Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
What is the relationship between entropy and work? Can someone explain the relationship between entropy and work? I've been reading my textbook and looking online but I feel like I'm missing something. Can someone explain it in layman's terms :)
| Let's frame this question in terms of a heat engine (Carnot engine).
Here is a diagram I made for a class when teaching this stuff.
Heat flow $\dot{Q}$ has an associated entropy flow $\dot{Q}/T$. The job of a thermodynamic engine is extract/filter as much useful work as possible from a flow of energy and entropy.
To answer your question in layman terms. Work is entropy free energy, it is what you have managed to extract/filter from a flow of heat by rejecting entropy. In this context entropy is a measure of inaccessible energy (i.e. a part of heat flow that cannot do work).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Where extra stored energy of Capacitor go ? Suppose We have a capacitor with capacitance $\ C $ and charge $\ Q $ . So total stored energy is $$ E=\frac{Q^2}{2C} $$ Now if I connect a capacitor with same capacitance parallel with it then current will flow until voltage across both capacitor become same and this case charges among them will be same . So each capacitor will contain $\ Q/2 $ charges. So energy stored in each capacitor will be $$ E'=\frac{Q^2}{8C}$$ . So total energy in 2 capacitor is $$E_{tot}=\frac{Q^2}{4C}$$ which is not clearly equals to $\ E$ . Where does the remaining stored energy goes ?? When does the loss occur ?
| Connecting two perfect capacitors like that would be like connecting two perfect but different voltage sources; you would get a hypothetical explosion.
In real life, every capacitor has inductance and resistance. So, as the current built up between the two capacitors, you'd heat up the wire between the capacitors as well as the capacitors themselves. If the inductance was high enough and the resistance low enough, you'd get oscillation that would be inevitably damped by the (always non-zero) resistance. In the end, you'd have somewhat warmer test equipment, and that's where your "lost" energy would be.
As a metaphor, consider a trough with a removable wall down the middle. Fill one half with water, and then calculate the potential energy. Then remove the wall, wait a bit, and calculate the potential energy again. You'll find that energy went away, because the average altitude of the water is lower. Where'd the energy go? It was dissipated into heat and sloshing sounds.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What does the g mean after the isotope given? I'm familiar with notation such as Sc-44m standing for the meta stable state of Sc-44.
What does Sc-44g mean? There are a few examples of this notation; here's one:
http://iopscience.iop.org/0031-9155/60/17/6847/pdf/0031-9155_60_17_6847.pdf
Many thanks
| The answer is already on page 2 of your link above:
"Among the large number of radionuclides of medical interest, Sc-44 is
promising for PET imaging. Either the ground-state Sc-44g or the metastable-state Sc-44m can be used for such applications, depending on the moleculeused as vector."
So the metastable state Sc-44m decays to the ground state Sc-44g.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why don't lakes have tides? There's a tidal effect that we can clearly observe in oceans, which is the effect of gravity from the Sun and the Moon. If gravity affects everything equally, why don't lakes have tides?
| You probably got voted down cause this can easily be google searched, but the simplest way to explain it is that a tide happens because the lunar tug on one side of the ocean is measurably more than on the other side of the ocean and as the earth rotates the tidal "bump" follows the moon so you get 2 high tides and 2 low tides a day.
A tide is effectively one very large wave. The distance from Peak (High tide) to Trough (low tide) is 1/4 the circumference of the Earth.
Lakes do have tides, but since all lakes are much smaller than 6,000 miles across, it's nothing like ocean tides. The wavelength still applies to all bodies of water, but the lake is so small compared to the wavelength that even a big lake would have a tidal rise and fall of maybe 1/3rd of an inch - too little for most people to notice.
Your bathtub also has tides, but now we're talking about maybe the height of a few atoms. Even the Mediterranean sea isn't large enough for significant tides. It has small ones.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
"answer_id": 0
} |
Books on waves with Fourier Transforms There are many waves and oscillations books out there that also include Fourier analysis but very few give the subject a thorough treatment, they just pass it in a few pages. If anybody has any sources(particularly books) that have Fourier analysis and particularly Fourier Transforms, I would appreciate if he could share his information with me.
Waves, Oscillations, Quantum Mechanics or Mathematics books are all ok if they have an intuitive Fourier analysis in them.
| The FourierTransform.com is a website maintained by an enthusiast. The site is not peer-reviewed, but it looks as though it might provide helpful explanations.
Here's a link which provides some basic introduction to the Fourier transform. And here is another link to class notes provided by Prof. Carlton M. Caves for an introduction to the Fourier transform.
Robert N. Bracewell, did some work on Fourier analysis regarding images. He authored two works on the subject:
Bracewell, R.N., The Fourier Transform and Its Applications (McGraw-Hill, 1965, 2nd ed. 1978, revised 1986)
Bracewell, R.N., Fourier Analysis and Imaging (Plenum, 2004)
Fredric J. Harris, Professor of Electrical Engineering at U.C. San Diego, wrote a paper on the use of windows with the discrete Fourier transform for harmonic analysis.
Professor Tom Körner is a mathematician and the titular Professor of Fourier Analysis at Cambridge University. He wrote an incredibly complete "introduction" to the subject - simply entitled "Fourier Analysis". The introduction of a review of the book states:
It is not an easy read - but it is really very good, and considered a classic.
These references range from the introductory to the specialized, but I hope you find something helpful in them.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/203382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is a relativistic quantum theory of a finite number of particles impossible? In Dyson's book Advanced Quantum Mechanics , he said "These two examples (the discovery of antimatter and meson) are special cases of the general principle, which is the basic success of the relativistic quantum theory, that A Relativistic Quantum Theory of a Finite Number of Particles is Impossible." However, when we calculate the Feynman diagram of particles collision process, the number of particles should be finite.
So my question is why a relativistic quantum theory of a finite number of particles is impossible. What does it mean explicitly? Does it mean that we need to consider infinite number of harmonic oscillators when we quantize free field?
| Because a pair of particle and anti-particle can be created from the vacuum, it means that infinite number of pairs of particle and anti-particle can be created from the vacuum. So when you consider relativistic quantum theory it's impossible to only consider a finite number of particles.
When you calculate Feynman diagram, you are actually only doing perturbation theory to some order. If you want to calculate the exact result by Feynman diagrams, then you need to consider infinite number of Feynman diagrams, because you can add loops to the diagrams. And it means that you need to consider infinite number of particles.
When quantize a field, we do have to consider an infinite number of harmonic oscillators. What do I mean by this is that if you look at the expression for the quantum field, say,a scalar field, you will find that it's a superposition of infinite creation operators and annihilation operators.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/203490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What do elements of CKM matrix imply? In CKM matrix, there are 9 elements, e.g. Vud = .974, Vus = .227 ,Vub = .004. The sum of these 3 elements is greater than one, so they cannot represent the probability of an up quark to transform in an interaction/decay into down, strange, bottom quark respectively by emitting W+ boson. Then what do these elements imply?
| https://en.wikipedia.org/wiki/Cabibbo%E2%80%93Kobayashi%E2%80%93Maskawa_matrix
$$ \begin{bmatrix} d^\prime \\ s^\prime \\ b^\prime \end{bmatrix} = \begin{bmatrix} V_{ud} & V_{us} & V_{ub} \\ V_{cd} & V_{cs} & V_{cb} \\ V_{td} & V_{ts} & V_{tb} \end{bmatrix} \begin{bmatrix} d \\ s \\ b \end{bmatrix} $$
The d,s,b quarks are eigenstates of mass. The d',s',b' states are states that W,Z leave within their own SU(2) doublets. For example, $W^+$ takes d' into u and nothing else. So, the CKM matrix V says d' is a linear combination of d, s, and b. The coefficients are "amplitudes" not probabilities. The probabilities are ${|amplitude|}^2$. V is unitary ($V V^{\dagger} =I$) to conserve probability. Therefore,
$$
|V_{ud}|^2+|V_{us}|^2+|V_{ub}|^2=1
$$
$$
.974^2+.224^2+.004^2=.999
$$
This is 1 within experimental error and can be used to argue that another generation of quarks is not needed. You can see all the experimentally measured CKM values with their errors at the above link. By calculating $V V^{\dagger}$ you can check for yourself how well conservation of probability is satisfied.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/204113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rockets and distance I am trying to create an equation which allows for me to change the aspects of the rocket so i can calculate the distance traveled vertically. My idea is for a rocket that only moves vertically; with this i can calculate the amount of time it would take to make it past the first Lagrange point. So far I have made an equation which calculates the acceleration of the rocket, this being:
$$
\frac{32T}{W_0+F-Bt}-\frac{Gm_e(W_0+F-Bt)}{r^2}
$$
$T$=Thrust
$W_0$=Initial weight
$F$=initial weight of fuel
$B$=Burn rate(lbs of fuel per second)
$t$=time in seconds
$G$=gravity constant
$m_e$=mass of earth
$r$=radius from earths center of gravity
Thank You for taking the time to read this.
| You are assuming constant thrust $T$ during flight, presumably until the rocket runs out of propellant.
You are also assuming (or neglecting) any air drag to be zero. The resulting drag force could be very significant at high speeds and assuming your rocket is launched from the Earth's surface.
With those limitations in mind the balance of forces of the rocket becomes:
$T=m(t)a+F_g$.
$a=\frac{T}{m(t)}-\frac{F_g}{m(t)}$.
Here $m(t)=W_0+F-Bt$ and $F_g=G\frac{m_em(t)}{r^2}$.
Substituting we obtain:
$a=\frac{T}{W_0+F-Bt}-G\frac{m_e}{r^2}$ (I'm assuming that by 'weights' $W_0$ and $F$ you mean masses).
So I don't know where your factor $32$ came from.
The total burn time is $t_b=\frac{F}{B}$.
Since as $a=\frac{d^2r}{dt^2}$, in principle we have a second order Differential Equation in $(r,t)$ which when integrated between $t=0$ and $t=t_b$ would yield the total distance traveled during thrust. Slightly rearranged it becomes:
$\frac{d^2r}{dt^2} + \frac{Gm_e}{r^2}= \frac{T}{W_0+F-Bt}$
But this is a non-linear Differential Equation and I don't believe it has analytical solutions, only numerical (iterative) ones.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What's the difference between the work function and ionisation energy? In a particular textbook, the work function of a metal (in the context of the photoelectric effect) is defined as:
the minimum amount of energy necessary to remove a free electron from the surface of the metal
This sounds similar to ionisation energy, which is:
the amount of energy required to remove an electron from an atom or molecule in the gaseous state
These two energies are generally different. For instance, Copper has a work function of about 4.7eV but has a higher ionisation energy of about 746kJ mol-1 or 7.7eV.
I've sort of figured it's because the work function deals with free electrons whilst ionisation is done with a valence electron still bound within the atom. Is the difference due to the energy required to overcome the attraction of the positive nucleus?
| Ionization energy is the energy required to remove the outermost shell electrons of an isolated atom of an element (gaseous phase). We define ionization energy when we have a single atom.
On the other hand, the work function is the energy required to remove the outermost shell electrons of a metal's surface atoms when they are involved in metallic bonding. The work function is the energy needed to take away METALLIC BONDED electrons. (It is obvious that all the outer shell electrons of a metal are involved in metallic bonding.)
The removal of the electrons from a sea of electrons needs less energy (generally) since they are attracted to many kernels at the same time, due to which the forces cancel out, resulting in free electrons.
Hence, the work function is generally lower than the ionization energy for a particular element.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
When does the concept of electric field in classical electrodynamics fail, and QED is needed? It is really hard to find reference to when the traditional concept of electric wave, especially TEM wave, fails, and needs to be replaced by quantum electrodynamics.
So when does the concept fail? At high frequencies of electric field?
| The failure of classical electrodynamics has to occur when we are talking about very low intensity light (such the wave is describing one photon) or when the frequency of the classical waves is very high (because a photon can decay to an electron-positron pair but an EM wave cannot).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Rectangular potential barrier Take the usual rectangular potential barrier, that is:
$$V(x)=0 \: \text{if} \: x<0 \: \text{or}\: \: x>a$$
$$V(x)=V_0 \: \text{if} \: 0\leq x \leq a.$$
I've looked at several notes and books and in everyone of them the books supposes, in the third region ($x\geq a$), a solution to the time-independent Schroedinger equation of the form:
$$\psi= Ae^{ikx}.$$
Advocating that "we don't want any wave travelling to the right". Now, it isn't immediately clear to me that the wavefunction can indeed be interpreted as a wave, and that is even travelling. What is precisely the boundary condition that permits me to say that the constant in front of the second exponential is $0$? If the wavefunction were normalizable then the usual condition of $\psi \rightarrow 0$ as $|x|\rightarrow \infty$ would do. But what in this case?
| To put it another way: typically with this sort of setup we want there to be an incoming wave from the left, $\psi_0 = e^{-i \omega t + i k x},$ which is partially "reflected" into an outgoing wave to the left $\psi_\ell = r e^{- i \omega t - i k x},$ and partly "transmitted" to an outgoing wave to the right, $\psi_r = t e^{- i \omega t + i k x}.$
The condition that these be travelling waves comes from the implicit $e^{-i\omega t}$ which is prefixing all of them, as a traveling wave has the form $f(x - v t)$ where $v$ is the velocity of the wave. The fact that this is the same for all three of these waves means that the scattering is elastic and is not consuming any energy, as it comes from the $i \hbar \partial_t \psi$ part of the wavefunction, enforcing that great Planck/Einstein law, $E = \hbar \omega.$ The $-i$ perfectly cancels the $i$, leaving just $\hbar \omega \psi = +\frac{\hbar^2 k^2}{2m} \psi = \frac {p^2}{2m} \psi.$
The condition that no wave is coming in from infinity on the right really is just saying "that's not what we're studying": we want only the left-hand-side to have an incoming wave so that we can study how the barrier reflects and transmits that wave. You could then have the barrier transmitting/reflecting two incoming waves, but it would be a superposition of their individual transmissions/reflections and would not be very interesting, once you've worked out the one-incoming-wave solution.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to Vary the wavelength of UV CFL? I have a $12$ $V$ $DC$ operated UV $[CFL]$(http://en.wikipedia.org/wiki/Compact_fluorescent_lamp) with $365$ $nm$ wavelength. I need to vary this wavelength in the $250-300-350-400-450-500$ $nm$. Please guide me regarding how to proceed with it.
| Mount the lamp on the outside of a really fast spaceship.
If you want a longer wavelength, fire up the engines of the spaceship and get it to travel away from you, so that the light is red-shifted.
If you want a shorter wavelength, do the same but with the spaceship travelling towards you, so that the light is blue-shifted.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help me understand Gauss law Suppose I have the following, the gaussian surface is the drawing in the middle. So charge enclosed is zero, and then eletric field must be zero since the area of the gaussian surface is not zero. But, clearly the eletric field is not zero in the middle, because if you put a charge there it will move. Why I'm getting the concept wrongly? Edit: you can consider the middle figure as a sphere.
| If the charge between the two surfaces is $0$ (zero) initially, that makes ${q}/{\epsilon} = 0$ and Gauss Law states that the Electric Flux is directly proportional to charge inside it, i.e.,
$\phi = \dfrac{q}{\epsilon} = \displaystyle\oint\vec{E}.\vec{dA}$ ,
Here $\vec{E}.\vec{dA}$ represents the dot product of the differential are and the electric field passing through it.
Even, if the flux is zero, $\displaystyle\oint\vec{E}.\vec{dA}$ becomes zero. Only if electric field and its area vector was always parallel i.e., always perpendicular to the surface, then the electric field can be claimed to be $0$ (zero).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Question Regarding torricelli's theorem/Law I recently studied about bernoulli's equation/principle.
After the derivation of the said equation , my book gave some applications of the principle, which include torricelli's theorem/law.
In deriving torricelli's law from bernoulli's principle, the pressure at the opening of the tank in which the fluid is contained , is said to be equal to the same pressure which is applied at the the top surface of the applied fluid , namely the pressure of the atmosphere.
But my book also states that the pressure drops (according to bernoullis principle) when the fluid passes through a narrow pipe or opening and its velocity increases.
So why does the pressure remain the same in this situation ? Why doesn't it change?
Any help would be much appreciated , THANKS.
Could you please answer in simple and easy to learn terms , Thanks AGAIN.
| I suppose that the best way to answer this question is by using an analogy. Take a glass of water(at room temperature) and place it in a refrigerator. What happens to the water? It cools down. Now you take this glass of cold water and keep it back outside. What happens now? The temperature of the water comes back to the room temperature. Why? Because the water is exposed to room conditions and hence through heat transfer, it's temperature rises up to that of the room temperature. Similarly, your fluid when flowing through a constricted passage(analogous to a refrigerator) experiences a drop in pressure(as your textbook states). But at the exit of the passage, it is once again exposed to atmospheric conditions and hence it's pressure rises back to the atmospheric pressure!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Where is the event horizon in a black hole? At the beginning I thought that the event horizon coincides with the surfaces, but then making a new name when you could just call it surface would seem a bit pointless.
Then where is the event horizon? Is it inside or outside the black hole?
Notice that I have a really basic knowledge about physics
| An event horizon is a "sphere-shaped surface of influence" of inescapable gravity from/towards the black hole. It is the point where the object (victim?) cannot overcome the gravity of the black hole, and will be sucked into it. This is also the point where, theoretically, you would vanish to an outside observer, since the light crossing the event horizon also will not escape to bounce back to the observer.
There is not one particular location/altitude for an event horizon, it depends on the size and mass of the black hole. The more the mass, the more the gravity, thus a larger event horizon.
Wikipedia has a nice writeup about them, although this article is more about event horizons in general, not just for black holes.
Edit: I use the term "Sphere of Influence" loosely -- you would enter the sphere of influence (meaning your motion would be affected by the black hole's gravity) before you would cross the event horizon -- you could return from the SOI, but you couldn't return once past the event horizon.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Stone dropped from a moving train
*
*This may look like a stupid question, but it is really getting to me. Imagine a train moving with an acceleration $a$, and a person drops a stone from the window. To an observer on the ground, the stone follows a parabolic path, as it is a projectile with initial velocity the same as the velocity of train when dropped. However, the person who dropped it sees that it falls down in a straight line. Why? Can someone explain the reason to me?
*What will the acceleration of the stone measured by an observer on the ground. I think it should be $a_{net}=\sqrt{g^2 + a^2}$, but I have a book which says it should be $g$. No explanation is provided. Help me out please.
| The path of the stone for the observer on the train should be a straight line as, when releasing the stone, it had the velocity same as train. However, air resistance may affect the stationary path. For the observer on ground it should be parabolic due to the action of gravity on the moving stone. The net acceleration of stone should be $g$ since after dropping it from the train it is only accelerated by gravity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
When I take a Gaussian surface inside an insulating solid sphere, why does the outer volume have no effect on the electric field? Say I try to find the magnitude of the electric field at any point within an insulating solid sphere. I know that in the case of a conductor, the electric field within it is 0. However, I have not learned anything about an insulator, so I assume that it would not be 0.
I used Gauss' Law and calculated the charge of the volume within the Gaussian surface, the radius of which is equal to the distance between the point of interest to the center of the sphere. So I got the right answer, but I want to know the physics behind it. Why does the remaining volume of the insulating sphere, which is just right outside the Gaussian surface, have no effect on the electric field at that point? Even to me, my question sounds flawed as I am pretty much asking why an insulator has no effect on an electric field. However, I just don't think it would be that simple.
| Who says the outside field doesn't effect.The Gauss's law gives the net field due to entire charges inside or outside the Gaussian surface only the charge taken is what inside.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the wavefunction of the Young Double Slit experiment? I have never seen the wavefunction for this experiment and would like to know how to derive it using the Schrodinger equation. I specifically want to see how the electron wave function leaves the source, then goes through the slits, and produces the characteristic diffraction pattern on the other end.
| Well, there are many things you could do. You could:
*
*consider two gaussian beams (the linked article is for electrodynamics)
*apply some paraxial approximation (which would be more appropriate to treat electrons with a high forward momentum)
*do a cheap/cheesy symmetric point source approximation using Green's functions.
I can do number three for you :)
If you take $\hbar=1$, $m=\frac{1}{2}$, then the equation in question becomes $i \dot{\varphi}+\nabla^2 \varphi=0$, which has a solution:
$$\left(\frac{a}{a+2 i t}\right)^{3/2} \exp \left( {-\frac{x^2+y^2+z^2}{2 a+4 i t}}\right) $$
Then you can add two of these point sources together and translate them:
$$\left(\frac{a}{a+2 i t}\right)^{3/2} \exp \left( {-\frac{x^2+y^2}{2 a+4 i t}}\right)\left(\exp\left( \frac{(z-h)^2}{2 a+4 i t} \right)+\exp\left( \frac{(z+h)^2}{2 a+4 i t} \right) \right) $$
The fact that these wave packets aren't moving is a bit of a cheat, but you can always "boost" to a moving frame by using the answer here: Galilean invariance of the Schrodinger equation (or if you're really on top of your quantum mechanics game you can apply the translation operator $e^{-i\hat{x}\cdot \hat{p}}$)
Voila, an appropriate wavefunction.
Here's an XZ slice of the initial condition $|\psi|^2$:
An XZ slice of $|\psi|^2$ at a later time and offset Y:
And an animation of $|\psi|^2$ (uploaded on imgur)
I used Mathematica to expand psi squared of the previous equation. You can see exactly where the interference comes in (the cosine term)
$$|\psi(x,y,z,t)|^2=\frac{\left(a^2\right)^{3/2}}{\left(a^2+4 t^2\right)^{3/2}} \cdot \left(\exp\left(-\frac{a \left(h^2+2 h z+x^2+y^2+z^2\right)}{a^2+4 t^2}\right)+\exp\left(-\frac{a \left(h^2-2 h z+x^2+y^2+z^2\right)}{a^2+4 t^2}\right)+2 \cos \left(\frac{4 h t z}{a^2+4 t^2}\right) \exp \left(-\frac{a \left(h^2+x^2+y^2+z^2\right)}{a^2+4 t^2}\right)\right)$$
So the oscillatory/important part is $\cos \left(\frac{4 h t z}{a^2+4 t^2}\right)$. This brings up the obvious problem with this approach: it doesn't directly provide the nice result that you usually want, relating the momentum of the particle to the "wavelength" of the interference pattern. The interference pattern reaches its maximum frequency at $t=\frac{a}{2}$, so I'll leave it as an exercise to the reader to see if there's a relation between the momentum ($\hat{p}^2$ maybe?), the de Broglie wavelength, and the usual peak/trough formulas diffraction formulas(this sort of thing)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Fluid speed and fluid density How does fluid density affect fluid speed?
Basically I am trying to figure out if, with all other quantities remaining constant, would an increase in fluid density cause the fluid speed to increase/decrease?
For example, would water and honey have different fluid speeds in a pipe, because their densities are very different? I know that:
$$Av = Av$$
and
$$P + ρgh + (1⁄2) ρv^2$$
But does an increase in density lead to an increase/decrease in fluid speed? How so?
| Continuity equation for $\textit{steady}$ flow, in which properties remain uniform over cross-section, is $\rho A v=$constant. If area remains constant along the flow then $v ~\alpha~ \frac{1}{\rho}$. For unsteady flow the statement is more complicated (read up compressible flows).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Coffee Straw Physics When I put my little, cylindrical coffee straw into my coffee, the liquid immediately rises about half a centimeter up the straw without provocation. This is also the amount of coffee that the surface tension of the coffee will allow to stay in the straw when removed from the liquid in the cup.
Keep in mind that all the while, the top end of the straw is open.
Why does the level of the liquid in the straw insist on being higher than the level of all the liquid in the cup?
| The liquid rises due to surface tension. In this case the adhesion between liquid and cup material Is higher than cohesion between liquid molecules. So it is higher than liquid in cup.
I think the liquid in the straw that remains after removing will be lesser than the liquid that you saw rising while in the cup. Try for transparent liquids. The liquid in the case you mentioned is purely due to adhesion and not surface tension forces.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Why don't non-Hermitian operators with all real-eigenvalues correspond to observables? Suppose you could construct an operator that was non-Hermitian but had all real eigenvalues or could at least be restricted in a way to create only real eigenvalues, why would this operator not correspond to an observable quantity?
| 1) If all the eigenvalues of an operator are real, then it is Hermitian. You can see this by writing the operator (call it A) in the eigenvector basis. Then A has all real eigenvalues along its diagonal and zeros everywhere else. Therefore, $A^\dagger = A$ which means it is Hermitian.
2) Many of the operators that we call "observables" are the generators of transformations. For example: J (angular momentum) is the generator of rotations $e^{i\theta J}$, P is the generator of translations $e^{ixP}$, and H is the generator of time translation $e^{itH}$. If these transformations are to be unitary to conserve probability, then the generator must be Hermitian. For example unitary means (where $\theta,x,t$ are real):
$$
(e^{ixP})^\dagger e^{ixP}=I
$$
$$
e^{-ixP^\dagger} e^{ixP}=I
$$
$$
-P^\dagger + P =0
$$
therefore P is Hermitian.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/207078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Physical interpretation of the creation operators in string theory? Is there any way to describe phsycially which each creation operator $a^{(i)+}_{n}$ in string theory does to the ground state string?
Here would be my guess (although it is likely to be totally wrong):
You can consider the ground state string as a string that is not moving. The creation operator $a^{(i)+}_{n}$ increases the oscillation harmonic in the direction $x^i$ by $n$. E.g. So that $a^{(1)+}_2$ acting on the ground state will do the following:
| The states of string theory are quantum states. They represent a "vibration" of the string in the same sense that a particle in standard QFT represents a "vibration" of the quantum field.
That is, they do not represent actual "physical" vibration at all. In particular, the states do not describe actual physical position, vibrations or whatever of the string. You may see the worldsheet as a propagator that turns states at one end into states at the other end - but these states live in Hilbert spaces associated to the strings, and do not describe anything about the string itself. Just like usual quantum field theory describes states in Hilbert spaces associated to spatial slices of spacetime, but that doesn't mean the spatial slice itself would behave/be formed in any particular way.
Furthermore, in some approaches to string theory, all you do is specify certain types of conformal field theories on the worldsheets, and the requirement is that their total central charge cancels to not have a Weyl anomaly. In this approach, you don't generally get the usual creation/annihilation operators, or it is at the very least not natural to interpret them as being related to any "vibration".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/207480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How big are clouds? How big are clouds? When I look up into the sky I have no frame of reference, so I don't know if they are 200 feet or 2 miles across. When I am in a plane looking out at a cloud, I try to use the wing as reference but I still don't have a good reference point, because the clouds are just a large white mass.
I realize that "cloud" is a very loose term, so interpret how you wish.
|
I don't know if they are 200 feet or 2 miles across
Clouds are fractal. Cloud particles can be a few dozens of micrometres, and big tropical cyclones can be thousands of kms across. That's a range of more than 10 orders of magnitude! That's why clouds are a pain to represent in models — it is simply impossible to have a physics-based model of a cloud with a domain large enough to represent a full cloud system.
Realistically, it is not possible to give a lower limit for the size of a cloud. When you "see your breath" on a cold day outdoors, that is fundamentally no different from the fog you see above a lake, which again is fundamentally no different from a large system. It's all liquid and solid particles floating in the air.
So, to answer your question: a cloud can be as large as thousands of kms across. There is no practically applicable lower size limit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/207632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 7,
"answer_id": 0
} |
Velocity of an Electron as it Passes through a Uniformly Charged Ring I've been presented with a problem in which an electron is placed a certain distance x from the center of a positively charged ring and allowed to move freely. The ring has a known charge density λ. I am tasked with finding the velocity of the electron as it passes through the center of the ring.
First of all, I know that when the electron is initially positioned it will have a certain electric potential energy and that as it moves in towards the ring that potential energy will be converted into kinetic energy until it reaches the center, at which point all of the potential energy will be converted into kinetic energy.
The best way that I can think of to find the initial potential energy is to first find the electric potential at that point and then translate that into potential energy via the equation $V = \frac{U_e}{q}$. However, I am not certain what charges should be used in each equation. I have tried using the total charge of the ring (extrapolated from λ) in the electric potential calculation, and the dividing the charge of the electron out of that expression to produce the electric potential energy. Which produces the following equations:
$$
V = k\int\frac{dq}{r} = k\int\frac{dq}{\sqrt{R^2 + x^2}} = \frac{kQ}{\sqrt{R^2 + x^2}}
$$
$$
U_e = \frac{V}{q_e} = \frac{kQ}{q_e\sqrt{R^2 + x^2}}
$$
$$
U_e = \frac{1}{2}mv^2
$$
$$
velocity = \sqrt{\frac{2U_e}{m_e}} = \sqrt{\frac{2kQ}{m_eq_e\sqrt{R^2+x^2}}}
$$
I'm told that this is an incorrect solution. Could somebody please explain to me where I went wrong?
| The first thing is that $U_e = q_e * V$. Then, electrostatic potential $V= \int {E.dr}$. For a ring, $E= \frac {kqx}{(x^2 + R^2)^{3/2}}$. The energy conservation otherwise is spot on.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/207727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Do relativistic events need to match if accounted for time dilation and length contraction? To explain the question let me give you a short example.
In the scenario there are two references frames A and B.
A consists of a x'=1 Ls (lightsecond) long pole in the positive x direction. At t=0 a flash is generated at its origin. 1s later the flash reaches the end of the pole.
B sees A moving with v=0.866c in the positive x direction. Due to length contraction, A's pole appears to only be 0.5 Ls long.
In B, 3.731s after A generated the flash the flash reaches the end of the pole, because:
$$x-ct=0 \quad with \quad x=v \cdot t+x'\sqrt{1-v^2/c^2}$$
$$(v \cdot t+x'\sqrt{1-v^2/c^2})-ct=0$$
$$(0.866c \cdot 3.731s+0.5)-3.731s \cdot c=0$$
So the flash reaches the pole's end after 1s in A. But from B's point of view it takes 3.371s.
Wouldn't this require a time dilation factor of 3.371? But the actual factor is $$\frac{1}{\sqrt{1-v^2/c^2}}=2$$
Based on a suggestion in a comment let me write out the problem more detailed:
For A: $$ct'-x'=0$$
$$t'=1s$$
$$c \cdot 1s - 1 (1c \cdot 1s) = 0$$
For B (for the formula of x see above):
$$ct-x=0$$
$$t = \frac{t'}{\sqrt{1-v^2/c^2}}=1s / 0.5 = 2s$$
$$c \cdot 2s - x \neq 0$$
| I think that your calculation is correct in that it would take $3.73$ seconds for the light pulse to reach the end of the pole according to B's perspective. However, that number is not the time dilation factor. For $v=0.866$, the factor should be equal to about $2 (= \frac{1}{\sqrt{1-(v/c)^2}})$.
So where did your reasoning go wrong? I think that the problem is your selection of a pole and then considering a light pulse in the positive x-direction. Why did you choose the positive x-direction? Why not send a light pulse in the negative x-direction instead? It should give the same answer if what you calculated was really the time dilation factor, right? But it doesn't. If you send a light pulse in the negative x-direction, you should that it takes less than 1 second for the light pulse to reach the end of the pole from B's perspective. Does that mean the time dilation factor is both less than 1 (for light pulses in the negative direction) and also greater than one (for light pulses in the positive direction)? Obviously, that's nonsense.
You can't use light pulses in either the positive or negative direction for your purpose because in addition to time dilation effects there is also a time contribution due to the fact that the far end of the pole is moving away (or moving towards) the light pulse. Try setting up your calculation so that instead you consider a light pulse moving in the y-direction (i.e., perpendicular to the direction of the motion of the light pulse and the observer B). That should give you the correct time dilation factor if you repeat your calculation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is the relation between energy levels of hydrogen atom in Bohr's solution to that of Dirac solution In Dirac solution for hydrogen atom, the energy levels are calculated as positive
\begin{equation}
E=\frac{mc^{2}}{R(t)\sqrt{1+\frac{z^{2}\alpha^{2}}{\left(n+\sqrt{\left(j+\frac{1}{2}\right)^{2}-z^{2}\alpha^{2}}\right)^{2}}}}
\end{equation}
, while in Bohr's model the energy levels are negative
\begin{equation}
E=\frac{-Ze^{2}}{8\pi\epsilon_{0}r}
\end{equation}
How are these two related to each other?
| I have found the answer myself here
Energy in Dirac model $E_d$ is related to energy in Bohr's model $E_b$ as
$E_b \approx E_d - m_ec^2$
where $m_e$ is mass of electron and $c$ is speed of light. The answer above is not useful.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is there a rigorous definition of 'much greater than'? I have encountered $\gg$ in many physics text books where it's used as a relation between constants or functions but in none of the text books I have read is it properly defined anywhere.
If $A \gg B$ where $A$ and $B$ are constants, or $f(x) \gg g(x)$ does this simply mean that $A \geq 10B$ and $f(x) \geq 10g(x)$ for all $x$?
Note: I'm asking this here because mathematicians don't use the much greater than relation. At least not that I know of.
| It is a symbol and an idea used in mathematics too. But the important part is just that $B$ is 'ignorable' relative to $A$. This depends on the level of precision that is being used experimentally. If you're working to a precision of 1 part in 100, then $B$ should not effect the answer to that level of precision. If you're working to 1 part in a million, then $B$ should not effect the answer to that level.
For example, we could say that general relativity is ignorable for putting a man on the moon, but not for running a GPS system. The more specific takeaway is to learn that there's no such thing as an exact number in physics. It's always $A\pm err$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 5,
"answer_id": 0
} |
How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles?
It would appear to a layman such as myself that these heavier unstable particles are just transient interplay of the stable forms.
| You're question is interesting because it is connected to the notion of elementary particle. As mentioned by anna v, the elementary particles (fermions) of the standard model have very specific properties under the symmetry of the standard model ($SU(2)_L\times U(1)_Y \times SU(3)_c$): they lie in the fundamental representation of the group, which in familiar language means that they are the building block of all other particles (baryons, mesons etc). This definition does not imply that elementary particles are necessarily stable. An electron, a muon and a tau (3 elementary particles, leptons) have exactly the same property under a transformation of the symmetry group. Nothing distinguishes one from the others except their mass. Heavier particles can decay into the lighter ones.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 7,
"answer_id": 4
} |
Are white noises always Markovian? Are white noises always Markovian? I am a bit confused about it. As white noise always has a constant power spectrum， its auto correlation function must contain a delta function of time. Thus the correlation time of the noise vanishes. But I don't know whether they can be called Markovian.
| Mathematically, the answer to your question is yes. The dynamics of a physical system that is driven by pure white noise, with constant power spectrum up to arbitrary high frequencies, will be perfectly Markovian.
However, as CuriousOne points out, it is essentially impossible to verify that a physical noise process is truly Markovian, because of the finite time or frequency resolution and range we can achieve with our measurements. Furthermore, many of the most common sources of noise are demonstrably not Markovian, e.g. thermal fluctuations at temperature $T$ must have a memory time on the order of $\hbar/k_B T$ in order to satisfy detailed balance. In practice, this often does not matter: as long as the correlation time of the noise is much shorter than the time scales of interest, then the noise can be treated as memory-less to a good approximation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why can't be the EPR experiment simplified? Alice measures the spin of her electron on the x axis. She now knows the spin value of Bob's electron on the x axis at time T0. Bob measures the spin of his electron on the z axis. He now knows the spin value of Alice's electron on the z axis at time T0.
The two meet up and speak, knowing both the x and z values of their electrons at T0, contradicting the uncertainty principle, which states that an electron "doesn't have" both the values at the same time.
There is obviously something wrong in what I wrote. What's that?
| Under your assumption of simultaneously well-defined x and z value, you reach predictions which are inconsistent with quantum theory. This is exactly what leads to Bell's inequality which is (experimentally!) violated by quantum theory, see https://en.wikipedia.org/wiki/Bell%27s_theorem or the explanation in Preskill's lecture notes (http://www.theory.caltech.edu/~preskill/ph229/notes/chap4_01.pdf, Section 4.2).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Models for populations of decay products I'm looking to create a population model for the specific nuclides in a neutron spallation source. The source is a target (Tantalum clad Tungsten) which is being bombarded with protons, and in turn is producing neutrons. The change in a specific population of a Nuclide is going to be dependent on the initial population, the half life of the specific isotope, the isotopes which decay into it, and other factors.
Have models like this been made before, and are there any publications on the matter which anyone knows of?
| These sorts of calculations are part of the design process for spallation sources and are done quite carefully. The tool of choice is MCNP, which has been extensively benchmarked over many decades. MCNP is maintained by folks at LANL, which happens to have a tantalum-clad tungsten spallation target; if you are on-site at LANL you should invite someone from that team to lunch and see if they can steer you in the right direction.
You should also look for papers (and their authors, who also like to eat lunch) describing the SNS mercury spallation target. Since the SNS is so much younger than LANSCE, the design computations were much more sophisticated. Likewise the ESS will have a helium-cooled tungsten target and is actively being designed at present.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does $v^2=v_0^2-2gh$ work if the positive axis is up and the initial velocity is down? In the situation where the positive axis is up, the acceleration due to gravity is $g$, and the velocity is represented by the equation $v^2=v_0^2 - 2gh$. This works great if the initial velocity is upwards. But say, what if the initial velocity is downwards?. While we all know intuitively that the acceleration due to gravity would increase the speed (and thus $v^2$ would increase), this equation says that the speed (and thus $v^2$) would decrease. How is this justified?
| You are looking at a specific application of a more general formula
$$v_q^2-v_{oq}^2=2a_q(q-q_o),$$
where
*
*$q$ is the coordinate direction,
*the $v$ terms are velocity components along the $q$ axis,
*$a_q$ is the constant acceleration component along the $q$ axis,
*and $q$ and $q_o$ are the positions along the $q$ axis, which match, respectively with $v_q$ and $v_{oq}$.
For constant acceleration along the $q$ axis between locations $q$ and $q_o$, this always works, regardless of the signs of $v_q$ and $v_{oq}$ (because both of those get squared). You must pay attention to the sign of the acceleration and the signs of the positions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Derivation of $E=h\nu$? I found this question here and the answers got me thinking.
Take a case where $\Psi(x,t)$ is the linear combination of two eigenvectors of a charged particle:
$\Psi(x,t)=c_1\psi_1e^{-iE_1t/\hbar}+c_2\psi_2e^{-iE_2t/\hbar}$.
At $t=0$ the wave function is:
$\Psi(x,0)=c_1\psi_1+c_2\psi_2$.
The probability density distribution is:
$P(x,t)=\Psi^*(x,t)\Psi(x,t)$.
$P(x,t)=|c_1\psi_1|^2+|c_2\psi_2|^2+c_1c_2^*\psi_1\psi_2^*e^{-i(E_1-E_2)t/\hbar}+c_1^*c_2\psi_1^*\psi_2e^{-i(E_1-E_2)t/\hbar}$.
So $P(x,t)$ is (obviously) not constant but contains elements that oscillate with angular frequency $\omega=(E_1-E_2)/\hbar$ and frequency $\nu=(E_1-E_2)/h$.
Does this explain Bohr's equation: $E_1-E_2=h\nu$?
| As pointed out in zeldredge's answer, what is missing is the connection to the photon. However, I do think you can fix this problem. If you couple a system that exhibits natural oscillations at certain frequencies to an external force that oscillates at some frequency $\omega$, then you get a resonance when $\omega$ matches one of the natural frequencies. Conservation of energy then yields that in a quantum mechanical treatment of the force field, the quanta that appear must have energies of the form $\hbar\omega$, which is indeed the case for a harmonic oscillator.
So, what one would be assuming here is that the electromagnetic field can indeed be described in terms of harmonic oscillators.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/209033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What will I see if launch a thing into a black hole? Suppose that I launch a thing into a black hole from a secure distance, this black hole is secure at 2 meters and is floating over my yard, doesn't matter.
What will I see?
Will I see that the thing increases their speed and falls quickly into the hole?
Or will I see that the thing decreases their speed and falls slowly, every time more slowly?
| You'll see the object at first accelerate towards the hole (under gravity) and then slow more and more as it approaches the event horizon. It will asympotically freeze in place at the event horizon and then gradually shift redder and redder until it disappears.
This is assuming that the black hole is big enough that the acceleration is similar across the body. For small black holes the tidal effects would rip the object up because of the great difference in acceleration between the parts of the object that are closer to the black hole and those that are further away.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/209159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sound when traveling faster than sound I was wondering, if I am running at the speed of sound while playing music on my iPod will I be able to listen to my iPod while running at the speed of sound? or we cant hear anything while running at the speed of sound.
| Well..
1) If you were running at the speed of sound, you probably wouldn't be for long. The human body isn't designed to handle those kinds of stresses.
2) Assuming you're listening to the iPod using ear buds (in your ear) You can probably think of the air between the seal on the ear bud and your ear drum as isolated from the air you're running through, so no. You could still hear your music.
It is important to note that the air turbulence will probably create an extremely loud sound at such a high velocity. So while you may be able to hear your music in theory, in practice the music would likely be drowned out by the sound of rushing air.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/209450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does an ion thrust engine consume more energy as it speeds up? This question goes to a very basic non-understanding of mine that I have had in the back of my mind for ages - I just read the following here:
ion thrusters are capable of propelling a spacecraft up to 90,000
meters per second (over 200,000 miles per hour (mph). To put that into
perspective, the space shuttle is capable of a top speed of around
18,000 mph. The tradeoff for this high top speed is low thrust (or low
acceleration). Thrust is the force that the thruster applies to the
spacecraft. Modern ion thrusters can deliver up to 0.5 Newtons (0.1
pounds) of thrust, which is equivalent to the force you would feel by
holding nine U.S. quarters in your hand.
So when it hits the top speed what is the bottle neck? The logical thing to me is that it takes more and more electricity to maintain the 0.1 pounds of thrust, but if this is the case, does this not violate the premise that you cannot tell how fast you are going without something to compare to? In other words, if I turn the engine on and then off again repeatedly, should I expect different results from one time to the next?
I know I'm confused about something very basic here - that's why I'm asking..
| That page is not well written.
The 90km/s speed is the exhaust velocity of the engine.
It is not the maximum speed of the spacecraft.
There is no maximum speed of the spacecraft, short of the speed of light.
They make the mistake again when they say:
"While a chemical rocket's top speed is limited by the thermal
capability of the rocket nozzle"
ADDED: Chris Drost makes a good point.
If the initial mass of fuel is about 99% of the vehicle's total mass, then it can get up to about 4.6 times the exhaust velocity, or about 400km/s, before it runs out of fuel.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/209828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Definition of a line charge with Dirac delta function Is the following statement correct for a line charge distribution $λ(x)$?
$$ρ(\mathbf r)=λ(x)δ(y)δ(z)$$
If yes - what does it say?
| The factor δ(y) indicates that the charge distribtution is non-zero only for y=0, i.e. on the zx plane; likewise, δ(z) that it is non-zero only on the xy plane. Thefore, the product δ(y)δ(x) Indiactes that the charge distribution in non-zero on the intersection of zx and xy planes, i.e. the x-axis. Then, the function λ(x) defines the actual form of the charge density on the x-axis. Not that, since, δ functions are distributions in the mathematical sence, this charge distrubtion makes sence only under 2-D integrals over y and z for a given x. Such a proccedure will reduce ρ(r) to λ(x).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/209922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to mix a drink with a non-standard phase of ice? Would it be possible to safely cool and drink a glass of water with anything else than the Ih form of ice?
Here and here you can see that some alternative forms of ice have a higher density than water, hence they would sink.
Would it be possible to have the coolest party ever, where the attendees would drink from glasses where the ice is sinking instead of floating?
| As you can see from the phase diagram plot in the first link you provided, the only other ice phase which is stable at atmospheric pressure is ice XI, and its density is about the same as that of the most familiar ice phase (ice Ih). The other denser ice phases that you see on the phase diagram are only stable at pressures significantly above 1 atmosphere. As far as I'm aware, none of those high-pressure phases of ice are metastable, so you would have no chance of synthesizing any of those high-pressure ice phases in a high-pressure device (e.g., diamond anvil cell, Paris-Edinburgh cell, etc.) and then trying to retain the phase as you download the pressure back to atmospheric pressure.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What has the potential energy: the spring or the body on the spring? Particles have gravitational potential energy due to its position in the gravitational field. We say the particle has potential energy and not the Earth (the body doing the work). Why is it not the same with a spring doing work on a body?
It is my understanding that we can define a potential energy function for all systems being acted on by a conservative force. Since the spring force is conservative, why can't we define a spring (elastic) potential energy for the body? Why is the potential energy defined only for the spring?
| Potential energy is just energy stored in a static state -without motion. So a spring can have potential energy, and so can a body attached to the spring that's in a gravitational field. So for this type of system (undamped harmonic oscillator in a gravitational field) potential energy is not strictly defined for the spring. If the forces are conservative and energy is trapped within the system, and not lost outside the system, the energy will continue to flow between potential and kinetic energy states, and the spring and body can share the potential energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why isn't the acceleration at the top point of a ball’s journey zero? When I shoot a ball vertically upward, its velocity is decreasing since there is a downward acceleration of about $9.8\,\mathrm{ms}^{-2}$.
I have read that at the top most point, when $v = 0$, the acceleration is still $9.8\,\mathrm{ms}^{-2}$ in the downward direction where $v=0$. That is, the acceleration is still the same.
But at the highest point, the ball is stationary, so it is not even moving. How can it accelerate?
| At the topmost point, the velocity vector is a null vector whereas the acceleration vector has constant magnitude $-9.8\,\mathrm{m/s^2}$ and constant direction downwards i.e. towards the centre of earth.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 2
} |
Can tidal forces significantly alter the orbits of satellites? I would assume that there are other larger, more significant, forces acting on artificial satellites, but can tidal forces drastically alter the orbit of a satellite over time?
I was thinking this could especially be an issue for a satellite in geostationary orbit, because they have to be extremely precisely positioned. However, I could see this being an issue for satellites in other orbits as well, just not to the same degree.
| Yes, of course the tidal forces affect the orbits. In the case of the Earth/moon
system, Earth's day used to be 18 hours, and when the tidal slowing got
to our current 24-hour solar day(23 hours fifty-odd minutes sidereal)
the angular momentum went into the Lunar orbit, and the moon is more distant
nowadays.
That 18-hour day was the status quo about 900 million years ago. The
moon's orbit is expanding about 3.8cm per year, and Earth's day is growing
at 2 milliseconds per century. Whether you call this orbital change 'significant'
or not, is a judgment call. I'd say it is significant, because the Moon
wouldn't be so near unless it was created or captured more recently than the
birth of planet Earth.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Conservation of momentum in a baseball Conservation of momentum: A thought experiment.
A baseball is placed on top of a baseball holder, the kind used to train young batters.
A batter hits the stationary ball perfectly horizontal, sending it flying through the air in a relatively straight line, during which time it will eventually fall to the ground due to Earth’s gravitational pull. Ignoring everything but momentum, the total distance traveled by the time it hits ground is, let’s say 200 feet.
Now, a second baseball is instead dropped down in front of the batter from a certain height, straight down, at which time the player again hits the ball squarely on its side, not at an up or down angle, but perfectly horizontal, sending it flying in a relatively straight line through the air.
Will this second ball travel the same distance as the first one, and if so, what happened to the downward momentum the ball obtained before it was hit by the batter? Alternatively, does the initial downward momentum in fact affect the distance traveled?
My friends say that the initial downward momentum is most likely converted into heat by friction with the bat, rendering it incapable of affecting the trajectory of the struck ball. I say the initial downward momentum of the dropped ball must be conserved and that it does in fact make the ball fall faster to the ground, thereby shortening the distance traveled. Any thoughts would be greatly appreciated.
| Provided the bat delivers exactly horizontal momentum impulsively to the second ball, it will not travel as far due to its initial downward velocity, as you say. Dissipating the downward momentum doesn't make much sense in the scenario you described.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How is normal force distributed along the surface of contact? Help me settle this argument.
A mass $m$ is placed on a thin diving board.
The base of the diving board has mass $M>>m$.
Does the board tip over?
I drew the following FBD and concluded there is no net torque.
My friend thinks the normal force $N$ will be applied at a different $x$-coordinate, perhaps $x=D_1/2$. This would result in a net torque. How can I prove him wrong using the laws of classical mechanics?
Of course, the normal force is not applied at a single point. It is distributed across the entire surface of contact. Is it possible to compute the force distribution $N(x)$? We could measure this experimentally by placing many small scales under the base.
| I think that you're making this problem more complicated than it has to be in order to simply determine if the assembly will tip over or not. You don't really need the spatial distribution of the forces being exerted by the table or ground on the assembly. All you need to note is that if the pivot point is at x=D1 then the ground will exert whatever counter-torque is needed in order to prevent the assembly from rotating counter-clockwise about the pivot point.
So all you need to do is to calculate the torques contributed by mass m and mass M about the pivot point. If the sum of those two torques acts in the counter-clockwise direction, then the ground will exert a counter-torque of the same magnitude but in the opposite direction to prevent the whole assembly from turning counter-clockwise. On the other hand, if the sum of the two torques from m and M acts in the clockwise direction, the ground plays no role in providing a counter-torque and the whole assembly tips over in the clockwise direction around the pivot point.
Calculating the total torque due to m and M should not be difficult. For the purposes of calculating the torque due to M, you can assume just a single force of magnitude Mg (where g is gravitational acceleration) acting downward at its center-of-mass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Power and frequency of photons and its relationship with biological safety I understand that x-rays are more dangerous than radio waves because they are of higher energy, since they have higher frequency. However, it’s less dangerous to stand near a radio station with a higher power output than to be near an x-ray machine with a lower power output. (This is what I’ve been told, anyway.) Why is that the case? Is power not energy per unit time? Should a lower power indicate lower energy (and therefore greater safety) regardless of the frequency of the wave?
| There are two types of energy involved, and the blurring of this distinction is cause of a huge number of misunderstandings.
Light comes in discrete packets called photons. The energy of each photon is proportional to the frequency of the light. On top of that, a light beam can have any number of photons in it, and this gives it its overall power. The power of the beam is the energy it transmits per unit time:
$$
\text{power}=\frac{\text{energy}}{\text{time}}=(\text{photon energy})\times\frac{\text{no. of photons}}{\text{unit time}}.
$$
It is perfectly possible for an x-ray beam to transmit as much power as a radio beam, simply by having lots of radio photons and comparatively few x-ray ones.
Biological damage, on the other hand, is slightly different. The biomolecules inside cells that suffer radiation damage interact with individual photons, one at a time.
*
*If the photon energy is small, the biomolecule can usually dispose of that extra energy in a safe manner, even if there are lots of small photons around - each photon gets handled in turn, unless there really are lots of photons around.
*If the photon energy is big, on the other hand, the biomolecule needs to handle a large amount of energy in a single go, and this usually makes it change its chemical state into a bunch of pretty reactive fragments.
This is why it’s less dangerous to stand near a radio station with a higher power output than to be near an x-ray machine with a lower power output.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Central force law An object has an orbit in polar coordinates as $r(\theta) = a\theta^2$ (where $a$ is constant).
Assuming the central force is directed towards the origin $r=0$, how can I know which central force law lead to such an orbit? And how to find $r$ and $\theta$ as function of time?
| Perhaps I can help.
For any object in orbit, the Earth exerts a force on the object and the object exerts a force on the Earth.
So we know from $F=ma$ that:
$$∑ F_θ=ma_θ$$
Using polar coordinates {r,θ}, this equation becomes:
$$ 0=2\left(\frac{dr}{dt}\right)\cdot \left(\frac{dθ}{dt}\right)+r\left(\frac{d^2θ}{dt^2}\right) \implies \frac{1}{r} \left(r\left(2\left(\frac{dr}{dt}\right)\cdot \left(\frac{dθ}{dt}\right)+r\left(\frac{d^2θ}{dt^2}\right)\right)\right) $$
$$ 0=\frac{1}{r} \left(2r\left(\frac{dr}{dt}\right)\cdot \left(\frac{dθ}{dt}\right)+r^2\left(\frac{d^2θ}{dt^2}\right)\right) \implies \frac{1}{r} \left(\frac{d}{dt}\left(r^2\dfrac{dθ}{dt}\right)\right)=0$$
Integrating yields: $$r^2\dfrac{dθ}{dt}=h$$ where $h$ is the constant of integration
Therefore:$$\dfrac{dθ}{dt}=\dfrac{h}{r^2}$$
This proves Kepler's Second Law that the aerial velocity of a particle subjected to central-force motion is constant.
Remember that this also accounts for orbital eccentricity.
I hope this answers your question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/211105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Where does the force appear when considering object interactions in another reference frame? Imagine I am sitting on an asteroid with my buddy and drinking a beer. When the bottles are empty we throw them simultaneously in opposite directions perpendicular to the asteroid's movement. What will happen?
From the logical standpoint and from momentum conservation, our velocity should not change - the total momentum of two bottles is zero in the asteroid's frame of reference.
Suppose somebody is watching the asteroid from another reference frame (velocity not equal to zero). According to Newton's second law, the force is equal to the change of momentum over time. The mass of asteroid was changed (remember the bottles). The momentum was changed ($M\times V$). Where is the force?
| The technical answer is that the force "comes" from your buddy and you, who spend (chemical) energy to throw the bottles.
If we consider an idealised scenario, though, where a system of three glued points breaks suddenly down, then I will argue that there is actually no change in the momentum => no need of force:
1) Before the break-down the system's momentum is:
$$\vec{P}(1) = \vec{P}_{A}(1) + \vec{P}_{b_1}(1) + \vec{P}_{b_2}(1) = (M_A + m_{b_1} + m_{b_2}) \vec{V}$$
2) Suppose now that the balls $b_1$ and $b_2$ (I suppose they have equal masses) are thrown at opposite directions relative to the asteroid, say $\vec{v}$ and $-\vec{v}$. Therefore:
$$\vec{P}(2) = \vec{P}_{A}(2) + \vec{P}_{b_1}(2) + \vec{P}_{b_2}(2) = M_A \vec{V} + m_{b_1} (\vec{V} + \vec{v}) + m_{b_2} (\vec{V} - \vec{v} ) = (M_A + m_{b_1} + m_{b_2}) \vec{V} = \vec{P}(1) \Rightarrow \triangle \vec{P} = 0$$
I've noticed that you worry particularly what happens with the asteroid (on its own). Then the answer is similar:
$$\vec{P}_A (1) = M_A \vec{V} = \vec{P}_A (2)$$
i.e. no change in momentum. I suppose your confusion comes from the fact that you initially consider the momentum of the asteroid as the total momentum $\vec{P} = \vec{P}_A + \vec{P}_{b_1} + \vec{P}_{b_2} \neq \vec{P}_A$ ;)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/211203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How does thin film interference work? So thin film interference is when light is half-reflected half-refracted then the reflected and refracted wave interfere with each other to produce another color.
What I don't understand is that the waves are off-sync; not on top of each other. So why do they interfere? And doesn't interference change only amplitude? Why does it affect the wavelength?
| Everything is explained in wikipedia at "thin film interference". What do you mean by "off-sync" ? In classical images like the wikipedia one (see below) only one ray is drawn, but in practice there are an infinity of parallele rays, so superimposition do occurs.
But if by "off-sync" you mean there is a phase difference in the "2" superimposed rays, this is the whole point: phases will positively or negatively (or intermediatly) interfere based on this phase difference. And since the phase corresponds to wavelength/offset, the phase difference varies with wavelength.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/211396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Can the mass of an object be changed by adding opposing magnetic fields? Apparently it does. Or is this voodoo physics. If not what is really happening in this case?
https://www.youtube.com/watch?v=8N2TS3VReTA Boyd Bushman Changed the mass of a black box object, rock. He took two black box rocks. One rock he added two magnets connected in oppositon S-N-N-S bolted together. And the other rock had no magnets. He then drop them both at same time from a 40 story building and the rock with magnetic fields in opposition canceled out the mass gravity to a small percent and fell more slowly then the rock with no magnetism. Real effect on not? Are there any field equations that tie gravity directly or indirectly to magnetism?
| It's obviously wrong: mass don't change. Now the effects of mass might be tilted by some other forces.
Moreover, the speed of free fall is not related to mass as well (at 1st order).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/211477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Feynman propagator for arbitrary values of the gauge parameter $\zeta$
For the choice $\zeta = 1$ the Lagrangian can be brought into a particularly simple form upon integration by parts in the action integral. Equation$$\mathcal{L}' = -{1\over4}F_{\mu\nu}F^{\mu\nu} - {1\over2}\zeta(\partial_\sigma A^\sigma)^2$$with $\zeta = 1$ can be transformed into$$\mathcal{L}' = -{1\over2}\partial_\mu A_\nu \partial^\mu A^\nu + {1\over2}\partial_\mu A_\nu \partial^\nu A^\mu - {1\over2}\partial_\mu \partial_\nu A^\nu$$$$= -{1\over2}\partial_\mu A_\nu \partial^\mu A^\nu + {1\over2} \partial_\mu [A_\nu(\partial^\nu A^\mu) - (\partial_\nu A^\nu)A^\mu].$$The last term is a four-divergence which has no influence on the field equations. Thus the dynamics of the electromagnetic field (in the Lorentz gauge) can be described by the simple Lagrangian$$\mathcal{L}'' = -{1\over2}\partial_\mu A_\nu \partial^\mu A^\nu.\tag*{$(*)$}$$
Later in the book I am reading, we have the following, where it's worked out for the case of arbitrary $\zeta$:
If the gauge-fixing parameter is $\zeta \neq 1$ the Lagrangian $(*)$ is changed to$$\mathcal{L}'' = -{1\over2} \partial_\mu A_\nu \partial^\mu A^\nu - {{\zeta - 1}\over2}(\partial_\nu A^\nu)^2.$$
To me, though, it is not so clear how this formula for $\mathcal{L}''$ comes from here in the case of arbitrary $\zeta$. Could anyone help explain?
| \begin{align}
\mathcal{L}&= -\frac{1}{4} F^{\mu\nu}F_{\mu\nu} -\frac{\zeta}{2}\left( \partial\cdot A\right)^2 \\
&=-\frac{1}{2} \partial _\mu A_\nu\partial^\mu A^\nu +\frac{1}{2} \partial _\mu A_\nu\partial^\nu A^\mu -\frac{\zeta}{2}\left( \partial\cdot A\right)^2 \\
&=-\frac{1}{2} \partial _\mu A_\nu\partial^\mu A^\nu +\frac{1}{2} \partial _\mu A_\mu\partial^\nu A^\nu -\frac{\zeta}{2}\left( \partial\cdot A\right)^2 \\
&=-\frac{1}{2} \partial _\mu A_\nu\partial^\mu A^\nu +\frac{1}{2} \left( \partial\cdot A\right)^2 -\frac{\zeta}{2}\left( \partial\cdot A\right)^2 \\
&=-\frac{1}{2} \partial _\mu A_\nu\partial^\mu A^\nu -\frac{\zeta-1}{2}\left( \partial\cdot A\right)^2 \\
\end{align}
where in the second line we just use the definition of $F^{\mu \nu}$, and in the third line we did integration by parts twice to swap places for the two derivatives
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/211799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Physics:Buoyant force and scale readings Sorry this might seem like a dumb question, but I'm having trouble understanding the concept behind buoyant force and scale readings.
Suppose I have a beaker filled with water, and the beaker is placed on a measuring scale.
a) If I then place a ball into the beaker, the ball floats, what will be scale reading be? I understand that Mg=Buoyant force, since the ball floats, they cancel out. Does that mean there's no change in scale reading?
b) If I place a ball and the ball submerges completely under the water, but does not touch the base of the beaker, what will the scale reading be? Now that Mg>Buoyant force, does the scale reading increase by the difference of the two? Mg-Buoyant force?
c) If I place a ball and the ball completely sinks under the water and touches the base of the beaker, what will be scale reading be? I understand that Mg>>>Fb, but why does Fb play no effect in the scale reading in this case?
| Buoyant force has no effect on the interaction between the beaker and the scales it stands on.
The scales will register a weight which is the sum of the weights of the beaker, water and ball.
Consider that if the ball is at least partially immersed so that some part is below the level of water, it has displaced some water and so raised the level of the water in the beaker.
The water pressure at the bottom is a function of the height of the water column above it and so is a higher pressure. Since pressure acts normally to the interface between water and beaker base it presses down on the base of the beaker which transmits this force to the scales.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/212299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Have David Wolpert's findings really “slammed the door” on scientific determinism? I recently read an article describing how mathematician/physicist David Wolpert's research closed the door on scientific determinism. I have huge doubts about the implied conclusion, considering the fact that a result like this would have significant implications philosophically, but I haven't seen his work discussed in philosophical circles (Wolpert first demonstrated this in 2008). His work is also cited in the Wikipedia entry for "Laplace's Demon."
If anything, I could see this result as having implications for the epistemology of determinism, as we might never be able to "know" that the world was indeed deterministic. But that is completely independent of whether or not the universe is ontologically deterministic. I'll mention that I am a strong proponent of causal determinism. Indeed I think true randomness is utterly absurd, as it would be almost akin to magic.
If anyone has any input on whether or not this result actually demonstrates that the world can't be deterministic, I'd be happy to listen and further question my own worldview. But at first blush I am taking this to be a wild exaggeration.
|
any input on whether or not this result actually demonstrates that the world can't be deterministic,
The universe can be deterministic. Full stop. And there can't be a way to show it isn't, since the determinism can itself apply to the methods you use to test it. So you shouldn't get super excited about the universe being deterministic if it is unfalsifiable.
You shouldn't get super excited about it not being deterministic either since there can't be evidence of that either.
Instead you can make much more precise theories that can be tested.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/212584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Specific heat capacity and temperature, 0 K? I've found similar threads like this, but with no clear answer. I understand that the specific heat capacity of a substance increases with temperature, because the vibrational nodes and rotational movements of the atoms are quantized (I assume that the increase is because the energy is distributed evenly over all degrees of freedom?).
What I don't understand is why the heat capacity goes to zero as temperature goes to zero. This comes out of the mathematics, sure - but what is the physical explanation here? If the translational movements are not quantized (or temperature-dependent, rather), what is explanation?
I hope my question makes sense.
| Short answer: at absolute zero there is one - and only one - energy state available to each particle. Any attempt to change the state of even a single particle introduces energy into the system and you're no longer at absolute zero. Put another way: if there were two energy states allowed for a particle, the number of available energy states would be N for N particles (any one of them might be in the second state), and the temperature, which is proportional to the logarithm of the number of available states, is greater than zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/212716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How can the atmospheric pressure be different in distinct points at the same altitude? From an hydrostatic point of view, the pressure in a fluid should be the same at the same depth/altitude.
Obviously, in our atmosphere that does not happen. I am guessing that the main reason is the fact that the atmosphere cannot be regarded as hydrostatic.
Is this the reason? How exactly can we explain these pressure differences?
I understand that a higher pressure region must have a higher density, and therefore it would take time for reducing such density gradient. But how fast is this? In the order of the speed of sound? Or it has nothing to do with it?
| The air moves in great swirls.
In places where the air is being warmed from below it moves up.
That causes air to be sucked in from below, and spread out at the top.
What it sees as the reason to be sucked in is a lower pressure pulling it.
When any fluid is pulled in to a center, its angular momentum is conserved (and it has plenty of that because it is spinning with the earth), so it spins faster.
(Coriolis force is another way to describe this.)
So, you have meteorological low pressure areas, where the air is spinning the same direction as the earth, only faster, and high pressure areas, which are the opposite.
So that's why you can see different pressures at sea level or any other altitude.
(By the way, a low atmospheric pressure at sea level causes the water itself to be pulled up, resulting in "storm surge".)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/212895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
For a diatomic molecule, what is the specific heat per mole at constant pressure/volume? At high temperatures, the specific heat at constant volume $\text{C}_{v}$ has three degrees of freedom from rotation, two from translation, and two from vibration.
That means $\text{C}_{v}=\frac{7}{2}\text{R}$ by the Equipartition Theorem.
However, I recall the Mayer formula, which states $\text{C}_{p}=\text{C}_{v}+\text{R}$.
The ratio of specific heats for a diatomic molecule is usually $\gamma=\text{C}_{p}/\text{C}_{v}=7/5$.
What is then the specific heat at constant pressure? Normally this value is $7/5$ for diatomic molecules?
| A diatomic molecule will have 7 degrees of freedom at high temperatures. However, the ratio of specific heats that you cited is for diatomic molecules around room temperatures, which have 5 degrees of freedom.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/212986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Relative Time from a far place I have read that whenever we see the sun we are seeing the sun as it was before 8 minutes ago. Meaning if the sun were to somehow go dark we would not know until 8 minutes after.
Now I became curious, say a civilization 4.7 billion light years away was looking at us through a very advanced telescope. Would they be seeing the Earth as it was 4.7 billion years ago (early earth formation) or would they be seeing us in our very selves right this second.
| It is like mailing a letter to a friend via cargo ship. The ship takes a week to cross the ocean and deliver the letter. If your friend is just now getting the letter, he is reading about how you were a week a ago.
It is the same with light. Light takes 4.7 billion years to travel a distance of 4.7 billion light years. The aliens would see what we were like 4.7 billion years ago.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What happens to Goldstone bosons in the Higgs potential after symmetry breaking? When the gauge symmetry of our Lagrangian breaks spontaneously through the Higgs mechanism, we usually find that $n$ Higgs degrees of freedom become massless through the vacuum expecation value (vev), where $n$ is the number of broken generators. This means if our original gauge group has $N$ generators, the group that leaves the vev invariant has only $N'=N-n$ generators. The $n$ broken generatore become massive through the Higgs vev and in turn we can see that the mass terms for $n$ Higgs degrees of freedom vanish if we put in the vev. Often one says the Goldstone boson get eaten by gauge bosons, which then become massive.
What exactly happens to these massless degrees of freedom in the Higgs potential after symmetry breaking, i.e. after we expand the Higgs fields about the vev? **The mass-terms= quadratic terms vanish as noted above if we put in the vev, but, in general, there are several quartic terms possible. These would describe interactions of the Goldstone bosons with the Higgs bosons and surely there must be a good way to see that these vanish, too or have no influence for some reason?
| First, note that, strictly speaking, there is no such thing as spontaneous symmetry breaking in Higgs mechanism. I mean, that below and under the Higgs scale (i.e., at scale, at which non-zero Higgs VEV appears) the lagrangian can be rewritten in a gauge invariant way. How is it possible? The answer is that there are different physical states (i.e., eigenstates of hamiltonian) below and under the Higgs scale. Under the Higgs scale eigenstates of hamiltonian form also representations of the gauge group. But below it eigenstates of hamiltonian don't form representations of the gauge group. These eigenstates are linear combination of transverse and longitudinal degrees of freedom.
So, in some sence, nothing happens with three scalar fields in Higgs doublet. They just don't appear as physical states below the Higgs scale.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Gradient, divergence and curl with covariant derivatives I am trying to do exercise 3.2 of Sean Carroll's Spacetime and geometry. I have to calculate the formulas for the gradient, the divergence and the curl of a vector field using covariant derivatives.
The covariant derivative is the ordinary derivative for a scalar,so
$$D_\mu f = \partial_\mu f$$
Which is different from
$${\partial f \over \partial r}\hat{\mathbf r}
+ {1 \over r}{\partial f \over \partial \theta}\hat{\boldsymbol \theta}
+ {1 \over r\sin\theta}{\partial f \over \partial \varphi}\hat{\boldsymbol \varphi}$$
Also, for the divergence, I used
$$\nabla_\mu V^\mu=\partial_\mu V^\nu + \Gamma^{\mu}_{\mu \lambda}V^\lambda = \partial_r V^r +\partial_\theta V^\theta+ \partial_\phi V^\phi + \frac2r v^r+ \frac{V^\theta}{\tan(\theta)} $$
Which didn't work either.
(Wikipedia: ${1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r}
+ {1 \over r\sin\theta}{\partial \over \partial \theta} \left( A_\theta\sin\theta \right)
+ {1 \over r\sin\theta}{\partial A_\varphi \over \partial \varphi}$).
I was going to try
$$(\nabla \times \vec{V})^\mu= \varepsilon^{\mu \nu \lambda}\nabla_\nu V_\lambda$$
But I think that that will not work. What am I missing?
EDIT: The problem is that the ortonormal basis used in vector calculus is different from the coordinate basis.
| This problem is really nicely adressed is Weinbergs Gravitation and Cosmology, chapter 4 ig I remember correctly. There is basicalky one issue which leads to confusion:
In physics orthogonal coordinates are used, for example spherical or cylindrical. This leads to a diagonal line element.
This allows to normalize the natural basis-vectors. So if the diagonal elements are called $h_i$ then a 'vector' V in physics is neither a covariant nor a contravariant vector, but
$V_j = h_j W_j$ with W a vector and no Einstein summation.
So to go from math to physics and back you have to keep track of the $h_i$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Thermal state vs equibilibrium state Could someone explain what's the difference between a thermal state and an equilibrium state? Or is it even the same?
| Thermal states are used to explain the thermal equilibrium state, where the system attains the same temperature as that of the bath and will not exchange any more heat with the bath. This situation can be explained using the quantum mechanical equivalent of the canonical ensemble (that represents possible states of the mechanical systems at thermal equilibrium at a constant temperature). That's why the density matrix is represented as Gibbs state $\rho= Z^{-1}e^{-\beta H}$.
On the other hand, equilibrium states can be thermal, chemical or mechanical equilibrium. This is a generalized context. for example, If the volume of the system is kept constant (quantum mechanically volume can be anything like the frequency of the harmonic oscillator or external magnetic field used to drive a quantum system etc.), then the system is in mechanical equilibrium, where, we can never extract any work from the system. At the same time, we can exchange heat in to/out of the system, which makes the system thermally out of equilibrium (cannot be a thermal state/Gibbs state).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Measuring the moment of inertia of a flywheel using simple pendulum motion I've seen a method for experimentally determining the moment of inertia of a flywheel and I'm not sure whats the reasoning behind it. You attach a small weight $m_1$ to the flywheel's edge, it's important that $m_1 \ll M $, $M$ being the mass of the flywheel, so we can approximate that the small mass does not change the MOI of the flywheel. You then move $m_1$ through a small angle and let it oscillate like a simple pendulum. We measure $T$, the period of oscillation.
Then $I$, the MOI of the flywheel along its perpendicular axis is:
$$I=\left( \frac{T^2 g}{4 \pi ^2R} - 1\right) m_1 R^2$$
Where $R$ is the radius of the flywheel and $g$ is the standard acceleration due to gravity.
I'm stumped as to why this works, the only thing I can identify is that $\frac{T^2 g}{4 \pi ^2R} - 1 \backsimeq 0$ because $m_1$ describes a simple pendulum, so $T^2=4\pi^2\frac{R}{g}$.
|
The force $m_1g\sin\theta$ provides the moment $Rm_1g\sin\theta$ ($R$ is the radius of the flywheel), so the equation of motion becomes:
$I\ddot\theta+Rm_1g\sin\theta=0$.
For small $\theta$, then $\sin\theta \approx \theta$, so we get:
$\large{\ddot\theta+\frac{Rm_1g}{I}\theta=0}$.
The solution of this classic DE is:
$\theta(t)=\theta_0 \cos(\sqrt{\frac{Rm_1g}{I}}t)$, with period $T$:
$T=2\pi\sqrt{\frac{I}{Rm_1g}}$, so that:
$\Large{I=\frac{Rm_1gT^2}{4\pi^2}}$.
I've seen a method for experimentally determining the moment of inertia of a flywheel and I'm not sure whats the reasoning behind it. You attach a small weight m1 to the flywheel's edge, it's important that m1≪M, M being the mass of the flywheel, so we can approximate that the small mass does not change the MOI of the flywheel.
The derivation above is for that case described in your question. But it doesn't correspond to the formula you gave.
For the formula you gave, the inertial moment of $m_1$ is actually taken into account as follows. The actual inertial moment, including that of $m_1$ is $I'$:
$I'=I+m_1R^2$.
All we have to do now is substitute $I$ with $I'$, so that:
$T=2\pi\sqrt{\frac{I+m_1R^2}{Rm_1g}}$, so that:
$\large{I+m_1R^2=\frac{Rm_1gT^2}{4\pi^2}}$
$\Large{I=\left( \frac{T^2 g}{4 \pi ^2R} - 1\right) m_1 R^2}$.
As in your formula, but only if you do allow for the effect of $m_1$ on the total moment of inertia.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Will this rope break due to the tidal forces or not? While I was thinking about how tidal forces can make objects float at the surface of a planet orbiting a massive object like a black hole, the fact that any material on the Earth isn't held together by gravity only, but also by chemical bonds which give it its tensile strength came into my mind.
The scenario I thought of is as following:
1- A 10 kg metal sphere is tied to a 1 meter long thin rope and this rope is nailed into the surface of the Earth.
2- The tensile strength of the rope is 10 N. (So the metal sphere has to accelerate at 1 $m/s^2$ to break the rope)
3- The Earth is orbiting a black hole at its Roche radius, and so our metal sphere on the surface of the Earth is effectively weightless and floating, but it is still held by the rope.
Here is a simple picture to summarize:
Now, if we move the Earth to orbit the black hole even closer until the tidal forces on Earth due to the black hole become $\Delta a$ = 10.8 $m/s^2$ and so the metal sphere is pulled by the difference between $\Delta a$ and the Earth's gravitational acceleration (9.8 $m/s^2$) which is 1 $m/s^2$ towards the BH, will the rope break or not ?
In other words, if a is the gravitational acceleration towards the BH, in order to break the rope, which one do we need ? :
1- $a_{Point A}$ - $a_{Point B}$ = 1 $m/s^2$
OR
2- $a_{Point A}$ - $a_{Point C}$ = 1 $m/s^2$
| Between points A and C
The earth can be considered as a wall and thus to break off the 'wall' a relative acceleration with respect to the wall is a must , If the tensile strength is 10N and mass 10 kg then the rope must move at a relative acceleration of 1m/s^2 or
Acceleration(a) - Acceleration(c) = 1
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to derive end-correction value relationship for open-ended air columns? According to Young and Freedman's Physics textbook, in open-ended air columns like some woodwind instruments, the position of the displacement antinode extends a tiny amount beyond the end of the column.
UCONN's website states that the end correction for a cylinder could be found by:
$$d=0.6 r$$
where $d$ is the end correction, the distance which the antinode extends beyond the end of the pipe and $r$ is the radius of the cylindrical pipe.
How does one derive that relationship?
| This is actually a fairly involved calculation that was done by Levine and Schwinger in 1948. If you are interested , the reference is H.Levine and J. Schwinger, "On the radiation of sound from an unflanged circular pipe", Physical Review 73:383-406
I'll not attempt to replicate that calculation here but will try to describe the main points.
The main factor leading to the end correction is the boundary condition at the end of the pipe. The continuity of air pressure and velocity at the end of the pipe requires that the mechanical impedance of the wave equal the acoustic radiation impedance of the end of the pipe. So, the acoustic radiation impedance at the end of the pipe determines the end correction on the antinode. The radiation impedance is not zero, but is the radiation impedance of the pipe end.
To calculate the radiation impedance of the pipe end, it's treated as an unflanged piston embedded in a plane. Also it's assumed that the wavelength of sound is much larger than the diameter of the pipe. The resulting radiation impedance has real and imaginary parts, and it's the imaginary part that leads to the end correction. Levine and Schwinger arrived at a value of 0.6133d for the end correction to the effective length of the pipe.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Why are HCP materials brittle while FCC materials are ductile? Why are hexagonal close packed materials brittle, While face centered cubic is ductile. Is it related to crystal planes?
| Yes and how close the planes are packed and of course their geometries.
See some good answers below.
Source : http://www.researchgate.net/post/What_actually_makes_a_material_ductile_or_brittle
We may understand brittleness/ductility of solids from its bonding nature. In every solid, the constituent atom/ions are held by primary bonds (covalent/ ionic/ metallic). When we apply stress, we deform the atom/ions from its lattice. If the material can accept the deformation by getting strained- we call it ductile.Ductile materials must have some mechanism to absorb the stress- forming defects in its lattice. Brittle materials can't create defect in its lattice to absorb stress, so it deforms upto certain stress then break suddenly.
Source https://www.physicsforums.com/threads/why-is-fcc-more-ductile-than-bcc.550403/
Crystalline structure is important because it contributes to the properties of a material. For example, it is easier for planes of atoms to slide by each other if those planes are closely packed. Therefore, lattice structures with closely packed planes allow more plastic deformation than those that are not closely packed. Additionally, cubic lattice structures allow slippage to occur more easily than non-cubic lattices. This is because their symmetry provides closely packed planes in several directions. A face-centered cubic crystal structure will exhibit more ductility (deform more readily under load before breaking) than a body-centered cubic structure. The bcc lattice, although cubic, is not closely packed and forms strong metals. Alpha-iron and tungsten have the bcc form. The fcc lattice is both cubic and closely packed and forms more ductile materials. Gamma-iron, silver, gold, and lead have fcc structures. Finally, HCP lattices are closely packed, but not cubic. HCP metals like cobalt and zinc are not as ductile as the fcc metals.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Friction of a rolling cylinder I was wondering why friction vectors are drawn differently regarding a cylinder rolling on a surface and a cylinder rolling down an inclined surface. Since friction is responsible for the rotational motion shouldn't it be always pointing in the same direction (given that the cylinder is rotating to the right)?
Here are two pics I googled so you can see what I mean:
|
Since friction is responsible for the rotational motion
That's an assumption. Rotational motion could be caused by something else (like a drive shaft, or an electric motor).
For a rolling object, what friction does is bind together the rotational motion and the lateral motion.
The second case is easier to analyze. The cylinder is on an incline. Gravity pulls it down the ramp (linear motion). Friction creates a torque to cause rotational motion.
The first case is a bit less obvious. Instead of something forcing the cylinder to move linearly, I assume that something is rotating the cylinder (a torque is being applied in the direction of rotation). This rotational acceleration is opposed by friction, which pushes the cylinder forward.
So the reason for the opposite directions for friction is that in one case linear motion is driving the rotation, and in the other case rotation is driving the linear motion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Can we predict throwing a dice? What happens if we throw a dice from same position, with same force, by creating a vacuum environment on earth? Will it be predictable now i.e. will the dice have same results all the time?
If answer to the question is no, I have another question, Why in quantum mechanics we say a particle for example electrons are in two states. We only get a probabilistic value of position, that does not mean, its in two states at a time. The way we see it may be destroying the probabilistic character.
| In a pure Newtonian model, you can indeed make a prediction of the outcome if you know the inputs. But in this case there is a good deal going on. Rotation of the dice, how the dice leave the throwing surface, linear velocity, angle of impact to the landing surface, the coefficient of restoration from the landing surface, etc. You have removed air resistance which is would be a very low impact to the physics model. The problem is that the results would be governed by Chaos theory than quantum mechanics. Even a slight change in any variable would cause completely different outcomes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Relation of orbital speed and eccentricity Earth's eccentricity is 0. 0167 and speed at perihelion is 30.3 Km/s and at aphelion 29.3 with a difference of +/- 1. 0164 wrt average orbital speed
*
*Is this a coincidence or are the variations of speed directly related to eccentricity?
*can we calculate the time elapsed from aphelion, knowing the eccentricity of the orbit?
| No it's not a coincidence.
The linear eccentricity, $c$, is the distance from the centre of the ellipse to either of the foci. This diagram shows an orbit with this marked - for clarity I've made the orbit very eccentric:
The eccentricity that you quote is defined as:
$$ e = \frac{c}{a} \tag{1} $$
where $a$ is the semi-major axis.
The lower diagram shows the Earth at its closest and most distant positions. These distances are:
$$\begin{align}
r_{\text{max}} &= a + c \\
r_{\text{min}} &= a - c
\end{align}$$
Conservation of angular momentum tells us that:
$$ r_{\text{max}}v_{\text{max}} = r_{\text{min}} v_{\text{min}} $$
and therefore the ratio of the velocities is:
$$ \frac{v_{\text{max}}}{v_{\text{min}}} = \frac{r_{\text{min}}}{r_{\text{max}}} = \frac{a-c}{a+c} $$
Since equation (1) tells us that $c = ae$ the above equation simplifies to:
$$ \frac{v_{\text{max}}}{v_{\text{min}}} = \frac{1 - e}{1 + e} $$
Now we use the binomial theorem to approximate $(1 + e)^{-1}$ as $1 - e$ and this gives us:
$$ \frac{v_{\text{max}}}{v_{\text{min}}} \approx (1 - e)(1 - e) \approx 1 - 2e $$
where I've dropped terms in $e^2$ and higher powers on the grounds that if $e$ is small higher powers will be much smaller and can be ignored.
So does this work? Well, if we substitute your figures for the velocities we get:
$$ \frac{29.3}{30.3} \approx 0.967 \approx 1 - 0.33 \approx 1 - 2 \times 0.165 \approx 1 - 2e $$
And that's why the ratio of the velocities is related to the eccentricity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Gauge theory for mathematicians? I'm looking for a textbook or set of lecture notes on gauge theory for mathematicians that assumes only minimal background in physics. I'd prefer a text that uses more sophisticated mathematical concepts like principal bundles and connections, and categorical language whenever convenient.
| I have been writing something in this direction in section 1 of the book Differential cohomology in a Cohesive topos (pdf). Have a look, just focus on section 1 and ignore the remaining sections on first reading.
The survey-part is presently also appearing as a series on PhysicsForums. See at Higher prequantum geometry I, II, III, IV, V and Examples of Prequantum Field Theories I -- Gauge fields, II -- Higher gauge fields.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Streamlines and line of flow of fluid particles
*
*Can the streamlines of a fluid particle show the position of the particle at a time(using the streamlines)?
*I know that streamlines cannot intersect because at a specific instant the particle reaching the intersection will have two different directions of motion (in the steady flow of particles), but can I say the same thing in turbulent flow for the line of flow of the particles, why or why not?
*Is it necessary that a streamline will be a perfect straight line(to a good approximation)? If no why is it called a line then?
| Two particle cannot overcome each other because if it crosses then the direction of particles will be two different tangents at 1point which is not possible
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Constructing Killing tensors from Killing vectors Background:
After reading about Carter constant and symmetries in GR, I became interested in Killing tensors.
I tried reading this paper by Alan Barnes, Brian Edgar and Raffaele Rani, discussing conformal Killing tensors. I have some trouble understanding the crux of the paper.
Question:
*
*Is there a general way to construct Killing tensors, if the Killing vectors are known?
*How would I do this?
*Are there any Killing tensors that can not be constructed from Killing vectors?
Initial guess/motivation for the question:
Initially, I thought Killing tensors could just be formed via $K_{\mu \nu}=k_\mu k'_\nu$, where $K_{\mu \nu}$=Killing tensor, $k_\mu,k'_\nu$=Killing vectors. After reading the above paper, I am no longer sure. The paper discusses conformal Killing tensors and vectors, which may be the source of my confusion.
| Killing tensors created by just product of two Killing vectors is only in the trivial case. In a non-trivial case this is not possible such as in finding the Carter constants. This is all I know as I'm also a beginner.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.
| In relativity (both special and general) one of the key quantities is the proper length given by:
$$ ds^2 = g_{\alpha\beta}dx^\alpha dx^\beta \tag{1} $$
where $g_{\alpha\beta}$ is the metric tensor. The physical significance of this is that if we have a small displacement in spacetime $(dx^0, dx^1, dx^2, dx^3)$ then $ds$ is the total distance moved. You can think of it as a spacetime equivalent of Pythagoras' theorem. The quantity $ds$ is an invariant i.e. all observers in any frame of reference will agree on the value of $ds$.
A metric is positive definite if $ds^2$ is always positive, and Riemannian manifolds have a metric that is positive definite.
However in relativity $ds^2$ can be positive, zero or negative, which correspond to timelike, lightlike and spacelike intervals respectively. It is because $ds^2$ can have different signs that manifolds in GR are not Riemannian but only pseudo-Riemannian.
Lorentzian manifolds are a special case of pseudo-Riemannian manifolds where the signature of the metric is $(3,1)$ (or $(1,3)$ depending on your sign convention).
If we take the metric tensor that corresponds to special relativity (i.e. flat spacetime) equation (1) becomes:
$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$
That minus sign on the $dt^2$ term means $ds^2$ can be negative as well as positive, so the manifold is pseudo-Riemannian, and the one negative and three positive signs on the right hand side make the signature $(3,1)$ so the manifold is Lorentzian.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Why, in quantum field theory, is $\hat{a}(p)|0\rangle=0$? My Quantum Field Theory lecturer just said that $\hat{a}(p)|0\rangle=0$ because the vacuum state contains no particles. Now, according to Wikipedia,
"according to quantum mechanics, the vacuum state is not truly empty but instead contains fleeting electromagnetic waves and particles that pop into and out of existence".
So is this argument incorrect?
| In field theory, there are two vacua. The non-perturbative vacuum $|\Omega\rangle$ and the vacuum of the free theory $|0\rangle$. The wikipedia article makes reference to $|\Omega\rangle$ in terms of $|0\rangle$ and its excitations.
The true vacuum is annihilated by the (dressed) annihilation operators, and can be thought of perturbatively in terms $|0\rangle$ plus fluctuations. Thus $|\Omega\rangle$ is not "empty" if we define as empty $|0\rangle$. But one should keep in mind that this is an artifact of the perturbation theory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Did I calculate this voltage correctly? I have just started to learn physics.
Please forgive me if I am completely wrong or something, I have just turned 14 haha.
I am trying to learn about how to work out voltage for my year 10 exams.
Here's what I have so far:
I got such a large value of 30000V so I think I might have made a mistake somewhere.
Any help would be greatly appreciated. Thanks :)
| There is an error in your multiplication.
$$
(9 \times 10^{9})(20 \times 10^{-9}) \neq 90\times20
$$
$$
(9 \times 10^{9})(20 \times 10^{-9}) = (9\times20)\times10^{0} = 9\times20 =180
$$
This means that your answer is off by a factor of 10. Without the error your answer should be 3kV.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Particle anti-particle annihilation and photon production This is just a conceptual question I guess. The annihilation of a particle with a finite mass and its anti-particle cannot lead to the emission of only one photon, and this is due to the conservation of energy and linear momentum.
However, how could this be shown in a mathematical way? Could this be done, perhaps, with a consideration of the four-vector momentum of the two particles?
| As dmckee says in a comment, the proof is ridiculously simple. Suppose we work in the centre of momentum frame so the total momentum is zero. The particle comes in with some momentum $p$ and the antiparticle comes in with the opposite momentum $-p$, and the two annihilate.
Suppose the annihilation produced a single photon. The momentum of a photon is:
$$ p = \frac{h}{\lambda} $$
but the problem is that the momentum of a photon is always $h/\lambda$. Unlike a massive particle a photon has no rest frame i.e. no frame in which it's momentum is zero. So creation of a single photon means the momentum would change from zero to $h/\lambda$ and momentum wouldn't be conserved.
For momentum to be conserved we have to create a minimum of two photons moving in opposite directions i.e. with momenta $h/\lambda$ and $-h/\lambda$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to start an artificial gravity? I understand how artificial gravity in space stations works. It is by normal force the wall exerts on the foot.
But I wonder how to start it in the first place. I just learned about centrifugation in a centrifuge. To start, the side-wall of the tube produce a tangential acceleration. Because of the inertia (tendency to go tangentially) of the material contained, normal force is thus needed to keep the material from going through the tube and keep it rotating in a circle.
But in the space station, there is no friction, so there is no way to create that tendency that produces the need for normal force in the first place.
| An astronaut dressed in a magnetic suit inside a room or environment of copper or similar element in characteristics of conductivity or alloy (copper chamber) is a human magnet (magnet man) than will face a force that is opposed to his displacement inside the "chamber". If one of six sides or of all the sides than compose the chamber, one of the sides is the based side, or floor than determines the position of contact of the feet and if this side is an ferromagnétic material that attracts the astronaut at time the isstronaut is subject to an attractive force towards "below" and a force (effect Lenz) forces him to develop a work "to "ascend" or to move inside the chamber and if the magnetic suit (magnetic vinyl or similar material) has major magnetism proportionally when it moves away from the feet, this is, the magnet is much stronger in them, shoulders and head, in a such way that the attraction towards "below" is similar in both ends and is distributed proportionally along the suit. Then we have a chamber of copper where the magnet man is attracted towards the base and it is difficult to him to displace, which looks like enough the gravity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
(How) can we determine mid-point on Earth's orbit? The eccentricity of earth's orbit varies with time, but at present time its eccentricity is roughly mean e (0.0167).
The position of equinoxes is far more complicated than I thought and it is not at mid-point, can someone explain how you determine with a certain accuracy the middle point of the ellipse (where $\lambda = \pi/2 $ (or $3/2 \pi$)?
Edit:
*
*is the formula given by Pulsar valid for any point of the ellipse (for example $\lambda = 2.4981 = 143.13°$ from perihelion? (143.13- 0.0167*.6?)
| If you just want to find the dates for various events, the the following link provides a list of sources for astronomical calculators of various types.
http://www.midnightkite.com/index.aspx?URL=Software
The US Naval Office also is a good source of this sort of astronomical information.
http://aa.usno.navy.mil/index.php
From the data you present in your graphic, I think it is fairly obvious that the motion of the earth around the sun is more complicated than you would expect, if modelled as a single body in the solar system, even over fairly short time scales. I'm no expert in this, but I guess that the moon has a large part to play in the year to year variations in the various dates you show.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Quantum Dot Confinement - Band Gap I understand that the quantum confinement effect is observed when the wavelength of an electron is smaller than the wavelength of a material. I have also read that Quantum Dots have energy band gaps, because of the semiconductor material that makes them. For an exciton to be created an electron must jump from the valence band to the conduction band, generally in semiconductors this occurs from light absorption. If electrons are not permitted to move, from the confinement effect then how are they going from the valence to the conduction band? Am I picturing this wrong in my head, or is just that even quantum dots have some free electrons?
| Confinement means that the possible electron states are discrete in energy, rather than continuous, NOT that the electron cannot move between these states. Creation of an exciton is an electron transition from a state in the valence band to one in the conduction band, induced by the optical field. This is similar to electron transitions in an atom.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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?
| This site: Bernoulli Equation also uses the term "Pressure Energy". The pressure energy per unite volume is measured in N x m / m^3 = N / m^2. So this pressure energy per unite volume is in fact a pressure. Instead of the word "pressure" you can use the expression "pressure energy per volume". They are equivalent.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 7
} |
Is work a form of energy or means of transfer of energy? What actually is energy? Is it a property associated with matter or just a number? By doing work are we changing the energy of the body or converting one form of energy into other which is already present.
| I suppose that one way to look at energy is that it is a convenient tool for book-keeping tool since it is a conserved quantity. It's actually amazing that there is something that we call "energy" which can exist in so many different forms (e.g., kinetic energy, gravitational potential energy, electrostatic energy, electromagnetic energy, etc.) but that if we sum up all of these individual energies then the total sum is always constant no matter what happens.
As for "work", that is typically defined as the energy associated with exerting a force 'F' on an object for a certain distance 'd'. So it's the energy Fd that is lost by whatever person, being, or contraption does the work and goes somewhere else. The energy can go directly into the object that is being subjected to the force F (say, by letting the energy Fd get transferred into increasing the kinetic energy of the object), but that's not necessarily the case. If the object is being pushed a rough surface, then all that Fd energy may instead get entirely transferred into heat energy associated with the friction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How does inhalation work? In school, we learn that during inhalation, the diaphragm expands, causing air to get sucked into our lungs. You can feel this suction by putting your hand over your mouth while inhaling.
Why is that? Does the expanded capacity of the lungs cause the air outside my body to rush into my body to, shall we say, keep the lungs full?
| You are correct that your chest muscles are in fact pulling the lungs "open," which creates a pressure differential and draws air into the lungs. When the muscles relax, the chest cavity collapses to its original state, expelling the air (not 100% of it!). You may have heard of a "collapsed lung" injury. What happens there is that the lung is ripped loose from the surrounding muscles, and thus cannot be reinflated.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Mercury-in-glass thermometer Question from my textbook: Jason says 'The mercury in the thermometer can be replaced by coloured water. The thermometer will function well after recalibrating using a similar method to calibrating a mercury-in-glass thermometer." Comment on his idea.
The answer is NO, but why?
Please use simple english to answer my question, as I am only grade 9, I haven't learnt much about physics, therefore I may not understand some technical words. Thanks!
| A mercury-in-glass thermometer is traditionally calibrated by measuring two temperatures: the freezing point of water (0°C), and the boiling point of water (100°C).
There's an obvious reason why this won't work for a water-in-glass thermometer: the water in the thermometer will freeze when when you stick it in the ice bath, and boil when you stick it in the boiling water, and in neither case will you get a good reading.
There's also a less-obvious problem with measuring the freezing point: even if you try to measure the height of the water column just before the water freezes, you won't get good information: water has a maximum density around 4°C, and expands as it cools from there to the freezing point. The height of the water column just before it freezes will be about the same as it was at 8°C.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why only higher energy photons can affect atoms even though their speed is same? I understood that a baseball moving at the cosmic speed limit can do lot of harm, but since photons are not comparable to a baseball so irrespective of their high speed they do not rip anything in their way.
Now I want to know that why every photon, whether its of RADIO WAVES, X-RAYS or VISIBLE RAYS have same speed, even if their energy is different? Is it only because photons (of different waves) do not have REST MASS so they can achieve the cosmic speed limit? But there is another fact that they do have momentum and momentum does increase with energy then why their speed does not change?
| It's because photons are massless and so they can and must move at the speed of light with any none zero amount of momentum. This means that even though they move at the cosmic speed limit they can have very little amounts of momentum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
High Power/Range Electromagnet Pretty new to electromagnetism but my team and I are working on a project so have some questions before we spend more money doing R&D. We are looking to create a fairly small 10-20lb electromagnet. What we need this to be able to do is create down force from a range of say 3-4 feet.
So when holding the magnet at around 3 feet it could create 20+ pounds of pull towards the metal sheet below(for a total pull of 40+lbs assuming the magnet itself is 20lbs). Just curious if this would be possible in the scale that is needed. We could pretty easily do the reverse and have the electromagnet at the bottom and the metal being pulled to the magnet but this is just not as practical for our project.
| The size of a magnetic field is on the order of the size of the magnet that creates it. You can focus the field so that it comes out one side of the magnet, but once the field emerges into air it will immediately spread out in order to minimize its energy. For example, it's impossible to make a magnetic jet that throws a magnetic field like a hose.
All that said, you only specified the weight of the electromagnet, not its size. So, if you had a large (but light) coil, several feet in diameter, you might be able to get a significant magnetic field at a distance of 3 feet. Doesn't seem likely to me that you'd get a 20 pound pull from it, though.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Complex scalar field theory For the complex scalar field theory
$$L = -\partial_{\mu}\phi^{*}\partial_{\mu}\phi - m^{2}\phi^{*}\phi + J\phi^{*}+J^{*}\phi,$$
*
*Why is there no factor of 1/2 in the lagrangian like in the real scalar field?
*Can we say $ Y = 0$ (renormalization) because we know the two-point function $<0|T\phi(x)\phi(x')|0> = 0$ and so $<0|\phi(x)|0> = 0$ is satisfied?
| Without the factor $1/2$ for a complex field all observables constructed out of the Lagrangian in the standard way vie Noether theorem, like the energy $H:= \int T_{00} dx$ or the momentum $P_i = \int T_{i0} dx$, turn out automatically to be the ones of a system of identical particles of two types, {\em proper particles} and {\em anti particles}. E.g.,
$$H = \int d^3k\: k^0 ( a^*_ka_k + b^*_kb_k)$$
This is the standard interpretation of the quanta associated to a complex, also known as charged, field. The presence of the factor $1/2$ would instead produce
$$H = \frac{1}{2}\int d^3k\: k^0 ( a^*_ka_k + b^*_kb_k)\:.$$
(I cannot answer the second question as I do not know what you mean by $Y$.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why is Gibbs Free Energy not equal to 0? Its definition itself makes it 0 Gibbs Free Energy is defined as
$$G = H - TS$$ and the change in $G,$
$$\Delta G = \Delta H - T\Delta S,$$ (isothermal and isobaric)
Now at constant pressure, $q = \Delta H$
and Entropy is defined as , $\Delta S = \dfrac{q}{T}$ which means that $T\Delta S = q.$
Thus shouldn't Gibbs Free Energy automatically be zero. Or is there something I'm missing, like a specific condition?
| The only common situation I'm aware of in which a system can undergo both an isothermal and an isobaric process at the same time is when the system is undergoing a phase change at constant pressure. For instance, we could be considering a system comprised of liquid water and steam in phase equilibrium.
For a quasi-static, isobaric process,
$$\Delta H = \Delta (U + pV) = \Delta U + p\Delta V = Q + W + p\Delta V = Q -p\Delta V + p\Delta V = Q,$$
and so the change in enthalpy during the process is exactly equal to the heating of the system. For a quasi-static, isothermal process,
$$Q = T\Delta S.$$
Therefore, for a process that is both isothermal and isobaric,
$$\Delta G = \Delta (H - TS) = \Delta H - T\Delta S = Q - Q = 0.$$
In other words, yes! - the Gibbs free energy of the whole system is constant during such a process, but note that this is a very special process. It is rare that a system can undergo both a process in which both the temperature and pressure can remain constant at the same time, but such a process$-$*if quasi-static*$-$is a process of constant Gibbs free energy.
So: as long as there is a system containing a single substance in two different phases, and the phases are in thermal, mechanical, and phase equilibrium, then any process occurring at constant pressure also occurs at constant temperature, and so $\Delta G = 0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why will you float in free fall with 0 relative acceleration? Imagine you are in an elevator that is falling freely to the surface of the earth. You begin to float. However you and the elevator have equal acceleration. Why is it that you begin to float with respect to the elevator?
|
Imagine you are in an elevator that is falling freely to the surface of the earth. You begin to float. However you and the elevator have equal acceleration. Why is it that you begin to float with respect to the elevator?
You do not really float, you simply travel at the same speed as the elevator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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?
| galaxy come in many different sizes: some of the small-er ones do rotate ["orbit"] around the edge of a large galaxy ... one can also visualize galaxy-clusters, in which the entire cluster rotates .....
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "48",
"answer_count": 3,
"answer_id": 2
} |
Why aren't all objects transparent? I know that for an object to be transparent, visible light must go through it undisturbed. In other words, if the light energy is sufficiently high to excite one of the electrons in the material, then it will be absorbed, and thus, the object will not be transparent. On the other hand, if the energy of the light is not sufficient to excite one of the electrons in the material, then it will pass through the material without being absorbed, and thus, the object will appear transparent.
My question is: For a non-transparent object like a brick, when the light is absorbed by an electron, it will eventually be re-emitted. When the light is re-emitted won't the object appear transparent since the light will have essentially gone through the object?
| You say:
For a non-transparent object like a brick, when the light is absorbed by an electron it will eventually be re-emitted.
but this isn't true. In a solid the excited state can decay by transferring energy to lattice vibrations instead of emitting a photon. This means the energy of the incident photon is converted to heat and the photon is lost forever.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 3,
"answer_id": 1
} |
What is the difference when we measure torque/angular momentum about a point and about an axis? When do we measure torque about an axis and when do we measure torque about a point? What's the difference between measuring torque about an axis or a point.
I tried searching this on google but did not get satisfactory answer.
| Everything in classical mechanics, momentum, angular momentum, torque, velocity etc. is measured about a point. Period. You can be sort of a Newtonian Nazi and complain that it is wrong to talk about torque about an axis and you'll be correct but here it means a completely different thing but in common language, we often make do with such words.
So, coming to torque about an axis, it means the component of a torque pointing across a fixed direction which is along a hypothetical thing, we call the axis of rotation. The torque too is measured about a point lying on that axis and it can be geometrically proved that the component of torque along this particular direction is equal for all the points lying on this particular axis.
So, in short, torque along an axis, means the part of total torque that is along that axis measured w.r.t. a point on the axis itself.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 0
} |
Sliding and rolling friction I heard that friction depends only on the normal force but not on the contact area. Let's take a cube and a sphere which are of same weight (then normal force will also same ) but the force needed to move these two objects is different, why?
| Whenever one applies a sideways force trough the center of gravity of an object, that force has two components: 1) a direct force that tries to overcome friction and slide the object, and 2) a torque that uses friction to produce a rotation of the object by lifting its center of gravity over the leading edge.
A short, flat object will tend to slide because the torque needed to raise the center of gravity over the long edge is large compared to the force needed to overcome friction; the same object place on its short edge will tumble, because the torque needed to raise the center of gravity over that short edge is comparatively small.
Thus rolling friction must always be less than sliding friction, by definition, because for an object to roll the counter-force of friction that leads to torque must be smaller than the counter-force of friction that must be overcome for a slide.
A wheel, then, is simply a special case in which (ideally) the center of gravity of the object never needs to be lifted, and so the friction needed to induce a roll (ideally) is vanishingly small.
therefore it is easier to move sphere than cube of same weight.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/219313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How viscosity and velocity relate in a pipe? Assume we have a pipe that there is a fluid stream in it. By increasing the velocity of the fluid the resistance will increase either (because of the viscosity I think). My question is how are these to parameters depending to each other and what is the formula?
| With hydrodynamics we normally find that at low shear rates the flow is limited by the viscosity of the liquid while at high shear rates it's limited by inertial forces and the viscosity doesn't matter. This is the case for flow in a pipe. At low flow rates the pressure drop $\Delta P$ is related to the flow rate $Q$ by the Hagen-Poiseuille equation:
$$ \Delta P = \frac{8\mu \ell}{\pi r^4} Q $$
where $\ell$ is the pipe length and $r$ is the pipe radius. So in this case the pressure drop is proportional to the viscosity $\mu$.
However at high flow rates the pressure drop is given by the Darcy-Weisbach equation:
$$ \Delta P = f_d \frac{\ell}{2r} \frac{\rho v^2}{2} $$
where $\rho$ is the water density, $v$ is the flow velocity and $f_D$ is a fudge factor called the Darcy friction factor. Note that in this case the pressure drop is independant of the viscosity.
At intermediate flow rates you get a complicated regime where the pressure drop has a sub-linear dependence on the viscosity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/219449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How does one build a framework of sensors from nothing? I would like to understand how you can go from nothing to a framework of sensors you can use to try out things in physics/thermodynamics. That is, everything has to be bottoms up, nothing premade allowed (except maybe things like a lighter, screwdrivers, saws, hammers, but nothing that is already a sensor. So using a multimeter is cheating, for example, unless you build it yourself). Is there any book on this?
|
How does one build a framework of sensors from nothing?
For example, you can build a primitive ampere-meter using a compass like in the picture.
If you really want to get as primitive as possible see: The Science Notebook. The site teaches you even how to build a compass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/219768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Bandwidth of laser pulses I am working on a practice test for my optics class. The professor does not provide the solutions for the test. I was just wondering if someone could tell me if I am doing this correctly because it seems kind of simple but I am unsure.
A laser emits short pulses of duration 14.0 ps separated in time by
1.0 ns at a wavelength of 1 um. From one pulse to the next, the timing of field oscillations is kept perfectly coherent and phase
stable using sophisticated feedback control.
What is the bandwidth of the source (in Hertz)?
I think the answer to this would be 1/t which is 1 /14 ps = 71 Ghz
I think it is just that simple but I want to know if I am wrong so I do not make that mistake on the actual test. Also just to clarify the bandwidth of the pulse is the range of frequencies in that pulse of light right?
| About 30 GHz,
according to this.
Check "Bandwidth-limited Pulses".
Because of Fourier transform, the product of the temporal duration and spectral width is ≈ 0.44 for Gaussian-shaped pulses. See also this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/219894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the direction of propagation of the natural light is perpendicular to the direction of electric and magnetic field making up natural light? I know that the direction of propagation of the wave(light) is perpendicular to the direction of electric and magnetic field in the situation of plane waves. And I want to know the relation between the direction of propagation and field in the situation of natural light.
| Your text is rather muddled, but to answer the question: the Poynting vector is normally in the direction of propagation, which is to say the E and B fields are perpendicular to the direction of prop. This is always true in a vacuum, but it turns out that in various materials, the Poynting vector can be off-axis.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/219978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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 distinction between these two equations is that: $p+ \rho u^2=constant$, is valid only for 1D compressible flow while, $p+(1/2) \rho u^2=constant$, is valid for incompressible flow.
The difference arises because of the coupling of continuity and momentum equation in compressible flow. This coupling is absent for incompressible flow. You can see this by a simple derivation from 1D Euler's equation. Euler equation is basically the momentum equation where the viscous forces are neglected.
$$ \frac{dP}{\rho}+ u du + gdz = 0 $$
Let's also neglect the body forces. Now this becomes:
$$ dP=- (\rho u) du $$
Where P is the pressure, $\rho$ is density and u is the 1D velocity. This is valid for both compressible and incompressible flows. For deriving the Bernoulli's equation, you simply integrate both sides.
$$ \int{dP}=- \int{(\rho u) du} $$
If the density, $\rho$ is constant, flow is incompressible, and you can take $\rho$ out of the integral sign to get your Bernoulli's equation.
$$ P= -\rho \int{u du} = -\rho \frac{u^2}{2}+constant$$
$$P+\frac{\rho u^2}{2}=constant$$
Now, consider the case of compressible flows. Here you cannot take the density $\rho$ out of the integral. Instead, you can use continuity equation for compressible flows, which says: $\rho u=constant$, to take $(\rho u)$ out of the integral. So, we have
$$ P= -(\rho u) \int{du} = -\rho u * u+constant$$
$$P+\rho u^2=constant$$
You see? We had to use the continuity equation in order to get this. This is the coupling I'm talking about. Even if you derive the 1D compressible flow equation using a control volume, using $P_1 , V_1 ; P_2 , V_2 $ at entry and exits, you'll still need to use the continuity equation $P_1 V_1 =P_2 V_2$ to get $P_1 + \rho {V_1}^2=P_2 + \rho {V_2}^2$. The derivation from Euler equation just helps to see the distinction very easily.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Power statement is valid for MW Or KiloWatts? If I can talk to someone and tell him that a new power plant inaugurated by Prime Minister will produce $60$ megawatts per hour, will it be true to use $\mathrm{MW}$ unit for Power?
| You're right that the unit "megawatt" is abbreviated MW.
However, as Aniket comments, watt itself means "energy per unit time", so saying that the power plant produces 60 MW per hour doesn't make sense.
In your comment, you question whether MW is a "basic unit". I'm not exactly sure what you mean by this, but the SI unit of power is watt, so if you want to express your statement in SI units, you should say
*
*"The power plant produces energy at a rate of 60 millon watt",
but either of the following statements are equally true:
*
*"The power plant produces energy at a rate of 60 MW".
*"The power plant produces energy at a rate of 60 megajoules per second".
*"The power plant produces energy at a rate of 80461.3 British horsepower".
*"The power plant produces energy at a rate of 44 million foot-pounds-force per second".
Being an astronomer I was brainwashed to cgs units, so I'd go for
*
*"The power plant produces energy at a rate of roughly $10^{15}\,\mathrm{erg}\,\mathrm{s}^{-1}$".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Force on a charged particle due to an uncharged infinite conducting plate How can I calculate the force on a charged particle due to an uncharged infinite conducting plate?
If there is a small object with positive charge placed above a metal plate, the object induces a negative charge on the surface of the plate facing the object. Let's call this surface as $S_1$.
Since the conductor was uncharged initially, and charges always reside on the surface of a conductor, $S_2$ gets a positive charge.
From the picture above, the particle would experience an upward force due to $S_2$ and downward force due to $S_1$. So is the force zero?
| This the kind of question that can be solved by the method of images.
Try placing a fictitious charge on the other side on the conducting plane. You should arrange it in such a way that the electrostatic potential is precisely zero on the surface of the conductor. If your case you put it at equal distance as the first but on the other side.
The physical interpretation is that the electrons in the neutral conductor will rearrange themselves because of the field from $q$.
Since the conductor is infinite the charges will not go from one side to the other. They will rather be brought in from infinity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a relationship between kinetic energy of emitted electron and photoelectric current? I know that photoelectric current is dependent upon intensity of incident light. But it should also be dependent upon kinetic energy of emitted electron because mathematically $I=Q/T$. So if kinetic energy is more then time becomes less. Please help and let me know if I'm right or wrong.
| You're wrong.
The current is limited by the number of electrons per second emitted from the metal surface. Once a steady state current is established the number of electrons per second received by the collecting electrode is the same as the number emitted per second from the metal surface and the speed the electrons travel from the metal to the electrode makes no difference.
However the speed of the electrons will affect the time the current takes to settle to a constant value after the light is turned on. When you trun the light on there is a delay before the collecting electrode starts registering a current, and the faster the electrons move the shorter this delay. Likewise the decay rate of the current when the light is turned off also depends on the electron speed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a way to make a heat pipe that can transfer heat downward? Is there a way to make a heat pipe that can transfer heat downward? As in a 10 to 20 ft vertical pipe with 100°f at the top and 50°f at the bottom.
| You can circulate chilled water through the pipe which should distribute the temperature. Rate of circulation, along with the thermal properties of the pipe and embed media will control the precise temperature gradient - if thats even important for your application.
Lastly, the problem of steady state temperature distribution in a metal rod is well understood and doesn't depend on the pipe orientation (up, down, left, right).
You can enter your specific numbers on this website:
http://demonstrations.wolfram.com/TemperaturePropagationInAThinSteelRod/
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compton effect in photo-electric? In photo-electric effect Einstein said that photons incidents on material and gives their energy which will gives kinetic energy to electrons. But i also want to know that why Compton's effect not works in this situation. In my view when photon incident on material it should eject a electron as well as a photon of less energy than incident photon.
| Compton scattering occurs on free electrons, i.e. not in a bound state. The equivalent to a Compton scatter would be a scattering of a photon off the field of a solid, momentum and energy balance happening collectively with the total mass of the solid. In the best case it would be a whole atom that the photon would scatter inelastically off, the atom taking the momentum balance and a reduced frequency photon leaving. This would give a continuous spectrum, no cut offs. This could happen only if the Thomson model of the atom were correct , the electrons freely distributed in a continuum . It would not give rise to the photoelectric effect, which gives a discrete electron momentum and implies quantized states for the electron.
In fact the photoelectric effect is one of the lynch pins of proof of the quantized condition of nature in the microcosm. The energy of the photon is constrained by the quantized binding of the electrons, as with a lower energy of the photon no electrons appear.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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?
| If you are indoors, then...
*
*Seal all cracks or openings in the ceiling and in the walls above the mid-point. This includes the gaps around junction boxes and between drywall panels. Caulk and duct tape are your friends here.
*Extinguish any open flames and get rid of any possible spark sources. No smoking either!
*Release a quantity of hydrogen equal to 1/2 the volume of the room (use more if the ceiling is higher than 10 meters). Hydrogen gas is colorless so you are not going to be able to do this by eye - get out your notepad and pencil to keep track of the total quantity of gas released! Keep in mind that the hydrogen will expand significantly when you release it from the bottles, so go slowly and don't over do it or you will likely suffocate yourself.
*Wait for a minute for all the hydrogen to rise and stabilize.
*Release the helium balloon. It should stop rising when it hits the interface between the air and the hydrogen and hover mysteriously 1-5 meters above the floor.
If you have problems with turbulence mixing the hydrogen and air layers, try to hold your breath and stop moving around so much. If there is still too much turbulence, evacuate the air in the bottom half of the room and replace it with sulfur hexafluoride.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 4
} |
Dimensions of $\phi$ in scalar field theory On Srednicki page 90-91 (in printed edition) he derives that $$[\phi] = \frac{1}{2}(d-2) \tag{12.10}$$ from $${\cal L}=-\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi -\frac{1}{2}m^{2}\phi^{2} - \sum_{n=3}^{N}\frac{1}{n!}g_{n}\phi^{n}$$ and the fact that $[\partial^\mu] = +1$ in units of mass. Can someone please elaborate how he gets to $$[\phi] = \frac{1}{2}(d-2)~?$$
| He uses that the action is dimensionless so that
\begin{align}
[ d^d x \left(\partial\varphi\right)^2] &= 0 \\
&=[d^d x]+2[\partial\varphi]\\
&= -d +2 + 2[\varphi]
\end{align}
using that $[dx]=-1$ and $[\partial\varphi] = [\partial] + [\varphi] = 1+[\varphi]$.
This gives $[\varphi] = (d-2)/2$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/221090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.