at focus, without any spherical abberation. How?
| I think that you have found from first principles that a hemispherical lens suffers from spherical aberration?
Later
Here is my attempt.
It might look a mess but I think that it does illustrate the fact that the wavefronts converge towards a region which is about half the radius of the mirror.
The radius of the mirror is 90 mm and the wavelets have a radius of 10 mm.
You start with a wavefront and draw wavelets centred on that wavefront.
You then construct the new wavefront and repeat the process.
The smaller you make the wavelets the more accurate will be the construction.
When you "know" what the new wavefront is going to look like then the size of the secondary wavelets does not matter as is shown in the right hand diagram.
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Conflicts between Bernoulli's Equation and Momentum Conservation? The well known Bernoulli's equation states that
$P+\frac{\rho V^2}{2}=c$
However, a simple momentum conservation considering $P_1$ and $P_2$ acting on two sides, and velocity changes from $V_1$ to $V_2$, yields
$P_1+\rho_1 V_1^2=P_2+\rho_2 V_2^2$, which differs from Bernoulli's by a coefficient $\frac{1}{2}$.
What is going on here? I understand the derivation of both, just want to know how to explain the conflict.
| The factor $\frac 12$ comes from the relation $\vec v \cdot\nabla \vec v = \nabla \frac{\vec v^2}{2} + (\nabla\times\vec v)\times \vec v$ in the momentum conservation equation $$\rho \left(\frac{\partial \vec v}{\partial t}+\vec v \cdot\nabla \vec v\right)=\vec g-\nabla p$$
(Sorry to post this as an answer, but I can't comment your post yet because of reputation)
EDIT: formalism as follows
*
*$\vec u \equiv \vec v$ is the velocity field
*$p$ is the pressure field
*$\vec g$ is the gravity field
If you develop the equation, assuming that the gravity field derives from a potential such as $\vec g=-\nabla \phi$, and that the flow is in steady state i.e. $\frac{\partial \vec v}{\partial t}=0$ then: $$\nabla\left(\frac{\vec v^2}{2}+\frac p\rho+\phi\right)+\vec \omega\times\vec v=0$$, where $\vec{\omega}\equiv \nabla \times \vec{v}$ is the vorticity operator
At equilibrium, $W=\Delta E_k + \Delta E_p$ (work of the pressure forces, kinetic energy and potential energy)
*
*$W=p_{1}A_{1}(v_{1}\Delta t) - p_{2}A_{2}(v_{2}\Delta t)$
*$\Delta E_k=\Delta m(v^{2}_{2}- v^{2}_{1})/2$
*$\Delta E_p=\Delta mgh_{2}- \Delta mgh_{1}$
| {
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What mechanism explains the effect of a hot surface on hydrogel balls? Adding little hydrogel balls to a hot surface has an interesting effect as seen in this video. Hydrogels (as the name suggests) contain a lot of water but rather than melt or burst, the little balls are seen to bounce up-and-down on the hot surface making a high-frequency squeaking noise.
My thought on this is that it is related to the Leidenfrost effect. At the hot surface, the membrane is quickly heated and perhaps some small amount of vapor escapes causing the balls to bounce. However, I don't think this would have enough force to have them bounce so high. Perhaps the elasticity of the hydrogel plays a part.
Does anyone have an alternative idea?
| Yes, the Leidenfrost effect plays a big part here. As the hydrogel balls aren't perfectly elastic (their restitution coefficient is smaller than one) they would quickly come to a halt but little jets of steam that are ejected by them when they hit the hot surface constantly provide extra upward force to compensate for inelastic losses. It's what keeps them bouncing.
The high pitch of the sound is related to the small size of the hydrogen balls.
| {
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On a molecular level, how does heat transfer take place? When an ice molecule hits a water molecule, the water molecule adds kinetic energy to the ice molecule. Why do the water molecules lose kinetic energy? In other words, why does heat like to go where there is less heat? I know thermodynamics tells us energy can't be created or destroyed. It doesn't really make intuitive sense, though. In the macroscopic world, if a ball hits a moving object in a non-opposing direction to the direction of force initially applied, the object's velocity actually increases, while the ball suffers from Newton's equal-and-opposite-reaction rule.
| Basically a "molecule" of water cannot heat up ice.
I think what you are trying to say is, how does heat transfer take place on a molecular level? If that's the case, then its something like this. In the interface between water and ice, water molecules are moving, while ice molecules are static. on contact, some molecules of ice acquire velocity (due to no binding forces in one direction, and cohesive forces towards liquid water molecules). as a result, surface molecules of ice start acquiring velocity, hence changes state from solid to liquid (simply, ice melts). and due to conservation of energy, an equivalent amount of kinetic energy (macroscopically, heat) is lost by water. hence water cools a bit.
Of course there may be better ways to explain this, but i tried my best. hope you understand :)
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Time dilation in the movie Interstellar I know that the science in movies is questionable and sometimes ridiculous but I would think this question would have been more obvious to the script writers. When they visited Miller's planet they were almost killed by a re-occurring tidal wave. In a few short minutes they ended up spending 23 years there. When they were still orbiting the planet before going down wouldn't they have seen multiple tidal waves occurring one after another? Therefore why even bother landing on the planet?
| In addition to what should be a miniscule difference in time dilation between the water planet's orbit and the water planet's surface, I still don't understand the tiny difference in subjective time separating the two landers' arriving on the water planet's surface.
If two identical ships start a journey in the same frame of reference, and then travel identical paths that end up in the same frame of reference, then the subjective difference in time separating the two ships should remain constant.
For instance, in an automobile race two cars may go faster in some parts of the track and slower in others, but the difference in departure/arrival times should always remains the same. If Car B starts the race 5 minutes after Car A, then Car B should also finish the race 5 minutes after Car A, no matter how twisty the track. Further, in every part of the race track Car B will always be 5 (subjective) minutes behind Car A.
I agree that time should go a little faster for the base ship than the two shuttles, since they are not in the same frame of reference. (Assuming the base ship is orbiting the planet which I am not sure on.)
| {
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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 $F_N$ is just the force between two surfaces. It's called "normal" because it acts perpendicular (normal) to the surfaces.
Gravitational force is completely unrelated. Gravity always acts with $F_g = -mg$. The minus sign indicates that the force points down.
These two forces often oppose each other, which is why $F_N$ OFTEN, BUT NOT ALWAYS, $=mg$. The sum of all y-components of forces must equal acceleration in the y direction (Newton's 2nd Law). For a book resting on the table, there is no acceleration in the y direction, and 2 forces acting: gravitational force and the normal force. Since $a_y=0$, $F_N+F_g=0$, and $F_g=-mg$, so $F_N=mg$.
Hairy details and garlic for the higher-ups:
"Down" depends on your coordinate system. It's more accurate to express gravitational force as a vector (although you'll have to decompose it sooner or later).
$|F_g|=mg$ only near the surface of the earth. For a more general relationship, use Newton's Law of Universal Gravitation.
Newton's 2nd Law actually states that the vector sum of all forces is equal to the mass times the acceleration of the object: $\sum \vec{F}=m \vec{a}$. This is actually 3 scalar equations: $\sum F_x=ma_x$, $\sum F_y=ma_y$, and $\sum F_z=ma_z$.
| {
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When sugar is added to water, how does the mass change, and how does that affect the water's density and boiling point? I can't find a good answer anywhere. How does the amount of sugar added change the boiling point, mass, and density of water? Does it affect the mass or the volume? Or both?
| As stated above, the mass of the whole system (sugar + water) doesn't change. In addition, with "ideal" mixing, the total volume of the water plus the total volume of the sugar equals the total volume of the mixture. However, this is not a sure bet, and there are many cases of a volume of one material mixed with a different volume of water, and the total volume is not the sum of the two. I'm not familiar with whether or not sugar mixed with water is ideal or non-ideal ... a Google search may have some hints for this.
The following link is a good start for this kind of data:
http://homepages.gac.edu/~cellab/chpts/chpt3/table3-2.html
Regarding boiling point, the boiling point increase (and the freezing point depression) depends only on the number of particles dissolved in the solution. The term for this is the "colligative" properties of these substances. Table sugar is a dimer, which means that it is composed of two 6-carbon sugars. It is easily possible that some of this dimer disassociates into its constituent parts when in solution, increasing the number of particles in solution, and hence, the boiling point. Rather than go through calculations to determine this, there should be published data that give the measured effect. This link should give you some good information:
http://kitchenscience.sci-toys.com/boiling_freezing_pressure
| {
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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?
| All of these oscillations have the same frequency, so the EM waves have the same wavelength.
Maxwell's equations are linear, and air is a linear medium, so EM waves don't interfere with each other. You can point a laser pointer across the room and it won't affect all of the other light rays flying around the room. So the incoming wave is not affected by anything created by the receiving antenna.
Any emission by the receiving antenna (scattering) is really small as compared to the wave coming from the transmitting antenna.
Things are a bit different if the scattering antenna is really close to the transmitting antenna. This is the basis of a beam antenna (https://en.wikipedia.org/wiki/Directional_antenna).
| {
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How to keep a helium balloon between 1 to 5 meters above ground? (without it being tied) I understand that helium balloons rise because their density is less than air, so they can rise up to a point where the air surrounding it has the same weight as the balloon.
I was thinking to fill it with something like half air and half helium. Will this work? If not, is there a way to do it?
| Ideal gases (of which real gases are only an approximation) obey the relationship $PV = const$ (at constant temperature). This won't work because pressure falls as altitude increases. As the balloon rises, the pressure outside the balloon falls, and it will expand, increasing its volume, and decreasing the density of the gas inside.
In the real atmosphere, the temperature also has ot be taken into account, but its effect should be roughly the same for both air and helium.
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Air flow speed and penetration abilities As I understand, two fans with the same diameter and with the same rotor speed(usually measured in RPM) can have different air pressure(usually measured in mmH2O) and air speed(usually measured in CFM) because of fan blades design. Higher air flow speed should provide worse penetration(for example, penetration to radiator with thin fin gaps) abilities. However, it is counterintuitive to imagine that air molecules with higher speed have smaller penetration ability than the ones with lower speed. Why is that so? Or is it because of complexity of air flow dynamics and air flow with slower speed has indeed for some (weird) reason better penetration abilities? At least I know that this is not because of Bernoulli's principle because this should apply only for tubes.
| The most likely explanation is that at lower speeds you get a more laminar flow. To quote the wiki article:
At low velocities, the fluid tends to flow without lateral mixing, and
adjacent layers slide past one another like playing cards. There are
no cross-currents perpendicular to the direction of flow, nor eddies
or swirls of fluids.[2] In laminar flow, the motion of the particles
of the fluid is very orderly with all particles moving in straight
lines parallel to the pipe walls.[3] Laminar flow is a flow regime
characterized by high momentum diffusion and low momentum convection.
| {
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Magnetic field at the center and ends of a long solenoid A long solenoid has current $I$ flowing through it, also denote $N$ as the turns per unit length. Take its axis to be the $z$-axis, by symmetry the only component of the magnetic field inside is $B_z$. Find the magnetic field at the center of the solenoid (on the axis). Also, find the magnetic field at the ends of the solenoid.
For the first part, since the solenoid is long we can approximate the magnetic field inside to be uniform and is given by $B_z = μ_0NI$, so we can say that the magnetic field at the center is also $μ_0NI$. I'm not sure if my argument is correct but based on my understanding, from the uniformity of the $B$-field inside, it should be the same everywhere inside. Can anyone kindly tell me if this is correct? Any suggestions and insights?
For the second part I don't have any idea on how to start.
| since you are concerned about a long solenoid, this problem has a very simple solution.
Suppose you have two identical long solenoids, each of them having magnetic field $B$ at the ends. You join them end to end, such that their magnetic moments are in same direction. Thus, at the junction the magnetic field adds up to $B+B=2B$.
But this junction is nothing but the mid point of another long sloenoid, with same value of N. Thus we get $μNI=2B$, or $B=μNI/2 $!
| {
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Can we find a vector if its dot product and cross product with another vector is given? If I have two vectors $\vec{b}$ and $\vec{v}$, and I know that
$$
\vec{b} \times \vec{v} = \vec{c}
$$
and
$$
\vec{b}\cdot\vec{v} = \lambda
$$
can I find the $\vec{v}$ vector in terms of the $\vec{c}$ vector, $\vec{b}$ vector, and $\lambda$? I have been struggling with for quite sometime. And I hope it's time I asked for help.
| Yes. If $\bf B \times V=C$ and ${\bf B \cdot V}=\lambda$, the BAC-CAB rule tells you:
$\bf B\times C=B\times(B\times V)=B(B\cdot V)-V(B\cdot B)$
So
${\bf B\times C}={\bf B}\lambda-{\bf V}B^2$
and
$${\bf V}=\frac{{\bf B}\lambda-{\bf B\times C}}{B^2}$$
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Conservation of energy and angular momentum I'm writing a java program to simulate the solar system. All planets are modelled as point masses. How do I check if my solar system is conserving energy? I'm not sure how to calculate the energy of the system at the start of the simulation, let alone at the end!
And given that my model is with point masses, presumably I can't calculate angular momentum at all?
| The conserved quantities in your problem are: total mechanic energy $T+U$, total linear momentum $\vec P$, and total angular momentum $\vec L$ of the $N$ point masses.
They are defined as:
$$E=T+U \quad\text{ total mechanical energy}$$
$$U=-G\sum_{i}^N \sum_{j}^N \frac{m_i m_j}{|\vec r_i -\vec r_j|}\quad\text{ total potential energy}$$
$$T=\frac12\sum_i^N m_i v_i^2\quad\text{ total kinetic energy}$$
$$\vec P=\sum_i^N m_i \vec v_i \quad\text{ total linear momentum}$$
$$\vec L=\sum_i^N m_i (\vec r_i\times \vec v_i) \quad\text{ total angular momentum}$$
where $G$ is the gravitational constant (the gravitational force between 2 masses is $F=G \frac{m_i m_j}{(\vec r_i -\vec r_j)^2}$).
Note that:
$m_i$, $\vec r_i$, and $\vec v_i$ are the mass, the position, and the velocity of the point mass $i$.
only the total mechanical energy $E=T+U$ is conserved, while the total kinetic and total potential energies are not conserved separately;
the total linear and angular momenta are vectors, while total energies are scalars;
you must include also the sun in the sums over point masses $i=1,...,N$ and $j=1,...,N$ in the definition of the energies and momenta.
A point mass has a well defined angular momentum $m (\vec r \times \vec v)$ which is non-zero if the mass is rotating around another mass (or any point in the space).
| {
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Infinite pulley system Infinite Atwood Machine Harvard Solution
Hi, I've been trying to solve this question for a while. I understand the first solution and also the solution to the second problem but I don't understand how to apply the second problem to solve the infinite pulley system.
In particular, I don't understand the last sentences:
Therefore, since $f N (x) → 3m$ as $N →∞$, our infinite Atwood’s machine
is equivalent to (as far as the top mass is concerned) just two masses, $m$ and $3m$. You can then quickly show that the acceleration of the top mass is $g/2$.
Note that as far as the support is concerned, the whole apparatus is equivalent to a mass $3m$. So $3mg$ is the upward force exerted by the support.
If the Atwood's machine is equivalent to just two masses $m$ and $3 m$, then wouldn't the value of acceleration be $2g$?
| No, an Atwood's machine with masses $m$ and $3m$ has acceleration of $g \frac{3m -m}{3m+m} = g/2$, as explained in https://en.wikipedia.org/wiki/Atwood_machine#Equation_for_constant_acceleration
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Principle behind electrostatic shielding? If we have a solid conducting sphere with charges around it, then the electric field inside the sphere is zero, otherwise the electrons of the sphere would not be in equilibrium as there would be a net force acting on it. However, if its a hollow sphere, then why does the electric field inside the hollow sphere have to be zero?
| Okay, draw a circle around the spherical void; done?
Then see is there any charge?
If not, there should not be any divergence of electric field, know that?
So, from Gauss Law, $$\text{div} \mathbf{E} \cdot dv= 0$$ which implies $$\mathbf{E}\cdot d\mathbf S= 0 \implies \mathbf E= 0$$ over that volume .
Edit:
Don't know what OP is trying to convey through his message but still let me elaborate a bit.
Suppose, there be an isolated conductor.
Initially, there are no net charges. So, the electric field inside & outside is null.
Then some, external charges are placed inside the conductor. They would feel force of repulsion among them & in order to minimise it, they spread over the surface in such a way that there is no net electric field inside.
Then the conductor is placed in some external electric field.
The field inside the surface of the conductor is the sum of the external electric-field & the field due to the charges residing on it. They arrange in such a way that they nullify the external electric field inside the surface.
This is indeed proved by the fact that the potential inside the surface needs to follow Laplace's equation. Now, a solution could be the potential of the surface itself which is equipotential.
But by Uniqueness theorem, we know there can be only one solution: one potential inside the surface.
This means potential inside the surface is same as that of the surface.
Now, electric field is the gradient of potential which implies it is zero inside the surface of the conductor.
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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?
| Only the force component that's perpendicular (90 degree angle) to the surface will create a pressure. Let's consider a toy car on a track that has a 45 degree incline. Gravity exerts a force on the car. We can split this force into two components: one that is parallel to the track and the other one is perpendicular to the track. Only the perpendicular force will create pressure on the track. The parallel will accelerate the car so it will start moving.
This becomes obvious if you consider the track on a 90 degree incline so it's vertical. The car just falls down. All force is used to accelerate the car and the pressure on the track will be zero.
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The anticommutator of $SU(N)$ generators For the Hermitian and traceless generators $T^A$ of the fundamental representation of the $SU(N)$ algebra the anticommutator can be written as
$$
\{T^A,T^{B}\} = \frac{1}{d}\delta^{AB}\cdot1\!\!1_{d} + d_{ABC}T^C
$$
where $\delta^{AB} = 2\text{Tr}[T^AT^B]$ is the normalization chosen for the generators (note that they are also chosen orthogonal), $d=N$ for the fundamental representation, and $1\!\!1_{d}$ is the $d$-dimensional identity matrix.
For the fundamental representation it seems possible to deduce this expression by arguing that the anticommutator is Hermitian and hence can be written in terms of the $N^2-1$ traceless generators and one matrix with non-vanishing trace.
Does this expression hold for a general representation of the generators? If yes please explain why and/or provide a reference.
The relevance in the above equation appears in trying to express a general product:
$$T^AT^B = \frac{1}{2}[T^{A},T^{B}]+\frac{1}{2}\{T^{A},T^{B}\}$$
where the commutator is already known as a consequence of the closure of $SU(N)$.
| For a general representation $t^{A}$ of the generators of $SU(N)$ it is possible to deduce the following form of the anticommutator
$$\{t^{A},t^{B}\} = \frac{2N}{d}\delta^{AB}\cdot 1_{d} + d_{ABC}t^{C} + M^{AB}$$
where
$$
\mathrm{Tr}[t^{A}t^{B}] = N\delta_{AB}\\
d_{ABC} = \frac{1}{N}\mathrm{Tr}[\{t^{A},t^{B}\}t^{C}]
$$
and the object $M^{AB}$ satisfies a number of properties
$$
\mathrm{Tr}[M^{AB}] = 0,\quad M^{AA}=0,\quad \mathrm{Tr}[M^{AB}t^{C}] = 0,\quad M^{AB} = M^{BA}, \quad (M^{AB})^{\dagger} = M^{AB}
$$
The second last property expresses the orthogonality of $M^{AB}$ to the generators $t^{A}$ showing that it is not contained in the algebra. In the case of the fundamental representation $M^{AB}=0$ as the degrees of freedom have been exhausted (or alternatively; the generators and the identity span the full space of Hermitian matrices).
In the case of the adjoint representations of $SU(2)$ and $SU(3)$ I performed an explicit calculation of $M^{AB}$, verifying the properties above.
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What is meant by the word "length" in definition of surface tension? Surface tension is defined as the force applied per unit length. What is that "length" belonging to? I can imagine force being applied per area but not length.
"It would take a force of 72 dynes to break a surface film of water 1 cm long." Film is two dimensional. Isn't it?
My textbook didn't describe that what is that length of? I have surfed for it but didn't found what that word exactly mean there.
| Image you have a U shaped piece of wire with a soap film inside it, and a straight piece of wire inside the U:
The red line is the stright piece of wire that is free to move in the U - in effect a 2D version of a piston.
The surface tension of the soap film $\gamma$ pulls on the wire so it produces a force $F$ on it where $F$ is given by:
$$ F = 2\gamma \ell $$
with $\ell$ being the length of the wire (the factor of two is because a soap film has two surfaces and each surface pulls on the wire).
This is why the surface tension is a force per unit length. If you consider the force produced by the surface on some straight element then the force is the surface tension multiplied by the length over which the surface tension acts.
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Earth as a conductor In a lightning rod or other safety devices, charge is directed towards the earth.
How the surface of the earth can be used as a conducting path despite being composed of material that is not a good electrical conductor?
| Small currents do pass for example when we want to create charges in a metallic sphere by induction.Earth has both insulating and conducting minerals like aluminium oxide,organic matter and aluminium iron etc respectively and moisture in the soil will do more good by dissolving ions.Talking about electric discharge from air to ground,Earth acts as a large capacitor with about 30,000 volts which acts on the broad clouds high up in the sky creating the discharge.
Also I would like to mention that a lightning rod is attached to a metallic plate placed deep inside the earth crust for such conduction,the metallic plate adjust to the potential created inside earth.This plate is mostly composed of soft iron coated with copper to prevent rusting.
When lightning occurs the charges accumulates in this plate which then gets dissipated to the ground as it has larger surface area in contact with it and thus there is less resistance.All the conducting materials that I mentioned in first paragraph helps in this dissipation.In earth many conducting paths can be followed up which also contributes in decreasing resistance just like parallel combinations.
| {
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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?
| It seems more complicated but I would just take the force which is pushing the water out (Gravity) minus the force by the difference in air pressure.
| {
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What is the root cause of elasticity of a material? I know that there exist some interatomic and intermolecular forces in the material but why does stretching a material will enhance attractive force over repulsive force and vice versa.
| The bonds between atoms will 'break' if they are pulled apart even a short distance; however, the angles between adjacent bonds can be distorted a fair amount. If the molecules have a zigzag chain structure, or better yet a helical chain structure, you can stretch the material more without breaking the bonds. If the chains are anchored near the ends (cross-linked) then the network can restore its original shape and the material is 'elastic'.
| {
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Moment of inertia of solid cube about body diagonal How do I find the above mentioned moment of inertia?
Steps I've tried:
1.) Triple integrations that proved to be to big.
2.) I noticed that the if we split a $2\times 2\times 2$ into individual $1\times1\times1$ components, the body diagonal of the $2\times 2\times 2$ either passes through or is parallel to body diagonals of the $1\times 1\times 1$ cubes.
If the moment of inertia of the $1\times 1\times 1$ about body diagonal be $I$, then the moment if inertia of the $2\times 2\times 2$ about its body diagonal will be $8I$ because it has 8 times as much mass. I though I could get an equation in terms of $I$ by equating $8I$ and moment of inertia of the individual $1\times 1\times 1$ cubes about the body diagonal of the $2\times 2\times 2$ using parallel axis theorem but it turns out to be greater than $8I$. Could someone resolve my mistake?
P.S: I am not familiar with the concept of tensors.
| One of the big tricks that you find helpful is that the inertia matrix of a sphere is of the same form of that of a cube. I'll describe what this means.
For a sphere, imagine the moment of inertia (MoI) about three perpendicular axes through the centre of mass. Each MoI will be the same due to symmetry. Now, rotate the three axes you were just using, and the MoI about each axis is still the same:
Now, the trick says you can do the same for a cube! So, you can easily calculate the MoI about an axis going through the centre of mass perpendicular to one of the faces of a cube. So you know the MoI about three perpendicular axes. Now, you can rotate the axes such that one aligns with the diagonal, like such:
And, like a cube, the MoI about the new axes will be the same as the MoI about the old axes, and so you can say: the moment of inertia about a body diagonal of a cube is the same as the moment of inertia about an axis through the centre perpendicular to one of the faces.
The reason this trick works is because a cube and a sphere have inertia matrices of the same form, and performing rotation operations on those matrices causes no change on either matrix, i.e. rotating axes does not change moment of inertia.
| {
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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 first expansion term is $\langle p_f | V |p_i\rangle$. If you assume energy and momentum conservation between initial and final state, namely $E_f = E_i$ and $p_f=p_i$ then the only possibility for the above to be non-zero is in correspondence of
$$
\delta(E_f-E_i)\delta(p_f-p_i)\langle p_f | V |p_i\rangle = \delta(E_f-E_i)\langle p_f | V |p_f\rangle.
$$
| {
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How do you determine if the spin is up or down? Fundamental particles such as quarks and leptons can have a spin either up or down. These spins are (obviously) opposite of each other. But what differentiates them? Let's say you examine a pair of electrons and you find out they're opposite (one up and one down). How do you know which one is up and which one is down? Do all up-spinning-particles in the universe point to an established direction? If not, how can we know the difference between an up and a down spin?
| There's no universal notion of "up" or "down", just like there's no universal notion of "left" or "right" in the universe.
Speaking of "spin up" and "spin down" involves two completely arbitrary choices - a choice of axis and then a choice of "up" and "down" along that axis. You can swap the meanings of "up" and "down", and you can choose another axis to speak of "up" and "down" along. It's just a choice.
As to how to distinguish "up" and "down": For example, if you have a constant magnetic field in a direction and put an electron into it, the electron states with spin "up" along the axis of that field will have a different energy value from the states with spin "down", since the magnetic field essentially couples to the spin "direction" through the magnetic moment. Which one of these you call "up" or "down" is still somewhat arbitrary, but you can distinguish them.
| {
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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?
| Black holes emit radiation through what is essentially a modified case of the Casimir Effect. That is what we know as Hawking Radiation. Most black holes would only emit in the microwave band, which is why a stellar black hole will outlive most of the matter in the universe. It also must be taken into consideration that they are already moving very quickly, as everything else is when taking the CMB as the frame of reference. All of that said, just due to the orders of magnitude we're dealing with, there is likely no way to effect the speed of evaporation due to hawking radiation. Black holes actually do emit more radiation as they gain mass, just as a smaller proportion of overall mass, because you are evaporating something with a bare minimum of one solar mass (smaller ones have never been observed in any way, though they could exist) with low-energy photons.
The pedantic answer, however, which is clearly what you're after, is yes. If you hucked the entire solar system at one for a tight gravitational slingshot, you might speed up the entire 10^80 year process of black hole evaporation by a couple of femtoseconds.
| {
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Photon absorption and Sight If light is made by emitting photons and photons are absorbed by our eyes to see light
then i have this question:
if there is one person in the room and he looks at a light source (L) with x photons emitted
then there is another room and there are 10 people who look at light source (L) which has same amount x photons.
then do the people in the room with 10 people see the light source x/10 darker as in 1/10 of the photons emitted are absorbed by each person?
basically i think my question boils down to can two people view the same photon of light??
| Two people cannot view the same photon of light, because it is destroyed in effect upon being absorbed. However this does not decrease the overall brightness of the light source if more than one person observes it because your eye was not going to capture those photons anyway. Think about it this way. The brightness of the Sun here on earth is not changed by the presence of other planets (that are not directly in front of the Sun), because the earth is in a certain path to capture a certain set of photons of the Sun. The same is true for your observers in the room.
| {
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Velocity-time-distance problem In my book the formula for the $y$-component of velocity during the upward projectile motion is given:
$$V_y=V_{iy}-gt$$
and next to it the formula for $y$-component of velocity during the downward projectile motion is given, differed only by a conjugate,
$$V_y=V_{iy}+gt$$
I think it must have been:
$$V_y=gt$$
since $V_{iy}=0$ at the maximum height of projectile. I took time arbitrarily same for both conditions i.e. $t$.
Am I right?
| It's customary to use small $v$ for velocity (but not mandatory).
Notations like $v_y$ and $v_{iy}$ make little sense without defining them. Here it implies the velocity component along the $y$-axis but without specifying this axis nothing makes much sense either.
So I'll define the $y$-axis as a vertical axis, pointing upwards. This sense is important too, as we'll define the scalar of any vector pointing in that direction as positive and the scalar of any vector pointing in the opposite direction as negative:
Having cleared this up, the expression:
$$v_y=v_{iy}-gt,$$
now makes perfect sense because $\vec{g}$ points downwards. The index iy of course refers to initial.
It's also the only expression (for speed) we need. Assuming $\vec{v_{iy}}$ pointed upward then $v_y$ is positive as long as:
$$v_{iy}>gt,$$
after which the object starts falling back to Earth and $v_y$ becomes negative in accordance with the convention agreed upon higher up.
Note also that if $\vec{v_{iy}}$ was pointing downwards (an object thrown downwards from a
building for instance) the expression for $v_y$ still returns the correct result.
The expression can be further generalised as:
$$v_y=v_{iy}+at.$$
For numerical computations, the correct signs then have to be assigned to the values of $v_{iy}$ and $a$, in accordance with the directions of the vectors $\vec{v_{iy}}$ and $\vec{a}$.
| {
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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?
| my profesor of structure of matter told us that is widely done in Russia, where there is a surplus of explosives and so the price is low.
The procedure consists in filling a container with graphite on the borders and with explosives in the center, and make the expolsive blow up.
This procedure will produce diamonds $100-200 \, \mu m$ thick, and not pure, so is suitable for industrial grade diamond production.
So if you was wondering to create a huge diamond for a ring to give to your gf at the science fair you're better choose another method ;)
| {
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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).
| If a conductor is ungrounded, then the charge on the conductor cannot change (unless touched by an electrically charged object).
Now, we know that every charge has electric field around it. So, if a charge is brought near an ungrounded conductor, (we are assuming that the conductor is uncharged), the will tend to develop electric field inside the conductor. But, we know that we cannot have any electric field inside any conductor, so the charges in conductor will align themselves so as to neutralize the electric field inside it. (But, the net charge inside the conductor is still zero).
| {
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Energy of classical ideal gas in the grand canonical ensemble The canonical partition function for an ideal gas is
$$
Z(N,V,\beta) = \frac{1}{N!} \left(\frac{V}{\lambda^3}\right)^N
$$
where $\lambda = \sqrt{\frac{\beta h^2}{2 \pi m}}$ is the thermal De-Broglie wavelength. It is straightforward to obtain
$$
\langle E \rangle = -\frac{\partial \log Z}{\partial \beta} = \frac{3}{2} N k_B T .
$$
From $Z$ the grand-canonical partition function is
$$
Q(\mu,V,\beta) = \sum_{N=0}^\infty \frac{1}{N!} \left(\frac{e^{\beta \mu} V}{\lambda^3}\right)^N = e^{\frac{e^{\beta \mu} V}{\lambda^3}} .
$$
The average particle number is
$$
\langle N \rangle = \frac{\partial \log Q}{\partial (\beta \mu)} = \frac{e^{\beta \mu} V}{\lambda^3} .
$$
To get the average energy we should do, substituting $\langle N \rangle$,
$$
\langle E \rangle = - \frac{\partial \log Q}{\partial \beta} = \frac{3}{2} \langle N \rangle k_B T
$$
but this is true only if we magically ignore the $e^{\beta \mu}$ factor when taking the derivative, otherwise there is an extra (nonsensical) term. I've checked a few sources and this is the accepted solution (after all, it must be this one to be consistent with the canonical ensemble result), although they mysteriously gloss over the issue, so I'm missing something. Thanks.
| You may take derivative by assuming constant fugacity i.e $e^{\beta\mu}$. Because the Main formula for calculation of partition function in grand canonical ensemble is $Q=\sum\,e^{-\beta(E-\mu N)}$. So when you want calculate mean Energy or $\langle E\rangle =\sum p_i E_i =\sum\frac{e^{-\beta(E_i-\mu N_i)}}{Q}E_i $.
We know that $p_i=\frac{e^{-\beta(E_i-\mu N_i)}}{Q}$. so you should take derivative of $\hbox{ln}(Q)$ in a way that for each term just $E_i$ come beside $p_i$ it mean you should take derivative in constant fugacity or constant $e^{\beta\mu}$
| {
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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)?
| No, it's not possible, sorry. This is because blurring (or more generally, convolution) is a lossy operation, meaning that information is lost when an image is blurred, such that it can never be completely retrieved. While there are ways to sharpen a blurred image, these are either very non-trivial or else they're only approximations - there's no way to sharpen an image such than when it is later blurred it will return to its original appearance.
From a quick skim of the article you linked to, I don't think there's any claim that what you suggest is possible. Could you have misunderstood something perhaps?
| {
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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 usual rule of thumb in flying is that ground effect starts to appear within about one wingspan of the ground. av8n always has a good intuitive explanation.
He explains it as a mirror-image aircraft beneath you.
In flight training, it becomes important in soft-field takeoffs, where you transfer weight to the wings at as slow an airspeed as possible and get into ground effect.
Then you accelerate to climb speed.
Aside:
In Lindbergh's flight from New York to Paris, at times he made use of ground effect to save fuel.
At a height above the water of 30 feet, more or less, its height is automatically stable, and it just putts along at low power like a sewing machine.
| {
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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?
| Infrared and radio radiation has the same origin as X rays. When a photon 'crashes' with an electron, that electron gains the energy from the photon (if the photon has high frequency it gains a lot of energy, if it has low frequency it gains little energy). Then, the electron goes to a more energetic energy level, farther away from the nucleus. Then, the electron sends off a photon and goes back to the original energy level. That applies to all kinds of radiation. X rays, infrared and radio radiation are all photons, of different frequencies.
| {
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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?
| Right Ascension and declination are angular measures, RA * 15 gives degrees &c.
Basically, RA is where the siderial zenith is at any time, declination is the distance from the zodiac. You multiply RA by 15, and get angles.
Neither of these are distance, you have to get a radial term from somewhere else. What they are are coordinates to find the points in the sky.
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Is moment of inertia numerically additive?
NOTE: The following argument is being made for square only, not any general shape.
We have this square plate:
From Perpendicular Axis theorem,
$$I_1 + I_2 = I_z \\ I_3 + I_4 = I_z$$
Also, $$I_1 = I_2 \\ I_3 = I_4$$
Therefore, $$2\ I_1 = I_z \\ \text{and }\ \ 2\ I_3 = I_z \ \ \ \ \ \ \ \ $$
Then we get $$I_1 = I_2 = I_3 = I_4 \\ I_1 + I_3 = I_z$$
Although we get this to be true numerically, My instructor told me that last equation is wrong even for square. He said that it doesn't follow any known theorem. But I think this reason is somewhat unsatisfactory.
Q Are the last two results really correct (for square) (at tensor level) ?
| The equations you derived, especially the last two are true only for the square shape. Because it has inherent symmetries as mentioned by ja72 and others. But your instructor told you that those equations were not true for squares also. He probably did some mistake. The most probable case is the following:
We know that moment of inertia actually depends not only on mass but also on distribution of mass around the considered axis. Now in your picture, clearly, the mass distribution around I1 is till distance a on both sides(assuming the length of the square to be 2a). But for I3 the distribution is till 1.4a. So how is it that they have same Moment of Inertia.
The answer to it is actually pretty simple. It's because the mass distribution is different in the two cases. In case of I1, the mass is equally spread throughout the distance, whereas in I3, the mass is more closer to the axis but decreases linearly with distance. That's why the two M.I. turns out to be same.
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Does Quantum Mechanics say that anything is possible? I may be incorrect in this, but doesn't Quantum Mechanics state that everything has a probability of occurring (with some probabilities so low that it will essentially never happen)?
And if I am correct in my thinking, could it mean that, quite literally, anything has a possibility despite a potentially low probability?
(A version of my idea, which I admit closely borders on fictional magic, would be:
It is completely possible to walk through a wall, but the probability is so low that you need to walk into a wall for eternity before it can be done.)
| You have probably heard a garbled version of Murray Gell-Mann's totalitarian principle:
Everything not forbidden is compulsory
In quantum mechanics some processes are forbidden usually because they violate conservation laws. This is what CuriousOne refers to in his comment. Gell-Mann's principle states that unless a process is expressly forbidden there is a non-zero probability that it will occur. That probability may be exceedingly small, but in principle if we wait long enough the process will occur.
As CuriousOne says, this does not mean anything is possible because many things are expressly forbidden.
Re walking through a wall: you and the wall are too large to be easily described using quantum mechanics. At this scale decoherence is effectively instantaneous and you need to take this into account when calculating the behaviour of the system. You cannot apply a simple tunneling calculation to get the probability that you will tunnel through the wall. having said this, I would guess the probability that you will pass through the wall is greater than zero, though by such a small amount that you would need to try repeatedly for many times the age of the universe for it to happen.
| {
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Lens combination Suppose there are two lenses touching each other as shown in the diagram below. the focal length of the lenses are $f_1$ and $f_2$ respectively. Now by using the formula for lens combination I obtained the final focal length as $$\frac{f_1f_2}{f_1+f_2}$$.
Now suppose I assume there is a very small distance between them say d,where $d\ll R$, and I calculate the focal length of the system I still obtain the same result. Now my question is whether I can conclude from this that when ever two optical devices(lens,prism,etc) are touching each other. Can I do the calculation by assuming there is a small distance between them?
| If there is a small gap between two optical components you have to do your calculations for a small separation. If the elements are touching each other then you have to set the separation $d$ to $0$.
If I understand your question correctly then the answer is yes, you can set the distance to $0$ before or after solving your system, due to continuity. Here, "before" means that you write your equations assuming that the components are touching each other, and "after" means that you account for a small gap, solve your equations and then set $d=0$ in the final result.
Click here to consider the case of 3 or more lenses, and try putting their separations to $0$.
| {
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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?
| Photons can be regarded as stable particles depending on perspective. In the Sun's core hydrogen atoms break down into a mixed plasma which includes protons and deuterium atoms. If these collide they form a helium-3 nucleus and a gamma ray is released. The gamma ray is a highly energized photon. It is very difficult for this photon to escape the Sun as it is surrounded by densely packed protons. It bounces around for a while (maybe a thousand or a million years) before it gets to the surface. Due to losing some energy from collisions with protons the photon will now be in the visible spectrum. Then if it is "aimed" towards Earth it will take about 8 minutes through the vacuum to get here but not in an exact straight line as space-time is curved by all of the matter nearby (planets etc). The photon does not move up and down with the waveform however. The photon is represented by a waveform because that it how we measure it. For us, with our measurements the photon is the waveform. One Wavelength = One Photon. The wave moves as one. If I threw a coat-hanger at you it would not take on the trajectory of its shape. The entire coat-hanger would move as one. Hope this breaks your perception that a photon travels along itself. You don't walk along yourself. All the best.
| {
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Operator Product Expansion I wonder why in OPE in CFT terms like
$$ \frac{:O(z) O(w):}{(z-w)^2} $$
occur, for example in the OPE of Energy-momentum tensor with itself:
$$T(z) T(w) = \frac{c/2}{(z-w)^4} + \frac{T(z)}{(z-w)^2} + \frac{\partial T}{z-w}$$
Here we have term $:\frac{T(z)}{(z-w)^2}:$, where for free scalar field $T(z) = :J(z)^2:$.
So my question is why such term $:J(z)^2:$ occurs, because all field operators are considered as correlation functions, so:
$$ < T(z) T(w) > = < : J(z)^2: :J(w)^2: >$$
And with help of Wick's theorem we can decompose this correlator of 4 field operators to product of correlators like $<J(z) J(w)> <J(z) J(w)>$ and no terms with "bare" operators and one correlator ($:J(z) J(w): <J(z) J(w)>$) occurs here.
EDIT: I think that when we write OPE we don't consider field operators as operators under correlation function. And when we evaluate this correlation, only correlation functions remain and no operators.
But then the question turns to be: Why do we need such OPE in terms of another operators? While evaluating correlation function of $T(z) T(w)$ all we need is the first term of OPE.
EDIT: Well, the answer is that we need OPE when we are interested in correlation functions with another operators, which are located far away from operators OPE is constructed for.
| The question (v4) is a bit unclear, so here are some general comments, which may help:
*
*In CFT, an OPE is calculated via a Wick-like theorem changing an (implicitly written, operator-valued) radial order ${\cal R}(\hat{A}_1\ldots\hat{A}_n)$ into combinations of c-number-valued contractions and operator-valued normal order $:\ldots:$. Wick's theorem is e.g. explained in this Phys.SE post.
*Such calculations are organized into zero contractions, single-contractions, double-contractions, and so forth. E.g. in string theory, the central charge in the $TT$ OPE is a result of double-contractions.
| {
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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?
| It quite simply means that the radial direction is orthogonal to the tangential direction.
The universe in which we live behaves, as far as we can tell and measure, as a 3D volume (time aside for now:). A 3D volume is defined by 3 orthogonal directions. We can choose to look at a 2D cross-section of this volume, and we get a 2D plain, in which, every point is defined by 2 orthogonal direction. One can show these directions can be X\Y, but also R\Teata. R is the radial direction, and Theata is the tangential direction.
Because these directions are orthogonal, BY DEFINITION, it's not hard to understand that if you measure the radial component of a purely tangential scale (the deliverance of the curl), of the other way around, you get nothing - because radialiry is not tangential, and tangentiality is not radial :)
| {
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Current from Middle Battery in a Two-looped Circuit
With this question, as with many tutorials of similar questions I’ve found online, my textbook only mentions three currents: $I_1$, which flows through the left loop from and to the 19 V battery, $I_2$, which flows through the right loop from and to the 19 V battery, and $I_3$, which flows through the middle section.
However, why can’t there be an $I_4$ which flows from and to the 12 V battery through the left loop, and an $I_5$ which flows through the right loop? In this diagram, it seems pretty clear that all the current comes from the 19 V battery. But about the current coming from the 12 V battery? Is there no current? (The subsequent analysis completely discounts the presence of any current from the 12 V battery, so I’m thoroughly confused.)
| Forget about the circuit details of the batteries and resistors for the moment and focus just on the circuit layout: You have three paths by which current can travel from the top to the bottom of the shown circuit. So you need three independent parameters, $I_1$, $I_2$, and $I_3$ to fully describe the current flow in this circuit. The directions of the arrows that you associate with $I_1$, $I_2$, and $I_3$ don't matter. If the arrow associated with, say, $I_1$ is drawn in the wrong direction then you will find that the current $I_1$ is negative when you work out the result, which would mean that the current if flowing in the opposite direction to that indicated by the arrow. So when you say "it seems pretty clear that all the current comes from the 19 V battery", that's actually not so. There is no assumption being made in the diagram that the directions of the current flows are determined by just the 19 V battery. Yes, the directions that the three arrows are pointing in are all consistent with a view of all the current being due to the 19 V battery, but that's not being assumed here. The current arrows can point in arbitrary directions. The only requirement is that you have to be consistent and keep the directions of all the arrows the same as you work though the problem.
| {
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What will happen to a human being exposed to Martian atmosphere? Mark Watney, in the movie The Martian, says that,
If the HAB breaches, I'm just gonna, kind of... implode.
The corresponding novel, by Andy Weir, says he will explode (as pointed out by @MikaelSundberg).
I think he will neither explode, nor implode, but simply die of cold and asphyxiation.
Can anyone scientifically explain what will happen?
PS: The HAB is a NASA designed habitat for humans on Mars.
| The Martian atmosphere is effectively vacuum. He would be unconscious in less than 20 seconds and the he will end up freeze dried.
| {
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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?
| When you are standing upright imagine a downward external force on your head and an equal magnitude upward force on your feet.
These two forces will compress you and you “feel” being compressed as a result of these two forces acting on you.
Imagine that you are being spun round in an apparatus as shown in @JohnRennie ‘s diagram.
In his example there is an external force on you due to the “floor”.
In order for all parts of your body to have an inward force so as to rotate your body must be compressed.
So you “feel” compressed and interpret this as being acted upon by two forces one of which is real and causing the centripetal acceleration, the force of the the floor on you, and another force which is not real, an outward force which is supposedly helping to compress you - a centrifugal force.
| {
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How to show period is defined by $T=dS/dE$ (V.I. Arnold Mathemtical Physics) I'm looking at a book by VI Arnold on mathematical physics and I've hit a roadblock pretty early on. I'll quote the question:
"Let $S(E)$ be the area enclosed by the closed phase curve corresponding to the energy level E. Show that the period of motion along this curve is equal to $T=\frac{dS}{dE}$."
Here the phase curve is a plotting the solutions to a system with one degree of freedom. An example of this would be the equation for simple harmonic motion (mass and spring constant k are set to one): $$\ddot{x}=-x$$
It's solution is defined by concentric circles about the origin, where each circle defines a particular energy level E. Anything that could possibly point me in the right direction would be very helpful.
| Notice that the "speed" in the phase space is given by
\begin{equation}
v_{\mathrm{ps}} = \sqrt{\dot{q}^2 + \dot{p}^2} = \sqrt{H_{p}^2 + H_q^2}.
\end{equation}
Here, $H=H(p,q)$ is the Hamiltonian, and $H_p$ and $H_q$ are shorthand for $\partial H/\partial p$ and $\partial H/\partial q$. Provided that there exists a closed constant-energy curve in the phase space corresponding to $H(p,q) = E$, the period is given by
\begin{equation}
T(E) = \oint_{\partial D(E)}\frac{dl}{v_{\mathrm{ps}}} = \oint_{\partial D(E)}\frac{dl}{\sqrt{H_{p}^2 + H_q^2}} ,
\end{equation}
where $dl$ is an infinitesimal length element along the curve. The symbols $D(E)$ and $\partial D(E)$ respectively denote the region bounded by the constant-energy curve and the curve itself.
Next, the area of the region $D(E)$ is given by
\begin{equation}
S(E) = \int_{D(E)} dpdq.
\end{equation}
One can make a change of variables from ($p,q$) to ($E,l$), where $l$ for a given $E$ parametrizes the constant-energy curve such that $dl$ is its infinitesimal length element. One should in principle calculate the associated Jacobian determinant. But an easy way to work around this is to first note that the area element of the phase space can be written as $dl_E\, dl$, where $dl_{E}$ is the length element in the direction perpendicular to a constant-energy curve. It is related to $dE$ as follows:
\begin{equation}
dl_{E} = \frac{dE}{|\nabla H|}=\frac{dE}{\sqrt{H_p^2 + H_q^2}}.
\end{equation}
Therefore,
\begin{equation}
S(E) = \int_{E_{\min}}^{E}dE^{\prime} \oint_{\partial D(E^{\prime})} \frac{dl}{\sqrt{H_p^2 + H_q^2}} .
\end{equation}
Comparing the expressions for $T(E)$ and $S(E)$ leads to the desired result.
| {
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Closing a switch in series with a capacitor Suppose we have the following circuit:
Such that for $t<0$ the switch M was open. If we close the switch at $t=0$ what will the voltage on the capacitor, $V_C$, be at $t=0^+$? What about $\dot V_C$ at $t=0^+$? Will there be a current passing through $R$ the moment the switch is closed?
I need to solve an ODE for a more complicated circuit which has this sub-circuit as a part of it, and I need initial conditions to solve for the ZIR case. I'm trying to figure out these initial conditions but I'm not sure. Here are my thoughts:
Before closing the switch, there will be a steady finite current $I$ in the circuit.The moment we close the switch, there can be no current passing through $R$, else there will be some finite voltage on the capacitor $\longrightarrow$ $\dot V_C$ will be infinite $\longrightarrow$ $I_C$=C$\cdot \dot V_C$ will be infinite which cannot be.
Therefore $V_C(t=0^+)=0$ and $\dot V_C(t=0^+)=I/C $. I'm not sure if what I said I correct, I mean we learned that if there is no impulse current (like Dirac's Delta function), the voltage on the capacitor will be continuous.Does this apply in the case too? I would really appreciate any help.
| There are some mistakes with your assumptions. When $t<0$ a current $I =V/R$ will be flowing through resistor and no current would be flowing through capacitor. As soon we close the switch the capacitor will get charged instantaneously (yes it could lead to $I = \infty$ at $t=0$ but it can be avoided if even a small resistor is placed between capacitor and voltage source. And we can never get the resistance of those wires down to $0$). After $t=0$ a potential difference will develop across capacitor due to its charge. But a same current $I = V/R$ through the resistor as to satisfy Kirchhoff's loop rule in loop containing resistor and source. And because charge on capacitor will be constant after $t=0$ so will be the potential difference so $\dot{V_c}$ will be zero, not infinite and so will be the current through capacitor.
| {
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Correct Yukawa Term with a SU(2) Higgs Triplet? Given $SU(2)$ doublet fermions $\Psi^1$ and $\Psi^2$ and a $SU(2)$ triplet Higgs $H$, how does the correct Yukawa term look like in tensor notation?
Schematically, we have
$$ 2 \otimes 2 \otimes 3 \stackrel{!}{=} 1$$
and
$$ \Psi^1 \otimes \Psi^2 \otimes H $$
A first guess could be $\Psi_i^1 \Psi_j^2 H_{ij}$, but we have $ 2\otimes 2 = 3 \oplus 1$ and therefore the naive tensor product $\Psi_i \Psi_j = M_{ij}$ is not an element of the $3$-dimensional representation, but belongs to the reducible $ 3 \oplus 1$. This means in abstract notation
$$ \Psi_i^1 \Psi_j^2 H_{ij} \hat = (3 \oplus 1) \otimes 3 \neq 1 $$
Therefore, we must first remove the singlet $1$ from the product $2 \otimes 2$ before we multiply the $\Psi_1 \otimes \Psi_2$ term with the triplet Higgs. In other words we must make sure our product $\Psi_1 \otimes \Psi_2$ yields an element of the $3$, because the only chance to get a singlet is by multiplying the triplet with another triplet $3 \otimes 3 = 1 \oplus ...$.
In tensor notation this last step, as soon as we have formed a $3$ from the two doublets, is ordinary matrix multiplication and taking the trace.
Two guess:
*
*Is there some additional tensor $\eta$ involved that makes sure we get an element of the $3$ from two doublets, i.e.
$$ 2\otimes 2= 3 \hat = \Psi^1_i \Psi^2_j \eta_{ijkl} ?$$
*
*Or do we simply need to remove the trace
$$ 3 = \Psi^1_i \Psi^2_j - \frac{1}{2} Tr(\Psi^1_i \Psi^2_j) ? $$
| The first thing to notice is that composing two spin 1/2 particles (doublets) symmetrically yields spin 1 (a triplet) and anti symmetrically, spin 0, a singlet. So to suppress the singlet component in the Clebsch -Gordan reduction, you just symmetrize. What you are seeking has already been written down in 1964, in the celebrated σ-model of Gell-Mann and Levy, in the context of isospin, for one isodoublet, namely $g\overline{\psi}(\sigma +i\vec{\tau}\cdot\vec{\pi} \gamma_5)\psi$, an SU(2) invariant.
You see that the scalar term is antisymmetric, by fermion interchange, but the pseudoscalar one is symmetric, so isotriplet, dotted onto the isotriplet of π s. So all you need is $i\overline{\Psi}^1 \vec{\tau}\cdot\vec{H} \gamma_5~\Psi^2$. But since I don't know the context of your construction, I cannot seek better γ-matrix implementations suitable for your application: the important thing is that they should result in a symmetric fermion bilinear.
| {
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What is electron? In quantum mechanics, we easily talk about some "particles" or "somethings" like electron and photon. Besides, in classical mechanics we talk about particles that have mass. As we know, one can formulate axiomatically the meaning of a particle in Newtonian mechanics. Is there any similiar approach to understand "what is electron" in quantum mechanics? Some people say the electron is something that has mass and charge, and we can talk about its wave function. Moreover, every "atoms" have natural number of electron. I guess I cannot understand what is electron.
| There was a lot of temptation and attempts to present the electron as a wave or as a wave packet. But the problem arose, because such waves and wave packets are quickly destroyed because of dispersion.
But there is a case where the dispersion disappears. This is the case when the group velocity of wave propagation becomes equal to the phase velocity.
However, the equations of quantum mechanics that existed up to now did not allow one to obtain such a wave without dispersion. Because the resulting dispersion relations did not have a point of intersection of the phase and group velocities.
Pay attention to the interesting work where this problem is solved.
http://vixra.org/abs/1710.0239
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Applying Kirchoff voltage law to a short circuit If you consider an ideal wire with no resistance that shorts an ideal battery, the only voltage drop that exists is the emf of the battery, with nothing to balance it.
Obviously in the real world such a scenario is impossible, for the wire will have some resistance, but in this ideal example, is KVL not violated?
| No, it doesn't break. All it means is that all the voltage is on the ideal wire. According to Ohm's law there will be infinite current to account for the zero resistance. That's what happens in a real circuit - if you short a battery the current is very high.
| {
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Expansion of wave function and energy in terms of small parameter In time-independent perturbation theory, the Hamiltonian is perturbed with a perturbation of the form $\lambda V$, and the eigen-energies and wave-functions of the unperturbed Hamiltonian are expressed as series in powers of $\lambda$:
$$E =E^{(0)} + \lambda E^{(1)} + \lambda^2 E^{(2)} +...,$$
$$\rvert n \rangle = \rvert n \rangle^{(0)} + \lambda \rvert n \rangle^{(1)} + \lambda^2 \rvert n \rangle^{(2)} +...$$
But why is the same parameter $\lambda$ used both in the expression $\lambda V$, and in the expansion? It makes sense that in the limit $\lambda \to 0$, $E \to E^{(0)}$ and $\rvert n \rangle \to \rvert n \rangle^{(0)}$. But is there a formal way to see why the expansion works generally? I feel like it is some sort of a Taylor expansion.
I am looking for a more formal reasoning as to why this expansion in powers of $\lambda$ would be appropriate in this correction. Maybe there's a way to derive it. Even the Taylor series for a function has a formal derivation; I am looking for a similar argument as to why the eigenkets and eigenvalues depend on $\lambda$ in a power series expansion, and not some other way.
| The perturbation is usually written in the form $\lambda V$ to make its dependence on a small parameter explicit. Clearly, the eigenstates and eigenvalues of your hamiltonian must depend on $\lambda$ as well and therefore one may hope that it can be expanded around $\lambda=0$ (yes, you can think of it as a Taylor expansion). The existence of such an expansion requires that the perturbed solutions are smoothly connected to the unperturbed solutions, which is an assumption made here. Perturbation theory may fail, e.g. if a phase transition occurs.
| {
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Sliding along a circular hoop: work done by friction
Assume a point object of mass $m$ slides along a hoop of radius $R$, starting from a position which makes 90 degrees with the line of radius connecting the center and the ground. Let the coefficient of kinetic friction between the hoop and the object be $\mu$. Assuming that the object starts at rest, what is the total work done by the friction when the object comes to the ground level?
My idea: the normal force at any instance is given by $$N=mg\sin\theta+\frac{mv^2}{R},$$ where $\theta$ is the angle between the radial line connecting the present position and the intital position of the object to the center of hoop. With this we have the frictional force as $$f_k=\mu\left(mg\sin\theta+\frac{mv^2}{R}\right),$$ so that the total work done by friction is $$W_k=\int_0^{\pi/2}\mu\left(mg\sin\theta+\frac{mv^2}{R}\right)R\mathop{\mathrm{d\theta}}.$$
The problem I am having is to figure out $v$ as a function of $\theta$, i.e $v(\theta)$. Any ideas?
|
Set up an equation of motion for the rotation of the mass around the centre-point.
$$\tau=I\alpha$$
Where:
$\tau=mg\cos\theta-\mu mg\sin\theta$
$I=mR^2$
$\alpha=\frac{d\omega}{dt}=\omega\frac{d\omega}{d\theta}$
So:
$$mg\cos\theta-\mu mg\sin\theta=mR^2\omega\frac{d\omega}{d\theta}$$
$$R^2\omega d\omega=g(\cos\theta-\mu \sin\theta)d\theta$$
Integrate between $0,\pi/2$ and $0,\omega$ to get an expression of $\omega^2$ in $\theta$. Then use $v=\omega R$.
Then calculate the gain in kinetic energy: $\Delta K=\frac{mv^2}{2}$ (*) and the loss in potential energy $\Delta U=mgR$. The difference between the two is the friction work.
(*) Or use $\Delta K=\frac{I\omega^2}{2}$.
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How does one show Maxwell's equations in vector calculus form describe the same motion in all reference frames? The covariant form of Maxwell's equations is Lorentz invariant.
$$\partial_{\alpha}F^{\alpha\beta} = \mu_{0} J^{\beta}$$
$$\partial_{\alpha}F_{\beta\gamma} + \partial_{\beta}F_{\gamma \alpha} + \partial_{\gamma} F_{\alpha \beta}=0.$$
However, in Griffiths book on electrodynamics, he says that Maxwell's equations in vector calculus form will allow us to predict the motions of stuff in a way that's the same in all reference frames, although what one considers a magnetic field in one frame may be considered a electric field in another.
Can someone show how to prove this?
| The "motion of stuff" is not the same in all reference frames.
Each frame has it's own coordinate system, and so the physical process you are looking at will be described by different variables. (This is true even when you think of Galilean coordinate systems).
What Griffiths means is that you can use the vector form of Maxwell's equations along with the relativistic equations of motion for the particles involved in any frame you like, and they will describe the same physical process (just from the point of view of two different frames).
A couple of points to aid understanding of this:
*
*If two reference frames are used to describe the same physical process then you should be able to transform the solution in one frame into the other frame and get the same answer.
*Maxwell's equations are Lorentz invariant (they "look" exactly the same when written down on paper!) even when they are written in vector form. To prove this you have to transform all variables in those equations (fields, current-density and charge-density) using the appropriate Lorentz transformations for those variables. The reason the tensor notation is used is because you never have to worry about proving this if the law can be written in tensor form you already know it's Lorentz invariant.
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Quantum circuit and Witness preserving amplification for QMA My question is about the witness preserving amplification scheme for QMA proposed by Marriott and Watrous in 2005 (see http://arxiv.org/abs/cs/0506068 and http://www.cims.nyu.edu/~regev/teaching/quantum_fall_2005/ln/qma.pdf).
The original (quantum verifier) circuit $A$ takes as input $|\psi\rangle \otimes |0^k\rangle$ (where $|\psi\rangle$ is the witness and $|0^k\rangle$ the auxiliary qubits) and it ouptuts (on its first row) the answer of the computation.
The amplifying circuit is defined using $A$ as follow:
So it repeats several times the following two measurements (see section 3 of http://arxiv.org/abs/cs/0506068, or http://www.cims.nyu.edu/~regev/teaching/quantum_fall_2005/ln/qma.pdf):
*
*apply $A$, measure if output qubit of $A$ is 1, apply $A^{\dagger}$
*measure if auxiliary qubits are $0$
What I don't understand is how the auxiliary qubits could not be equal to 0. The initial state is $|\psi\rangle \otimes |0^k\rangle$ and then $AA^{\dagger} = I$, so we should always have state $|\psi\rangle \otimes |0^k\rangle$ at the second step (and so always obtaining Yes at this step)?
| Since we measure the output in step 1, we do not apply $AA^\dagger$, but rather
$$
A\vert0\rangle\langle 0\vert_1 A^\dagger
$$
or
$$
A\vert1\rangle\langle 1\vert_1 A^\dagger\ ,
$$
depending on the measurement outcome,
which is generally not equal to the identity.
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Do free-electron lasers actually lase? Free-electron lasers are devices which use the motion of highly energetic electron beams to produce bright, coherent radiation in the x-ray regime. More specifically, they start with a high-energy electron beam and feed it into an undulator, which is an array of alternating magnetic fields designed to make the electron beam move in a 'zigzag' path with sharp turns on either side, emitting synchrotron radiation during each turn.
The radiation thus produced is added up over successive turns, and it is produced coherently via self-amplified spontaneous emission.
One common question frequently posed of this setup is: is this actually a laser? That is, does it have population inversion, and is the radiation actually emitted via stimulated emission in some suitable understanding of it? People in the know tend to answer in the affirmative, but I've yet to see a nice explanation of why - and definitely not online. So: do FELs actually lase?
| If you use "population inversion" as an essential part of the definition of what a laser is, then you're right it's not a laser. But that doesn't deny that the properties of light can be just light from any "laser." So, this is a bit of semantics issue, and less about the physics.
Some other "non-laser" coherent light sources: Optical Parametric Oscillator, laser without inversion, etc
| {
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"timestamp": "2023-03-29T00:00:00",
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How exactly does a bulb light up?
A typical value for the electron drift velocity in a copper wire is $10^3\ \mathrm{m\ s^{-1}}$. In the circuit below, the length of the copper wire joining the negative terminal of the batter to the lamp is $0.50\ \mathrm{m}$.
(i) The switch S is closed. Calculate the time it would take for an electron to move from the negative terminal of the battery to the lamp.
(ii) The lamp lights in a time much less than that calculated in (e)(i). Explain this observation.
In the second part, I can't imagine the situation. I reckon that not all electrons travel with a drift velocity, so the faster ones reach the bulb and make it glow.
But how exactly does this lighting happen? The electron comes in contact with the circuitry and lights up the bulb, or is it because of the electric field?
| A single electron takes some time to move from the battery to the bulb but the lamp lights up faster than that. The reason is due to the fact that it is not the electrons travelling from the battery that light up the lamp when it is first lit, rather due to nearby electrons. The electric field is set up almost instantaneously in the circuit due to movement of electrons from their initial position all over the wire,(at the speed of nearly c, depending on the medium) and the electrons nearer to the lamp pass through the circuit lighting it. So even if the drift velocity of the electrons is so small, the lamp gets lit up.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Negative powers of operators This may sound like a strange question, but just to be sure: Suppose I have a general Hermitian operator in Hilbert space whose action on an eigenvector is given by $R|r\rangle = r|r\rangle$. Then, I assume that the following is true:
$\frac{1}{R}|r\rangle = \frac{1}{r}|r\rangle$ and similarly for other powers $R^{-2}|r\rangle = r^{-2}|r\rangle$
Does this follow immediately from the action of the operator, as in the case $R^2|r\rangle = RR|r\rangle = r^2|r\rangle$ or does this have to be defined?
| If you have $R|r\rangle=r|r\rangle$ then you have, where $I$ is the identity operator,
$|r\rangle=I|r\rangle=R^{-1}R|r\rangle=rR^{-1}|r\rangle$ and you immediately have
$R^{-1}|r\rangle=\frac{1}{r}|r\rangle$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to move a bubble which is trapped by the capillary pressure? I have a question about how to move a trapped bubble in a tube.
If we assume to have a horizontal tube, with water on each side of the bubble. The point to the left of the bubble is point 1, while the point to the right is point 2.
The capillary pressure equation is: $\Delta P_{cap}=\frac{2\cdot \sigma \cdot cos(\theta)}{R}$
Where $\theta$ is the contact angle, $\sigma$ is the interfacial tension between the gas buble and the water, while $R$ is the radius of the tube.
Since the bubble is trapped, I assume that pressure in point 2 ($P_2$) is larger than the pressure in point 1 ($P_1$). In order to accomplish this, the contact angle between the bubble and point 2 must be smaller than the contact angle betweenthe bubble and point 1, or the radius of point 2 must be smaller than the radius of point 1:
$$P_2-P_1=\frac{2\sigma cos(\theta_1)}{R}-\frac{2\sigma cos(\theta_1)}{R}=?\tag1$$
My question is therefore how we can get this bubble moving by applying more force on the left side (I want to move it towards the right)? If we apply more force from the left side, won't that increase the pressure $P_2$ the same as $P_1$, because of the capillary pressure, so that $P_2$ will always be larger than $P_1$ (if not the contact angle changes),and the bubble cannot be moved with more pressure?
| The stuck bubble problem happens when there is static equilibrium between the circumferential contact forces of surface tension and the forces due to the pressure difference upstream and downstream of the bubble. The gas pressure inside the bubble is uniform and between the upstream and downstream liquid pressures. In that equilibrium state there is a liquid flow from upstream to downstream of the bubble.
There are three things you can do to unbalance the equilibrium and get the bubble to move: 1. increase upstream liquid pressure. This will initially cause the bubble size to decrease but eventually the contact surface tension will become less than the pressure force. 2. Reduce the downstream pressure. This will initially cause the size of the bubble to expand, but eventually the contact forces will become smaller than the pressure force. 3. Finally you could add a surfactant to the upstream liquid flow. This would reduce surface tension when it reaches the bubble and unbalance the equilibrium.
There may also be another way. The stuck bubble problem is one of laminar flow. If you somehow create pulsatile flow and achieve the Womersly frequency the flow becomes turbulent. This may also unbalance the equilibrium.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why is the Fourier transform more useful than the Hartley transform in physics? The Hartley transform is defined as
$$
H(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty
f(t) \, \mbox{cas}(\omega t) \mathrm{d}t,
$$
with $\mbox{cas}(\omega t) = \cos(\omega t) + \sin(\omega t)$.
The Fourier transform on the other hand is defined very similar as
$$
F(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty
f(t) \, \mbox{exp}(i \omega t) \mathrm{d}t,
$$
with $\mbox{exp}(i \omega t) = \cos(\omega t) + i \sin(\omega t)$.
But although the Fourier transform requires complex numbers it is used much more in physics than the Hartley transform. Why is that? Are their any properties that make the Fourier transformation more "physical"? Or what is the advantage of the Fourier transformation over the Hartley transformation?
| *
*On the set ${\cal L}_1(\mathbb{R})$ of integrable functions $f:\mathbb{R}\to \mathbb{C}$, we can define the sine and cosine transforms
$$\tag{1}\left\{\begin{array}{ccc} ({\cal C}f)(\omega) &:=& \int_{\mathbb{R}}\! dt ~\cos(\omega t) f(t),\cr
({\cal S}f)(\omega) &:=& \int_{\mathbb{R}}\! dt ~\sin(\omega t) f(t). \end{array} \right.$$
*Next, consider the reflection
$$\tag{2} ({\cal R}f)(t)~:=~f(-t),$$
and note that
$$\tag{3}\left\{\begin{array}{ccc}
{\cal C}{\cal R}f &=& {\cal C}f, \cr
{\cal S}{\cal R}f &=& -{\cal S} f.\end{array} \right.$$
*Similarly, we can define the complex Fourier transform
$$\tag{4}\left\{\begin{array}{ccccc} ({\cal F}f)(\omega) &:=& \int_{\mathbb{R}}\! dt~ {\rm cis}(-\omega t) f(t) &=& ({\cal C}f)(\omega)-i({\cal S}f)(\omega) ,\cr
({\cal F}{\cal R}f)(\omega) &=& \int_{\mathbb{R}}\! dt ~{\rm cis}(\omega t) f(t)&=& ({\cal C}f)(\omega)+i({\cal S}f)(\omega), \end{array} \right.$$
or the Hartley transform as
$$\tag{5}\left\{\begin{array}{ccccc} ({\cal H}f)(\omega) &:=& \int_{\mathbb{R}}\! dt ~{\rm cas}(\omega t) f(t)&=& ({\cal C}f)(\omega)+({\cal S}f)(\omega) ,\cr
({\cal H}{\cal R}f)(\omega) &=& \int_{\mathbb{R}}\! dt ~{\rm cas}(-\omega t) f(t) &=& ({\cal C}f)(\omega)-({\cal S}f)(\omega). \end{array} \right.$$
*Clearly, the three integral transforms (1), (4) & (5) are bijectively related on ${\cal L}_1(\mathbb{R})$, and it becomes just a matter of convenience which one to use. DanielSank's answer already lists some of these conveniences.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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Why doesn't gravity mess up the double slit experiment? So let's say you are doing a double slit experiment. Also, let's use electrons.
My question is, won't the gravity of the electron affect the earth, thereby causing it decoherence and its wave function to collapse (or for MWI, entanglement and loss of information to the environment, preventing interference)?
The reason why I think this would happen is because you could tell which path the electron took based off its tidal effects on the earth: everything is a detector.
| "everything is a detector"
This can't be true, or else there would be no such thing as persistent entanglement.
As @Conifold points out, the electron's charge should be a far more potent source of environmental disturbance, anyway. Why doesn't the charge of the electron leave a trace as it passes through the slits - some persistent disturbance of the charged particles that make up the atoms that make up the filter?
The answer must be that the coupling in both cases (electromagnetic, gravitational) genuinely does not cause decoherence. In the case of gravitation, I would think it's just the extreme weakness of the interaction. In the case of electromagnetism, I'm not so sure.
This answer is a placeholder, written in a rush. I will return to it and improve it a few days from now, if no-one has written a better one.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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Why there aren't gluons with charges like $b b$ or $r r$? So my principal question is the one in the object above. But then I have another question related.
Supposing I have a $bg$ gluon.
How may I write it in the octet basis?
Thank you for your help!!
| Gluons carry color in color-anticolor pairs and form an octet and a singlet representation as seen below:
As you see the octed is filled with a color and an anticolor. The anti has to be a different color otherwise the gluon would be color neutral, and no strong force would be transferred.
Since to be exchangeable as gauge bosons in the strong interaction gluons have to be part of an octet , these are the acceptable combinations of color/anticolor. A bit similar to the fact that a photon has to be a is singlet, carries no charge.
There exist no free quarks or gluons to be able to set up states with higher color "spins" the way one can do with spin and isotopic spin.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why is the speed of sound lower at higher altitudes? At sea level the speed of sound is 760mph, but at altitudes like the Concorde would fly at (55,000ft) the sound barrier is at 660mph, so 1000th slower. Does it have to do with lower pressure?
| it has to do with the temperature lapse with altitude. since the speeed of sound is related to temperature by:
$a = \sqrt{\gamma RT}$, where $\gamma$ and R are gas properties and T is temperature
and the temperature profile follows (generally) like the left of these three plots:
The area of interest for airliners is in the lowermost region where the temperature is steadily decreasing with altitude (at a rate of ~6.5K/km)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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