Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Do moving singularities crack/tear space-time? Question: The tip of a crack in a continuum like glass is a singularity.
If we
A) set QM and Planck's volume arguments aside and stick to GR only
B) assume that at least some* blackholes contain a singularity and
C) assume black holes move in space-time,
then shouldn't it follow that the hole punctured in space-time due to the singularity is cutting/cracking/tearing space-time as it moves along?
Does the space-time get sewn back up magically?
Should it not leave a trail of damage behind?
| You are thinking classical physics + General Relativity.
In the centre of a black hole is a gravitational singularity, a one-dimensional point which contains infinite mass in an infinitely small space, where gravity become infinite and space-time curves infinitely, and where the laws of physics as we know them cease to operate. As the eminent American physicist Kip Thorne describes it, it is "the point where all laws of physics break down".
In this sense, one can define the singularity as a trajectory in the space time reference frame of the earth, it is a moving point, but "crack" attributes characteristics to space time that it does not have. There are continuous distortions of space time due to the existence of the singularity and these have to be explored in the framework of General Relativity. It means that as t changes the functional forms in space of the distortions of the singularity change . There is no permanent space medium that would keep a record of the "passing" of the singularity. In the absence of mass/energy, space time is flat.
Current research is attempting to quantize gravity, which means that singularities become blurred, due to the uncertainties introduced by the probabilistic nature. Have a look at the current Big Bang model to get an idea of what happens to singularities when quantization of gravity is assumed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The counterfactual possibility to explain discrete energies in Thomsons plumpudding model In the german edition of "Tipler: Physics for scientists and engineers" there i a small text about the plum pudding model of the atom. i give a direct translation:
"Thomsons plumpudding model of a atom: The negatively charged electrons are embedded in kind of liquid or porridge [...]. For that kind of configuration, one can compute the resonance frequencies of the electron oscillations. According to the laws of classical physics, such an atom should emit light with this frequency, but Thomson couldn't find such a configuration, for which the computed frequencies correspond to the observed specral lines."
My (counterfaactual) question is, what is meant by "configuration" of electrons? And how could a oscillating electron emit descernible lines as the acceleration - the cause of radiation - would be continuos and not discrete or so.
|
how could a oscillating electron emit descernible lines as the acceleration - the cause of radiation - would be continuous and not discrete or so.
The membrane of a speaker, playing a pure sinusoidal musical note, moves back and forth with a continuous acceleration. Despite this, the sound it produces has a single frequency.
In a similar manner, the electron that oscillates with a given frequency around an equilibrium position, generates an electromagnetic wave that is continuous in time and space but has a definite frequency $f_0$. The human eye is sensible just to the frequency (color) of light. The eye can not follow the time varying amplitude of the light waves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Visualising gas temperature and gas pressure Gas pressure is created when gas molecules collide with the wall of the container creating a force.
Gas temperature is a measure of how fast the molecules are moving / vibrating.
However, they both seem to be concerned by "kinetic energy" of the molecules, or in other words, the "collision" they impose on the target.
How do we visualize the difference between pressure and temperature of gas?
Is there any obvious difference between the two?
The same question in another form:
*
*A gas is hot when the molecules collided with your measuring device.
*A gas have high pressure when the molecules collided with your measuring device.
So, what is the difference between the two "collisions" in the physical sense and how do we visualize the difference?
For Simplicity,
How can a Hot gas be Low Pressured? ( They are supposed to have High Kinetic Energy since it is Hot. Therefore should be High Pressured at all times! But no. )
How can a High Pressured gas be Cold? ( They are supposed to collide extremely frequently with the walls of the container. Therefore should be Hot at all times! But no. )
|
*
*A gas is hot when the molecules collided with your measuring device.
Not quite. Gas heats your measuring device when the collisions are mostly such that the colliding gas molecule has more kinetic energy than the colliding measuring device molecule.
It's instructive to think colliding molecules as sumo wrestlers: The molecule which has more momentum wins the bout, the winner does work on the loser by throwing him. Winner loses energy, loser gains energy.
The above rule works for straight head-on collisions. For other kind of collisions there are different rules. For example a molecule that experiences a collision on its rear gains energy. And a molecule with lot of kinetic energy rarely experiences rear collisions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
} |
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?
| There are plenty of satellite galaxies orbiting larger galaxies. The question is how long are you willing to wait for an orbit?
The Milky Way has a mass $M$ of something like $6\times10^{11}$ solar masses, or $10^{42}\ \mathrm{kg}$. The small Magellanic Cloud is at a distance $R$ of $2\times10^5$ light years, or $2\times10^{21}\ \mathrm{m}$. A test mass orbiting a mass $M$ at a separation $R$ will have a period of
$$ P = 2\pi \sqrt{\frac{R^3}{GM}} = \text{2 billion years}. $$
Such a system could undergo at most $7$ orbits in the entire history of the universe. The universe isn't old enough for the nearest major galaxy to have completed a single orbit around us at its current separation.
Even if you did wait long enough, galaxies aren't particularly good at holding their shape. If you put them in a situation where gravity is strong enough to bend their path into a closed orbit, odds are they will also be tidally torn apart by that same gravity. And we see this all the time, as for example with the Mice Galaxies:
| {
"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": 1
} |
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?
| For an object to be transparent, the light must be emitted in the same direction with the same wavelength as initially. When light strikes a brick, some is reflected in other directions, and the rest is re-emitted in longer, non-visible wavelengths. That is why a brick is opaque to visible light.
Some materials we consider transparent, like glass, are opaque to other wavelengths of light. Most window glass these days, for example, is coated with infrared- and ultraviolet-reflective films to increase insulative capacity. You can see through these fine with your eyes, but an infrared-based night vision system would see them as opaque objects. Another example is that most materials are transparent to radio waves, which is why both radio broadcasts and radio telescopes are so successful.
| {
"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": 0
} |
FWHM increase with energy (gamma spectra) Below I have two plots from a gamma spectrum which I've been analyzing. The first plot is between a low energy range, the second between a significantly higher energy range. It is clear that the FWHMs (Full Width Half Maxima) of the peaks in the spectrum increase with energy. This is apparently on each of the spectra I've seen and seems to be an integral property of all of them.
Generally speaking, the peaks/Gamma lines in a Gamma spectra are wider at higher energies.
What is the exact reason for this? I do assume intuitively that there would be more inherent variance in the energy of peaks at higher energies. Whether this is a correct assumption is part of my question. Is the increase in width to do with error in the measurement of a particular Gamma? Do the Gammas at higher energies genuinely have a larger fluctuation in their energy? Is it a combination of both?
| The gamma photon energy is a function of the energy levels in the nucleus. There is some uncertainty in these levels. However, the data that you present in the chart is the amount of energy deposited in a detector, where the uncertainty is determined by the uncertainty in the interaction of the photon with the material of the detector and the processing of that physical interaction. At low energies, the photoelectric effect (absorption) dominates; at mid energies, the Compton effect (scattering) dominates.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
What is the relation between Nuclear magnetic resonance (NMR) and electron paramagnetic resonance (EPR)? It seems to me that the basic principles are exactly the same, right? Then I am puzzled that the former was awarded a nobel prize while the later not.
I noticed a similar question here What's the difference between NMR and EPR?
It seems that the difference is purely quantitative, in the frequency.
| I would say the maths and equations are pretty much identical except in H NMR you would use the gyromagnetic ratio for a proton, while in EPR you use the data for an electron. Both are spin 1/2 systems.
In terms of medical imaging it is easier to pick H2O via pulse NMR (rather than continuous field i.e what chemists do for molecules etc) than observe free radicals. However there are plenty of free radicals in the body. In fact most of the metalloproteins that use copper can exhibit free radical chemistry - or redox cofactors for electron transfer chemistry to name but two.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/219236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Time dilation on Satellites due to GR I am trying to determine the time dilation onboard a satellite (say GPS @ 20,000km) w.r.t an observer on the earth. I have already determined the special relativity component using:
$$
t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
And I got the correct answer for the time dilation simply due to relative motion (7 microseconds after 24 hrs). Im not sure if determining the component due to GR is this simple, but my first attempt was to evaluate it using the equation to determiine gravitational time dilation outside a non rotating sphere in a circular orbit using the Schwarzschild metric.
$$
t' = t\sqrt{1-\frac{3GM}{rc^2}}
$$
I dont seem to be getting the correct answer (45 microseconds after 24 hrs). Any ideas?
| The equation you quote:
$$ t' = t\sqrt{1-\frac{3GM}{rc^2}} \tag{1} $$
gives the time relative to an observer at infinity. You want the time relative to an observer on the Earth's surface. You need to calculate:
$$ t_\text{satellite} = t\sqrt{1-\frac{3GM}{r_\text{satellite}c^2}} $$
and:
$$ t_\text{Earth} = t\sqrt{1-\frac{2GM}{r_\text{Earth}c^2}} $$
where $r_\text{Earth}$ is the radius of the Earth and $r_\text{satellite}$ is the radius of the satellite's orbit (measured from the centre of the Earth). The relative time dilation is then the ratio of these two times.
Note that equation (1) combines the special and general relativity contributions to the time dilation i.e. it includes both the gravitational time dilation and the effect of the orbital velocity. The observer on the Earth's surface isn't in a circular orbit so the equation is slightly different (a factor of 2 in the square root rather than 3).
Incidentally, when the gravitational field is weak (like the Earth's) we can use the weak field approximation for time dilation:
$$ \frac{dt_B}{dt_A} = \sqrt{1 - \frac{2(\Phi_A - \Phi_B)}{c^2}} \tag{1} $$
The quantity $\Phi_A - \Phi_B$ is the difference in the Newtonian gravitational potential energy between $A$ and $B$, and $dt_B/dt_A$ is the time dilation of $B$'s clock relative to $A$'s clock.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/219573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the number of turns in a loop part of the magnetic flux? In all both my physics textbooks, the number of loops $N$ is left out of the equation for magnetic flux $\phi_B = \int_s \vec B \cdotp\hat n dA$ and only when calculating the Emf induced by an inductor is the number of turns taken into account.
That leads to the question : does the number of loops in a coil effect the magnetic flux?
Consider the following setup. You have a coil of 5 turns/loops placed perpendicularly to a magnetic field. There is no current going through the coil. With the magnetic field, area of the coil, and position of the coil kept constant, the number of loops/turns in the coil is increased to 50. Is an Emf generated across the coil?
In other words, if you were to change the number of loops in a coil, would the magnetic flux change (thus inducing an Emf)?
| Yes, if you go around N times then the emf will be N times greater than if you went around once.
Why? Suppose one loop bounds a surface with area A, then N loops will bound a surface with area perpendicular to the field NA. Imagine you're looking directly through the solenoid and the surface it bounds is made out coloured glass, how many layers of glass are you looking through? (Answer: N)
Why don't your books include it? Actually they do! It's the the "S" term, i.e. the surface that your integrating over.
What about the E field? The E field doesn't change because although your emf increases by N, the tangential distance of the curve you integrate over also increases by N. The electric field doesn't care what curve you integrate over, or if you call it one big wire or many loops. The electric field is going to do what it wants.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/219881",
"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.
| I think I figured it out. Bernoulli's assumption is incompressible flow. The equation yielding from momentum conservation always holds. When velocity is low (incompressibility holds), the two equations yields similar results.
| {
"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": 1
} |
Does the second law of thermodynamics imply a spacetime beginning of the universe? Recently I have been studying thermodynamics and I noticed a article by a religious person which says that the second law of thermodynamics proves that the universe had a beginning. A spacetime one. He says that the universe should be in complete disorder if it was eternal so the second law proves a finite universe. Simply trying to prove the universe began to exist. I tried to do some of my own research but I reached dead ends so I came here for help from you guys. Any thoughts on this topic?
| No. A low entropy "initial state" could be the result of a so-called anthropic fluctuation in a (past) eternal universe.
Fluctuations about equilibrium could, fortuitously, create the initial conditions for life as we know it. This was proposed by Boltzmann and his assistant Schutz in the late 19th century, though ultimately deemed unsatisfactory by a succession of physicists, notably Eddington and Feynman. Critics stressed that in an anthropic fluctuation, one should expect maximum entropy compatible with life (this is related to the Boltzmann brain paradox). This, or something like it, remains, however, a logical possibility.
This discussion assumes some level of ergodicity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Working of electric-tester How does the circuit gets completed when we put an electric-tester glow when we put it in the +ve terminal, when actually the electrons flow from the -ve terminal ?
| The electric tester displays a light when electrons flow through it. When you touch a negative terminal electrons flow from the terminal through the tester and into you. When you touch a positive terminal electrons flow from you through the tester and into the terminal. Either way the light illuminates.
The current flowing is so small that you don't notice it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Work Done On Circular Motion I am trying to find the total work done on a ball, $m=0.8kg$, tied to a rope of r=1.6m length and swung in a vertical circle.
I understand the total work done by both the tension in the string and gravity is 0 for a complete circle, but I am having trouble trying to find the work done by gravity when the ball is at its lowest point and at its highest point, work done during a semicircle.
I know that as the ball swings it constantly changes direction and therefore the $\theta$ between $F_g$ and $\Delta s$ changes.
So there is a varying theta that is associated with both $F_g,\Delta S$
I know that $$W_{g}=\int_0^\pi -F_g \cdot dl$$
If $\Delta S = r\theta$
then
$dl=ds=rd\theta$
and $F_g=mg$
$$W_{g}=\int_0^\pi -mgrcos\theta d\theta $$
$$W_{g}=-mgr[sin(\pi)-sin(0)]=0J$$
I don't understand what I am doing wrong the answer is -25.1J. It makes sense that the answer is 0 because when the ball is at the bottom and at the top, $\Delta s$ and $F_g$ are perpendicular to each other and the total work increases until $\frac{\pi}{2}$ then decreases resulting in a 0J.
So what am I doing wrong?
Here is the solution:
I found out that $\Delta s = 2r$ since it started at $y_o=0,y_f=2r=3.2m$
There for $W_g=F_g\Delta s cos\theta=3.2mgcos(180)=-25.1J$
Okay so since the actual solution was so much easier, what exactly did I do wrong with my initial thought of trying to solve it?
And ironically if we do:
$$W_{g}=-2mgr\int_0^{\frac{\pi}{2}} cos\theta d\theta $$
$$W_{g}=-mgr[sin(\frac{\pi}{2})-sin(0)]=-25.1J$$
|
If $\Delta S = r cos\theta$
then
$dl=ds=rd\theta$
and $F_g=mg$
$$W_{g}=\int_0^\pi mgrcos\theta d\theta $$
If you're taking the angle from the center of the circle (which you are, since you said that $\Delta S = r cos\theta$, then the initial position of the ball is $-R$, since displacement is a vector quantity (and the final position of the ball is on the opposite side of the center of the circle), and hence your angle must vary from $-\frac {\pi}{2}$ to $\frac {\pi}{2}$ instead of $0$ to $\pi$, which gives you the required answer.
It makes sense that the answer is 0 because when the ball is at the bottom and at the top, $\Delta s$ and $F_g$ are perpendicular to each other and the total work increases until $\frac{\pi}{2}$ then decreases resulting in a 0J.
This logic is flawed, because the vertical direction of displacement (which is along the direction of gravity) remains the same (downwards) when the ball is moving from top to bottom, and hence the work done by the Gravitational force is always positive and on the body always negative, during this course.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Meaning of components of Maxwell's stress tensor $T_{ij}$ I am learning about Maxwell's stress tensor and what I understood is that the components, say $T_{ij}$ is something like a force parallel to the $j$th-direction acting on the surface with its normal in the $i$th-direction.
I was working on a problem which is to find the net force on the upper hemisphere of a uniformly-charged solid sphere of radius $R$ and charge $Q$.
Calculating the force using Maxwell's stress tensor and symmetry arguments(ignoring $F_x$ and $F_y$), I got
$$F = \int{T_{zz}da_{z} + T_{zx}da_{x} + T_{zy}da_{y}}$$
Then came the confusion. When calculating just the $\int{T_{zz}da_{z}}$ part, I got 0. Which meant $F_z$ arises only from shear forces $T_{zx}da_{x} + T_{zy}da_{y}$. I cannot visualize how this is possible given a $T_{zx}$ acting along $x$-direction give rise to a force in the $z$-direction and same for $T_{zy}$. What did I understand wrongly here?
| To find the total force in the z axis you should sum over the z vector embedded in the field's matrix, which is the
The integral should be (for the net force in the z-axis):
$$ F_{z} = \sum_{i = 1, j = 3}^{i=3} T_{i}^{j} \cdot \hat{n}dS $$
With
$$ T_{ij} = \left( \begin{array}{ccc}
xx & yx & zx \\
xy & yy & zy \\
xz & yz & zz \end{array} \right) $$
with n is a unit normal vector and dS is some area element, in the case of a sphere it would be:
$$ S = 4 \pi r^{2} $$
$$dS = 8 \pi r dr $$
$$ r = \sqrt{x^{2} + y^{2} + z^{2}} $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
When sugar is added to water, how does the mass change, and how does that affect the water's density and boiling point? I can't find a good answer anywhere. How does the amount of sugar added change the boiling point, mass, and density of water? Does it affect the mass or the volume? Or both?
| The mass doesn't change at all, it will be just the sum of the water mass and the mass Added, what happens is the change of density because the mixture, in general the molecules get closer to each other ( through the intermolecular forces) and, this way, the volume become lower to the same mass quantity, what increase the density by the equation
$$
\rho = \frac{m}{V}
$$
Another effect because the changing of the intermolecular forces is the increasing of boiling temperature. It happens due the increasing of the intensity of the intermolecular force after the mixture, this way you need more thermal energy to break the bounds, and so, you need more heat to do that.
I really hope it helps you!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/220734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.
| In some cases it does eject a photon with a lower wavelength, if it did not do this then the laws of conservation of momentum would not be supported thus disproving many aspects of modern physics. The problem with this is without the experimental evidence or data, it is hard for someone to calculate or predict the new photons wavelength, let alone detect it. The photon would be emitted with such a low energy that the problem would be "How would you detect it?"
You could test the discrepancy of how much energy the electron should have compared to what it does have but even that would be difficult.
| {
"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": 0
} |
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?
| You might be interested to have a look at my answer to Why do helium balloons rise and fall? where I answer a closely related problem.
The change in bouyancy with pressure depends on how rigid the skin of the ballon is. If the skin is very rigid, i.e. the balloon volume doesn't change as the external pressure changes, then the bouyant force is proportional to the external pressure:
$$ F \propto \frac{P}{T} $$
There will be a pressure at which the bouyancy equals the balloon mass, and in principle the ballon will hover at this pressure. The trouble is that (as others have said in comments) the pressure changes only very gradually with height so the range of heights over which the balloon has effectively neutral bouyancy will be quite large. The balloon motion will be dominated by random air currents and I suspect you have no chance of getting it to hover at a selected height between one and five metres.
Incidentally, if the ballooon skin is very flexible the bouyancy is less affected by pressure and the problem becomes even harder. In the limit of a completely flexible kin the bouyancy is unaffected by pressure and you stand no chance of making the ballon hover at a selected height.
| {
"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": 3
} |
Magnetic field at the center and ends of a long solenoid A long solenoid has current $I$ flowing through it, also denote $N$ as the turns per unit length. Take its axis to be the $z$-axis, by symmetry the only component of the magnetic field inside is $B_z$. Find the magnetic field at the center of the solenoid (on the axis). Also, find the magnetic field at the ends of the solenoid.
For the first part, since the solenoid is long we can approximate the magnetic field inside to be uniform and is given by $B_z = μ_0NI$, so we can say that the magnetic field at the center is also $μ_0NI$. I'm not sure if my argument is correct but based on my understanding, from the uniformity of the $B$-field inside, it should be the same everywhere inside. Can anyone kindly tell me if this is correct? Any suggestions and insights?
For the second part I don't have any idea on how to start.
| Notice, the magnetic field at some internal point on the axis of a solenoid is given by general expression
$$B=\frac{\mu_0 NI}{2}(\cos\theta_1-\cos\theta_2)$$
where, $\theta_1$ & $\theta_2$ are the angles between axis & the lines joining the extreme-points of both the ends of solenoid to the concerned point.
1) magnetic field at the center of a long solenoid is given by setting $\theta_1=0$ & $\theta_2=\pi$ $$B=\frac{\mu_0 NI}{2}(\cos 0-\cos\pi)=\color{blue}{\mu_0 NI}$$
2) magnetic field at the end of a long solenoid is given by setting $\theta_1=\pi/2$ & $\theta_2=\pi$ $$B=\frac{\mu_0 NI}{2}(\cos \pi/2-\cos\pi)=\color{blue}{\frac{\mu_0 NI}{2}}$$
Thus, the magnetic field at the center of a long solenoid is two times the magnetic field at the ends .
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/221086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute affect of a shower on density altitude As a pilot I have a basic understanding of density altitude, how temperature affects the effective air pressure:
I noticed recently that I have difficulty breathing when I take a shower in Santa Fe, NM, which is at an altitude of 8000 feet. According to the chart above if the shower is 105F / 40C then the effective altitude is actually 12,300 feet due to the hotness of the air. However, this does not take into account the higher humidity of the shower which also has an effect. How can I compute the density altitude inside the shower accounting for humidity as well?
| To account for the humidity, you need a chart (or equation) that relates the effective altitude to the humidity of the air. For example, if the chart you show here is for 50% humidity, then since the air density increases(lower altitude) as the humidity of the air increases, at 100% humidity, the effective density altitude will be lower. However, if the chart is for 100% humidity, then you already have the correct chart to obtain the effective altitude of your shower!
In conclusion, you need to know the humidity of the given chart, and the variation of air density with humidity, to be able to calculate the effective density altitude.
Regarding your breathing problem, I suspect that it is more likely due to the fact that you are now breathing water vapor in together with the air, thus less oxygen per volume taken in per breath.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/221301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What's the bubble's wall made up of in false vacuum decay? It is well known that for some kind of double well potentials, there are two minima with one is unstable called the false vacuum while the other stable one called the true vacuum. The tunneling is allowed by quantum mechanics which is also called the false vacuum decay followed by the nucleation of bubbles. Colemann proved that the energy mostly locate on the bubble's wall. Now comes the question, what's the bublle's wall made up of? Since we always consider Higgs field, are they Higgs particles?
Inspired by the following discussions, I now know that there are no real particles associated with a static field. Because from the classical point of view, a real particle is an oscillation mode which we call plane wave. The static field may be treated as the a sea of virtual particles from the point of view of Fourier transformation. So are the domain walls.
| That depends on what you mean by "made up of". Is static electric field "made up of" photons? Is the stream of water "made up of" waves?
To me these are semantic games that have no connection to reality. But if you are willing to answer "yes" to the questions above, then you can safely say that those domain walls are "made up of" Higgs bosons (and also the longitudinal components of W and Z while we are on it).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/221409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Infinite pulley system Infinite Atwood Machine Harvard Solution
Hi, I've been trying to solve this question for a while. I understand the first solution and also the solution to the second problem but I don't understand how to apply the second problem to solve the infinite pulley system.
In particular, I don't understand the last sentences:
Therefore, since $f N (x) → 3m$ as $N →∞$, our infinite Atwood’s machine
is equivalent to (as far as the top mass is concerned) just two masses, $m$ and $3m$. You can then quickly show that the acceleration of the top mass is $g/2$.
Note that as far as the support is concerned, the whole apparatus is equivalent to a mass $3m$. So $3mg$ is the upward force exerted by the support.
If the Atwood's machine is equivalent to just two masses $m$ and $3 m$, then wouldn't the value of acceleration be $2g$?
| Here's a simple diagram of the Atwood machine you describe.
For reference:
$F_{total} = m_{total} \cdot a$
Let us call $g$ the gravitational acceleration, $m$ the mass of the less massive "block," and $a$ the acceleration of the "duel-block-system."
$$F_{total} = m_{total} \cdot a$$
$$a = \frac{F_{total}}{m_{total}}
= \frac{F_{m} + F_{3m}}{3m + m}
= \frac{-mg + 3mg}{4m}
= \frac{2mg}{4m}
= \frac{g}{2}$$
For more complicated Atwood machines, use a similar method.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/221484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Principle behind electrostatic shielding? If we have a solid conducting sphere with charges around it, then the electric field inside the sphere is zero, otherwise the electrons of the sphere would not be in equilibrium as there would be a net force acting on it. However, if its a hollow sphere, then why does the electric field inside the hollow sphere have to be zero?
| $E$ is necessarily zero inside hollow sphere because,
Inside hollow sphere
$Q = 0$
from gauss’s law
$$\phi=\frac{Q}{\epsilon}$$
$$E.A = \frac{Q}{\epsilon}$$
Since,
$E.A = 0$
$E = 0$ or $A = 0$ but $A\ne0$ so $E = 0$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/221618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can the hole concentration of cuprate superconductors generally be determined from lattice parameters? As it is known, in YBCO superconductors, lattice parameters are strongly dependent on hole concentration, so hole concentration can be determined by measuring the c-axis parameter, see e.g. "Oxygen determination from cell dimensions in
YBCO superconductors" by Benzi, Bottizzo and Rizzi.
Is it similar for other cuprates, LSCO (La2-xSrxCuO4) or LBCO, or are lattice parameters independent from doping level?
| Here is paper where lattice parameters dependence on x is shownn:
https://journals.aps.org/prb/pdf/10.1103/PhysRevB.49.4163
This graph I extact from this paper:
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/221771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The anticommutator of $SU(N)$ generators For the Hermitian and traceless generators $T^A$ of the fundamental representation of the $SU(N)$ algebra the anticommutator can be written as
$$
\{T^A,T^{B}\} = \frac{1}{d}\delta^{AB}\cdot1\!\!1_{d} + d_{ABC}T^C
$$
where $\delta^{AB} = 2\text{Tr}[T^AT^B]$ is the normalization chosen for the generators (note that they are also chosen orthogonal), $d=N$ for the fundamental representation, and $1\!\!1_{d}$ is the $d$-dimensional identity matrix.
For the fundamental representation it seems possible to deduce this expression by arguing that the anticommutator is Hermitian and hence can be written in terms of the $N^2-1$ traceless generators and one matrix with non-vanishing trace.
Does this expression hold for a general representation of the generators? If yes please explain why and/or provide a reference.
The relevance in the above equation appears in trying to express a general product:
$$T^AT^B = \frac{1}{2}[T^{A},T^{B}]+\frac{1}{2}\{T^{A},T^{B}\}$$
where the commutator is already known as a consequence of the closure of $SU(N)$.
| I am not sure what you are asking. For every antisymmetric d-dimensional matrix $T$ you can extract the trace part and hence have
$$T=\frac{I}{d}\cdot tr{T}+(T-tr{T}).$$
You can check out that the first term is really the trace part and the second term is traceless. So in your equation, it is simply a definition of $d_{ABC}$. Note that $tr{{T^{AB}}}=2C\delta^{AB}$ in your equation. More generally, for any d-dimensional matrix T, you can have
$$T=\frac{1}{2}({T+T^{T}})+\frac{I}{d}\cdot tr\left({\frac{1}{2}(T-T^T)}\right)+\left(\frac{1}{2}(T-T^T)-\frac{I}{d}\cdot tr\left({\frac{1}{2}(T-T^T)}\right)\right)$$
with the symmetric, the trace and the traceless antisymmetric parts correspondingly.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/221851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Induced EMF dependent on terminal wire connection? Figure(a):
When a conductor moving inside magnetic field, of a given length at a certain velocity the induced EMF is:
$$\epsilon = vBL$$
However, what if we changed the position where the bottom wire is connected to the wire like so:
Is the induced EMF now:
$$\epsilon = vBL_2$$
I'm not sure how that can be true, when the conductor's length has not changed just the position of where the circuit wire is connected "shortining" the current medium(or path) I agree, however, how would it change the induced EMF? The charges are still at the top & bottom of the conductor, would the terminal wire's connection reduce the induced EMF?
| The magnetic flux intercepted by the conductor will be reduced by connecting the wire at intermediate point of conductor because the effective length of the conductor will now $L_2$ & induced E.M.F. is given as $$\epsilon =vBL_2$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/221986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Null geodesic equations If one is constrained to the $xt$ plane, one can define the intersection with that plane of the null hypersurfaces originating at some point $P$ as
$$ g_{tt} \frac{d P^t}{d \lambda}\frac{d P^t}{d \lambda} + g_{xx}\frac{d P^x}{d \lambda}\frac{d P^x}{d \lambda} = 0,\tag{1}$$
$$ \sqrt{ \frac{g_{xx}}{-g_{tt}}} \frac{d P^x}{d \lambda} = \frac{d P^t}{d \lambda}. \tag{2}$$
It is not clear that a curve satisfying this equation will also satisfy the Geodesic equation:
$$ \frac{d^2 P^{\alpha}}{d \lambda^2} = - \Gamma^{\alpha}_{\beta \gamma} \frac{d P^{\beta}}{d \lambda} \frac{d P^{\gamma}}{d \lambda}.\tag{3}$$
What would be a reasonable approach to show that the first equation is also a geodesic?
| Comments to the question (v3):
*
*In 3+1D, the 1D-intersection of a null-hypersurface with the constraints $y=0=z$ does not need to locally be a geodesic, as simple counterexamples show.
*The analogous 1+1D question is more interesting: In 1+1D, the eq. (1) is locally always a non-affine parametrized geodesic. It does not have to satisfy the affinely parametrized geodesic eq. (3). For a proof, see my Phys.SE answer here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/222118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Relativistic acceleration in sinusoidal electric field Consider a relativistic charge $q$ moving with an oscillating electric field $E_z$ with phase velocity $v_p=c$ in direction $\hat{z}$ (e.g. radially polarized laser coprogating with electron). What is the energy gain of this charge as a function of time?
I set this up from the relativistic force
$$F=\frac{dp}{dt}=\frac{d}{dt}(\gamma m \dot{z})= qE_0 \sin{((c-\dot{z})t k)}$$
where $t$ is time, and $k$ is the usual wave number $k=2\pi/\lambda$.
My confusion arises from the $\dot{z}$ on the RHS. I don't have much experience with differential equations, and so I wonder if it is necessary to actually write it as $\dot{z}=\int_0^t dt' \ddot{z}$ or if the LHS' attribution of $\ddot{z}$ automatically leads to an appropriate $\dot{z}$?
Thank you for any help.
After some thinking the equation written above is actually wrong. It should indeed be written like so:
$$
F=\frac{dp}{dt}=\frac{d}{dt}(\gamma m \dot{z})= qE_0 \sin{(k(ct-\int_0^t\dot{z}dt' +z(t=0))}
$$
I would appreciate help solving this equation if someone has experience. TY!
| Since I can't comment yet, here some thoughts to your initial question:
A changing electric field will always generate a magnetic field due to Maxwell:
$$\nabla \times {\bf H} = \varepsilon_0 \frac{\partial {\bf E}}{\partial t}$$
So you can't consider an electric field changing with time seperately! And you want to consider a particle moving with $\gamma>1$. The equation of motion for an electron in an em-field is
$$\frac{{\rm d}}{{\rm d}t}(\gamma m_e c_0^2)=-e({\bf E + v\times B})$$.
The contribution of the magnetic field to the force is suppresed by a factor of $1/c_0$ against the electric field contribution, i.e. $B_0=\frac{E_0}{c_0}$. But this means, that the contribution to the force ${\bf v \times B}$ is not negligible, when $v\sim c$. If you want to neglect the magnetic field, you either have to consider a constant electric field, or a particle with $\gamma \ll 1$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/222204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Does the proton wobble due to the electron's orbit in a Hydrogen atom? In a hydrogen atom the nucleus only has one proton and no neutrons so the electron to nucleus radio is higher for hydrogen than that for any other atom. Does the orbiting electron induce a wobble on the proton similar to the way a star wobbles due to the orbit of planets. Different forces aside. Larger atoms would be more electrically symmetrical with more electrons.
| Perfunctory quantum comment #1: "orbital", not "orbit". There are no little balls moving in a circles in there. These things are quantum objects.
But the sort answer is "yes", the proton has a non-zero momentum distribution that mirrors the electron's. Now, because the proton is nearly 2000 as massive, the proton's position wave-function is nearly 2000 more compact than that of the electron, and its extent can be ignored for many purposes.
However, it matters when comparing the spectroscopy of protium (Hydrogen-1) to that or deuterium or tritium: the changed reduced mass results in very slightly different energy levels.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/222435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Exact meaning of radial coordinate of the Schwarzschild metric In this answer as well as on Wikipedia the radial coordinate of the Schwarzschild metric is described as follows:
...the r co-ordinate is the value you get by dividing the circumference of the circle by 2π.
This circumvents (no pun intended) the problem of talking about the path outwards from the singularity, but what about the circle itself: does it have the length we would measure when flying around it (however this is done) or is it a virtual circle assuming flat space, i.e. no mass at the center.
| The circumference is the distance you would measure if you laid out a (very long!) tape measure along a circle centred on the black hole. Or if piloted your spaceship round the black hole it's the distance your navigation computer would record.
Extracting these distances is less work then you might think. Suppose we use polar coordinates so $r$ is the distance from the black hole, $\theta$ is the latitude and $\phi$ is the longitude. The metric is then:
$$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{1-\frac{r_s}{r}}+r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$
Suppose we trace out a circle round the equator so $dt = dr = d\theta = 0$ and $\theta = \pi/2$, then the metric simplifies to:
$$ ds^2 = r^2d\phi^2 $$
or:
$$ ds = rd\phi $$
As we go round the equator $\phi$ goes from zero to $2\pi$, so integrating to get the distance travelled gives:
$$ \Delta s = r\int_0^{2\pi} d\phi = 2\pi r $$
To show that the $r$ coordinate is not simply the radial distance let's use the same technique to measure the distance from $r = R$ to the singularity $r = 0$. We'll use a radial line at constant $t$, $\theta$ and $\phi$, so $dt = d\theta = d\phi = 0$, and substituting these into the metric gives:
$$ ds = \frac{dr}{\sqrt{1-\frac{r_s}{r}}} $$
so the proper distance from $r=R$ to the centre is:
$$ \Delta s = \int_0^R \frac{dr}{\sqrt{1-\frac{r_s}{r}}} $$
which is most definitely not just $R$. I won't evaluate the integral here because I've already done it in my answer to How much extra distance to an event horizon?. The result isn't particularly illuminating.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/222550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Velocity of a leak in a closed water tank Bernoulli's equation states
$P_1+{1\over2}\rho v_1^2+\rho g h_1 = P_2+{1\over2}\rho v_2^2+\rho g h_2$
In a classic "water tank with an open top and a leak" scenario, "point 1" is the surface water in the tank, and "point 2" is the leak. The equation could be rewritten for $v_2$ as
$v_2= \sqrt{2g\Delta h}$
This is simple, but suppose it involves a tank where the top is closed off. The above simplification will no longer yield the correct velocity.
How do I apply Bernoulli's equation for "water tank" scenarios in which the water tank is closed?
| If the leak is closed, the velocity goes to zero. So instead of the water down near the leak having extra flow velocity ($v_2 > v_1$), it has extra pressure ($P_2 > P_1$).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/222656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Chiral current VEV below the QCD scale Let's have pure QCD. I know that after spontaneous symmetry breaking quark bilinear form are replaced by their averaged values:
$$
\bar{q}_{i}q_{j} \to \langle \bar{q}_{i}q_{j}\rangle \approx \Lambda_{QCD}^3, \quad \bar{q}_{i}\gamma_{5}q_{j} \to \langle \bar{q}_{i}\gamma_{5}q_{j}\rangle \approx 0
$$
What can be said about VEVs of $\partial_{\mu}\bar{q}_{i}\gamma^{\mu}\gamma_{5}q_{i}$,
$$
\int d^4x d^4y\langle 0|T\left(\partial^{x}_{\mu}\bar{q}_{i}\gamma_{\mu}\gamma_{5}q_{i}(x))(\partial^{y}_{\nu}\bar{q}_{i}\gamma^{\nu}\gamma_{5}q_{i}(y))\right)|0\rangle?
$$
An edit. It seems that the second correlator is zero in momentum space for $k \to 0$, since no massless states couples to correlator $\Pi^{\mu \nu}(k) \equiv \int d^{4}x e^{ikx}\langle 0|T(J^{\mu}_{5}(x)J^{\nu}_{5}(0))|0\rangle$ in QCD.
| By the chiral anomaly equation
$$
\partial^\mu \bar{q}_f\gamma_\mu\gamma_5 q_f
= \frac{N_f}{16\pi^2}
\tilde{G}^a_{\alpha\beta}G^{a\,\alpha\beta}
$$
this correlator is proportional to the topological susceptibility
$$
\chi_{top}=
\frac{1}{V}\frac{1}{(16\pi^2)^2}\int d^4x \int d^4 y \; \langle T\,
\tilde{G}^a_{\alpha\beta}G^{a\,\alpha\beta}(x)
\tilde{G}^b_{\gamma\delta}G^{b\,\gamma\delta}(y)
\rangle
$$
The topological susceptibility is zero if one of the quarks is massless, but it is non-zero in general, and of $O(\Lambda^4_{QCD})$ in pure gauge theory or the large $N_c$ limit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/222775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Moment of inertia of solid cube about body diagonal How do I find the above mentioned moment of inertia?
Steps I've tried:
1.) Triple integrations that proved to be to big.
2.) I noticed that the if we split a $2\times 2\times 2$ into individual $1\times1\times1$ components, the body diagonal of the $2\times 2\times 2$ either passes through or is parallel to body diagonals of the $1\times 1\times 1$ cubes.
If the moment of inertia of the $1\times 1\times 1$ about body diagonal be $I$, then the moment if inertia of the $2\times 2\times 2$ about its body diagonal will be $8I$ because it has 8 times as much mass. I though I could get an equation in terms of $I$ by equating $8I$ and moment of inertia of the individual $1\times 1\times 1$ cubes about the body diagonal of the $2\times 2\times 2$ using parallel axis theorem but it turns out to be greater than $8I$. Could someone resolve my mistake?
P.S: I am not familiar with the concept of tensors.
|
If the moment of inertia of the 1x1x1 about body diagonal be I, then the moment if inertia of the 2x2x2 about its body diagonal will be 8I because it has 8 times as much mass.
This is the source of your confusion. The moment of inertia of a solid, uniform density cube about any axis that passes through the center of the cube is $\mathrm I = \frac 1 6 {ml}^2$, where $m$ is the mass of the cube and $l$ is the length of any one of the cube's sides. Since the mass of a solid, uniform density cube is given by $m={\rho l}^3$, another way to write the moment of inertia for such a cube is $\mathrm I = \frac 1 6 {\rho l}^5$. This means that the moment of inertia of your 2x2x2 cube will be 32 times that of the moment of inertia of your 1x1x1 cube.
You will get exactly the same result (a factor of 32) if you consider that 2x2x2 cube to consist of eight 1x1x1 cubes and apply the parallel axis theorem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/222905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do opaque materials have an index of refraction? The index of refraction defines how much light is bent as it passes through an object, right? So how can opaque objects, which don't transmit light, have a value for the index of refraction? (e.g., Blender Docs lists the IoR of steel as 2.5)
| Index of refraction refers to the speed of light in a material, which comes up when determining how much light is reflected vs. refracted. In an opaque material the refracted light is absorbed, but the intensity of reflection still depends on the illumination angle.
For example, light reflected at Brewster's angle is completely polarized in the plane of the surface, because the angle between the reflected and refracted light is 90 degrees. The effect is easy to see in the polished floor of a long hallway if you have polarizing sunglasses. If you can find some physical polished steel and it has refractive index 2.5, you should see completely polarized light reflected at a 22 degree angle from the horizontal.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/223048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How do you determine if the spin is up or down? Fundamental particles such as quarks and leptons can have a spin either up or down. These spins are (obviously) opposite of each other. But what differentiates them? Let's say you examine a pair of electrons and you find out they're opposite (one up and one down). How do you know which one is up and which one is down? Do all up-spinning-particles in the universe point to an established direction? If not, how can we know the difference between an up and a down spin?
| Let's say you examine a pair of electrons and you find out they're opposite (one up and one down). How do you know which one is up and which one is down?
There is no way of knowing, or more accurately, there is no answer to your question. The wave function of the state you describe is a super position of the two options. There is not even any way of distinguishing between the two (they don't have individual names).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/223326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can we speed up the evaporation of black holes manually by accelerating it? If we throw an object to pass near a black hole, to bypass it, it will change the speed of the black hole, just like gravitational assist for a space probe. Does an accelerating black hole evaporate faster because:
*
*When object accelerates, mass increases
*When mass increases, gravity increases
*When gravity increases, the black hole collects virtual particles more rapidly
Is above true?
| The question related to speeding up evaporation of Black Holes manually has no basis ever in the science of Physics of the universe. The General Theory of Relativity had predicted the presence of Black Holes as regions in space, in which space-time distorted in such a way that nothing, not even light can escape.
However, the characteristics of these Black Holes are subjected to evaporation, destruction of the inner parts of information, and depletion of radiation that caused confusion in the world of physics, reference to the Theory of Quantum Mechanics. The Laws of Quantum Mechanics are clear and straight forward and state:
a) Black Holes cannot evaporate.
b) Radiation of Black Holes never stops.
c) Black Holes have an infinite life, in other words are eternal.
As a conclusion, the chaos between the Theory of Relativity and the Theory of Quantum Mechanics started forty years ago and still going on, that created a dilemma of having Black Hole information paradox and Black Hole radiation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/223449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How can I prove that D'Alembert operator is invariant under Lorentz transformations? I'm currently taking a course on Classical Electrodynamics and I'm trying to prove that the D'Alembert operator ($\square=\eta^{\mu \nu} \partial_{\mu} \partial_{\nu}$) is invariant under Lorentz-like transformations.
My professor does this using an argument to show that the variation of the operator must be zero, but I don't see where he specifically uses the fact that the transformation must be Lorentz.
I thought at first of proving it by applying the transformation and showing that the operator stays the same.
What he does is the following:
He shows that $[\delta, \partial_{\mu}]=\delta \omega_{\mu}^{\nu} \partial_{\nu}$ and then showing that $[\partial, \square]=0$.
Why is the first fact so?
| Let $\square = \eta^{\mu\nu}\partial_{\mu}\partial_{\nu}$ and let $\Lambda$ be any operator leaving the metric invariant, i. e. $\Lambda\eta\Lambda^{-1}=\eta$.
From the above it follows that the components of a $(2,0)$ tensor transform with twice the matrix $\Lambda$, whereas the vector components transform with $\Lambda^{-1}$, namely (I suppress all the indexes for the sake of the notations, but they are obviously summed over):
$$
\square' = \eta'\partial'\partial' = (\Lambda \Lambda\eta)(\Lambda^{-1}\partial)(\Lambda^{-1}\partial) = \eta\partial\partial = \square
$$
summing over the corresponding indeces.
For the transformation laws of tensors you may want to check this other answer of mine.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/223563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Deriving the equation of motion for a rigid system I want to derive equation of motion for the system shown in picture. How do I choose a generalized coordinate in order to calculate kinetic and potential energy of the system? I need the aforementioned quantities to implement Lagrange equations for determining the equation of motion. How many DOFs does the system has?
| Consider the net forces $F_x$, $F_y$ and moments $M_A$ at a point A, let's say the corner of the beam
now consider the center of mass at a point C located at $(c_x,c_y)$ relative to A. This location changes in each time frame because it is measured from an inertial reference frame (and not in the body frame).
The three degree of freedom of the body can be the position of A $(x_A,y_A)$ and the orientation angle $\theta$. The equations of motion from a Newtonian point of view are
$$ \begin{align}
F_x & = m \ddot{x}_A - m c_y \ddot{\theta} - m c_x \dot{\theta}^2 \\
F_y & = m \ddot{y}_A + m c_x \ddot{\theta} - m c_y \dot{\theta}^2 \\
M_A &= (I_C+ m (c_x^2+c_y^2)) \ddot{\theta} + m (c_x \ddot{y}_A - c_y \ddot{x}_A )
\end{align} $$
Where $m$ is the total mass, and $I_C$ is the mass moment of inertia about the center of mass. (See Derivation of Newton-Euler equations of motion for a derivation of the equations of motion not on the center of mass in 3D).
What you are interested in the motion due to the net forces at each frame which is
$$\begin{pmatrix} \ddot{x}_A \\ \ddot{y}_A \\ \ddot{\theta} \end{pmatrix} =
\begin{vmatrix} \frac{1}{m} + \frac{c_y^2}{I_C} & - \frac{c_x c_y}{I_C} & \frac{c_y}{I_C} \\ - \frac{c_x c_y}{I_C} & \frac{1}{m} + \frac{c_x^2}{I_C} & -\frac{c_x}{I_C} \\
\frac{c_y}{I_C} & - \frac{c_x}{I_C} & \frac{1}{I_C} \end{vmatrix} \begin{pmatrix} F_x \\ F_y \\ M_A \end{pmatrix} + \begin{pmatrix} c_x \dot{\theta}^2 \\ c_y \dot{\theta}^2 \\ 0 \end{pmatrix} $$
The last part that is needed is to expressed the forces as a function of the degrees of freedom. For small movements you can estimate
$$ \begin{pmatrix}
F_x = -k_2 x_A - c_1 (\dot{x}_A-L \dot{\theta}) \\
F_y = -k_1 y_A - c_2 (\dot{y}_A) - m g \\
M_A = - c_1 L (\dot{x}_A-L \dot{\theta}) - c_x m g
\end{pmatrix}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/223857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve for $v_2$ where $mv_1^2 + MU_1^2 = mv_2^2 + M U_2^2$ and $MU_1 - Mv_1 = MU_2 - mv_2$ by eliminating $U_2$? I was trying to solve the head on collision slingshot problem where the rocket moving with speed $v_1$ approaches a planet which is moving with speed $U_1$. I wanted the final speed of the rocket ($v_2$). $U_2$ is the final speed of the planet. Mass of planet is $M$. Mass of rocket is $m$. So I made two equations-
$$ M(U_1)^2 + m(v_1)^2 = M(U_2)^2 + m(v_2)^2 $$
$$ M(U_1) - m(v_1) = M(U_2) - m(v_2) $$
However, I am unable to eliminate $U_2$ to get ($v_2 = 2U_1 + v_1$) as the answer by also taking $\frac{m}{M} = 0$
Note :- This is a head on u-turn slingshot.
| I am getting $ U2=-U1-v1-v2 $ (Remember, these are added according to vector rules)
writing the two equations as,
\begin{equation}
M(U1)^{2}-M(U2)^{2}=m(v2^{2})-m(v1^{2})
\end{equation}
\begin{equation}
\implies M(U1-U2)(U1+U2)=m(v2-v1)(v2+v1)
\end{equation}
\begin{equation}
M(U1-U2)=m(v1-v2)
\end{equation}
Divide last two equations to get the relation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/223941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many particles can a particle accelerator accelerate at once? Does a particle accelerator lose its accelerating effectiveness as the number of particles being accelerated increases? According to Wikipedia, the mean acceleration of a proton in the Large Hadron Collider is 190,000,000 g's. Could the LHC accelerate one gram worth of protons at the rate while using the same amount of power?
I suspect not, but I don't know why.
| In general the total voltage $V$ seen by a particle (for a 1 TeV proton must be 1 trillion volts), multiplied by the beam current $I$, gives you the beam power:
$ P = VI $. So in principle you may have some trade off between the two of them, but often there are other limitations both to $V$ and to $I$. The designs of most of the existing machines have stretched these limits as much as possible (either because of cost or physical/technological problems) so it will not be immediately possible trade $V$ for $I$ or vice versa.
For instance at the LHC the energy is limited by the maximum bending field that the magnets are capable to produce, so even accelerating a single proton, you wont be able to push it beyond the nominal energy (currently 6.5 TeV). On the other hand you cannot arbitrary increase the current as collective behaviours of the beams may lead to instabilities and beam losses. Moreover, due to field quality issues, the LHC cannot even accept particles at an energy lower than 450 GeV.
So basically the number of particle that an accelerator can accept is the design one. Most of the time, at an operating machine, it cannot be easily increased even giving up something else (a counter example is the fully-loaded linac of CTF3). During the design phase you can optimize it, but this may lead to completely different design! Have a look at the (experimental) fusion reactors: those are basically very-low energy and quality accelerators (they just need to overcome the Coulomb barrier and particles scatter everywhere) which however store a huge amount of particles. Their designs do not share much with the high energy accelerators.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/224105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is there a relation between complexity of a system and entropy? Disclamer: I'm not a physics professional, so pardon me if the question is stupid/incomperhensible/generally doesn't make sense. And I've googled it, but didn't find an answer.
Getting to the point, I would love to know if a bigger, more complex system experiences higher levels of entropy, compared to a simpler and smaller system?
Thanks in advance to anyone who would enlighten me on the subject.
| Complexity behaves as the time derivative of entropy. In a closed system, entropy and complexity increase together initially, in other words the greater the disorder the more difficult it is to describe the system. But things change later on. Toward the end, as entropy approaches its final maximum where there is also maximal disorder, complexity diminishes.
See my publication: https://doi.org/10.1016/j.techfore.2021.121457
or pre-print: https://osf.io/6nwf9/
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/224201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is it possible that a person with myopia will see a blurry picture as normal? I am trying to process an image in good quality to appear blurred to a normal person and good to a person suffering from myopia
as seen in this source.
Is it possible that a picture that is blurry will appear normal to a person suffering from myopia (farsightedness)?
| I think that you can do something like that, but you have to exploit a caveat.
Myopia affected people have the same depth of field of normal-viewing people, with the exception that instead of going from tens of cm to infinity, it goes from few cm to tens of cm. This forces myopics to go close to the objects in order to put them on focus, but also allows them to resolve much more details. It is basically like having the macro turned on in a camera.
So you have to start from an image rich of tiny details which in reality could not be resolved by the human eye at a standard distance of ~30 cm. Then they would appeared blurred to a normal viewing person, while a myopic, being able to go much closer, may see them.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/224351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
How do I figure out the totally airborne height for a given machine? Technically "airborne" can just mean to move through the air, but I would like to know how high you have to be before you are entirely supported by air in a helicopter-like machine, as opposed to benefiting from the reaction from the earth (or whatever platform you are taking flight from). I am effectively asking for an equation that will tell me how high I have to be before the effect demonstrated in the following image is effectively zero.
I assume that this has something to do with mass, but am unclear on how to proceed beyond that.
| The ground effect is only present in winged craft because in a flight with wings there is a much high pressure under the wings when closer to the ground this increases the normal lift effect. In propeller lifted craft the effect still applies because the propellers are just air foils twisted ,but they are also smaller which means less surface area to have an effect on. The smaller the surface area the close to the ground they have to be and Most craft like helicopters have there propellers on top of the craft for stability thus making to far away from the ground to have the ground effect apply. Also you should check out back spin lift driven craft.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/224513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Elliptical orbit changing as a star's mass increases I'm studying Kepler's Laws, specifically the orbit of the Earth around the Sun. I know that if the Earth was more massive, the orbit would not be significantly affected. If the Sun was more massive, I know the velocity of Earth's orbit around the Sun would increase, but how would the shape of the orbit change? This is a theoretical case, neglecting conservation of mass and assuming the Sun's radius and volume don't change.
| If the existing orbit were already circular then any changes like you describe would immediately result in an elliptical orbit. After that the orbit would tend toward a circular orbit. How long that takes depends on the total tidal forces. A rigid body would take longer than a pliable one as tidal forces are exchanged.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/224703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How many fixed points does a Kelvin scale have? I have a book that says:
In the absolute Kelvin scale, the triple point of water is assigned the value of 273.16 K. The absolute zero is taken as the other fixed point.
But, then another section in the same book says:
On the Kelvin scale, the lower fixed point is taken as 273.15 K and the upper fixed point as 373.15 K.
What does this all mean? Can anyone clarify?
All I want to know is: How many fixed points does a Kelvin scale have?
| A linear scale for measuring something - temperature, pressure, whatever - has to have two fixed points, because two points are necessary and sufficient to define a line.
If a scale is defined with more than two fixed points, then calibration using any two of them is expected to produce the same result (up to experimental error), and which two are used is a matter of convenience. The SI definition of Kelvin only has two fixed points (0K and 273.16K = the triple point of water), but the ITS-90 thermometer calibration standard adds 13 more, because most thermometers don't include both 0K and 273.16K in their range, and it's useful to have fixed points in the middle as well as near the ends of a thermometer's range, to verify linearity.
Your book seems to be making a distinction between "Kelvin" and "absolute Kelvin". There is no such distinction; they are the same thing. I think it is (sloppily) trying to emphasize the historical development of the temperature scale, which started out defined as "Celsius shifted so that zero lies at absolute zero" and thus had Celsius's fixed points embedded in its definition, plus absolute zero.
Note also that "the boiling point of water" is deliberately not used in either the SI definition or ITS-90 specification, because it depends on the ambient pressure (whereas "the triple point of water" fixes the ambient pressure).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/224958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Magnetic field due to stationary electric dipole As we know from Maxwell's 3rd equation the magnetic field is given as
$$\nabla \times \mathbf{E} = - \frac {\partial \mathbf{B}}{\partial t}$$
Now, if we consider an electric dipole which is stationary, there will be electric field lines
like this:
I want to know what will be the curl of this electric field, if it exists. And will there be a magnetic field associated with this electric field too?
| Since everything is stationary, all time derivatives are zero, and there are no currents. Now look at the relevant equations. The curl of $\mathbf{E}$ you wrote yourself: $\nabla \times \mathbf{E} = -\partial \mathbf{B} / \partial t$. And will there be a magnetic field? Well, let's look at the relevant equations:
$$\nabla \cdot \mathbf{B} = 0$$
$$\nabla \times \mathbf{B} = \mu_0 \mathbf{j} + \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t}$$
If all time derivatives and currents are zero, what happens to $\mathbf{B}$?
Edit: I want to add a couple of things based on the comments. First, while intuition is definitely a great thing to have, it can fail. It doesn't come by itself; it has to be learned. Math, however, doesn't lie. So if the equation says $\nabla \times \mathbf{E} = -\partial \mathbf{B} / \partial t$ and you know $\partial \mathbf{B} / \partial t = 0$ because everything is stationary, well, then the curl of $\mathbf{E}$ is zero and there's nothing you can do about it.
But how does this fit with the picture? A non zero curl doesn't just mean that the lines curve. Rather, it means (roughly) that the lines are going around some place, and they must be going all the way around. This is most easily seen in the magnetic field of a wire. The field has a non zero curl right at the wire, and you can see that it goes around the wire. This can be justified more rigorously by Stoke's theorem, which says that the circulation of the field around a closed curve is equal to the surface integral of the curl on a surface which has the curve as its border.
Take your picture, draw a closed curve somewhere, and sketch the electric field at different points along the curve. If you take a line integral around it, along half the curve the field points one way and along the other half it points the other way, so the contributions cancel and the integral is zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/225121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does Quantum Mechanics say that anything is possible? I may be incorrect in this, but doesn't Quantum Mechanics state that everything has a probability of occurring (with some probabilities so low that it will essentially never happen)?
And if I am correct in my thinking, could it mean that, quite literally, anything has a possibility despite a potentially low probability?
(A version of my idea, which I admit closely borders on fictional magic, would be:
It is completely possible to walk through a wall, but the probability is so low that you need to walk into a wall for eternity before it can be done.)
| The short and direct answer to your main question is no, quantum mechanics does not say that anything is possible.
In regards to the "walking through a wall" question, the answer is not as you say, "walk for eternity." All you have to do is live long enough for the wall to crumble, then you walk through. This would be much shorter than eternity!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/225218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 3
} |
Is time created by movement through a higher dimensional object? In a comment I read on this page, someone mentions a theory where time is created by moving through and slicing "moments" of a higher dimensional object. For analogy, a 2-D creature living on a plane that moves through a 3-D object may experience the different slices of the 3-D object as "the passage of time" in their 2-D world.
*
*Is this idea well developed in physics?
*Can you share the name of this idea so that I can find it on google?
*Also, could dark matter play a role in this process?
| I don't think time is even a well defined concept and how it behaves conceptually can vary between different theoretical frameworks. Asking if time is created or explained by something in particular is assuming a standard definition already exists.
So it would probably be better to just ask if time can be defined as slices of a higher dimensional object.
I would say that it can. But whether or not that definition satisfies you is another question. (Since it's still not clear if such a definition could explain change in general, e.g. what governs the change between slices?)
The idea is used in physics, you can look up the concepts of world lines, world sheets, world volumes, etc..
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/225302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Intuitive explanation for subsonic Fanno flow In most situations in physics, the effect of kinetic friction is to reduce the macroscopic kinetic energy of a system and convert it into heat, thereby increasing its temperature.
but in the case of subsonic Fanno flow, the opposite happens: temperature decreases and velocity increases. I know the mathematical theory behind this, but can someone explain how friction is playing its dissipative role over here? There is the familiar effect of entropy increase but I'm also looking for a answer to how reduction of temperature and increase in velocity can have the above effect which seems quite counter intuitive to me.
| Friction still dissipates kinetic energy from the fluid. However, since we are assuming that the flow is adiabatic, heat is not transferred out of the system and is instead fed back into the fluid.
To conserve total energy, this thermal energy can either increase the temperature and decrease velocity or decrease velocity and increase temperature.
The choice is governed by the second law of thermodynamics: total entropy must increase.
As a thermodynamic state, entropy can written in terms of two other states.
While it may be intuitive that a process that increases temperature increases the number of microstates (and therefore increases entropy), we also need to consider how the other states are affected.
For an ideal gas, the speed of sound $a$ is related to temperature by
$$
a^2 = \gamma R T.
$$
Using the definition of the Mach number $M = u/a$, we can get the following relationship between the flow velocity and temperature
$$
\frac{\mathrm{d}T}{T} + \frac{\mathrm{d} M^2}{M^2} = \frac{\mathrm{d}u^2}{u^2}.
$$
Since we are also assuming that the flow rate is constant, the continuity equation reduces to $\mathrm{d}(u \rho) = 0$.
This assumption fixes the relationship between flow velocity and density
$$
\frac{\mathrm{d}\rho}{\rho} + \frac{1}{2} \frac{\mathrm{d}u^2}{u^2} = 0.
$$
So as the velocity increases, the density must decrease in order to maintain flow rate.
The change in entropy is then determined by the rate of temperature change and the rate of density change, both of behave differently at different Mach numbers. For subsonic flows, the combination that will result in increasing entropy is to drop temperature and density.
Let me know if this is still too hand-wavy, I can dig up some old notes and go over it step by step.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/225386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
What does it mean that the Rutherford's cross section is infinite? I'm studying elastic scattering and I read that the Rutherford's differential cross section is defined as:
$$\left( \frac{d \sigma}{d \Omega} \right)_R = \frac{Z^2}{4} r_o^2 z^2 \frac{(m_ec / \beta p)^2}{\sin^4(\theta/2)}$$
And the total cross section is defined as:
$$ \sigma_T = \int_0^{2 \pi} d \phi \int_0^\pi \frac{d \sigma}{d \Omega} \sin \theta d \theta$$
For the Rutherford's cross section the total cross section equals infinity ($(\sigma_T)_R = \infty$).
What does this mean physically? The cross section is the probability of interaction per unit of surface. Does this mean that we are considering as an interaction even when the particle continues its path as if nothing happens (i.e. $\theta=0$)?
| That the total cross section is infinite just means that every charged particle that passes by the (bare) nucleus is scattered to some extent. This is a consequence of the Coulomb potential being long range. Classically, it is sufficient for the potential to be nonzero for all radii in order to have an infinite total cross section. Quantum mechanically, if the potential approaches zero rapidly enough, the total cross section will be finite. However, the Coulomb potential goes to zero slowly at infinity, and the resulting Rutherford cross section (which is correct both classically and quantum mechanically) has an infinite total cross section.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/225715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Current from Middle Battery in a Two-looped Circuit
With this question, as with many tutorials of similar questions I’ve found online, my textbook only mentions three currents: $I_1$, which flows through the left loop from and to the 19 V battery, $I_2$, which flows through the right loop from and to the 19 V battery, and $I_3$, which flows through the middle section.
However, why can’t there be an $I_4$ which flows from and to the 12 V battery through the left loop, and an $I_5$ which flows through the right loop? In this diagram, it seems pretty clear that all the current comes from the 19 V battery. But about the current coming from the 12 V battery? Is there no current? (The subsequent analysis completely discounts the presence of any current from the 12 V battery, so I’m thoroughly confused.)
| *
*If only the 19 V battery was present, there would be currents in all branches, right?
*If only the 12 V battery was there, there would also be currents (but of different values) in all branches, right?
When both the 19 V and the 12 V batteries are present, they both contribute to all branches. The $I_1$, $I_2$ and $I_3$ in the diagram are the combined resulting currents, not only those coming from one battery.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/225815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
What will happen to a human being exposed to Martian atmosphere? Mark Watney, in the movie The Martian, says that,
If the HAB breaches, I'm just gonna, kind of... implode.
The corresponding novel, by Andy Weir, says he will explode (as pointed out by @MikaelSundberg).
I think he will neither explode, nor implode, but simply die of cold and asphyxiation.
Can anyone scientifically explain what will happen?
PS: The HAB is a NASA designed habitat for humans on Mars.
| The Martian atmospheric pressure is approximately equal to $0.6\%$ of Earth's mean, at sea level, mostly consisted of $CO_2$ ($98\%$).
This is equivalent of pressure at altitude of approximately $17 km$ in the Earth atmosphere, with boiling point $30 ^oC$.
The pilots use oxygen masks at altitudes $> 4km$, so even if Mars's atmosphere was entirely of $O_2$, people wouldn't survive.
Regarding the temperatures on Mars, they vary from $-150 ^oC$ to $+20 ^o C$.
Consequently, sure death by asphyxiation and depending on the location and season, instant or later freezing, which will be accelerated by the low boiling point, especially if there is no suit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/225910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Which elementary particles are behind magnetic field, similar as photons behind radio waves? I see, there are photons behind radio waves.
As Wave–particle duality said: the radio waves are waves and at the same time are fluxes of particles called Photons.
I'm wondering, what is behind magnetic fields?
Means, magnetic fields - are waves too. Shall this waves consists of photons too? Then how does photons moves by such trajectories??
| First. Eletric and magnetic field is dependent of the inertia frame. Only when we see this two fields together, that we have an physical object living in space-time, being independent of the inertial frame. Actually, when we do this we learn what is the truly symmetries of our space-time. Einstein is the one that discover that ;).
Radio waves are compound by electric and magnetic field. One create the another through space and time. See Maxwell's equations that describes this process.
Let's go to you question:
Behind the electromagnet (EM) field we have photons. But this is in some way a quantum relationship. We can say that photons and EM field is two faces of the same thing, as a generalization of wave-particle duality. In some cases we may have a bunch of photons propagating through space building an EM wave. In other cases we may have a state builded by a lot of superpositions of photons created from the vacuum propagating through space and annihilated to the vacuum, that constructively build an stationary EM field state. Actually the vacuum itself is build in that way. This process is more clear when you understand Feynman diagrams, that is a perturbative application of Path integral formulation of Quantum Mechanics.
This is quantum mechanics, so don't think that this particles are actually there. Only if you put an delicate detector you would measure they. But in quantum mechanics before the measure we have only potentialities of possible outcomes and they probabilities. And after the measurement you destroy the field state or vacuum state, in case of local measurements.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/226025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Decomposition of Spring Can a spring respectively its stiffness in a 2D-system be decomposed into its x- and y- components (stiffnesses)?
Assuming a spring located in an 2D coordinate system with positions p(x,y), length l and angle alpha, describing the orientation of the spring inside the system. Can the spring be represented by its xx, xy, yx and yy components?
Imagine an object which interacts with the spring not along its main direction, but along the global x-axis. Is it possible to model the interaction between spring and object by the decomposed xx-component of the spring?
Many thanks in advance
| If you are asking about the stiffness ( or spring constant $k$ ) , then obviously it cannot split into x and y components, because $k$ is a scalar constant. Suppose in your example, if the spring , at an angle $\alpha$ from x axis in $2D$ travels a distance $r$ then the displacement and force can be split into x and y component. That is, now the equation of such system is $$F=-kr$$
this can be split into
$$F_x=-kx$$
and $$F_y=-ky$$
as $k$ is a scalar. You can connect these three equations using
$$F=\sqrt{F_x^2+F_y^2}\\
r=\sqrt{x^2+y^2}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/226112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Dirac equation, $\alpha_i$, $\beta$ hermitian The argument I've seen is the one given here: http://epx.phys.tohoku.ac.jp/~yhitoshi/particleweb/ptest-3.pdf under (3.10):
$$H=\vec{\alpha}\cdot(-i\vec{\nabla})+\beta m$$
$H$ is hermitian, $-i\vec{\nabla}$ is hermitian, so $\vec{\alpha},\beta$ are hermitian.
This is not convincing, because a hermitian operator being a sum of two operators does not imply that the two are also hermitian, for example in the harmonic oscillator, the position and momentum operators can be written as a sum or difference of the annihilation/creation operators, which are non-hermitian.
So is the presented argument wrong? Why are the $\alpha_i, \beta$ really hermitian?
| I think its important to say that the Dirac equation cannot be derived using standard QM: it can be motivated, but ultimalely the definitive argument for/against its correction is a matter of experimental confirmation.
This means that the arguments that the link gives you in favour of the equation are just formal: they might seem very logical and unavoidable, but if the resulting equation doesnt work, its wrong.
I think its important to remember this, as we novice physicists tend to forget/ignore the infinite number of proposed equation, which felt very logical at the time, but soon ruled out by experiments. Dirac equation remained because it works. $\boldsymbol \alpha,\beta$ are hermitian because the resulting equation works.
That being said, we can give some arguments as to why $\boldsymbol \alpha,\beta$ cannot be non-hermitian: if we want a Hamiltonian like
$$
H=\boldsymbol\alpha\cdot\boldsymbol p+m\beta
$$
to be hermitian, then we are forced to take both $\boldsymbol \alpha$ and $\beta$ hermitian. One way of proving this is to see that $H$ must be hermitian regardless of the value of $m$. In particular, $H$ must be hermitian for massless particles, so
$$
H_\text{massless}=\boldsymbol\alpha\cdot\boldsymbol p
$$
must be hermitian. But as $\boldsymbol p$ is hermitian, the three matrices $\boldsymbol \alpha$ must be hermitian. As $\boldsymbol \alpha$ is independent of $m$, $\boldsymbol\alpha$ must also be hermitian in the massive case:
$$
H=\boldsymbol\alpha\cdot\boldsymbol p+m\beta
$$
Finally, as we already proved, the first term in the expression above is hermitian, so if $H$ is to be hermitian, the so must the term $m\beta$. This proves that $\beta$ must be hermitian.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/226526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Reconciling De Broglie wavelength with relativity I have trouble understanding what a De Broglie wavelength means when differing frames of reference are taken into account.
Observers in differing frames of reference will see a particle's wavelength as being different. I think I understand a matter wave as similar to a probability wave, but what impact does it have on my understanding when probability becomes relative? Does this mean different observers have a different probability of finding a particle at a position?
This might seem silly, but I am at rest relative to myself, do I have a definable wavelength from my own frame of reference even though observers will see a matter wave?
| The De Broglie wavelength corresponds to a free particle's momentum, $\bf p=\hbar \bf k$, and a particle's observed momentum obviously depends on it's velocity relative to the observer.
The localization probability, on the other hand, is given by the wave function $\psi$ as $|\psi|^2$. But for a free particle $\psi \sim e^{i(\bf k \cdot\bf x -\omega t)}$ and the localization probability is uniform in space (albeit infinitesimal), regardless of the observer's reference frame.
In the particle's rest frame the wavelength becomes infinite (zero momentum) and the wave function simply oscillates in time as $e^{-i\omega t}$.
For wave packets that are no longer uniformly distributed throughout 3D-space, the localization probability will look different for different observers, but this is no different then a localized light pulse looking different in different frames.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/226634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does moving air have low pressure? According to Wikipedia lift in an aircraft is due to an area of low pressure formed above the wings of an aircraft due to the fast moving air there. So why exactly is an area of low pressure created due to fast moving air?
| I don't know if it helps or not but give it a try
For this to understand let us consider a room in place of a hose say of length l in which a ball is bouncing up and down and colliding elastically with the ceiling and the floor continuously.
Now if we increase the speed of the ball in this upward or downward direction using some techniques, the ball will strike the ceiling and the floor with a greater momentum and thus the change in momentum will also increase and hence will apply a greater force and hence pressure on the ceiling and the floor will increase.
But what if we push the ball in the horizontal direction ?
If we give it a appreciable horizontal speed , it will rarely strike both the walls before coming out of the room and hence we can conclude that the pressure on the ceiling and the floor has decreased with increase in kinetic energy.
The above statements are taken from this answer of mine.
And I think that because of more hitting of molecules (present below the paper) than that above it, the paper goes up.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/226700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 11,
"answer_id": 5
} |
Is all of a black hole's mass situated arbitrarily close to the event horizon? Forgive me if I'm thinking about black holes in completely the wrong way, but since time dilation increases to arbitrarily large amounts the closer you get to the event horizon of a black hole, wouldn't that make it impossible for anything to cross it, since it would require an infinite amount of time?
Is every single black hole in existence just an infinitely thin sphere of infinitely red-shifting matter around some kind of physics-breaking empty space that we'll never know anything about because it's impossible to get to? And if that's the case, does the "singularity" at the center of a black hole actually exist, or is it just a center of mass?
| From what I know of black holes, you may get a few theories from this question, but the truth is we don't know presently. Singularities such as this are complete mysteries. We may never know, but as long as we survive our own self destruction, I believe we will figure it out eventually. :)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/226808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If I toss a coin, vertically, on the surface of Mars, will it land back in my hand? When I toss a coin in Mars, is the planets atmosphere rare enough that I'd rotate with the planet (at its angular velocity), but not the coin?
| The coin will come back to your hand just like it would on the earth. The effect of atmosphere is negligible comparing to the coin's inertia, so the horizontal position of the coin relative to your hand will hardly be affected. The rareness of the atmosphere will only affect the vertical motion of the coin, like how quickly the coin will fall into your hand.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/226882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 7,
"answer_id": 1
} |
Will physics ever be able to answer the question: "What caused the universe to come into existance"? Or is this just the flip side of a metaphysical question "why does the universe not exist", if the opposite was the case and nothing existed i.e. a universe devoid of all energy, matter forces, particles of any kind, massless or otherwise.
| There used to be a belief that eventually we would discover/create a Theory of Everything (TOE) that would explain why all the constants have the value they have and why the universe and laws of physics could be no other way than they are.
I doubt whether anyone still believes that is possible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/226964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does quantum randomness measurably affect macro-sized objects? I understand that while it is believed that there is no true randomness on the macro scale, there is true randomness on the quantum scale. A previous theory that quantum processes could be determined through "hidden variables" has been disproven (through polarizing photons and radioactive particle decay), confirming that true randomness does exist.
Now for my question. Does quantum randomness measurably affect the macro scale such that true randomness actually does exist outside quantum mechanics, or will rolling a die in identical conditions always yield the same result even after factoring quantum randomness?
| The stochastic features of QM could leave, in principle, a "trace" at the macroscopic level.
This is because not every $\hslash$-dependent family of quantum states yields, in the limit $\hslash\to 0$, a completely deterministic classical state (phase space point). As a matter of fact:
Every possible classical phase-space probability distribution can be obtained from some suitable quantum configuration, in the classical limit.
In other words, in the classical limit the quantum non-commutative probabilities (states) become, in general, classical probabilities in the phase space (statistical states). There are quantum states that in the limit become points of the phase space (not surprisingly, this is the case for the squeezed coherent states of minimal uncertainty), but these are only special cases.
This classical statistical description, that emerges from quantum states in the classical limit, can be seen as a "trace" of the probabilistic nature of the underlying quantum theory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/227243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sliding along a circular hoop: work done by friction
Assume a point object of mass $m$ slides along a hoop of radius $R$, starting from a position which makes 90 degrees with the line of radius connecting the center and the ground. Let the coefficient of kinetic friction between the hoop and the object be $\mu$. Assuming that the object starts at rest, what is the total work done by the friction when the object comes to the ground level?
My idea: the normal force at any instance is given by $$N=mg\sin\theta+\frac{mv^2}{R},$$ where $\theta$ is the angle between the radial line connecting the present position and the intital position of the object to the center of hoop. With this we have the frictional force as $$f_k=\mu\left(mg\sin\theta+\frac{mv^2}{R}\right),$$ so that the total work done by friction is $$W_k=\int_0^{\pi/2}\mu\left(mg\sin\theta+\frac{mv^2}{R}\right)R\mathop{\mathrm{d\theta}}.$$
The problem I am having is to figure out $v$ as a function of $\theta$, i.e $v(\theta)$. Any ideas?
| You are almost there. All you need to do now is realize that the velocity can be deduced from the kinetic energy, which is the difference between the potential energy lost and the work done by friction. And you have expressions for both of those. See if that gets you there.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/227354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How is $ \left(1-\frac{p^2}{2mE}\right)^{3N/2-2} =\; \exp\left(-\frac{3N}{2}\frac{p^2}{2mE}\right)\;?$ How is
$$ \left(1-\frac{p^2}{2mE}\right)^{3N/2-2} = \exp\left(-\frac{3N}{2}\frac{p^2}{2mE}\right)$$
(Karder, Statistical Physics of Particles, Page 107)
in the large $E$ limit. Here $N$ is particle, of the order of $10^{23}$, $E$ is the total energy.
I roughly guess that it should be $\exp(-\frac{p^2}{2m})$ since both $N$ and $E$ can be treated as infinitely large.
Update: a hint to solution is provided in the comments.
| You can use the approximations
$$1+x \simeq e^x$$ and
$$(1+rx) \simeq (1+x)^r$$
You can obtain
$$(1+(-p^2/2mE))^{3N/2-2}$$ which can be approximated in the $N\gg1$ limit as $$(1+(-p^2/2mE))^{3N/2}$$ which is approximately equal to$$(1+\frac{3N}{2}(-p^2/2mE))\rightarrow \exp [1 + (-p^2/2mE)]^\frac{3N}{2}$$ by using the first foruma above. Here $\frac{3N}{2}$ is considered as a constant number much greater than 1, probably the ${3N/2-2}$ is the correction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/227654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Velocity of liquid molecules in turbulent flow I was solving some questions when I came across this:
what is the velocity of liquid molecules in contact with the walls of the tube? and the answer was given that it can have any velocity and then no further explanation. I tried to search from different sources but could not find any reason. My doubt is that by any velocity does it mean it can have zero velocity too? and even maximum velocity?
| In laminar flow, fluid moves in defined layers. The boundary layer closest to the wall moves with least magnitude of velocity. The direction of its velocity is the direction of flow through the tube.
In turbulent flow, molecules become disorganized and can move in eddy currents that swirl in any direction, including against each other, regardless of distance from the wall. The velocity of molecules in turbulent flow is not constrained by distance from the wall, although the link I cited says that "a general specific feature of the near-wall turbulent flows is the presence, on the wall, of a thin viscous sublayer, wherein molecular viscosity forces are dominant and the velocity distribution is linear."
Perhaps the answer you cite was trying to illustrate that neither direction nor magnitude of velocity of any one molecule in turbulent flow at the wall is predictable.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/227768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
About the ratio of the density of dark energy at the time of cosmic microwave background emission to the current density of dark energy In a question, I am given the current densities of dark energy, dark matter and normal matter and am asked to find the ratio of density of dark energy at the time of CMB and now.
The answer is 1. Is this a simple calculation or is it something we are expected to know?
| Assuming dark energy is a cosmological constant and not quintessence or something more complicated, it is a constant and does not change as the universe expands. Well, if it changed it would be a cosmological variable not a constant!
For more on this type of calculation you should read Pulsar's answer to Equation for Hubble Value as a function of time as this describes the principles involved.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/227860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Do free-electron lasers actually lase? Free-electron lasers are devices which use the motion of highly energetic electron beams to produce bright, coherent radiation in the x-ray regime. More specifically, they start with a high-energy electron beam and feed it into an undulator, which is an array of alternating magnetic fields designed to make the electron beam move in a 'zigzag' path with sharp turns on either side, emitting synchrotron radiation during each turn.
The radiation thus produced is added up over successive turns, and it is produced coherently via self-amplified spontaneous emission.
One common question frequently posed of this setup is: is this actually a laser? That is, does it have population inversion, and is the radiation actually emitted via stimulated emission in some suitable understanding of it? People in the know tend to answer in the affirmative, but I've yet to see a nice explanation of why - and definitely not online. So: do FELs actually lase?
| A free-electron laser (FEL) is a parametric amplifier, which operates by transferring energy to the output signal (photon pulse) from an oscillator (electron bunch running down a long undulator magnet). An electron bunch is accelerated to relativistic energies and sent through a periodic magnetic structure (undulator) where transverse oscillations and interference produce synchrotron radiation enhanced at specific wavelengths. The intensity of this radiation scales with the number of electrons in the bunch. Photons co-propagate with the relativistic electrons and if the undulator is long enough, induce significant energy modulation in the electron bunch, leading to a periodic density modulation of the electron cloud (microbunching). The resulting microbunches behave like giant charged particles, and emit photons proportional to the square of their total charge at wavelengths longer than the bunch length. So a free-electron laser is not a laser in the usual meaning of the acronym but a parametric amplifier that produces coherent radiation pulses.
Note that because of the need for microbunching for an FEL to operate, there is no FEL that can produce truly continuous photon beams as a function of time today.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/227960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 5,
"answer_id": 1
} |
How exactly does a bulb light up?
A typical value for the electron drift velocity in a copper wire is $10^3\ \mathrm{m\ s^{-1}}$. In the circuit below, the length of the copper wire joining the negative terminal of the batter to the lamp is $0.50\ \mathrm{m}$.
(i) The switch S is closed. Calculate the time it would take for an electron to move from the negative terminal of the battery to the lamp.
(ii) The lamp lights in a time much less than that calculated in (e)(i). Explain this observation.
In the second part, I can't imagine the situation. I reckon that not all electrons travel with a drift velocity, so the faster ones reach the bulb and make it glow.
But how exactly does this lighting happen? The electron comes in contact with the circuitry and lights up the bulb, or is it because of the electric field?
| We must understand that the internal electromagnetic forces inside the wire constrains the charge movement.
It is not so different from a rigid body as a rocket or a ship. The source of power is at the rear in both cases, but the rear part can move only if all body moves. A ship can have a slow velocity, but as soon as its rear part starts to move, its front part (say 100 m ahead) also moves.
It is not possible for individual electrons to flow through the wire unless some of its fellows ahead also moves. Otherwise an excess of negative charge would be built in regions of the wire.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to move a bubble which is trapped by the capillary pressure? I have a question about how to move a trapped bubble in a tube.
If we assume to have a horizontal tube, with water on each side of the bubble. The point to the left of the bubble is point 1, while the point to the right is point 2.
The capillary pressure equation is: $\Delta P_{cap}=\frac{2\cdot \sigma \cdot cos(\theta)}{R}$
Where $\theta$ is the contact angle, $\sigma$ is the interfacial tension between the gas buble and the water, while $R$ is the radius of the tube.
Since the bubble is trapped, I assume that pressure in point 2 ($P_2$) is larger than the pressure in point 1 ($P_1$). In order to accomplish this, the contact angle between the bubble and point 2 must be smaller than the contact angle betweenthe bubble and point 1, or the radius of point 2 must be smaller than the radius of point 1:
$$P_2-P_1=\frac{2\sigma cos(\theta_1)}{R}-\frac{2\sigma cos(\theta_1)}{R}=?\tag1$$
My question is therefore how we can get this bubble moving by applying more force on the left side (I want to move it towards the right)? If we apply more force from the left side, won't that increase the pressure $P_2$ the same as $P_1$, because of the capillary pressure, so that $P_2$ will always be larger than $P_1$ (if not the contact angle changes),and the bubble cannot be moved with more pressure?
| From an engineering POV you could use sound, either in the audible or ultrasonic range to disrupt the adhesion. A transducer could be placed in the liquid at one end of the tube and arranged so some sound travels along the tube. If that is not possible them maybe attach a transducer to the outside. All experimental though. One of those "try it and see" answers with no theory to back it up.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Why can the spin of a relativistic particle not be orthogonal to its momentum? I have read that the 3-momentum of a relativistic particle cannot be orthogonal to its spin 3-vector. When thinking about how the spin vector transforms when the particle approaches light speed, it seems clear that it cannot be orthogonal to the boost direction, but I'm wondering what is the exact mechanism that prevents it.
An example for clarification: Let's take an electron at rest in the lab frame, which has been put in a spin-up eigenstate ($z$-axis). Then, a potential is added, such that the electron gains some $x$-momentum.
Since the spin was pure $z$ axis and the momentum pure $x$, some relativistic process must change the spin vector to prevent it from being orthogonal to the momentum.
EDIT: I'm starting to think that this must be answered with Dirac spinors. The $x$ momentum must have some effect on the spin vector direction. But I cannot see the quantitative answer, so I'm still looking for an answer.
| I think you are just referring to a simple kinematical phenomenon. Consider a massive particle carrying a $3$-vector $\vec{S} \equiv (s_x,s_y,s_z)$ in its rest frame, which exists just because the particle is massive.
This vector may describe the spin, but not necessarily it. Now suppose that the particle travels with a speed $v$ along the direction $z$ if referring to the laboratory reference frame whose spatial $z$-axis is collinear with the one of the rest frame (up to a pair of internal rotations the decomposition theorem of Lorentz group guarantees that this is a standard situation).
The vector $\vec{S}$ defines a four vector $S^\mu$ whose components in the laboratory frame are $$\left(\frac{vs_3}{c\sqrt{1-\frac{v^2}{c^2}}},s_x,s_y, \frac{s_3}{\sqrt{1-\frac{v^2}{c^2}}}\right)\:.$$
You see that, among the last three spatial components, the component parallel to the momentum (in the direction $z$) $\frac{s_3}{\sqrt{1-\frac{v^2}{c^2}}}$ dominates with respect to the normal components $s_x,s_y$ as $v \sim c$.
In the limit $v=c$, if one assumes that something like the parallel component exists, the normal components cannot be defined at all. However the rest frame itself do not exist, so this case cannot be easily treated and another approach is more suitable, the one arising from the classification of unitary irreducible representations of Lorentz group. In a sense, however, the remnant of the spatial components of $S^\mu$ in the laboratory frame, in this limit case is the so called helicity, which could be viewed as always parallel (or antiparallel) to the $3$-momentum (see ACuriousMind's answer).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Direction of static friction in inclined rolling motion I can't seem to understand in which direction static friction faces for inclined plane motion with rolling motion. This considers rolling motion without slipping, how do i find the direction of the static frictional force?
| The way i ended thinking about it is as follows: I convinced myself friction is forcing opposing motion. Now imagining a ball rolling down an incline, I considered the point of contact of the ball with the incline, call it x. This point intends to move in a direction that is opposed to the general translational motion of the ball as a whole (to be more precise, you have to consider infinitesimally small change in time from the moment x is in contact with the incline).
Hence the force of friction is opposed in direction to x's motion thus, (finally) the force of static friction is parallel to the general direction of translational motion of the ball.
Note that it is "static friction" because otherwise x would remained in contact with the incline as the ball translates downward, which we refer to as "slipping".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to determine the radius of curvature of a convex lens? Suppose there is an equi-convex lens made of glass which has a focal length ($f$) of 30cm. Then, can we not say that the radius of curvature, $R$ of the lens is twice the focal length, i.e. $R = 60cm$?
Why do we need to use the Lens Maker's Formula for the same, which in fact gives a different result :
$$
\frac{1}{f}= \frac{\mu_2 - \mu_1}{\mu_1}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
$$
For $f=30 cm$, $\mu_1 = 1$, $\mu_2 = 1.5$, $R_1 = R$ and $R_2 = -R$, we get :
$$\frac{1}{30}= 0.5\times\frac{2}{R}$$
Or, $R=30cm$
How can one explain this? Also, how can the focal length be equal to the radius of curvature?
| Intuitively, the radius of curvature has to depend on the index of refraction of the glass. If the index were $1$, the lens would have no effect at all. If the index were very high, say $10$, it would not take much curvature to get a given f'ocal length. Clearly we cannot just say the radius of curvature is twice the focal length.
You have applied the lens maker's formula correctly to your problem. There is no problem with $R=f$, in fact that is always true for $\mu_2=1.5\mu_1$. We are assuming a thin lens in this formula, so the diameter of the lens must be small compared with $R$. If the focal length is $30$ cm and the diameter of the lens is $1$ cm the thickness is twice the height of a circular segment. Given $R=30, c=1$ we have $h=R-\sqrt{R^2-(\frac c4)^2}=30-\sqrt{900-\frac 14}\approx 0.004$ so the lens is about $1$ mm thick
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How is the perceived quality of sound measured and analyzed? I am doing a physics experiment researching the effect of temperature inside the clarinet pipe on the quality of its sound. I know that the speed of sound is directly proportional to temperature of the medium it propagates in, I can measure decibels, frequencies, air pressure, and have theoretical knowledge on the production of resonance in open pipes. But how can the perceived quality of sound be measured and analyzed?
| This question is "turned upside down". The proper direction of research is:
$$
I \ can \ describe \ the \ percieved \ quality \rightarrow \ Let's \ seek \ for \ its \ physical \ (objective) \ description
$$
not the other way.
At this point the question suddenly becomes very broad and will be closed probably. Generally: any measurements of intensity, pressure etc. will not provide you any "quality" as a physical quantity, at least without listening tests and a great deal of psychoacoustics.
There are some phenomena that would certainly lower the quality as unwanted impedance discontinuities in the clarinet tube, but you should study some literature first. This is, of course, a studied phenomenon. Try The Physics of Musical Instruments by Fletcher and Rossing and maybe Signals, Sound and Sensation by Hartmann.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Why is the speed of sound lower at higher altitudes? At sea level the speed of sound is 760mph, but at altitudes like the Concorde would fly at (55,000ft) the sound barrier is at 660mph, so 1000th slower. Does it have to do with lower pressure?
| From a non-technical viewpoint, I would say that the simplest way I understand this is the following one.
Yes, it has to do with pressure. Actually, it might be easier to think that it has to do with density. Consider the dominoes effect. If the dominoes are more apart from each other, that is, if the density of dominoes is lower, it will take longer for a certain domino to "communicate" to the next one that it has suffered a mechanical push from the previous one. Analogously, if the density of the air is lower, the propagation of the mechanical pulse will take longer, that is, its speed will be lower.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Hamiltonian of a quantum harmonic oscillator On page 286-287 of Nielsen Chuang's Quantum Information and Quantum Computation (10th edition) book, the Hamiltonian for a quantum harmonic oscillator is approximated as $H=a^\dagger a.$ What are the assumptions involved in such an approximation and why is this approximation needed?
| I have not looked at the book , however the sense in which it is an approximation is that it is neglecting the constant term
The Hamiltonian of a SHO is ,
$H= (a^{\dagger}a + 1/2)\hbar\omega$
This means that the ground state energy of the SHO is $1/2\hbar\omega$. This is what is being neglected since it is only a constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/228915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
visible light spectrum Why do we see black objects? Colors of objects are formed when the spectrum of that color is reflected. Example Green objects are green because they reflect the green spectrum of light, red objects are red because they reflect red spectrum of the visible light and white objects because they reflect all the visible spectrum of light. When an object absorbs all the visible spectrum of light, it becomes black. But if nothing is reflected from the object, shouldn't it be invisible instead of black?
| As long as there is a discontinuity in the index of refraction ( real or imaginary parts) then there will be a reflection from the "black" object. How your eye-brain perceives it is a different matter..
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
What is the metric tensor for? I am wondering how to use the metric tensor, in practice? I read the book and done the exercises in A student's guide to vectors and tensors by Dan Fleisch. The concept of a tensor and their applications are well defined.
In that book, is explained how to get the metric tensor for coordinate system transformation, such as from spherical coordinates to ordinary Cartesian coordinates or even from cylindrical coordinates to Cartesian coordinates; which are easy to obtain, given enough practice. But what do such metric tensors mean (in practice), how does one use such a tensor in an actual math/physics problem?
| The metric is a rank two tensor that defines several features of a differential manifold. It defines how to relate changes in distance to changes in coordinates, how to take take the inner product of two vectors. Given an inner product we have a way of measuring angles.
More indirectly, the metric describes geodesics in a manifold, the shortest distance between two points. Just as the shortest distance between two points on the surface of a sphere is not a straight line, this is also the case in a general non-flat space. Speaking of, the metric contains information about the curvature of a manifold as well as how coordinate basis vectors change within the manifold.
Given its relation with the inner product, the metric provides a means to generalize Gauss' and Stoke's Laws to higher dimensions.
This is an excellent introduction to these topics:A First Course in General Relativity by Schultz
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What came first, the Universe or the Physical laws that govern the Universe? This sounds like the Egg and the Hen question but I am curious about this. If universe came first and created physical laws for itself, then what created the law or the principle as a consequence of which the universe came into existence in the first place? And if there where pre-existing physical laws that governed the big bang or whatever the origin of the universe was, then where did those laws come from and what were they a part of? If we assume that creation and destruction of universe is cyclic and the same laws are carried onto the next creation and destruction cycle then shouldn't the law which is governing this cycle be a consequence or a part of some bigger something (like a mega-verse). Whichever the case, we again come down to same basic question as in the title.
Thanks.
| As far as I understand, Physics is not able to awnser this question, because the physical laws we use to describe the Universe are not valid up to the exact event of the Big Bang (the Big Bang is said to be a singularity of spacetime).
Physics attempts to describe the Universe at a moment when it already existed, but does not states causes for its existence itself.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Kinetic energy of a rotating object in an exercise, a linear molecule is being subject to a force applied on the edge in its axis. Then $K_1=\frac{1}{2}mv^2$, all is well.
Then in the second point of the exercise, the force is applied on the same edge but in an orthogonal direction to its axis. Then the molecule begins to rotate. So its kinetic energy is composed of two terms: $K_2=\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$,$\omega$ being the angular velocity of the rotation. The thing is that the linear velocity is the same as before and the correction says that $K_2=K_1+\frac{1}{2}I\omega^2$. But how come the same force can give two different energies to the molecule? I thought that $v$ would decrease in the second case, because of the apparition of the angular velocity $\omega$ so that the energy would be conserved. So in the second case the molecule goes as fast as before but in addition it rotates on itself?
| The energy of the system is not only proportionate to the force, applied to it, but is actually the work done by that force ($F\Delta s$) on the path that your system has travelled ($\Delta s$). In the first case the molecule has only the translational motion. In the second case, in addition to translational motion there is also a rotation, thus the path is NOT the same and the energy is NOT the same. The key-point here is that the energy depends also on the path, not only to the applied force.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What's the reason double-slit experiment can't be explained by edge effects rather than quantum interference? Say we had exactly this...
But instead, it was a PING PONG GUN (imagine as table tennis players use to train),
throwing out PING PONG BALLS. The two slits are say 20 cm wide, and the observing screen is say 5m distant.
If the ball goes through the EXACT MIDDLE of a 20cm slit, it will travel in a perfectly straight line and make a "dot" on the observing screen.
If the ball travels nearer and nearer to the left or right edge of a slit, the flight path will bend slightly towards that side. For example, due to electrostatic force (rather like how a vertical pour of water from a faucet will bend slightly as your hand approaches).
Note that this is not some sort of fantasy; you could very easily organise for the ball path to bend slightly when near an edge, using either electrostatic force, magnetic force, aerodynamic factors or other forces, with the correct material of balls and slits (substitute small metal balls and slits of magnetic material .. whatever).
Indeed, you could trivial arrange so that precisely this famous image
is the outcome.
This is the "trivial mechanical bending" explanation of "all this interference pattern stuff".
Can you help me understand in a clear way, What is the explanation of why this is not at all the explanation?
| Your explanation makes no sense. To see why, suppose you have two slits and you record a particular interference pattern as a result: a series of light and dark bars. If you then cut an additional pair of slits half way between the first pair of slits, the resulting pattern may have some dark bars where formerly there were light bars. The only way this can be explained is if there is something going through the additional slits that deflects the light that would have hit the bars that were light in the two slit experiment. A full explanation of this point can be found in "The Fabric of Reality" by David Deutsch, Chapter 2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
What would be the view like from inside a black hole looking towards the event horizon? Ignoring the fact that we would be torn apart by gravitational gradient and assuming we get some time to make some observations before hitting singularity, what would we see looking towards the event horizon or in any other direction away from the singularity?
| There are actually some nifty simulations that show what you would see:
http://jila.colorado.edu/~ajsh/insidebh/intro.html
(Had to post as 'answer' because I don't have enough reputation to comment)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/229868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
Why wavefunction is sometimes multiplied by the radius to get probability density? When solving 1d particle in a box, the probability density is said to be proportional to $|\psi|$, but when solving 3d orbitals, the probability density is said to be proportional to $|\psi|^2 r^2$. Why this difference?
| It's not "multiplied by $r^2$ to get the probility density". The issue is that the volume element in spherical coordinates is
$$ \mathrm{d}V = r^2\sin(\theta)\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi$$
and since the probability to find a particle in a subspace $X\subset \mathbb{R}^3$ is
$$ P(X) = \int_X\lvert \psi(r)\rvert^2\mathrm{d}V$$
by definition of a probability density, the quantity $r^2\lvert \psi(r)\rvert^2$ is what behaves like the "normal" probability density in flat coordinates: The probability to find the particle between $r_1$ and $r_2$ is proportional to $ \int_{r_1}^{r_2} r^2\lvert \psi \rvert^2\mathrm{d}r$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Derive drag coefficient of plate There is any analytical way to derive drag coefficient of flat plate aligned perpendicular to the flow? Wikipedia says it's between 1.98~2.05 but I want to get this value in calculation, not experimental value.
| Not really. For laminar flows, the solution will be the Blasius solution, but the solution is still numerical. There are analytical functions that can approximate it fairly well.
For turbulent boundary layers, there is even less hope for an analytical solution. The flow is non-linear and time dependent, but mean equations can be found. These will still need to be solved numerically though.
Although the flow configuration seems like it is one of the most basic possible, the Navier-Stokes equations admit relatively few known analytical solutions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why are four-legged chairs so common? Four-legged chairs are by far the most common form of chair. However, only three legs are necessary to maintain stability whilst sitting on the chair. If the chair were to tilt, then with both a four-legged and three-legged chair, there is only one direction in which the chair can tilt whilst retaining two legs on the ground. So why not go for the simpler, cheaper, three-legged chair? Or how about a more robust, five-legged chair? What is so special about the four-legged case?
One suggestion is that the load supported by each leg is lower in a four-legged chair, so the legs themselves can be weaker and cheaper. But then why not 5 or 6 legs? Another suggestion is that the force to cause a tilt is more likely to be directed forwards or sideways with respect to the person's body, which would retain two legs on the floor with a four-legged chair, but not a three-legged chair. A third suggestion is that four-legged chairs just look the best aesthetically, due to the symmetry. Finally, perhaps it is just simpler to manufacture a four-legged chair, again due to this symmetry.
Or is it just a custom that started years ago and never changed?
| The four-legged chair is the chair with the minimum number of legs (for whatever form of the sitting surface) to let you move in two perpendicular directions with equal effort. Leaning back, leaning forward, leaning right, leaning left.
One leg will make you tumble
Two legs will do that too
Three legs are coming closer
Four legs are the best for you
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "265",
"answer_count": 12,
"answer_id": 11
} |
Total number of primary maxima in diffraction grating I am trying to determine the total number of primary maxima that can be observed when light of wavelength 500 nm is incident normally on a diffraction grating, with the third-order maximum of the diffraction pattern observed at 32.0 degrees.
Rearranging the diffraction grating formula for maxima number ( $m$ ):
$$ m= \frac{d \space \sin \space\theta_\text{bright}}{\lambda} \, . $$
I can get the right answer if I let $$\theta = 90 ^\circ \, .$$ However, I do not understand why this angle value is used.
| The diffraction pattern is essentially infinite on the screen on which it appears. We regard the location on the screen with a single coordinate, $\theta$, which is the angle between the perpendicular line stretching from the center of the grating to the prime maximum (we can call this point the center of the diffraction pattern). If you imagine opening this angle slowly, you can see that the point in which the line touches the screen moves further and further away from the center of the pattern. The more you move towards $\theta=90^o$ the point on the screen gets infinitely farther from the center of the pattern. So, $\theta=90^o$ is the upper bound for the angle $\theta$. I wrote upper bound and not the maximal value since reaching $90^o$ in $\theta$ is equivalent to reaching infinite distance from the center of the pattern on the screen, which is of course not possible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can one write down a Hamiltonian in the absence of a Lagrangian? How can I define the Hamiltonian independent of the Lagrangian? For instance, let's assume that i have a set of field equations that cannot be integrated to an action. Is there any prescription to construct the Hamiltonian of a such system starting from the field equations?
| The field equations must be conservative in a fairly precise sense in order that this can be done in a physically appropriate sense.
Then there are several Hamiltonian approaches to field theory: the De Donder-Weyl formalism and the multisymplectic formalism. Although both formalisms can accommodate Lagrangians, the can also be understood without any Lagrangian, in a purely Hamiltonian form. Both formalisms can be made fully covariant.
This holds for classical fields. How to quantize a theory in one of these formalisms is very poorly understood.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/230934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
} |
Why are these equations valid despite seemingly inconsistent units? I am having quite a difficult time in trying to understand what units are used in this paper and how to convert things to SI. For example, look at equation (1):
$$T_M \approx 1500 \rho^{1/3}\ \mathrm{K}\tag{1}$$
It seems to be showing that temperature is measured in units of $\mathrm{g\,cm^{-3}\,K}$. Then look at equation (2),
$$T_M \approx 2800 \rho^2\ \mathrm{K}\tag{2}$$
which seems to be showing that temperature is measured in $\mathrm{g^2\,cm^{-6}\,K}$. Equation (10) doesn't make sense with these either:
$$\sigma \approx \frac{5\times 10^{20}\rho^{4/3}}{T(1 + 3x)}\mathrm{esu}\tag{10}$$
How are these consistent?
| The units are not consistent. Or in less precise terms, wrong.
Here's the only way I can think of for this to make some sense: just after equation (1), the paper says
...where $\rho$ is the density in $\mathrm{g\,cm^{-3}}$.
My guess is that they intend you to take $\rho$ as a pure number. For example, if the density is $0.1\ \mathrm{g\,cm^{-3}}$, then you should take $\rho = 0.1$. But that's inconsistent where the part just above equation (2) where it says
For $\rho \gtrsim 0.4\ \mathrm{g\,cm^{-3}}$...
which requires that $\rho$ actually have units in it.
I suppose the intent is that you always consider $\rho$ to be either $\text{density}$ or $\frac{\text{density}}{\mathrm{g\,cm^{-3}}}$, as needed. One would hope that, especially in modern times, this sort of sloppiness with units becomes increasingly rare, because it is confusing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
When sunlight bounces off the Earth, why isn't the entire spectrum reflected rather than just the infrared portion? I've read that greenhouse gases absorb and reemit sunlight, and that the infrared portion is what bounces off Earth back to space. When sunlight bounces off the Earth, why isn't the entire spectrum reflected rather than just the infrared portion?
| The reflectivity of the atmosphere, and of the surface itself, is strongly wavelength-sensitive. So while some percentage of any given wavelength is reflected -- and some percentage is absorbed rather than transmitted, the variation over wavelength is what leads to the somewhat misleading statement you refer to. Here's an example of atmospheric absorption, as can be seen at wikipedia
There are also curves of reflectance. $transmittance+absorptance+reflectance = 1$, in case you were wondering :-) .
The reason all this matters is that shorter-wave energy, e.g. visible and some UV, that is absorbed either in the atmosphere or by the ground, is re-emitted at different wavelengths in accordance with black-body theory. In general this leads to a lot of IR-radiation, so if the atmosphere is reflective at these wavelengths, the energy is retained rather than re-emitted to space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the minimum force required to move this block Please don't report. It's not a homework question. Yesterday on my physics test there was this question. there is a block of mass $m$ connected to a spring as shown in the figure. the spring constant is $k$ and the friction coefficient between the block and the floor is $\mu$. they have asked what is the minimum horizontal force $F$ applied as shown in the figure so that the block starts to move. I answered $\mu mg$ considering the whole spring mass system as a single system of mass $m$. but a friend said afterwards that it would be $\frac12 \mu mg$ because the force is being applied on the spring and if the force elongates it by $x$ length, $$kx=\mu mg$$ and, $$Fx=\frac12kx^2$$ so solving, $$F=\frac12µmg$$ . Is he right ? please explain in detail why he is right or wrong. And please point out the problem in my thinking if I am wrong.
what he is saying is that he is equating the increase in potential energy to the work done by the force on the spring. I don't understand his point.
| What would the force be if there was no spring? Each side of the spring feels the same force - so if you put a black box around the spring and only saw the string "going in" and a string "coming out" of the box, with the same tension on each, the force needed to move the box would be the same.
This means your approach is correct.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Would you save energy by heating the air in a shower stall so that you could use colder water? It is refreshing to take a cool shower in hot weather. And for the sake of discussion, lets assume that one should be "comfortable" with temeratures when taking a shower.
Considering that the vast majority of the heat from a shower is lost down the drain, would it not be more energy efficient to heat a confined space (shower stall) to a temperature that would allow you to shower with cool water rather than hot?
How hot would it have to be in the room for 70 degree water to be "nice." 60 degree?
And since all of the energy used to heat the air in the space would remain in the space/room, would you use less energy to take a comfortable shower doing it that way?
| According to a steam shower vendor, a 10KW unit is required to provide enough steam at 118 deg.F in a 6'x8'x 8' enclosure. The timer runs for 20 minutes (about right for a nice shower), so figure the unit would use 3.3KW and 2 gallons of water to make the steam room hot and steamy.
According to numbers pulled from all over, figure an average 20 min shower uses 50 gallons of water (416.5 lbs.), and that in my area that water comes out of the ground at 55 degrees F. To heat that water to 110 deg.F would take approx 22,910 BTU.
Converting KW to BTU (and here are the limits of my understanding - please correct my conversions/assumptions as needed) it looks like 3.3KW is about 11,260 BTU.
If this is all reasonable, then heating the shower to 82.5 deg.F would require 11,500 BTU, which, when added to the 11,260 BTU for the steam would total 22,760 BTU - or slightly less than just running the hot shower by itself. If you could go with heating the shower water even less,then the savings would appear to increase.
I am fully aware that this answer is full of broad assumptions, simplifications, and perhaps misunderstandings.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Charge inside a charged spherical shell
*
*If I were to put a negative charge inside a negatively charged spherical shell, will it move to the center?
*Electric field inside the shell due to the shell is zero (Gauss's Law), would that mean the charge inside the sphere faces no force?
But, that doesn't make intuitive sense to me. If the negative charge was near the walls of the sphere, wouldn't the charges on the near wall push the negative charge to the centre as the force due to the charges on the wall closest to it is higher than that form the walls further away from it.
*What about in the case of a ring? Will the charge move towards the center?
| How about this?
1) There is a charged spherical shell. The origin of the sphere must not have any electric field due to symmetry.
$${\bf E}({\bf 0}) = {\bf 0}$$.
2) Now take a point from the to the origin at $\bf r$. Due to symmetry of the problem the electric field has to be radial (points away from the origin), but can (still) have a magnitude $A_r$.
$${\bf E}({\bf r}) = A_r(|{\bf r}|) \hat {\bf r}$$
3) Apply the differential form of Gauss Law (with knowledge, that there is no charges, thus the charge density needs to vanish)
$$ \nabla \cdot {\bf E} = \rho({\bf r}) = 0$$,
to this function in spherical coordinates (https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates), yielding
$$ \nabla \cdot {\bf E}(r) = \frac{1}{r^2} \frac{\delta (r A_r(r))}{\delta r} = 0.$$
This indicates that $A_r(r)$ i.e. the electric field has to be constant wrt r.
Now let's combine the result from section 1) that the electric field is zero at the origin, and from 3) that the electric field is constant to yield that the electric field has to vanish in the whole system.
To gain further insight, consider how Gauss law is connected to conservation of field lines, and consider that in spherical symmetry.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Could a hydrophobic surface increase a liquid's resistance to compression/displacement? Imagine a quantity of an aqueous (yet slightly viscous) solution is resting on a hydrophobic surface with a contact angle around 100°.
A downward force is then applied as a (repellant) surface is lowered onto it, in order to spread the liquid out and reduce the height $h$ to some value, call it $h_1$.
Is it possible that as $h$ decreases, a shear friction force between the hydrophobic surface and the small amount of liquid (say 100$\mu$m thick) would be significant enough such that the force needed to continue displacing the liquid would be large?
| I guess that depends on what qualifies as a large force, but I don't think that the shear forces will be significant.
Consider the Couette flow, the shear stress and thus the viscous force along the boundary scales with the velocity. As velocity increases so too does the viscous force. In your problem, the magnitude of the shear stress is related to the rate at which $h$ decreases. Higher stresses can be achieved by decreasing $h$ more rapidly. However at some point, this might result in a break up of the droplet.
Having a hydrophobic surface makes it easier for the contact line to move, and thus easier for the droplet to spread. In this case having a hydrophilic surface would increase the fluid's resistance to spread, however the total contact line force is probably still small as it scales with contact line length.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Dependence of average speed of molecules of gaseous mixture We know that the average speed of gases in a single gas chamber is given by $\sqrt{8RT/\pi M}$
where R is universal gas constant,T is temperature,M is molar mass of gas.
But what if we mix two gases in any ratio say 1:1 and then try to find average speed of anyone of the gases.
Will the both gases have have same average speed or different?If same ,how will it be calculated and if different will it be given by same above formula?
| If the two gases are NOT interacting, then you can treat them independently and use your formula for each of them, weighting the result according to the mean speed, using your formula, will be
$$v=f \sqrt{8RT/\pi M_1}+(1-f)\sqrt{8RT/\pi M_2}$$
where $f$ is the fraction of gas number 1 and $M_1$ and $M_2$ the two molar masses - so you can solve it easily. However, in general, if the two gases are interacting you might a different formula for the mean speed of each of them, or, in other words, your formula for the mean speed is not valid anymore and depends on the interaction itself.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Why is the "expansion postulate" a postulate of quantum mechanics? I'm currently reading the following set of lecture notes on quantum chemistry, which includes the so-called "expansion postulate" as a fundamental postulate of quantum mechanics: "The eigenfunctions of a linear and hermitian operator form a complete basis set."
How is this a postulate of quantum mechanics, rather than a provable mathematical property of linear hermitian operators? Isn't the expansion postulate just a consequence of the spectral theorem for an infinite-dimensional Hilbert Space?
If the eigenfunctions of a linear, hermitian operator forming an orthogonal basis is not a mathematically provable fact about wavefunctions, wouldn't that necessarily imply that there exist hermitian, linear operators on wavefunctions whose eigenfunctions do not form an orthogonal basis?
| It's a postulate, not because it's a self-consistent mathematical property, but rather because it is an assumption about how the physical world may, or may not, be described.
There are many many different types of linear vector spaces other than a Hilbert space, so to make things easy they make an assumption and run with it until proven otherwise.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/231924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can energy be negative in a finite square well? Say if the potential $V(x) < 0$ in the well but the sides or the scattered states its zero potential, anyways
*
*How is that the energy in the well is less than zero?
*Is it because the potential is less than zero?
| Energy or the value of $V(x)$ negative means it is a bound system. Think of it in this way, if a particle is free and has no kinetic energy and potential energy then it's total energy is zero. If this particle is not free or otherwise is bounded by a negative potential well then it's potential energy is $-V$. You have to give the same amount of energy, in this case $+V$ to make it free. Then it's total energy will be $-V+V=0$ and it will go free.
To summarize a particle in a potential well $-V(x)$ means it is bounded by the well and has $V$ amount of energy less than of what it would need to become free.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Backyard experiments to falsify the Flat Earth theory I recently became aware that the flat Earth theory still exists in the 21st century, and has colored the views of a friend of mine. Roughly speaking, the tenets are:
*
*The Earth is a flat disk, with the south pole blown up into a circular "ice wall" where one would expect Antarctica to be.
*
*The sun and moon are either spheres or disks floating above the earth and moving in a spiral pattern with the seasons.
*NASA is part of a conspiracy to conceal the truth, and sends us animations and faked photo shoots.
The Flat Earth Theory is a scientific theory, in the sense that it makes falsifiable predictions about the universe that can be seen to match observation (or not). What are some good arguments or backyard experiments that could convince a layman that in fact the Flat Earth Theory is false?
A similar question and useful related reading is What is the simplest way to prove the Earth is round?, but it focuses more on the theoretical and conceptually simple side, applicable to the proverbial "numskull cousin". This question is more focused towards convincing a doubtful scientific person using (preferably low-tech) experiments and observations.
| Foucault Pendulum is a great example.
The original purpose of this experiment was to prove that earth rotates relative to the stars and not the other way around, yet nevertheless it proves that the earth rotates in a way which contradicts the "flat earth" theorem.
This experiment can be easily recreated at home, if you don't live close enough to equator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 11,
"answer_id": 9
} |
Mass of proton vs mass of nucleus I have just started reading nuclear physics.I know that the sum of masses of the quarks is less than the proton or neutron itself as a whole .
But why is it that the sum of the masses of the nucleons(protons and neutrons) is more than the nucleus itself?
whats the difference between the two cases?
| The difference comes from the kind of force that holds the constituents together. The force on the quarks in the nucleon is the color force, (one of the four fundamental forces), between nucleons, it is the residual color force, which appears as the strong force that binds the protons and neutrons in a nucleus.
The color force and the fact that the coupling constant is 1 in the color interactions between quarks and gluons is what creates the nucleons, protons and neutrons, with the exchange of gluons in a "bag" where a sea of quarks antiquarks and gluons binds everything together to the mass of the nucleon. Within a nucleon the color force acts as a spring , binding the constituents together. The invariant mass of the whole depends on the sum of the four vectors of the numerous temporary constituents and thus to a very small degree on the masses of the three constituent quarks which define the baryon quantum number.
The force between protons and neutrons is a spill over force, similar to the van der Waals forces that hold molecules in structures , it is attractive still but it is short range . It has been modeled by Yukawa as pion exchange with an exponentially falling effect. When the protons/neutrons are close enough, within the range of the Yukawa force, they can be bound: As it is attractive it has energy levels in which protons and neutrons can be bound giving up energy with the special relativity effect of the sum of the bound masses to be less than the free ones.
The sketch is an attempt to show one of many forms the gluon interaction between nucleons could take, this one involving up-antiup pair production and annililation and producing a π- bridging the nucleons.
(Feynman diagrams for strong interactions have to be taken with care, as with a coupling constant of 1 , perturbative expansions, which is what Feynman diagrams are about, do not converge. New methods of calculations have been found.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does potential difference or electric field change with distance between parallel plates? Say you have a set of parallel plates, one is positive and one is negative, if you change the distance between them would electric field strength change or potential difference, given the equation
$E=dV/dx$
From pure intuition, I think electric field strength will change as the field line bulges out from the sides, and as the plates are far enough from each other, each plate can be treated as a point charge and the system becomes a dipole, and E decreases at 1/r^3:
picture: (sorry I didn't upload the picture here in case I violate copyright)
http://www.electrobasic.com/uploads/3/2/3/4/32342637/9893979_orig.jpg
But I read from this post:Why does the potential difference between two charged plates increase as they move further apart?
and it says E must remain constant, and I am not sure why that is, and I do not see how simply moving the plates apart will increase the potential difference, as
$V=kQ/r$
and as r increases, v should decrease, and I don't understand how simply moving them apart will store energy in the system.
So why must E remain constant and V increase as the plates are separated further?
Thanks.
| If you treat parallel plates as infinite sheets, then you get a constant field in between that depends only on charge density. This isn't exactly true in reality, since real plates aren't infinite. But it gives a good estimation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Why is meteor speed what it appears to be? Is the speed of a meteor through our sky because of the speed of the earth's axis rotation, or because the meteor is speeding towards us at that speed?
| Meteors are essentially bits of rock that are independently in orbit around the Sun and which cross the Earth's orbit. If the Earth happens to be there at the same time then it will enter the Earth's atmosphere and we will see a meteor.
The velocity of meteors is related to how fast they we going in their orbit around the Sun, combined with how fast the Earth is going around the Sun. In addition, the Earth's gravitational field will accelerate the meteors as they approach the Earth.
Objects falling from infinity towards the Earth could reach a free-fall velocity as high as 11.2 km/s. The orbital velocity of the Earth in its orbit around the Sun is around 30 km/s. A meteor initially in orbit around the Sun however could have a velocity as high as 42 km/s at the distance that the Earth is from the Sun and that could be directed with the Earth's motion or against it.
The velocities of meteors as they enter the Earth's atmosphere can vary from $30 + 42 = 72$ km/s for a head-on collision with a meteor in a very long period orbit, to as little as $11$ km/s for something "approaching Earth from behind" with an initially small closing velocity, because it is in a very similar orbit to the Earth to begin with.
The axial rotation of the Earth (0.5 km/s) is a negligible contributor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why do materials show plastic behaviour for large stress? As the stress is increased, the strain increases proportionally up to elastic limit and the material regains its original dimension within elastic limit. When the stress is increased further the material shows a plastic behaviour. What change in the internal structure causes the transformation from elastic to plastic behavior?
| Plastic behaviour is characterised by there being permanent (non-reversible) deformations. In terms of molecules held together with springs (the bonds) in plastic deformation the springs are broken and the bonds then might the be between different molecules.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/232713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Torque: The T-shaped stick problem This is a problem that I have been unable to solve for some time.
Imaging a T-shaped stick, as shown in below image, which do not deform in any appreciable way and has pivot point at the tail of the "T".
There are two questions in my problem, the first is: given that the "T" is symmetrical, how would applying two units of force on one of the "T"'s horizontal appendage differ from two one unit of force applied on both appendages in opposite direction, in terms of the torque received at the pivot?
(See image)
The second question is: how would the situation differ if the "T" is "italicized", but the forces are exerted perpendicular to the appendages? see image:
What is a systematic way to explain and describe the torque received at the pivot in terms of the lengths L and H and possibly the angles between them?
| The torque is done with a cross product and thus is going to be:$$\tau = |\vec F| |\vec r| \sin\theta$$where $\theta$ is the angle between the position vector $\vec r$ (which points from the pivot to the place where the force is applied) and $\vec F$ (which points however the force points).
You can also express this as $x F_y - y F_x$ and since your forces point in the $x$-direction, the only term is $-y F_x$. The following effects hold:
*
*As long as the pivot is fixed, there is no torque difference between your two scenarios. The only difference is that when you only push with the $2~\rm N$ force on the top, the pivot needs to push back with a $2~\rm N$ force to be fixed in place -- so if you press too hard you could hypothetically break the pivot or the beam between, whereas the balanced $-1~\rm N$ at $+y$ and $+1~\rm N$ at $-y$ do not require any constraint force from the pivot.
*The only difference when you "italicize" is that the sides come closer to the center-line of the T, as long as your forces remain in the $x$-direction. In fact if you lengthen the sides slightly so that these $y$-coordinates remain the same, there is no difference! As we can see that this lowers the "effort arm" of the lever that you're using, the torque will lessen if you "italicize" the T without thus lengthening it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Coefficient of friction and practical experience of sliding The classical model of friction has a coefficient of friction depend only on the materials, but not area, and the force proportional to the normal force and coefficient of friction. So a given object on the same surface has the same friction whether it is supported by full bottom area or small legs as long as the materials are the same.
However every child knows that on a slide one goes faster if one lays down on their back compared to sitting on their butt. The slide is obviously still the same and since jackets usually extend below butt, the other material is also the same. So the friction should be the same as well, but it clearly isn't. So what is going on here?
Note: I mean typical stainless steel or fibreglass laminate slide, not ice, which is soft enough to complicate the matter further.
| In laboratory experiments with bricks, the sliding angle comes to be the same irrespective of the orientation of the brick. However, when it comes to a child sliding there are some extra factors. The child usually is not sitting at normal to the plane especially at large angles of sliding due to the fact that he can topple more easily than slide. He wants his COM nearer to the slide in order to prevent the toppling and so lays on his back. If he stands to the normal, then the free body diagram states that the acceleration of his centre of mass is dependent on the frictional force. So along with acceleration, he also topples.
If a standing child starts to topple, his feet won't move much, and therefore he won't 'slide', although his COM definitely would and therefore would land on his head. Thus in order for his fun and security, the child prefers to lay on back instead of doing anything else. The jacket may not play a big role here. The coefficient of friction should remain the same and not dependent on the surface area. I encourage you to try the same thing too. :P
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/233094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.