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What is the point of a voltage divider if you can't drive anything with it? The voltage divider formula is only valid if there is no current drawn across the output voltage, so how could they be used practically? Since using the voltage for anything would require drawing current, that would invalidate the formula. So what's the point; how can they be used?
| As an alternative to using a high-input-impedance device like a comparator or op-amp attached to the division point, one can instead use a low-impedance device that draws zero current at its operating point. For example, a Wheatstone Bridge connects a galvanometer (a sensitive low-impedance current detector) between two voltage dividers.
Describing how to balance a Wheatstone Bridge, to find the value of an unknown resistance Rₓ by the equation Rₓ = (R₂/R₁)·R₃, the Wikipedia article mentioned above says:
R₁, R₂, and R₃ are resistors of known resistance and the resistance of R₂ is adjustable. The resistance R₂ is adjusted until the bridge is "balanced" and no current flows through the galvanometer ... At this point, the voltage between the two midpoints ... will be zero.
When that voltage is zero, no current will flow, and both dividers will produce voltages at their true ratios.
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Discrete Harmonic Oscillator matrix representation of $x$ for Quantum Simulation (The paper I'm referring to in this question is "Quantum simulations of one dimensional quantum systems")
I've been trying to understand the paper above, specifically on constructing a matrix representation of the position operator, $\hat{x}$, in discrete real space (Equation (11)).
In analogy with the CV QHO, we define a discrete
QHO by the Hamiltonian
$$H^{\text{d}}=\frac{1}{2}((x^{\text{d}})^2+(p^{\text{d}})^2). \tag{10}$$
The Hilbert space dimension is $N$, where $N\geq 2$ is even
for simplicity. $x^{\text{d}}$ is the discrete "position" operator
given by the $N\times N$ diagonal matrix
$$x^{\text{d}} = \sqrt{\frac{2\pi}{N}}\frac{1}{2}
\begin{pmatrix}
-N & 0 & \dots & 0 \\
0 & -(N+2) & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & (N-2)
\end{pmatrix}, \tag{11}$$
I'm quite a bit lost on how this matrix is derived. Since we are in the basis of real space, I expect that the matrix should be diagonal (as it is). My guess is that the basis of real space that we are in is really the basis of Hermite Polynomials: the diagonal entries are the entries that would satisfy something along the lines of:
$$ \hat{H} H_n(x) = a_{nn}H_n(x)$$
where $a_{nn}$ is the diagonal entry in the $n$th row and column, and $H_n(x)$ is the $n$th Hermite polynomial.
I'm not entirely sure if this is proper thinking, so any insight would be greatly appreciated!
| I think this is it: It seems like we can treat $X$ as the Fourier transform of $p$ to explain the factors.
| {
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Relative Velocity of two particles If two particle are neither approaching towards nor receding away from other then their relative velocity is non zero.
How is this possible??
| The answer lies in your question itself. The two particles neither approach each other i.e. their separation isn't decreasing, nor are they moving apart i.e. their separation isn't increasing
In other words with respect to both the particles, the separation between them is not changing. Relative velocity is given as the rate of change in relative separation (with respect to the particle) with respect to time, and as the relative separation is a constant, it's rate of change is zero i.e. the relative velocity is zero.
Note only their relative separation matters, if both the first particle moves 5 meters in a second and so does the second particle, then their separation hasn't changed, thus the two particles could be moving, but with respect to each other the relative velocity is zero.
Edit: I didn't read the question properly, as it asks how there is a relative velocity between two objects, even though their separation is neither increasing nor decreasing. Well Velocity is a vector, so it can have both magnitude and direction, so it is possible that the separation doesn't change but their direction does change continuously, for example, two objects in circular motion, their relative velocity at every instant is non zero, despite their relative speed being zero
| {
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Enthalpy during compression of water What happens to the temperature of water when compressed?
Enthalpy $H = U + PV$.
$H$ is conserved in a closed system. By which I mean adiabatic and negligible external work applied.
We compress a litre of water to 10 bar (say). This requires negligible work because water is almost incompressible. But $P$ goes up, $V$ hardly changes, so $PV$ goes up, so $U$ should go down. It should get quite a bit colder.
But is that correct? Because very little work was done, yet $PV$ changes quite a lot, and it would produce a lot of cooling.
| For control masses at constant volume it is common to analyse them in terms of the internal energy; the enthalpy is commonly used for reversible isobaric control mass problems (in which the change of enthalpy is equal to the heat flux), and for open systems (such as flowing fluids). The reason for the latter can be seen in derivations given elsewhere; the related concept is sometimes termed flow work.
I'm not sure where your notion that the enthalpy is conserved in a closed system comes from (especially as you don't specify for what kind of process).
Returning to your problem: from the fundamental relation for the internal energy:
$$\mathrm{d}U = T\cdot\mathrm{d}S - P\cdot\mathrm{d}V$$
If the fluid were truly incompressible and the system was initially at equilibrium, pressurisation by application of an external force, with no heat input or output, would cause no change of the internal energy (no work has been done and the system's thermodynamic state has not changed; note that $P$ is independent of $u$ and $v$ for an incompressible fluid, as $v$ is a constant, so from the equation above we have that $u$ is solely temperature-dependent). A mechanical analogue is the zero compression work done to a spring of infinite stiffness when the applied force is varied. If the fluid were compressible, the previous answer shows that a temperature rise would be expected.
| {
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Why didn't heavier elements settle at the core of the solar system? As the solar system formed, why didn't all of the heavier elements such as iron, collect where the sun is leaving the lighter elements in the outer solar system?
| This question is tantamount to asking why do stellar systems, galaxies, and even planet-moon systems have a spatial extent as opposed to forming centralized blobs of decreasing density radially. The answer lies in the fact that all these objects have a rotational motion that spreads the mass outward, with the occasional accretion of heavier elements happening further away from the center of rotation. The chaotic nature of that accretion generates "islands" along the radius where heavier elements can indeed form enough critical mass (and thus rotational inertia) so as to not end up at the center.
| {
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Derivation for optical path length and the origins of the formula So I've learnt that the formula for optical path length is $OPL = ns$, where $n$ is refractive index of the medium and $s$ is its geometrical length, the problem is i cant really get around this formula like where does it come from? I tried deriving it but couldnt get anywhere.
I tried adjusting the last equation using snell's law, got nowhere.
| The quantity $ns$ has a physical meaning. I think that's what you're trying to figure out. If light travels a distance $s$ through a medium with refractive index $n$ in a time $T$, then the distance light would travel in the same time $T$ in a vacuum would be $ns$.
$$T = \frac{s}{v} = \frac{ns}{c}$$
and so the distance travelled in vacuum in this time is
$$cT = ns$$
| {
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Different variations of covariant derivative product rule This is a follow-up question to the accepted answer to this question: Leibniz Rule for Covariant derivatives
The standard Leibniz rule for covariant derivatives is $$\nabla(T\otimes S)=\nabla T\otimes S+T\otimes\nabla S$$
so for $T\otimes\omega\otimes Y$ this would translate to $$\nabla(T\otimes\omega\otimes Y)=(\nabla T)\otimes(\omega\otimes Y)+T\otimes(\nabla\omega\otimes Y)+T\otimes(\omega\otimes\nabla Y).$$
My question is: given a vector field $X$, how do I get from the above that $$\nabla_X(T\otimes\omega\otimes Y)=(\nabla_X T\otimes\omega\otimes Y)+T\otimes\nabla_X\omega\otimes Y+T\otimes\omega\otimes\nabla_XY$$ as written in that answer?
| Are you just rearraging the backets? If so remember that
the temsor product is defined to be associative: $ (a\otimes b) \otimes c= a\otimes (b \otimes c)$, so we can write eiher form as simply $a\otimes b \otimes c$.
If you are referring to replacing $\nabla$ by $\nabla_X$ remember that $\nabla$ is always $\nabla_X$ for some $X$. I.e. $\nabla_\mu\equiv \nabla_{\partial_\mu}$
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What is the maximum deflection angle of a pendulum in a car, when the car, initially at rest, accelerates suddenly? I was doing Kleppner-D.-Kolenkow-R.J. and I came across the following problem:-
A pendulum is tied vertically to a car at rest, the car suddenly accelerates at a rate A. Find the maximum angle of deflection $\phi$ through which the weight swings.
MY TRY: I saw the solution of this problem in the book which uses car's frame of reference, which was fairly simple.
I tried to do it in the ground frame of reference.
Deflection of the pendulum will be maximum when the angular velocity of the mass hung to pendulum relative to the hanged point will be zero, hence the velocity of mass relative to the car, perpendicular to the string is zero. But the constraint of a taut string doesn't allow velocity of mass relative to the car along the string also. So, velocity of mass relative to car is zero at the point of maximum deflection.
I have the following two tools to solve the problem:-
*
*Apply Work energy theorem to the mass.
*Use the string constraint i.e. the acceleration of the mass and the topmost point along the string will be equal at any instant i.e. $T-mgcos(\theta)=masin(\theta)$
The tension force and gravity are only two forces acting on the mass. But, how could one find the work done by tension on the mass in the journey from $A$ to $B$. Any hint would be a great help!
| Work done by tension in car's frame of reference would be zero. Then we can apply conservation of energy where initial and final kinetic would be zero(since at start it didn't have any energy) then use pseudo force's work equal to work done by gravity. We get answer $$\theta = 2\arctan(\frac{a}{g})$$ (where $\arctan$ is $\tan$ inverse function).
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Relativity without constancy of light speed Using homogeneity of space, isotropy of space and the principle of relativity (without the constancy of light speed), one can derive:
$$x' = \frac{x-vt}{\sqrt{1+\kappa v^2}}$$
$$t' = \frac{t+\kappa vx}{\sqrt{1+\kappa v^2}}$$
$\kappa = 0$ denotes Galilean and $\kappa < 0$ denotes Lorentz Transformation.
What does $\kappa > 0$ denote? Is it physically possible? I was told that it is self-inconsistent. Can somebody help me with a proof of this?
| As said in the answer of @m4r35n357 it is the case of Euclidean geometry. To see this, look at the transformations, that preserve the distance : $$ds^2 = dx^2 + dt^2$$
Among with the translations there are also rotations:
$$
\begin{pmatrix}
t^{'} \\
x^{'}
\end{pmatrix} =
\begin{pmatrix}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix}
\begin{pmatrix}
t \\
x
\end{pmatrix}
$$
Look, for example, at the first row. After defining $v = \tan \theta$, it is actually the $t$ transformation, with $\kappa = 1$:
$$
t^{'} = \frac{t + v x}{\sqrt{1 + v^2}}
$$
This geometry is obtained from the Minkowski space by Wick rotation $t \rightarrow i t$. It is mathematically consistent and alright, however, our world is described by the metric (in the flat case), with the minus sign between $dx^2$ and $dt^2$.
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Will objects that go beyond the cosmological event horizon eventually vanish from sight? Suppose a galaxy is headed beyond the cosmological event horizon. Photons it emits now will eventually reach us, but there is a point at which photons will no longer be able to reach us.
Supposing that a finite number of photons are emitted, this would seem to imply that the galaxy will eventually vanish from sight. But my understanding is that instead we would see the galaxy asymptotically slow down as we see more and more redshifted light from it. Why is this?
| Sources that say that the light is redshifted into the indefinite future are talking about classical cosmology and not considering quantum effects.
Quantum mechanically, as you say, the galaxy emits only finitely many photons before crossing the horizon and so there should be a last photon – although I should qualify that in a couple of ways. First, unless you know all future measurement outcomes, the disappearance still takes forever in the sense that you can never be sure that you've seen the last photon; another one could arrive at an arbitrarily late time. Second, even if you do know all future measurement outcomes, the horizon emits Hawking/Unruh radiation, and because of boson indistinguishability I'm not sure that there's any way even in principle to unambiguously identify the last galaxy photon amidst that noise.
| {
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Why can't photons cancel each other? The textbook argument against photons canceling each other draws upon the conservation of energy. Does this mean that energy conservation is a "stronger" principle than superposition? Waves in other media than the EM field, e.g., sound or water, do cancel out---presumably by passing on their energy to some other degree of freedom (e.g., heat). Could this imply that EM waves don't have any alternative channel to pass on the destructed energy and thus can't cancel out?
| All waves traveling through a medium don't cancel out. Sound waves, water waves, waves in a rope, etc. pass each other and travel further after they have passed. They don't (or almost not) exchange energy with the medium (like being converted to heat, though there is dampening). Two oppositely traveling waves may seem to vanish for a moment, but the medium contains the kinetic energy of the waves.
The same holds for classical em waves, although they don't travel in a medium. A circular em wave traveling outward from a center and a circular wave moving toward the same center will pass each other and continue their journey, without losing energy because they can't lose energy to the vacuum.
So they don't cancel out (they interact and travel along as if they didn't have encountered each other) because they don't have a channel to lose their energy.
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Screened Coulomb potential in metals One of the reasons why we can neglect electron-electron interactions in metals is the fact that their coulomb interaction is screened. I'm confused about the nature of this screening. In the literature the process is usually described like this:
If we bring an additional charge inside a neutral metal, the coulomb potential of this charge will be screened due to electrons accumulating (or rejecting) at the charge. This leads to the Thomas-Fermi potential
$$ \phi(r) =\frac{e}{r}\,e^{-r/r_{TF}}.$$
But if we consider the potential between electrons in the neutral metal, the screening must have a different origin because the electron density is homogeneous. The above mentioned approach can't be the reason can it?
I think the only way for the electron-electron interaction to be screened in the neutral metal is by the presence of the positively charged ions. If we consider a small volume in the metal, the total charge inside will be zero so the long range potential falls off faster then $\frac{1}{r}$. It appears the literature claims that the other electrons are responsible for the screening but I can't see how this is possible.
Edit:
Maybe I should rephrase the question. In the literature, it seems the mechanism behind the screening is the dynamics of the electrons, rather then the presence of the ions, so my question is
Q: How can this possibly be true, given that the electron distribution is homogeneous?
| Maybe not very useful regarding you question, but other way to see why e-e interactions in a metal can be neglected is by comparing typical kinetic to potential energy.
$ E_F = \hbar^2k_F^2/2m$ and $U_{e-e} = e^2/{4\pi\epsilon_0d}$.
Using the free electron relation $k_F = (3\pi^2n)^{1/3}$ and $d^{-1} \approx n^{1/3}$ you get
$E_F/U_{e-e} \approx 5 a_B n^{1/3}$
where $a_B$ is the Bohr radius in SI units. When the potential energy is equal to or larger than the kinetic energy it is also known as the Mott criterion.
The insight of appx is that counter-intuitively, as the density increases e-e interactions are negligible.
| {
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Terrell-Penrose effect and surface reflectance All explanations of Terrell-Penrose effect seem imply that the effect makes some of the back-facing walls of a passing-by object visible. You can see some examples of those in many different references. However, from my understanding many of those sources assume that the light reflect or emitted from the surface travels in all directions, which is not at all how surface reflectance works.
Even for a perfect Lambertian reflector the outgoing light will travel only in the direction of the hemisphere around the surface normal – none of the light is reflected "into" the wall. How would it be possible to see any light reflected from a back-facing wall if that light was never reflected in the observer's direction in the first place?
I can convince myself of being able to see a wall that is facing at most 90° away from me, based on the assumption that I'd see photons emitted from it in the direction parallel to the surface. How could I ever observer photons from a wall that's facing more than 90° away? Some of the references clearly show walls that were originally facing away from the observer as visible, like the wall with 4 dots on the moving dices:
Are those mistaken?
| Concentrate on the nearest top die on the right. The reason you can see the "four" face is that you have already passed the "one" face of the die! It is just that aberration has distorted the image to place it in front of you.
There is nothing more to "Terrell Rotation" than this simple fact.
You can find a more complete discussion here.
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What is the purpose of the roll maneuver on a rocket? I have watched several videos on both rocket launches, ballistic missile launches, even a really cool one by Northrup Grumman on launching a nuclear missile.
(https://www.youtube.com/watch?v=HWZXinRwCaE)
I understand why missiles have a pitch maneuver. I have watched videos such as (https://www.youtube.com/watch?v=kB-GKvdydho) that state that they roll to cancel out the difference to match its azimuth. But could not a second pitch maneuver actually accomplish as well? The Falcon 9, and Electron don't need to do a roll maneuver, and the Falcon Heavy cannot due to only having 2 boosters attached that have to stay perpendicular to the ground.
Why do non-boosted or basically round rockets, need to do one?
| When I was teaching physics, one of my students was an intern at NASA. I asked him the specific question of why the space shuttle performed the roll maneuver. He talked to NASA personnel, and reported to me that the maneuver was performed in order to turn the shuttle antenna towards the ground so the astronauts could maintain communications with mission control.
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Why is maximal kinetic energy lost in a perfectly inelastic collision? A perfectly inelastic collision is one where both of the colliding objects stick together and move as one.
My question is, why, of all possible combinations of final velocities that conserve momentum, does this one lead to the greatest loss in kinetic energy?
One reasoning I found was that this is the only combination in which the total kinetic energy of the system becomes 0 in some frame of reference (com frame). But just because the KE is 0 in some frame doesn't mean that it is the least possible in every other frame, does it?
| If you don't want calculus and need a physical interpretation, here's one:-
Meaning of inelastic collision is some energy of collision transforms into potential energy, either by changing shape or heat or sound etc.
So, in a perfect inelastic collision, the maximum amount of energy is converted into potential energy. And by conservation of energy, maximum kinetic energy is lost.
(Considering environment to be a part of system.)
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What is the difference between the specific heat capacities of water under isobaric and isochoric conditions Can the difference of specific heat capacity of water under isochoric and isobaric conditions be explained in terms of the internal energy of the system? Most of the videos I have watched base their explanation in terms of ideal gases. I guess its something to do with the fact isochoric conditions mean all the heat energy provided goes to the internal energy of the molecules. I also have the graphs of the specific heat capacities plotted against time
| In general it is the same idea as with ideal gases. This here is not what is formal answer, because specific heat is generally defined with entalpy and internal energy. This is rather the explanation, why there is a difference.
In order to change volume $V$ when the pressure is constant, some work $A$ has to be provided. In differential case (very small change): $dA=pdV$.
From conservation of energy we can than determine that:
$$
dQ=dW+dA
$$
$Q$ is internal energy of a system, and dW is energy added, and A is work done by the system.
So we can denote specific heat as $dQ/dT$:
$$
c_p=\frac{dQ}{dT}=\frac{dW+pdV}{dT}
$$
and
$$
c_V=\frac{dQ}{dT}=\frac{dW+pdV}{dT}=\frac{dW+p\cdot0}{dT}=\frac{dW}{dT}
$$
You can see from here, that $c_p$ is greater than $c_v$. The relation between this two depends on equation of state and it can be quite ugly for liquids. But in general when we have isobaric conditions, some added energy is converted into work needed, to change the volume of system.
If you are not familiar with differentials $d$, they are just very small changes $\Delta$.
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Confusion in definition of emf The emf of a cell is defined as the work done per unit positive charge in taking it around the complete circuit of the cell (i.e. in the wire outside the cell and the electrolyte within the cell). But Kirchoff's Second Rule states that the work done in moving a charge around a closed loop is zero. How then do we get a nonzero value of the emf?
|
Electromotive force is the electrical work done to move a unit positive charge from the positive electrode to the negative electrode. It is not about moving the charge through the whole circuit. It is basically the potential difference between the electrodes when no current is flowing.
But according to Kirchoff’s Second Rule, in a loop you reach the same point where you had started from; so the potential difference between the initial and final points is zero, and the net work done in this case is 0. This is different from what we do while understanding the concept of emf!
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Black Hole Formation If you plot a space-time diagram of an object falling through the event horizon of a black hole, and draw the past and future "light cones" of the object at every point, wouldn't the point infinitely to the event horizon have a light cone which allows light being radiated by the object to reach an observer outside the event horizon at time = infinity? (at the point when the object touches the event horizon, a radiated photon will never reach the observer outside the event horizon) If so, then why don't we, outside the event horizon of black holes being formed by the collapse of stars, observe light from the collapsing star (AKA the light from a supernova forming a black hole) forever? Please let me know if I'm wrong.
| You are not wrong. If we could live forever, and if we could observe indefinitely small light energies, we would observe the light from material falling into to a forming black hole forever. Because time appears to stop at the Schwarzschild radius, an issue is raised as to whether a singularity can actually form. In 1939 Julius Robert Oppenheimer and one his students, Hartland Snyder, published the seminal paper on gravitational collapse to a black hole (Oppenheimer J. R., Snyder H., 1939, On Continued Gravitational Attraction, Phys. Rev. 56, 455). They concluded that, from the point of view of an exterior observer, “it is impossible for a singularity to form in a finite time.”
The things usually called black holes in modern terminology are stable state solutions of Einstein's equation assuming boundary conditions which never physically appear in practice. As described by Oppenheimer and Snyder, these solutions never occur in our universe.
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Why is the electromotive force (EMF) highest in the loop when $\theta$ = $90$? My question is in regard of the following snippet provided by my textbook.
So why is the electromotive force (EMF) highest in the loop when $\theta$ = $90$ or $270$?
So the magnitude of the induced EMF will be determined by the rate at which the loop is rotating, according to Faraday's Law. EMF will be maximum when the rate of change of flux is at maximum.
But why does this means that the loop has moved to a position parallel to the magnetic field and the flux through the loop is zero? Since there is no magnetic field penetrates the area at that instant shouldn't there be no current? In turns, shouldn't there be no magnetic field?
| *
*The level magnets are exherting their normal N - pole and S - pole magnetic fields.
*As the loop wire moves into 90*, the electric current going through the loop is also generating a secondary electro-magnetic field.
*The interaction and clash between the 2 magnetic fields is the highest as the loop wire electromagnetic field is NEAREST the block magnets at 90*.
*The magnetic interaction WEAKENS as the DISTANCE becomes longer
when the loop MOVES AWAY from the 90* POSITION.
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How is melting time affected by flow rate and temperature of surroundings? Suppose you have a solid sphere of m, where m is an element with freezing point of 0 degrees Celsius.
In one scenario, you place your sphere in a (“static”) 25 degree Celsius environment and measure time, t, until melting. The sphere is fixed and cannot be displaced.
In the other, you place your sphere in environment with temperature, T, and with constant flow rate, v. Again, you measure time, t, until melting.
What is the equation that would relate the two scenarios? In other words, at what temperature and flow rate would time required for melting in the second scenario equal time required in the first?
| For the first case the differential equation for evolution of temperature of the sphere
$$
m * C_p * \frac{dT_m}{dt} = h_{nat} (T_{amb} - T_s) \\
$$
$$
\begin{array}
\text{where} \\
m & \text{mass of of the sphere} \\
C_p & \text{Specific heat of the solid} \\
T_m & \text{Mean temperature of the sphere} \\
T_s & \text{Surface temperature of the sphere} \\
T_{amb} & \text{Ambient temperature} \\
h_{nat} & \text{Heat transfer coeff. (natural convection)} \\
\end{array}
$$
The above combined with internal transient conduction equation for the sphere
with thermal conductivity (k)
$$ \frac{\partial T}{\partial t} = k \nabla ^2T $$
should provide necessary equations to determine the temporal and spatial variation of the sphere over time. I have omitted other gory details of boundary and initial conditions here. Under certain conditions, one can omit the above equation and assume that sphere temperature is uniform. (high thermal conductivity and and small heat flux at the surface of sphere)
Now it is possible to evaluate the second case, simply replacing the $h_{nat}$ with appropriate forced convection heat transfer coefficient. In general for air forced convection heat transfer coefficient is proportional to $v^{0.8}$
| {
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Can a single-slit experiment demonstrate the particle nature of light? Young's two-slit experiment is generally credited for demonstrating the wave nature of light. But what about a similar experiment with just one slit? My understanding is that this will create an interference pattern. Shouldn't that be enough to demonstrate light's wave nature? Perhaps the technology available at the time wasn't good enough to create interference, or perhaps there's a plausible wave explanation?
| Actually it proves the wave theory. If the slit is narrow enough, then the light would diffract, which cannot be explained using particle nature, rather wave nature has to be used to explain diffraction, so it would actually prove the wave nature of light.
| {
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Why does energy come in the form of packets? Photons are the packets of energy released by continuous oscillation of charges.
But I have some questions about this.
Since the electrons oscillate regularly while transitioning between orbitals then why is the energy released in the form of packets and not continuously in the form of waves ? What causes this discontinuity in released energy by electrons ?
A physical reasoning will be more appreciated than a mathematical one.
Note : By the word continuous , I am referring to the fact that two consecutive photons have time gap between their emissions while waves are just continuously produced with no gaps at all.
| The straight answer is that nobody knows why. We only know how.
By how I mean that we have accurate methods to predict experimental numbers. We can solve wave equations with advanced methods, as in quantum chemistry, and add on QED radiative corrections, for example.
So we know how but not why wave equations account, and very accurately so, for the behaviour of discrete particles.
I think it is important that we lay our cards on the table and that students are not trained to think that we understand QM fully.
| {
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Why is QED renormalizable? My understanding of renormalizability is that a theory is renormalizable if it the divergences in its amplitudes can be cancelled out by finitely many terms. I see that by adding counterterm (in the MS-bar scheme)
$$L_{ct}=-\frac{g^2}{12\pi^2}\left(\frac{2}{\epsilon}-\gamma+\ln4\pi\right),$$
the one-loop divergence of QED can be made finite. However, I do not see how this makes QED renormalizable? Surely as we work with diagrams with more loops, we will get more counterterms - given that we can have diagrams with arbitrarily many loops, do we not need an infinite number of counterterms to cancel these out?
| QED has only a finite number of irreducible divergent diagrams. The main notion of divergence of a diagram is power-counting: The term every diagram represents has the form of a fraction like
$$ \frac{\int\mathrm{d}^n p_1\dots\int\mathrm{d}^n p_m}{p_1^{i_1}\dots p_k^{i_k}}$$
and you can compute the difference between the momentum power in the numerator and denominator and call it $D$. Heuristically the diagram diverges like $\Lambda^D$ in a momentum scale $\Lambda$ if $D > 0$, like $\ln(\Lambda)$ if $D=0$, and is finite if $D < 0$. This can fail - the diagram can be divergent for $D < 0$ - if it contains a smaller divergent subdiagram.
If you work out the general structure of $D$ for the diagrams of QED, you should be able to convince yourself that QED has only a finite number of divergent one-particle irreducible diagrams. That cancelling the irreducible diagrams is enough to cancel iteratively the divergences in all higher-order diagrams containing them in arbitrary combinations to all orders is a non-trivial statement sometimes called the BPHZ theorem, whose technical meaning - though not by this name - is explained by the Scholarpedia article on BPHZ renormalization.
| {
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When did the Big Bang happen? Did the Big Bang happen at a point? goes through the fact that the Big Bang happened everywhere at the same time. John Rennie's answer explains this as being a consequence of all points in space beings squished into a single point, so technically the Big Bang happened everywhere. But, when we talk about relativity, we also use a fourth dimension of time. So, at the Big Bang all points in space, as well as all points in time should be squished (for lack of better terms) into a single point. So did the Big Bang happen at happen 'every-time'?
Edit: The age of the universe goes into how the true age of the universe can be determined as time flows in different ways for different observers. I am asking whether the Big Bang happened at all points in time together.
| No, the Big Bang did not happen at all points in time. It happened at one point in time, approximately 13.8 billion years ago. As you already know, the Big Bang was not an explosion in space, it was an explosion of space. However, the same is not true of time. (But if you're interested in the idea of an explosion encompassing all of space and time, then I'd recommend watching season 5 of modern Doctor Who. It's great. :)
| {
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Why does the spring constant not depend on the mass of the object attached? It is said that:
$$ F = -m\omega^2 x = -kx, $$
so $k=m\omega^2$. Since $k$ is the spring constant it doesn't depend on the mass of the object attached to it, but here $m$ signifies the mass of the object. Then how is $k$ independent of the mass attached?
| $\omega$ isn't a constant of the spring, but it actually depends on the mass you attach to the spring. $\omega$ refers to the frequency of oscillation of the attached mass. The formula for $\omega$ for an attached mass $m$ is $\sqrt{\frac{k}{m}}$, where $k$ is the spring constant. If you use $\omega=\sqrt{\frac{k}{m}}$ in the formula, $m$ cancels out leaving only $k$
| {
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Phase velocity in monatomic chain When considering a one-dimensional monatomic chain of atoms (identical masses $m$ & spring constant $\kappa$), one finds the following dispersion:
$$ \omega(k) = \sqrt\frac{\kappa}{m}\cdot\left|\sin\left(\frac{ka}{2}\right)\right|\, ,$$
which is $\frac{2\mathrm{\pi}}{a}$-periodic. So wavewectors higher than $\mathrm{\pi}/a$ do not provide new physical behaviour.
However, when computing the phase velocity, one finds:
$$ v_p = \frac{\omega}{k} = \frac{1}{k}\sqrt\frac{\kappa}{m}\cdot\left|\sin\left(\frac{ka}{2}\right)\right|\, .$$
This means that the phase velocity goes like a sinc, which is not periodic; wavevectors outside the first Brioullin zone yield a much lower phase velocity.
How is this possible? Is there a good reason to consider only the first Brioullin zone for the phase velocity? Or are there other errors my calculation?
| The phase velocity is kind of meaningless outside the first Brillouin zone. The phase velocity is the speed that the "crest" of a wave travels, but outside the first Brillouin zone, the wavelength is less than the spacing between atoms, so there aren't really crests; most "crests" occur in the gaps between the atoms where there is nothing to displace, so the crests are kind of mathematical artifacts.
While you can define a continuous function for the displacement of the atoms from their equilibrium position $u\left(x, t\right)$ for the wave, that doesn't mean that the wave is really continuous; the wave only has a meaningful displacement at the $x$ positions where there are atoms. So, some of the intuition coming from waves in a continuous medium doesn't really apply.
| {
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Why do electrons flow in the opposite direction to current? I'm 15 and just had a question about physics and electric fields.
I've read that electrons flow in the opposite direction to current. Isn't current the flow of negative charge and therefore the flow of electrons?
Or are they referring to conventional current?
| By 'current' we normally mean electrical (also called conventional) current, i.e. the amount of electrical charge that is passing a given point in the circuit at any given time.
If a positive charge carrier (say, a 'hole' in a semiconductor) passes from left to right, then that counts the charge $q$ to the current, over the time $\Delta t$ that the process takes.
On the other hand, if a negative charge carrier (most often, an electron) passes from left to right, that means that the total charge on the right is becoming more negative and the total charge on the left is becoming less negative, i.e., it means that electrical charge is moving from right to left. Thus, the electrical current goes from right to left, oppositely to the velocity of the electron.
| {
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Continuity equation in quantum mechanics - Verification Sakurai 2.7.30 I am trying to verify Equation 2.7.30 from Sakurai's "Modern Quantum Mechanics" 2ed. The bottom line of my question is:
$?? \psi^{*}\vec{A}\cdot\nabla\psi+\psi\vec{A}\cdot\nabla\psi^{*}=0 ?? $
And here is how I arrived at it. Below, all equations straight from Sakurai have (==) and all of my intermediate equaitons just have (=)
$$\frac{\partial \rho}{\partial t}+\nabla\cdot \vec{j}==0 \ \ $$
$$\rho=={\left| \psi \right|}^2 ,\ \ \ \ \ \vec{j}==\frac{\hbar}{m}Im\left( \psi^{*}\nabla\psi \right)-\frac{e}{mc}\vec{A}\left| \psi \right| ^2==\frac{i\hbar}{2m}\left( \psi\nabla\psi^{*}-\psi^{*}\nabla\psi \right)-\frac{e}{mc}\vec{A}\left| \psi \right| ^2 $$
$$\frac{1}{2m}\left[ -i\hbar\nabla-\frac{e}{c}\vec{A} \right]\cdot\left[ -i\hbar\nabla-\frac{e}{c}\vec{A} \right]\psi+e\Phi\psi==i\hbar\frac{\partial}{\partial t}\psi=\frac{1}{2m}\left[ -\hbar^2\nabla^2+\frac{i\hbar e}{c}\left( \nabla\cdot\vec{A}+\vec{A}\cdot\nabla \right)+\frac{e^2}{c^2}A^2 \right]\psi+e\Phi\psi $$
$$ \frac{\partial\rho}{\partial t}=\psi\frac{\partial}{\partial t}\psi^{*}+\psi^{*}\frac{\partial}{\partial t}\psi =\frac{i\hbar}{2m}\left( \psi^{*}\nabla^2\psi-\psi\nabla^2\psi^{*}\right)+\frac{e}{mc}\nabla\cdot\vec{A}\left|\psi\right|^2+\frac{e}{2mc}\left( \psi^{*}\vec{A}\cdot\nabla\psi+\psi\vec{A}\cdot\nabla\psi^{*} \right)$$
$$ \nabla\cdot\vec{j}=\frac{i\hbar}{2m}\left( -\psi^{*}\nabla^2\psi+\psi\nabla^2\psi^{*} \right)-\frac{e}{mc}\nabla\cdot\vec{A}\left|\psi\right|^2-\frac{e}{mc}\left( \psi^{*}\vec{A}\cdot\nabla\psi+\psi\vec{A}\cdot\nabla\psi^{*} \right) $$
In the last two equations, I have lined things up so it is easy enough to see that everything cancels except $ \frac{e}{2mc}\left( \psi^{*}\vec{A}\cdot\nabla\psi+\psi\vec{A}\cdot\nabla\psi^{*} \right)-\frac{e}{mc}\left( \psi^{*}\vec{A}\cdot\nabla\psi+\psi\vec{A}\cdot\nabla\psi^{*} \right) $
Which can be remedied if
$?? \psi^{*}\vec{A}\cdot\nabla\psi+\psi\vec{A}\cdot\nabla\psi^{*}=0 ?? $
| There should be a factor of 2 in the equation $d\rho /dt=...$ since $ \nabla (\bar{A} \psi) + \bar{A} \nabla\psi = 2 \bar{A} \nabla \psi $ since we can impose $\nabla \bar{A}=0$. Hope it can be helpful :)
| {
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Is my understanding of how a force is applied through a string correct? Let there be a situation where a force 'F' is acting on one end of an in extensible string which is connected to a box named 'A' resting on a friction less horizontal surface through the other end.
The Force F is transmitted through the string and acts on box 'A'. As a reaction to this box 'A' pulls the string toward itself with force 'F'.
In the FBDs we can see that the string is balanced by the forces but the box is not, so the box starts accelerating towards the right. As the box starts to accelerate, for a very small moment there is slack in the string which makes the tension force 0 for a brief moment.Then the force F being applied on the string is used to make the string taut again and tension forces again begin to accelerate box A.
This keeps repeating itself.
Is this how it works or there something wrong?
| There is a mistake in modeling the system. Indeed, the string is not a rigid body and you can't use the rigid body equations of motion and to be expecting that they will work. You consider, for example, in place of the string, a rigid thin (metallic) rod. In this case, you can consider the box and the rod as a unique rigid body on which the only acting force on x direction is the one you have drawn in your fist picture. So the whole body moves with a constant acceleration on x direction. While, how I said, if you want to consider the string, you have to use proper equations that take into account the physical properties of the string and which will lead to a realistic result.
| {
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Reason for peaks in graph of binding energy per nucleon A similar question was asked before, but it asked for a different thing. My question here is: What is the reason for spikes in this graph? The graph initially has spikes and then shows a constant decrease. Is it related to something called magic numbers as it is seen in multiple of 4?
| It has to be with the pairing term. Nature likes even-even pairs of nucleons. I mean, an even number of protons and an even number of protons. The reason is ultimately related to spin couplings.
So, odd-even pairs are more or less over the curve. Even-even isotopes, like $C^6$, or $O^18$, are especially stable. On the other hand, odd-odd pairs are especially unstable, but there are only 4 stable nuclei which are odd-odd.
Edit:
So, odd-even pairs are more or less over the curve could you elaborate this point?
Okay, I'll ellaborate.
Let's take the liquid drop model, which is empirical, but explains quite good what is happening. It has 5 parameters, though.
Let $B=B(Z,A)$ be the binding energy of the nucleus. The more energy, the more stable. Because that's the energy you have to overcome if you want to separate the nucleus.
The liquid drop model stablishes
$$B(Z,A)=a\cdot A -b\ A^{2/3} - s \frac{(A-2Z)^2}{A} \ -d \frac{Z^2}{A^{1/3}} - \delta\frac{Z^2}{A^{2/3}} $$
That's the function that fits the curve you are showing, with
$a=15,835 MeV; \quad b=18,33 MeV; \quad s=23,20MeV; \quad d=0,714 MeV$
The first term (a) is due to volume. It is responsible that the curve saturates at a certain value. $B/a=cosnt$ for large $A$.
The second term is due to the surface. Sicne small $A$ have much more surface, they are more unbounded. HEnce the stong decay at the beginning.
The 3rd term (s) is due to symmetry. Note that the fraction contains $N-Z$. If there is a big unbalance of nucleons, the nucleus will be unstable.
And the next one (d) is the Coulomb's repulsion. Check that $R\propto A^{1/3}$.
And what about $\delta$? Well,
$$\delta=\begin{cases} +11,2 MeV & if\ even-even \\ 0 & if\ odd-even \\ -11,2 MeV & if\ odd-odd \end{cases}$$
So, the curve that is usually plotted is the function without this delta-term. That's why I say that that "odd-even nuclei are on the curve". However, even-even will be above and odd-odd will be below.
In your curve, they have joined thereal nuclei, which is good.
But since this function $B(Z,A)$ is defined by parts, it is easier to represent it without the delta term. The curve is like this much more smooth. IT's like the "mean curve". Then, if we add the delta, we find more peaks. That's what I meant.
| {
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Energy Conservation in Rolling without Slipping Scenario
A solid ball with mass $M$ and radius $R$ is placed on a table and given a sharp impulse so that its center of mass initially moves with velocity $v_o$, with no rolling. The ball has a friction coefficient (both kinetic and static) $μ$ with the table. How far does the ball travel before it starts rolling without slipping?
The solution I found starts by setting up a conservation of energy and setting $v = rw$:
$$ \ \frac{1}{2}m v_o^2 = \frac{1}{2}m v^2 + \frac{1}{2}Iw^2 \to v_o^2 = \frac{7}{5}v^2 \quad{(1)}$$
It then goes on to say $W = \Delta K_{rotation}$ and solves for $D$ :
$$ \ \int_{0}^{D} F_{f} dx = μmgD= \frac{1}{2}Iw^2\quad{(2)}$$
There are a couple of things I do not understand about this approach. How does $(1)$ account for the loss of energy due to the friction force which is causing the rotation and the slipping that occurs before it starts rolling purely? Second, how does $(2)$ account for the change in center of mass velocity? Wouldn't $W = \Delta K_{rotation} + \Delta K_{transitional}$ ? I am most likely misunderstanding something and help is greatly appreciated.
| *
*since it's rolling without slipping the friction (static) won't affect the energy , so yes energy is conserved.
*here is a good demonstration https://www.youtube.com/watch?v=hxa6jAYA980 about similar case , after applying delta k= w ( transltaion )the forces you have are( only weight because again the friction you have is static
while applying delta k= w for rotation you have one torque which comes from friction
| {
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Young Modulus in Finn's Thermal Physics In Finn's Thermal Physics (equation 2.4), the Young modulus $Y$ of a stretched wire with tension $F$ is given to be
$$Y = \frac{L}{A} \left( \frac{\partial F}{\partial L}\right)_T$$
However, usually the Young modulus is defined to be the ratio of stress and strain, specifically
$$Y = \frac{\sigma}{\varepsilon} = \frac{F L_0}{A \Delta L} = \frac{FL_0}{A(L-L_0)}\implies dF = \frac{AY}{L_0} dL$$
this would suggest that the relationship given in Finn's textbook should actually be
$$Y = \frac{L_0}{A} \left( \frac{\partial F}{\partial L}\right)_T$$So I wondered why they have used $L$ instead of $L_0$ in the definition. Is it a mistake? Thanks!
| When assuming infinitesimal strains, one can use either $Y=\frac{L}{A}\left(\frac{\partial F}{\partial L}\right)_T$ or $Y=\frac{L_0}{A}\left(\frac{\partial F}{\partial L}\right)_T$, depending on what's convenient; the difference is negligible. The former is related to the definition of the true strain $e$ from $de=dL/L$ (so that $e=\ln(1+\Delta L/L)$) and the latter to the definition of the engineering strain $\varepsilon=\Delta L/L$. The difference between the true strain and engineering strain is, of course, also negligible in this context.
| {
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Does Energy & Momentum also Dilate & Contract respectively? Does energy and momentum also dilate and contract as time and length do respectively, since energy and time and momentum and length are complementary quantities both in relativity & QM?
| Energy and momentum together form a "four vector" just like time and space do. So we find that $(E/c,p_x,p_y,p_z)$ transforms the same way that $(c t, x, y, z)$ transforms. However, time dilation and length contraction are more specific.
Time dilation is the time in a "moving" frame for a clock which is not moving in the "rest" frame. So it is the transform $(c \Delta t,0,0,0) \rightarrow (c \gamma \Delta t, \gamma v \Delta t,0,0)$. This would correspond, for energy and momentum to a case where an object has (rest) energy but not momentum. In other words $(E/c,0,0,0) \rightarrow (\gamma E/c,\gamma v E,0,0)$. So indeed there is energy dilation.
Length contraction is the length in a "moving" frame for a rod which is not moving in the "rest" frame. So you would start with $(0,\Delta x,0,0)$. For energy and momentum this would correspond to $(0,p_x,0,0)$. However, this does not make sense for momentum because this would represent a particle with momentum but no energy. Such a particle would have an imaginary mass, which is not possible. So there is no direct equivalent of length contraction for energy and momentum, i.e. no "momentum contraction".
That said, together energy and momentum do transform as time and space so in general $(E/c,p,0,0) \rightarrow (\gamma E/c + \gamma v p, \gamma p + \gamma v E/c^2,0,0)$. It is only because there is no such thing as a particle with momentum but not energy that there is no direct equivalent for "momentum contraction".
| {
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Why do fluids not accelerate? A fluid flowing in a horizontal pipe must be flowing at a constant velocity because of the conservation of mass.
However, considering how there would be a pressure and hence force acting behind the fluid, for it to have a constant velocity, there must be an equal force slowing it down (depicted as $F?$).
I can't see a force that would be as big as the driving force. Can someone explain to me what this force is and how it's created?
| Once into the pipe, there is no net acceleration (before it leaves the pipe), $ma=0$. But you have significant viscous sheering. This creates resistance to flow. Similar to drag. Viscous friction.
The only force driving flow from point A in the pipe to nearby point B in the pipe is the pressure gradient.
$$F= A ~\Delta P = A ~(P_A-P_B)$$
For lower viscosity that will be a lower pressure gradient. That pressure drop (force) increases with length and viscosity and velocity, and decreases with diameter (for a given volumetric flow $Q$ it does. I.e. yes it’s higher at the same $V$ and higher $d$, but lower at the same $Q$ and higher $d$ - point being a big pipe resists less for same volumetric flow than a little pipe).
But no, if it’s not going real fast, you don’t need a lot of pressure-force to drive flow. Depends on factors mentioned.
| {
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Artificial Muscle | Is Electromagnetic Attraction Better then Maxwell Stress? I was watching a TED-talk on artificial muscle, HASEL, where the inventor demos that an insulated oil in the presence of electric potential field gets displaced due to induced Maxwell stress. In other words, electric potential gets converted directly to displacement, in contrast to electromagnetic actuation that convert electric potential to magnetic field, which in-turn produces motion due to field interaction.
Question: Is HASEL more efficient compared to electromagnetic actuators given then fact that electric potential is directly converted to motion, instead of intermediary potential (magnetic field) which in-turn convert motion in the case of electromagnetic actuators? The intermediary magnetic field also generates heat.
| Short answer "No" HASEL is not more efficient then electromagnetic actuaturs.
The following is taken from their paper:
Peak specific power during contraction of the two-unit actuator was 614 W/kg; specific work during contraction was 70 J/kg (fig. S12) (24). The measured peak specific power is double that of natural muscle and comparable to values for silicone DE actuators (26). Thermally activated coiled polymer fiber actuators (49.9 kW/kg)
We performed a full-cycle analysis of actuator efficiency using force
displacement and voltage charge work-conjugate planes (fig. S4) (24).
Conversion efficiency was 21%
while for electromagnetic motors has its “peak” specific power ranging from 1.1-3 kW /kg and efficiency of around 60%
| {
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Can rolling without slipping occur without friction? If a body is rolling without slipping is it necessary that there is friction acting on it ? I encountered a question in which there is a spherical body and a force is being applied on its top point ...so if there is only force then it should do translation motion only .. If there is friction also then it then only it can rotate with translation am I right?
| If there is no frictional force between an object and the surface it is moving on, then there will be no relationship (or connection) between the rate (or direction) of rotation and the translational velocity. If a horizontal force is applied to the top of a sphere on a friction-less surface, it will cause (independently) both translation and rotation. If there is friction, then the two motions will be related.
| {
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Energy transfer between oscillators Suppose I have two mechanical oscillators $a(t), b(t)$, coupled through the interaction $V_\text{int} = \mu^2 a(t) b(t)$. Is there a simple way to express the rate of energy transfer from $a$ to $b$ using only $V_\text{int}$? Something like $\partial_t V_\text{int}$ would have the correct units, but it is symmetric in $a$ and $b$, and therefore cannot represent energy transfer from $a$ to $b$. Something like $a\partial_t(\partial_a V) - b\partial_t(\partial_b V)$ is antisymmetric under $a\leftrightarrow b$, but I can't figure out how to justify this expression.
| The answer is actually extremely simple. The power transfer between oscillators is just the time derivative of the work done on $a$ by $b$, minus the work done by $b$ on $a$
\begin{align}
P &= \frac{dW_{a\to b} - dW_{b\to a}}{d t}\,,\\
W_{a\to b}-W_{b\to a}&=\int{\rm d}a \partial_b V_\text{int} - \int{\rm d}b \partial_a V_\text{int}\,.
\end{align}
Changing integration variables from $a$ and $b$ to $t$ using the chain rule
\begin{align}
{\rm d}a = \frac{da}{dt}{\rm d t}\,,\hspace{1cm} {\rm d}b = \frac{d b}{dt}{\rm d}t\,,
\end{align}
we arrive at the following expression for the power
\begin{align}
P = (\dot a\partial_b - \dot b\partial_a)V_\text{int}\,,
\end{align}
regardless of the form of $V_\text{int}$.
| {
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Do elementary particles last forever? I have heard that not even black holes last forever, because of Hawking radiation. But what about elementary particles? Will an electron, for example, exist for all time?
| The electron, an elementary particle, is the least massive carrier of negative EM charge currently known. If it would decay, it would involve the production of lower mass particles (such as the neutrino), but all known particles with lower rest mass have no EM charge.
https://en.wikipedia.org/wiki/Electron
Thus, during the hypothetical decay of the electron as you say, the EM charge would have to vanish, which would violate conservation laws.
https://physicsworld.com/a/electron-lifetime-is-at-least-66000-yottayears/
Thus, in the standard model, we consider the electron as a fundamental particle that, in isolation, cannot decay.
| {
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Probability of measuring state $|+\rangle$ and state $|-\rangle$ given a state and a basis I am given a basis $|+\rangle = \frac{1}{\sqrt 2}(|0\rangle + |1\rangle)$ and $|-\rangle = \frac{1}{\sqrt 2}(|0\rangle - |1\rangle)$ and i am given a three qubit state $|\phi\rangle = \frac{1}{\sqrt 3}|1\rangle |0\rangle |1\rangle + \frac{2}{\sqrt 3}|0\rangle |1\rangle |0\rangle $
What is the probability of measuring state $|+\rangle|+\rangle|+\rangle$ and what is the probability of measuring state $|-\rangle|-\rangle|-\rangle$?
I know how to express the state $|0\rangle $ and $|1\rangle $ in the basis above, that would be $|0\rangle = \frac{1}{\sqrt 2}(|+\rangle + |-\rangle)$ and $|1\rangle = \frac{1}{\sqrt 2}(|+\rangle - |-\rangle)$
but when attempting to do $|\left(\langle+|\langle+|\langle+|\right)|\phi\rangle|^2$ I can't really get anywhere.
How do I solve this?
| Step 1: Notice that $\vert 0\rangle = \frac{1}{\sqrt{2}}(\vert +\rangle + \vert - \rangle)$ and $\vert 1\rangle = \frac{1}{\sqrt{2}}(\vert +\rangle - \vert - \rangle)$.
Step 2: Expand $\vert\phi\rangle = \frac{1}{\sqrt 3}\vert 1\rangle \vert0\rangle \vert1\rangle + \frac{\sqrt{2}}{\sqrt 3}\vert0\rangle \vert1\rangle \vert0\rangle$ using the substitution in Step 1.
Step 3: The squares of the coefficients in your expansion in Step 2 should add to 1. The squares of the coefficients of the $\vert +\rangle\vert +\rangle\vert +\rangle$ and $\vert -\rangle\vert -\rangle\vert -\rangle$ states are the probabilities you desire.
The statement below is incorrect. $\vert 0\rangle$ and $\vert 1\rangle$ are as written above, not as you have written them.
I know how to express the state $|\phi\rangle $ in the basis above,
that would be $|0\rangle = \frac{1}{\sqrt 2}(|+\rangle + |-\rangle)$
and $|1\rangle = \frac{1}{\sqrt 2}(|+\rangle - |-\rangle)$
| {
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Why are voltage and volt both are denoted by $V$? Why are voltage and volt both are denoted by $V$? Won't it cause confusion?
| Volt is the unit of measurement of Voltage. There's no possible confusion. The voltage is the physical quantity that we measure, using a multimeter for example, and the result of that measurement is given in Volts.
If you have a 9V battery you're implicitly saying that if you measure the voltage across the leads of the battery, you'll measure 9 Volts.
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Transformers: How does current in primary coil change? I was doing a question on transformers and found this really confusing question:
A 100% efficient transformer converts a 240V input voltage to a 12V
output voltage. The output power of the transformer can be a maximum
of 20W. The output is connected to two 0.30A bulbs in parallel. One of
the bulbs fails. How does the current in the primary coil change?
What I did:
Since I have been given power and voltage I thought it would be helpful to figure out the current, so I did that which was fairly easy
20W/12V = 1.67 A
the output is connected to 2 0.3A bulbs so if one fails that's only 1 0.3A bulb
What I am stuck on
Now, this is where I kinda fall apart since I don't know where to go, I thought I would need to figure out some sort of ratio due to the transformer rule of the ratio of coil turns is equal to the ratio of voltages, but I don't really see where I would get information of the number of coils turns, however, my intuition is telling me that the current would increase...
Am I missing something that I haven't calculated from the question?
| The 20 W is a maximum rating for the transformer and is of no concern unless it is exceeded. At 100% efficiency the power in equals the power out. In this case the power out (7.2 W) gets cut in half, so the current in (0.03 A) will also be cut in half.
| {
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The Cause of lightning and charge distribution in clouds I was going through this video about lightning and I couldn't understand some points .
1: What caused the water molecules in ice crystals to be arranged in that specific pattern i.e. having positive charges at its boundary and negative in the inner region ?
2: Also WHY do the crystals break (which created or may have led to the distribution of charges in the way given in the figure) ? (Since the outer regions have positive charges two crystals approaching each other should finally come at rest before striking each other and no collision should occur).
I have also referred to the Wikipedia page and still have the same confusion . What is the cause of distribution of charges in saich a pattern (as shown in the figure) ?
| Assuming that the lightning occurs at the boundary or near boundary between the region of sky experiencing cold weather and the region of sky experiencing warm or hot weather, winds from the warmer side tends to move to the cold side. I'm the same manner, wind from the cold side moves towards the warmer side. When these winds (cold and warm) meet, they tend to collide which due to charges carried by them produces "sparks" due to friction between them. These sparks are seen as lightning.
It is important to understand that these winds are the ones driving clouds thus it could be the reason you have lightning on a cloudy day.
| {
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Are distributions of position and momentum assumed to be independent in quantum mechanics? Given a wave-function of a single particle we can calculate probability density for positions. We can also calculate probability density for momenta. Are these probability densities assumed to be always independent?
Or, in other words, if we measure position and momentum of a particle (for example electron in hydrogen being in the ground state), should we expect that these two random quantities independent?
| Any real world position measurement result also implicitly includes a momentum measurement. Why? Any measurement of x results in some psi(x), its Fourier transform is psi(p), both are measurement results and quantum states, they are simply represented in different bases and have inversely related widths, neither width can be zero. If the position measurement result is accurate (ie, psi(x) is narrow), the momentum measurement is commensurately not accurate (ie, psi(p) is wide). This is the stipulation of the Heisenberg uncertainty principle, it distinctly does not say x and p cannot be measured simultaneously, it simply says the accuracies of the simultaneous x and p results are inversely related, and neither width can be zero, as the product of delta x and delta p must be greater than or equal to hbar/2. This product cannot be zero, so neither delta can be zero. Thus, ideal quantum pure states, and single real numbered eigenvalues, are not physically realizable by any means whether manmade or natural. So every physical realization of a position measurement is also a momentum measurement.
Re independence of x and p, a free particle at position x’ may have any value of p independent of x’, and a free particle with momentum value p’ may have any value of x independent of p’. So x and p may be considered independent, while their wave function widths (in their respective eigenvalue domain spaces) are inversely related, as any Fourier transform pairs.
| {
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Why does metal change its color under polarized light? I've taken two photos of a metal in an experimental setup.
The first image shows the metal illuminated by a halogen-lamp from above. The second image shows the same metal illuminated by the same lamp but there are two additions: There is a linear-polarizing filter in front of the lamp and one in front of the camera.
I was expecting all specular reflection to be eliminated and only the diffuse reflection to be visible. But as you can see, the metal seems to have changed its color as well. Why is that? Shouldn't the metal still appear to be yellowish?
I've made two more photos of a similar object; the setup is the same.
The object appears to be blue when the polarizers are added to the setup.
Both metals are anodized aluminum.
I've done the same "experiment" with wood, plastic, and fabric; they don't appear in a different color as the aluminum does. I've also tried out white paper to see if white-balance might be the cause: no difference, the white paper stays white.
| Let's look at the frequency distribution of a halogen lamp:
You can see that all frequencies are present (hence the white color). The temperature of the burning lamp is about $800$ Kelvin. If you polarize the beam nothing, in particular, should happen because the photons have a random distribution of polarization. The intensity should obviously get less (with a polarization filter).
Did you change the settings of the camera? I think you have to (why did you use two filters, by the way?).
I can think of nothing else to conclude that the metal reflects the light in such a way to produce your observation(s). Paper reflects in a completely different manner. Try using another metal.
See also this article. Maybe you have already seen it, maybe not, but for sure it contains information for your project.
| {
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Maximizing entropy with Lagrange multipliers This is a problem I saw in a stat mech textbook, and I think it is a fairly common problem.
Given the entropy function: $$S = - \sum_{i=1}^N p_i \log p_i$$
Maximize $S$ subject to constraints:
$$ \sum_{i=1}^N p_i = 1 \\ \sum_{i=1}^N p_i e_i = c$$
It was suggested to solve this problem using Lagrange multipliers. So this is how I went about it:
$$L(p,\lambda, \mu) = - \sum_{i=1}^N p_i \log p_i - (\lambda \sum_{i=1}^N p_i -1)- (\mu \sum_{i=1}^N p_i e_i - c) $$
$$\frac{\partial L}{\partial p_k} = -(\log p_i +1) - \lambda - \mu e_i = 0$$
A little arithmetic gives:
$$p_i = e^{-\lambda - \mu e_i -1}$$
Then I used the above constraints to solve for $p_i$.
$$\sum_{i=1}^N p_i = \frac{\sum e^{-\mu e_i}}{e^{\lambda+1}} = 1 \implies e^{\lambda +1} = \sum e^{-\mu e_i} $$
And
$$\sum_{i=1}^N e_i p_i = \frac{\sum_{i=1}^N e_i e^{-\mu e_i}}{e^{\lambda +1}} = c$$
Since I am not sure how to solve this final constraint and get a value for $\mu$, I said
$$p_i = \frac{e^{-\mu e_i}}{\sum_i e^{-\mu e_i}}$$
My question is, how do I solve for $\mu$?
| You cannot solve for $\mu$ unless you know the $e_i$'s. But that shouldn't bother you, because in the context of the canonical ensemble, $\mu$ is defined to be the (inverse) temperature, and all quantities are written in terms of it. $c$ is the expected energy for a given $\mu$.
| {
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What is the centripetal force when instead of a mass point we have a physical rotating body? I was wondering what is the centripetal force of a body rotating in a circular motion. I know that the centripetal force of a point mass is $mv^2/r$. I only have done an introductory physics class so I can not find the answer.
| Centripetal force is the force which keeps a body on a circular path. It is not a new force.
Any force that acts towards the center of that circular path is your centripetal force. For example in case of earth and sun , the gravitational force is the centripetal force on the earth and it is just
$ mg = \frac{mv^2}{r} $
Where $r$ is the distance between the center of mass of the revolving body and the point about which it is rotated. Changing shape will only affect this distance and nothing else.
We actually don't need to prove that changing shape i.e. center of mass will affect the formula , the formula is itself defined to be force between the center of masses. So changing shape will affect the distance between the center of masses. For example , if a hemispherical or a triangular object is being rotated with a string then we will have to use the extra distance between the center of mass and the point where the string is attached to the body i.e.
$T = \frac{mv^2}{r + d_{centre of mass }}$
, Where $T$ is the tension force and $r$ is the distance between the fixed point and the point where body and string are attached.
And since $d_{center of mass}$ will be different for different shapes the force will be different.
| {
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Application of Noether Theorem I attempt to understand one of the examples of the application of Noether theorem given in Peskin & Schroeder's An Introduction to Quantum Field Theory (Page no. 18, Student Economy Edition). The relevant portion of the text is given below.
If I understand the derivation and the corresponding discussion here properly, then it was assumed that the Lagrangian density $\mathcal{L}$ satisfies the Euler-Lagrange equation:
$$\frac{\partial\mathcal{L}}{\partial\phi} = \partial_{\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right].$$
My Confusion: I don't see how $\mathcal{L} = \frac{1}{2} (\partial_{\mu} \phi)^2$ satisfies the Euler-Lagrange equation. Because on the left hand side, I get $\frac{\partial\mathcal{L}}{\partial\phi} = 0$, and on the right hand side, I get $\partial_{\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right] = \partial_{\mu} \partial^{\mu} \phi.$ If the given $\mathcal{L}$ doesn't satisfy the Euler-Lagrange equation, then how Peskin & Schroeder's formulation can be applied to this case? What am I missing here?
| For what it's worth, it is very important at which stage one uses Euler-Lagrange (EL) equations in an application of Noether's (first) theorem. Noether's first theorem has 2 sides:
*
*Input: A global off-shell$^1$ (quasi)symmetry. Here one should not use EOM. (An on-shell symmetry is a vacuous notion, because whenever we vary the action $\delta S$ infinitesimally and apply EOM, then by definition $\delta S\approx 0$ vanishes modulo boundary terms.)
*Output: An on-shell continuity equation. Here one should use EOM. (If it happens to hold off-shell as well, it is because the global symmetry is part of a bigger local/gauge symmetry. See Noether's second theorem and e.g. this Phys.SE post.)
--
$^1$The words on-shell and off-shell refer to whether the Euler-Lagrange (EL) equations (=EOM) are satisfied or not.
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Why do electric field lines curve at the edges of a uniform electric field? I see a lot of images, including one in my textbook, like this one, where at the ends of a uniform field, field lines curve.
However, I know that field lines are perpendicular to the surface. The only case I see them curving is when drawing field lines to connect two points which aren't collinear (like with charged sphere or opposite charges) and each point of the rod is collinear to its opposite pair, so why are they curved here?
| I have taken your image and created a few additional field lines at one end of the plates in the first diagram below.
When you come to the ends of the plates, the field starts to resemble that associated with two point charges instead of a sheet of charge. The second diagram below shows the field lines between two point charges. Note that as you move away from the two point charges an equal distance apart, the lines look like those at the ends of your parallel plate capacitor (curved lines). Towards the center between the charges, the field lines start to look straight and evenly spaced (parallel lines).
Hope this helps.
| {
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Would an atomic bomb detonate a uranium stockpile? If a uranium atomic bomb directly hit a stockpile of weapons grade uranium, would the chain reaction also detonate the stockpile?
what about a stockpile of nuclear reactor fuel rods?
what about a stockpile of various nuclear weapons?
what about a plutonium bomb or a hydrogen bomb?
what about all possible permutations of these?
| If the chunk of fissile material that the atomic bomb explodes close to is subcritical then the neutron flux from the bomb will trigger fissions and energy release in that material but as soon as the neutron burst is over, those fissions will die out and stop; no chain reaction will result. The material may melt down and fly apart in response but it will not explode like a real atomic bomb; this condition is called a fizzle.
It is possible that one bomb exploding may set off the ignition sequence in a warehoused warhead but warheads are equipped with elaborate safety interlocks to prevent this from happening. This is true for fission and fusion bombs.
| {
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Particle as wave, stable? I've started reading about the wave-particle duality but, after a few steps, reached a dead end:
*
*Schrodinger equation solutions for a free particle is a sum of terms of the form:
$$\psi(\mathbf{r}, t) = Ae^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$$
however, a single element of this form can not normalize, thus, can not exists alone. That is, a particle must be an addition of several terms, a wave packet. (TODO: verify a Gaussian wave packet normalizes :-).
*The restriction:
$$ \omega = \frac{\hbar k^2}{2m} $$
applies to previous wave function. That means that each component of the wave packet has a different propagation speed. As consequence, the particle spreads.
The question: particles tends to disperse (dissolve) ? If so, how to explain the stable existence of protons, fermions, ... ?
| The solutions of the Schroedinger equation (SE) are not "particles." They are wave functions. (More precisely, a wave function can be written as a normalizable superposition of the solutions of the SE.) The modulus square of a wave function gives a probability distribution to observe a particle at a specific point. This is the essence of the wave-particle duality.
| {
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A photon scatters an electron at an angle... Does it imply electron having an area greater then the photon's? Even we don't know much about scattering areas of photons and electrons does the fact that a photon scattering an electron at an angle mean that the photon cross-section area hits only a small lateral area of the electron causing it to move at an angle. If the photon cross-section area was greater than the electron's one would the electron be scattered only in the direction of the photon incoming motion because the whole its area would be uniformly objected to photon's pressure? So should the electron have a finite volume?
| I think the question is based on a very mechanistic idea of scattering. IN reality what is important is energy and momentum conservation. Assuming for simplicity a non-relativistic electron, this means
$$
\frac{p_i^2}{2m} + \hbar ck_i = \frac{p_f^2}{2m} + \hbar ck_f,\\
\mathbf{p}_i + \hbar\mathbf{k}_i = \mathbf{p}_f + \hbar\mathbf{k}_f.
$$
To satisfy the momentum conservation (i.e. the second equation) the initial and final momenta of either the electron or the photon do not have to be parallel. In fact, there are simply many more possibilities to satisfy both conservation laws when they are not parallel, i.e. when photon and electron are scattered at an angle.
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Kinematics: rest and motion A ball is dropped from a height of 19.6 m above the ground. It rebounds from the ground and raises itself up to the same height. Take the starting point is the origin and vertically downward as the positie X-axis. Draw approximate plots of a versus t graph. Neglect the small interval during which the ball was in contact with the ground;The acceleration changes from 9.8m/s2 to -9.8m/s2, but answer given in my reference book H C Verma pg:44 shows a straight line drawn in +ve axis and a small gap at t=2 . it is wrong right?
| The acceleration points downwards when the object is moving freely, no matter if rising or falling (its gravity what accelerates it, and gravity points to the ground, doesn't it?) The only time when acceleration is not $g$ is when the object is not falling/rising freely, this is when it is in contact with the ground.
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Will a plastic feel less heavy when I put it in a bucket of water and carry it? If I'm carrying a bucket of water in one hand and a piece of plastic in the other, and then I decide to keep the plastic in the bucket of water (it floats). Will it feel less heavy in the second case?
I think it will feel the same because it's mass adds up to the bucket's mass and will be pulled by gravity with the same extent. But somehow I can't get my mind off from the fact that it's weight is already balanced by the up-thrust.
Is there a simple way to explain how this works? It would be clearer if you helped me with some free body diagrams or an analogy or something simple.
| Fluids, like all objects, obey Newton's Third Law. This means that any upward buoyant force exerted on the plastic by the fluid has a counterpart downward force exerted on the fluid by the plastic. The force you must exert on the bucket & the fluid to keep it from falling is therefore increased by the amount of the buoyant force exerted on the plastic. If the plastic is in equilibrium, this is just equal to the weight of the plastic. Thus, the total weight doesn't change when you put the plastic in the water.
Your idea in the comment is basically correct. The situation is not fundamentally different from putting an object on top of another object that's sitting on a scale. When you do this, the reading of the scale will increase by the weight of the top object. The explanation is essentially the same as given above; just replace "buoyant force" with "normal force".
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Force exerted by blocks on an angled rail I'm trying to figure out the force exerted on blocks positioned on an angled rail due to gravity. This is the scenario I have (apologies for the poor graphic):
where each block (red square) is equipped with a wheel (black circle) which is fitted on a rail (black lines). The top section of the rail is angled at 50 degrees (or just $\theta$) and the bottom section is vertical. I'm trying to figure out on the force exerted on the bottom wheel (green arrow). Here is what I have so far:
We have 3 forces: 1) The force exerted by the top block parallel to the rail is $mg\sin(\theta)$; and 2) The force exerted by the middle and bottom blocks are $mg$. I end up with the following vector triangle:
$R$ is the force exerted on the wheel. Am I missing something?
Any advice appreciated.
| I think the force exerted on the bottom wheel is just the vertical component of your force diagram i.e. $2mg + mg \sin^2 (50^o)$.
The top block exerts a force $mg \sin (50^o)$ on the middle block along the line of the upper part of the track, but the horizontal component of this is opposed by an equal and opposite horizontal force from the lower part of the track. This leaves its vertical component, which is $mg \sin^2 (50^o)$. The middle block can only exert a force on the bottom block along the line of lower part the track, which is vertical, so there are no horizontal forces on the bottom block.
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How do forces 'know' they need to move when a system is in motion? I am curious as to how forces move when a system is in motion. This was never fully explained in my physics classes at university. Let me explain:
I understand the Newtonian (classical) physics that there are equal and opposite forces in play. So when I am standing the force I am exerting on the ground due to gravity is balanced by an upward force from the ground. However, when I lift one foot (say the left foot) the force from my body is now transferred through the right foot. However, where did the upwards force that was under my left foot go?
I assume the upward force 'moved' to balance the increased force exerted by my right foot. I can understand that it general, except for one point. How did the upward force 'know' that it needed to move - and, secondly, where it needed to move to?
This same question can be applied to many dynamic situations of motion, such in a moving vehicle. (I can think of many other examples as well).
I had one physicist trying to explain it to me but, I admit, I lost his explanation when he went down the quantum mechanics rabbit hole. Is there a classical explanation as to how forces know when and where to move when a system is in motion?
| Assumptions
Energy is conserved
Static forces are all relative potential energy.
Zero motion (macro) is a system in equilibrium.
Change in position required human energy, but net forces (due to F=mg) never changed.
Only the % of total force shared by each foot changed.
Mass did not change
gravity, g did not change.
Human energy was dissipation raising one leg as heat but no sweat.
| {
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Properties of conductors If there is a cavity inside a conductor and a charge is placed within it then what exactly happens?
I've read in one book that the charge in the cavity induces a charge (the induced charge is opposite in nature obviously) on the outer surface surface of the cavity. Here is the exact picture. (It is from a pdf which I downloaded from the internet, by MIT). Here is the link https://www.google.com/url?sa=t&source=web&rct=j&url=http://web.mit.edu/sahughes/www/8.022/lec05.pdf&ved=2ahUKEwjQio-zt-XrAhXBjOYKHencDu0QFjABegQICxAH&usg=AOvVaw02X1KKVhOE12qLIW7Rw9FN
But when I read it from another book it said that the inner surface cannot be charge free and if a charge +Q is placed in the cavity, there must be a charge -Q on the inner surface of the conductor. Here is the picture in my book.
So which diagram and explanation is correct?
Also, in the second picture why is there +Q charge on the outer surface of the conductor?
Please help me
|
So which diagram and explanation is correct?
Both of them are correct. They just are drawn in different ways.
Figure 1 shows how the positive charge in the middle attracts electrons in the conductor around it. The minuses, representing a negative charge (electrons), are drawn onto the conductor. It leaves out the fact, that by having the electrons move, there will be less electrons on the outer surface of the conductor.
In Figure 2 the electrons that are attracted to the positive charge in the center are indicated by minus symbols that are outside the conductor. That also is a perfectly reasonable representation. Figure 2 also is more detailed, as it points out that there will be a positive charge (fewer electrons) on the outside of the conductor.
| {
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What does $\hat{\phi}$ mean in cylindrical coordinates? When talking about the unit vectors in cylindrical coordinates, $\hat{\phi}$ often comes up. However, I cannot find a straightforward meaning for it. However, I do know that it is perpendicular to $\hat{\rho}$. How is that significant?
| If you're giving a thumbs-up with your right hand with your fingers loosely curled, and your thumb is pointing along $\hat{z}$, then your other fingers are curled along $\hat{\phi}$.
| {
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Is it possible to bend light without changing its color? It seems to me that whenever you change the direction of a wave it also affects frequency. Would this not also be true of light waves bending from, for example, gravity?
| No, it doesn't, as the comments point out.
You might possibly be confusing the change in wavelength when the light is inside the "bending material" such as a glass lens. However, upon exiting the lens -- or the localized gravitational field -- the light, now moving in a new direction, retains its vacuum wavelength. The wavelength change is entirely due to the effective index of refraction in the region or medium in question.
You might want to take a peek at the discussion in this physics.SE question.
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Charge on the plate of the capacitor I was reading parallel plate capacitors in my book today and there I saw that the battery provides +Q charge to positive plate and -Q charge to negative plate? But we know that only electrons can flow so the battery should provide only -Q charge right?
Is it that, the electrons flow from the negative terminal of the battery to the negative plate and since the conductor is seperated by a dielectric medium, here (air) the negative charge will pile up at the negative plate, but since the seperation is less this will induce a positive charge on the other plate of the capacitor and since positive charge can't flow so it will stay there only. Also, the negative charge is provided by this positive plate only, which is then transferred by the battery to the opposite side of the capacitor or the negative plate? Is this explanation correct?
Also, My teacher said that the battery provides energy to the charges but my book says that the battery provides the charges, I'm totally confused.
Please help me.
| You have got the right idea. When a circuit consisting of a battery and capacitor is switched on, the electrons from the negative terminal start accumulating on the capacitor plate. The negative capacitor plate induces an equal and opposite charge on the other capacitor plate. Together, an electric field is created and there is a new potential difference in between the plates. This potential difference serves to counter the emf provided by the battery.
The electrons from the positive plate go into the positive terminal and the battery is slowly consumed. This occurs until the enduced potential difference perfectly balances the battery. When this happens, we say that the capacitor has become charged and $\frac12CV^2$ worth of energy has been used up by the battery in doing so. The battery provides the potential difference which allows for the flow of charge. Both the textbook and your teacher are correct.
| {
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Electric field inside charged non-conducting spherical shell In class we had an exercise, where for a non-conductiong spherical shell given a potential inside and out we had to find a charge distribution and E field inside and out, with a charge distribution $\sigma =Q\cos\theta$, so the sphere is kind of like a charged dipole, where positive charge is on the upper and negative charge on the lower side since $0<\theta<\pi$. And we calculated that $E_{\text{inside}}=0$, however using Gauss law, if $\rho$ inside is $0$ then the field inside should also be zero.
However the teacher said that that holds only for spherically symmetric situations, and in this situation $\bf E$ field is not zero inside, and I dont understand that, can someone explain why is that.
Thanks.
| For any closed surface, Gauss's law states
$$ \oint \textbf{E} \cdot d\textbf{a} = Q_{enc} / \epsilon_0 $$
Applying the divergence theorem to the left side of the above leads to
$$ \int_{volume} ( \nabla \cdot E ) \ dV = \oint \textbf{E} \cdot d\textbf{a} $$
and rewriting the right hand term in terms of the charge density $ \rho$
$$Q_{enc} / \epsilon_0 = \frac{1}{\epsilon_0} \int \rho dV $$
Equating both sides leads to
$$ \frac{1}{\epsilon_0} \int \rho dV = \int_{volume} ( \nabla \cdot E ) \ dV $$
Which then you can cancel to be:
$$ \frac{1}{\epsilon_0} \rho = \nabla \cdot E $$
In spherically symmetric situations, you can just assume that E only depends on r such that the other terms in the divergence of E drop out: $$ \nabla \cdot E = \frac{1}{r^2} \frac{d}{dr} (r^2 E_r) $$ which then leads to the typical differential equation such that $ \frac{1}{r^2} \frac{d}{dr} (r^2 E_r )= \rho \frac{1}{\epsilon_0}$ in a sphere.
In this case, you cannot assume that E solely depends on r, as your teacher implies.
Inside the sphere, it would be true that $\rho =0$, but it is no longer true that E is independent of $\theta $ or $\phi $, and thus these terms would need to be included in the differential equation as well.
PS:
I don't know what level you're at-- I don't know if your teacher wants you to work with the differential equations like this, but this at least should explain why it's more complicated than that.
| {
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Why are the capacitors in this circuit in parallel but not in series?
In the circuit, the capacitors are said to be connected in parallel. Why is that so?
Edit: The switch will be closed and C2 is fully charged by C1 and no more current will flow between C1 and C2. The question asks for the voltages and charges hold by C1 and C2. In the solution, it is mentioned that C1 and C2 are connected in parallel (V1 = V2), which is the part I don't quite understand.
| I assume that this is an example where one charged capacitor charges another after the switch ic closed.
The main use of assigning the labels series or parallel to capacitors (and other circuit elements) is to decide which combination rule to use to find the effective capacitance of a number of capacitors.
The derivation of such combination rules have assumptions in them: the magnitude of the charge on a capacitor plate is the same as the charge on the plate of another capacitor to which it is connected in series and the potential difference across capacitors connected in parallel is the same.
So considering the final state of your circuit after the switch is closed which of the series or parallel conditions is going to be satisfied?
| {
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Expansion in flat spacetime I have been studying Raychaudhuri equation and focusing theorem related to it. Focusing theorem says that if the strong energy condition is satisfied and rotation tensor vanishes $\omega_{ab}$=0 then rate of expansion is negative. Frobenius theorem for timelike vector says that timelike geodesic is hypersurface orthogonal iff $\omega_{ab}$=0.
I was wondering to apply this in flat spacetime but I can't find any suitable timelike geodesic in flat spacetime which would be hypersurface orthogonal and d\theta /d\tau is negative. Can anyone help with this?
If I have any such geodesic and as in flat spacetime Riemann curvature tensor would be 0 therefore only expansion term and shear tensor term would be left in Raychaudhuri equation which can be found through simple computation and hence focusing theorem could be satisfied in flat spacetime.
| In flat spacetime, let $(t,x,y,z)$ be a global inertial cartesian coordinate system. Then the lines with $x,y,z$ fixed are orthogonal to the hypersurfaces of constant $t$.
This is the simplest case possible: trajectories of infinitely many inertial observers in rest relative to each other.
If you apply Raychaudhuri to this example, you will find $\dot{\theta}=0$ since $\theta_0 = 0$ initially, and thus $\theta = 0$ for any $t$.
| {
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What is the polarization of an EM wave after it suffers scattering? If a linearly polarized classical monochromatic electromagnetic radiation undergoes a scattering, does the scattered electric field have the same polarization as the incident electric field? I am looking for an answer (or deduce the conclusion mathematically) from classical electromagnetic theory of scattering.
| This is a huge subject in radar and it has a vast literature but there are no simple answers besides that depolarization (i.e., the generation of the orthogonal polarized reflected/scattered wave) is dependent on frequency, incident angle, statistics of the reflecting surface, etc. I only mention here one application, namely meteorological radar, where the transmitted signal is linearly polarized but the receiver measures both polarizations of the reflected wave, this is especially important to detect/classify rain.
| {
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Earth doesn't have seafloor craters. Does it mean water has been on Earth since long time before in the past? Is it true that Earth didn't get its water by cometes impacts as there are no subocean craters so it seems water acted as a comete bumper long before in the past?
| Because the sea floor is continually being renewed through spreading and subduction, most of the sea floor is no older than something like ~250 million years. Also note that meteor impacts were much more common longer ago than that than they are today, which means that almost all of the marine impact craters have been wiped clean off the bottom long ago.
| {
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Why is the mean density the same for all nuclei? Tell me if this is a correct theory? So the radius $R$ of the nucleus is directly proportional to $A^{1/3}$ (the nucleon number).
As $$V = \frac 43 \pi r^3,$$ this makes $V$ directly proportional to $R^2$. Also, as the nucleon number increases, the mass also increases and as the masses of protons and neutrons are similar you could say that the mass of the nucleon is directly proportional to the nucleon number.
If you put all of this together, you get the mass of the nucleon being directly proportional to the volume where the constant is the density.
Thus, that is why the density is constant for all nuclei?
| Yes, the nuclear density is approximately constant with nucleon number. Expanding on Vadim's answer, perhaps I can explain why this is somewhat intuitive.
Nucleons are held together by the residual strong force, which is attractive between nucleons and mediated by massive pions. As as result, it has limited range such that nucleons only feel the pull of their nearest neighbours (and to a much lesser extent their next-to-nearest neightbours). Consequently, we expect that we can keep adding nucleons to the nucleus without it increasing in density, as the inner nucleons do not attract the outer nucleons, which would in turn lead to a more compact (and hence more dense) nucleus.
Now in practice, one should expect that the nuclear density will in fact decrease slightly with radius, as the repulsive (electromagnetic) force between protons in the nucleus has an infinite range.
If instead the nucleus were made of nucleons held together by an infinite range, attractive force, then we would expect the density to increase with nucleon number.
| {
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How is it possible for the same force to do different amounts of work in two different inertial frames? Consider an object which has been given a speed $v$ on a rough horizontal surface. As time passes, the object covers a distance $l$ until it stops because of friction. Now,
Initial kinetic energy = $\frac{1}2mv^2$
And final kinetic energy is zero. Therefore, work done by friction on the object is equal in magnitude to $\frac{1}2mv^2$.
Now here is the part that I found weird: Consider another frame moving with a speed $v_0$ in the same direction with respect to the ground frame. Now, kinetic energy of the original object with respect to this new frame is $\frac{1}2m(v-v_0)^2$.
And, the final kinetic energy is equal to $\frac{1}2mv_0^2$.
So this means that the work done by frictional force, in this case, will have a magnitude of $\frac{1}2m[(v-v_0)^2-v_0^2]$, which is obviously different from the value which we get with respect to a stationary frame.And this part seems very unintuitive to me. How is it possible for the same force to do different amounts of work in two different inertial frames? (I would consider it unintuitive even if we consider non inertial frames, after considering pseudo forces).
And if we were to do more calculations based on the two values of the work done by friction, we would land on different values of some quantities which aren't supposed to be different in any frame. For example, the coefficient of friction would be different, as the amount of frictional force is constant, acting over a distance $l$. We can say that Work done by frictional force is $\alpha$$mgl$, where $\alpha$ is the coefficient of friction and $g$ is the acceleration due to gravity. We can clearly see that $\alpha$$mgl$ equals two different values.
So, is this just how physics works, or is there something wrong here?
| The coefficient of friction is the same in both cases. You have assumed the distance traveled to be the same in both cases, which is why you are getting different values for $\alpha$. Your other questions have been cleared in many answers above, so I just wanted to mention this point.
| {
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When is a quantum state pure and when mixed? Every definition of the two is always very abstract to me. Like, A pure state is located on the surface of the bloch sphere while the mixed state is somewhere within.
First of all, what is an intuitively definition?
And second of all, how do you practically recognize whether a given state is mixed or pure?
| A pure state is a ket in Hilbert space. A mixed state is a probabilistic mixture, which cannot be described by a single ket. The importance of the concept is illustrated in Bell tests, which show that quantum mechanics (i.e. the result of calculation using Hilbert space) cannot be replicated by classical probability theory. In practice, I do not know of any other examples giving a clear empirical distinction. I don't find it helpful to worry about it too much. The concept of mixed states was due to von Neumann, who was more concerned with matter of principle than practicality.
| {
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Why are there fewer lines on the absorption spectrum than on the emission spectrum for some gases? I don't know if this is right, but I think that it is because:
*
*when the gas is heated, the electrons get 'excited' and move up to an energy level (lets say an electron moves up 4 energy levels).
*Then as they move down, they emit all 4 photons of different wavelengths which are then shown on the emission spectrum.
*when the star's light goes through the gas, the gas absorbs specific wavelengths which causes their electrons to get 'excited'. This time, if the electrons move up 4 energy levels with this one specific wavelength they have absorbed, then you can't see the 4 other wavelengths that were emitted?
Thus, 4 lines are missing from the absorption spectrum?
I don't know if this is correct...
Thanks for any help.
| I found the answer!
So the continuous spectrum is formed is usually formed from a heated body (i.e a heated filament or a star) and when the continuous spectrum passes through a cooler gas, the electrons in the gas absorb the energy of a photon with a specific wavelength (which is why is shows black lines against a continuous spectrum).
*
*The electrons then get excited and move up an energy level. As they move down, they emit photons with energy of the discrete energy levels of an atom.
In the emission spectrum, the electrons in the energy levels usually start at random energy levels and so there is more of a variety of wavelengths that could possibly be emitted.
Whereas in the absorption spectrum, there are a few lines missing because most electrons start from ground state, meaning that there are less options of energies that a photon can be emitted at.
| {
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Magnetic, geographic and geomagnetic poles
Can someone explain me in simple terms where and what are the magnetic, geographic and geomagnetic poles?
Some sites say that magnetic north pole is in the south and thus it attracts the south pole of the magnetic needle while some say the north pole of the needle points south. In each place, I seem to get a different angle between the magnetic and geographical axis (10°, 11°, 17° etc.). And I don't get clearly what the geomagnetic pole is.
I'm utterly confused, please help me out.
This question slightly addresses my doubt but the answers contradict each other.
Thanks in advance.
| Although an answer has been accepted, there are lots of errors and ambiguities in the above answers. First of all, when you talk about getting a local angle between the magnetic and geographic axis, that is the magnetic declination. At most points on Earth, your compass does not point to any of the North geographic pole, the North Magnetic Pole or the geomagnetic North pole.
The three kinds of poles would coincide if the source of Earth's field were a very tiny but powerful bar magnet with its South pole pointing at geographic North. A field with this geometry is called a dipolar field (see Magnetic dipole). If the bar were tilted away from the rotation axis, then clearly magnetic and geographic North would not coincide, but everywhere on Earth our compasses would point to the North magnetic pole.
In reality, Earth's magnetic field is very complex, but it has modest deviations from a dipolar field. If you measure the field all over Earth and find the best fitting dipole field, its North pole at Earth's surface is the geomagnetic North pole.
If the field were dipolar, then at the geomagnetic North pole, a compass needle that could rotate in all directions would point straight down. This is referred to as the North Magnetic Pole. Since it is not, the North Magnetic Pole is at a different location. Also, the deviations from a dipolar field are complex, so as you move a compass around Earth, the declination changes. Finally, the field is always changing, so the declination is changing everywhere, as are the magnetic poles. Fortunately for navigation, these changes are slow.
| {
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Understand the total angular momentum in 2 cases I'm a little bit confused with how to understand the relationships between spin angular momentum $S$, orbital angular momentum $L$, and the total angular momentum $J$ by visualizing those quantities. Classically, if we consider the motion of the Earth, I think $S$ is its rotation, $L$ would be its orbit around the sun, and $J$ is the combination of both. However if we consider the electron spin, I've seen the following diagram to illustrate their relations:
I'm wondering is there a way to understand the Earth's total angular momentum from this diagram? If not, can we draw a different diagram to show their relations?
| In this picture, both orbital and angular momenta are treated as vector. As the third case illustrates, it is perfectly possible for the sum of two vectors to be smaller in length than the length of both constituents in the sum.
This third figure would correspond roughly to the case where the rotation of the Earth about its axis is reversed compared to what it is now, so that the Sun would rise in the West. Then the $\vec S$ for Earth would be mostly antiparallel to the orbital angular momentum of Earth about the Sun.
The big difference of course is that, for the “quantum vectors”, only some projections about $z$ are possible, and only some lengths of the resulting vectors are possible so $\vec L$ and $\vec S$ must combine in specific ways so that their projections and lengths add up correctly.
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Will energy in an LC circuit become 0 if its is disconnected when the capacitor is fully discharged? In a recent test I had a question in which there was an LC circuit with an inductor a capacitor and a switch. According to the answer key
If switch is opened when capacitor is fully charged energy of LC system remains same.
If switch is opened when capacitor is fully discharged energy of LC system becomes 0.
I can understand the first one but not the second one. The answer keys to this particular exam do tend to be wrong once in a while so I thought I'd get a second opinion.!
|
If switch is opened when capacitor is fully charged energy of LC
system remains same.
If it's a series LC circuit, then the energy will be $E=\frac{CV^2}{2}$ all residing in the electric field of the capacitor. If it is a parallel LC circuit of an ideal inductor and capacitor, the energy will remain the same but it will be exchanged between the electric field of the capacitor and magnetic field of the inductor at the resonant frequency.
If switch is opened when capacitor is fully discharged energy of LC
system becomes 0.
The only place where energy, if any, can be stored is in the magnetic field of the inductor, $E=\frac{LI^2}{2}$ and that would require current in the circuit prior to the switch opening. Once the switch is opened, current cannot flow. The energy stored in the inductor will be dissipated in an arc between the switch contacts as they open, caused by the high voltage induced in the inductor in response to the attempt to disrupt the current by the switch.
Hope this helps.
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Difficulty with Kallen-Lehmann spectral functions The following is for $D=4$. The correlators at a fixed point are power laws of the form $x^{-2\Delta}$, where $\Delta$ is the scaling dimension. Suppose I wish to find the nature of the spectrum at the fixed point, for which I calculate the spectral function $\rho(p^2)$ which is defined so that $$\langle\phi(x)\phi(0)\rangle=\int \frac{d^4p}{2\pi}^4e^{-ipx}\rho(p^2)$$
Now, for $\Delta=1$, I expect this to be the same as that of a fundamental scalar field, with $\rho(p^2)=\delta(p^2)$.
$\Delta=2$ should correspond to a composite operator of 2 massless fields, and thus I expect $\rho(p^2)=\int d^4k\delta[(k-p)^2]\delta(k^2)$ and so on.
However, I am unable to derive these relations formally. Any help would be appreciated.
As an example of the kind of attempts I made-none of which I claim to be rigorous in any way-note that the problem reduces to finding the fourier transform of $\frac{1}{x^2}$ and $\frac{1}{x^4}$. I tried introducing a regulator to control $x\to 0$, but got nowhere. Another approach was to call $\int d^4x e^{ipx}\frac{1}{x^4}=f(p)$, and find the differential equation for $f(p)$ by differentiating both sides with respect to $p$ until the LHS reduces to something like $\int d^4p e^{ipx}=\delta(x)$. It is likely that this is infact the right way to proceed, but I'm a little lost and frustrated by what should have been a simple calculation.
| I answered that in
For which values of lambda the euclidean two-point function $(p^2 +m^2)^{-\lambda}$ is reflection positive
In the notations of that other question, take $m=0$ and $\lambda=\frac{D}{2}-\Delta$ where
$D=4$ is the dimension of spacetime. Note that the condition $\lambda\le 1$ amounts to the unitary bound $\Delta\ge\frac{D-2}{2}$.
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Why is uranium's atomic weight listed as 238.02891, higher than almost all known isotopes? The vast majority of uranium is U-238, and most of the rest is U-235, U-232, U-234, etc....
So how can the averaged atomic weight be a little over 238?
| The isotope mass of uranium-238 is 238.05078826 amu, which is greater than 238. So it's not surprising that the standard atomic mass (averaged over relative abundances of all isotopes) is greater than 238 amu as well.
In fact, the masses of most heavy nuclei are greater than 1 amu per nucleon. This is because the binding energy per nucleon is lower for larger nuclei, and so the mass defect is smaller.
Note that both the mass of a bare proton and a bare neutron are slightly greater than 1 amu (1.0073 amu and 1.0087 amu respectively), and the masses displayed in isotope tables include the masses of the electrons as well (0.00055 amu). So the only reason that any nuclides have a mass below 1 amu per nucleon is because they are more tightly bound and their mass defect is correspondingly larger.
| {
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Ordering ambiguity in the Feynman propagators obtained using Wick's theorem Applying Wick's theorem to a string of four field operators, $\phi_a\equiv\phi(x_a)$:
$$T(\phi_1\phi_2\phi_3\phi_4)=\{...\}, \tag{1}$$
we obtain several terms, three of which are fully contracted fields: $$\phi_1^{\bullet}\phi_2^{\bullet}\phi_3^{\bullet\bullet}\phi_4^{\bullet\bullet},\quad \phi_1^{\bullet}\phi_2^{\bullet\bullet}\phi_3^{\bullet}\phi_4^{\bullet\bullet},\quad \phi_1^{\bullet}\phi_2^{\bullet\bullet}\phi_3^{\bullet\bullet}\phi_4^{\bullet}. \tag{2}$$
Where I have given contracted fields the same number of dots. Each contracted field gives the associated Feynman propagator: $\phi_1^{\bullet}\phi_2^{\bullet}\equiv D_F(x_1-x_2)$.
My question is, when we have terms with more than one contraction, which propagator goes first? Based on what I am reading in Peskin and Schroeder we order them according to the ordering of the left-most contraction arm, however the book only demonstrates this for terms with four operators (so far), and I am unsure if this relation holds for terms with more fields.
| Mmm... For bosonic fields isn't the propagator symmetric? If so there is no ordering issue.
For fermionic fields you get a determinant or a Pfaffian instead of a hafnian or a permanent, just start with things in their original order and count the number of interchanges as you move them into adjacent pairs: a minus sign for each interchange.
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Will back emf supports flux change? In an inductor, back emf is induced as there comes flux change. It is said that back emf decreases as time passes (when inductor is just connected to DC source).
When back emf decreases current increases in inductor. Current increases flux associated with inductor.
Back emf comes obeying Lenzs law. (Back emf is to fight flux change but why it is supporting flux change by decreasing...) Decreasing back emf is again increasing flux change. Then why and how back emf reduces?
|
Then why and how back emf reduces?
The back emf reduces as the current increases, only because the inductor also has some parasitic resistance, and/or you've connected it in series with a resistor, so that some of the applied voltage is required to maintain the current through the resistance and isn't available to drive further increases in flux (with the associated back-emf).
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Proof of a uniqueness theorem in electrostatics I am trying to understand problem 3.4 in Griffiths' Introduction to Electrodynamics:
Prove that the field is uniquely determined when the charge density is given and the potential $V$ is specified on each boundary surface. Do not assume the boundaries are conductors, or that $V$ is constant over any given surface.
I've tried and this is my approach:
Let there exist two potentials for the given constraints as $V_1$ and $V_2$.And let $V_{3} \equiv V_{1}-V_{2}$.
Both of the potentials$V_1$ and $V_2$ obey :$\nabla^{2} V=-\frac{\rho}{\epsilon_{0}}$. So $\nabla^{2} V_{3}=\nabla^{2} V_{1}-\nabla^{2} V_{2}=0$.
Also as the potential at every boundary is specified, then at the boundaries $V _3=0$ .
So we have$\nabla^{2} {V_3}=0$ and
$V_{3}=0$ at boundaries.
Since $V_3$ is a harmonic function so we can deduce from above that $V_3=0$ everywhere and hence $V$ is specified uniquely which implies that the field is unique as well because$-\nabla V=\vec{E}$. Which Completes the proof.
The textbook has a different and more difficult proof, and I feel that I'm incorrect. Please hint me how to do this problem or where I'm wrong. I don't have any teacher to go to. Thank you
| I think that you are assuming that $\nabla^2\phi=0$, $x\in \Omega$, and with $\phi=0$ on the boundary $\nabla\Omega$ then $\phi$ must vansish everywhere. This is kind of begging the question because it assumes that solutions to Laplace are unique. One can prove this, though, by
$$
\int_\Omega |\nabla \phi|^2 d^dx = \int_\Omega (-\phi \nabla^2 \phi ) d^dx + \int_{\partial \Omega} \phi ({\bf n}\cdot \nabla \phi)d|S|\\=0+0
$$
As $|\nabla \phi|^2\ge 0$ the only way that the LHS can be zero is if $\nabla\phi=0$ everywhere, so $\phi$ is constant. Then, as $\phi$ is zero on the boundary, it must be zero everywhere .
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Trouble solving partial differential equation with Laplacian squared I am working in extensions of General Relativity Theory and at the moment of taking the Newtonian limit of this extension theory (essentialy, mathematically speaking, this is just linearizing the field equations obtained via the variational principle, but this is not important) I arrive to the following partial differential equation:
\begin{equation}
\nabla^2 h+b\nabla^4h=\alpha \delta^3(\vec r)+\beta \dfrac{1}{r}e^{-r/\gamma}.
\end{equation}
Here, $\nabla^2$ is the Laplacian operator, $\nabla^4=\nabla^2\nabla^2$ is the ''squared'' of Laplacian operator, $\alpha,\beta$ and $\gamma$ are just real constants, $\delta^3(\vec r)$ is the three dimensional Dirac delta, $r$ is the variable and $h(r)$ is the function we're solving for (physically, it is basically the Newtonian potential).
I am having a lot of trouble to solve this differential equation, I tried to solve it by Fourier transforms but I'm not capable to find the analytic expression for $h(r)$. Even though, I know the solution must be what in physics we call Yukawa-like potentials, i.e, the solution must be of the form $\dfrac{k_1}{r}$ plus terms like $\dfrac{k_2}{r}e^{-mr}$.
Can someone help me out to find the solution?
| Take the Fourier transform of each side with
$$
h(x) = \int \tilde h(k) e^{-ikx} \frac{d^3k}{(2\pi)^3}
$$
so that
$$
\nabla^2 h(x)= \int \left\{-|k^2|\tilde h(k)\right\} e^{-ikx} \frac{d^3k}{(2\pi)^3}, \quad etc.
$$
Here $|k^2| = k_x^2+k_y^2+k_z^2$.
As (I think!)
$$
\frac{e^{-m|x|}}{4\pi r}= \int e^{ikx}\frac 1 {|k^2|+m^2}\frac{d^3k}{(2\pi)^3},
$$
we get
$$
(-|k^2|+ b |k^2|^2) \tilde h(k) = \left(\alpha +\beta \frac{1}{k^2+m^2}\right) .
$$
This leads to
$$
\tilde h(k)= \left(\alpha +\beta \frac{1}{|k^2|+m^2}\right) \left(\frac {1}{|k^2|-b |k^2|^2}\right)\\
= \left(\alpha +\beta \frac{1}{|k^2|+m^2}\right) \left(\frac {1}{|k^2|(1-b |k^2|)}\right).
$$
Now invert the FT to get
$$
h(x)= -\int \left(\alpha +\frac{\beta}{|k|^2+m^2}\right)\left(\frac{e^{-ikx}}{|k|^2(1-b |k|^2)}\right) \frac{d^3k}{(2\pi)^3}.
$$
Resolve the various factors using partial fractions to get a sum of Yukawas.
I have not tried doing the latter, so I have not yet seen your divergence problem.
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Is string theory the boundary theory of M-Theory? Looking at various AdS/CFT correspondences, we find that some (n-1) dimensional field theories on the boundary of $AdS_n$ with $N=\frac{8}{n-3}$ supersymmetries are equivalent to M-Theory in $AdS_n \times S_{11-n}$.
(e.g. for $n=7$ we get 6D $N=(2,0)$ superconformal CFT.)
Setting $n=11$, seems to suggest that there is a 10 dimensional N=1 supersymmetric theory on the boundary of $AdS_{11}$ that is equivalent to M-Theory in $AdS_{11}$.
The natural theory in 10 dimensions with $N=1$ supersymmetry is superstring theory (or one of them).
Therefor could 11D M-Theory and 10D superstring theory be related by the AdS/CFT correspondence? If not what 10D theory living on the boudary of $AdS_{11}$ would be equivalent to M-Theory on $AdS_{11}$?
Therefor is superstring theory not a theory of the bulk, but actually a theory of what happens on the surface of a black hole? i.e. string theory is just the theory of the hologram?
| No, AdS/CFT relates a theory of quantum gravity in d dimensions to a QFT without gravity in d-1 dimensions. M theory and string theory are both quantum gravity theories.
The holographic dual to M theory is a type of CFT called ABJM theory.
However, there is a case unrelated to holography where your suggestion is valid in a different sense. In heterotic M theory, aka Horava-Witten theory, we imagine M theory with two 1+9D "end of the world" branes and the 11th dimension as a distance between the branes. The M2 branes are then open tubes ending on the branes. The endpoints of these M2 branes are now 1D closed strings, and the theory restricted to one or the other 1+9D brane is a heterotic string theory. In this sense, a string theory lives on the boundary of an M theory, but this is not a holographic boundary.
| {
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Doubt on net acceleration during non-uniform circular motion During non-uniform circular motion, the direction of net acceleration is not in the direction of the centripetal acceleration, then why does a particle still move in a circular path, please explain.
|
During non uniform circular motion,the direction of net acceleration
is not in the direction of the centripetal acceleration, then why does
a particle still move in a circular path,please explain..
Assuming by "non uniform circular motion" you mean a change in speed of the particle moving in a circle, then it is because the centripetal acceleration depends only on the magnitude of the tangential velocity (the speed of the particle), not on the rate of change of speed of the particle, or change in tangential velocity (tangential acceleration). The following explanation is offered:
For circular motion there are two types of possible acceleration: centripetal and tangential.
Centripetal acceleration, $a_c$, is the acceleration towards the center of the circular path. It is always present and it is what keeps the particle in circular motion. It is due to a centripetal force. In the case of a car, the centripetal force is the static friction force between the tires and the road and directed towards the center of the circular path. The centripetal acceleration depends on the magnitude of the tangential velocity $v_t$ (the speed of the car or its angular velocity, ω, in rad/s) and the radius $r$ of the circular motion according to,
$$a_{c}=\frac{v^{2}_t}{r}=rω^2$$
The tangential acceleration $a_t$ results from the change in the magnitude of the tangential velocity. An object can move in a circle and not have any tangential acceleration simply because the angular acceleration $α$ (rad/sec$^2$) is zero because the object is moving with a constant angular velocity ω ($\Delta ω =0$). In the case of a car in circular motion this is the acceleration due to braking or increasing the speed of the car due to the static friction force between the tires and the road in the tangential direction.
$$a_{t}=rα=\frac{\Delta ω}{\Delta t}$$
Hope this helps.
| {
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How does information of the parent neutron star get encoded on the event horizon of a black hole? I was watching a video on a typical black hole formation from a neutron star and it suggested that the event horizon appears instantaneously at the surface of the star even as the stellar matter inside starts to converge into a singularity.
I have also read that the information that gets encoded on an event horizon is because of the fact that all the infalling matter gets sort of smeared on the black hole's surface from the point of view of an outside observer. So what happens to the information of all the original stuff in the neutron star if the stuff is inside the event horizon to begin with and so never gets smeared over it?
( And if we say that the event horizon starts out small from the center of the neutron star then shouldn't Hawking radiation destroy or at least impede the formation of the nascent and tiny black hole? )
| The event horizon is by definition the boundary between the black hole interior and exterior in spacetime. Any matter that starts outside the hole and ends inside it must cross the horizon at some point by definition, whether it's "founding" matter or matter that falls in later.
The horizon doesn't appear instantaneously at the star surface. It starts as a set of points that has no interior and encloses no space (possibly but not necessarily a single point), and grows outward at the speed of light.
When the event horizon appears and starts to grow, there is much less spacetime curvature locally than there would be in the vicinity of a small black hole of that size, so there's no reason to expect it to immediately evaporate by Hawking radiation.
| {
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Why does homogeneity imply $d P^{\mu} / dX^{i} = 0$ along geodesic? I'm reading the Cosmology lecture notes Daniel Baumann and there they describe the path of a freely-falling particle along a geodesic, which is denoted by the curve $X^{\mu}(\tau)$, $\tau$ being proper time and $\mu$ denoting a spacetime coordinate. The geodesic equation is given by
$$ \frac{d^2 X^{\mu}}{d \tau^2} = - \Gamma^{\mu}_{\alpha \beta} \frac{dX^{\alpha}}{d\tau} \frac{dX^{\beta}}{d\tau},
$$
with $\Gamma^{\mu}_{\alpha \beta}$ the Cristoffel symbols. Defining $U^{\mu} := dX^{\mu}/d\tau$ and $P^{\mu} := m U^{\mu}$, they say that due to the homogeneity of the universe, one has $d P^{\mu}/dX^{i} = 0$ (the $i$ representing a spatial coordinate). This seems to be equivalent to saying that $\frac{d^2 X^{\mu}}{dX^{i} d \tau} = 0$, however I don't see how this follows from homogeneity. I would say $X^{\mu}(\tau)$ is only a function of $\tau$, hence the derivative would vanish. However, it seems that this reasoning is too simple and there is more at play here. What am I missing in my understanding?
| First, $\frac{d}{dX^{i}}= 0$ can be considered as "generator" of displacement in $X^i$ direction so the statement $d P^{\mu}/dX^{i} = 0$ implies the velocity/momentum is independent of displacement along the $X^i$ axis, which is just the definition of homogeneity. I used quote when saying generator since Hamiltonian formulation of GR is bit involved. But the essence is what I just said.
Second, the fact that $P^{i}$ is only a function of $t$ or $x^0$ comes not just from homogeneity but also from isotropy of space. It's another way to say Hubble's law while excluding some parts.
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In what case can we describe forces by potential? Let's consider a particle in an $N$-dimensional space and let's assume that acceleration of this particle depends on its position. So, one can say that we have an $N$-dimensional vector field in an $N$-dimensional space (a mapping from position to acceleration)
What allows us describe this "acceleration field" as gradient of some potential?
Should dependency of acceleration on position have a special form to make it possible to introduce a potential?
| If the force at $\vec r$, $\vec F(\vec r)$, is the gradient of a potential i.e. $\vec F = - \nabla \phi$ then $\nabla \times \vec F=0$. And vice versa, if the curl of $\vec F$ is zero everywhere then there is a potential field $\phi( \vec r)$ such that $F= -\nabla \phi$. See this Wikipedia article for more details.
| {
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Does light have a mass? Could you please argue with work and gravity? Or however you like. I just don't get it. Thanks.
A physics professor talked about if light had a mass, it should do work ($W=N\cdot m = \frac{kg\cdot m}{s^2}\cdot m$) to get out of the gravitational field. But because we somehow can see it doesn't, light doesn't have a mass.
| Does light have a rest mass? No, the rest mass of a photon is zero. Is light affected by gravitational fields? Yes, because it has energy. In particular, light moving against a gravitation field does work and loses energy - we can see this because the light is red-shifted.
| {
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Sand leaking out of bag and skater "A person skating on a frictionless icy surface is holding a sandbag. The sandbag has a small hole at the bottom, from which the sand starts to leak. As the sand leaks from the sandbag, the speed of the skater..."
The answer was that the speed of the skater stays the same. I thought that the speed of the skater must increase because now the mass of the sandbag is less than the mass before it leaked. What is wrong with my intuition here? The system I'm considering is the skater and sandbag together.
$p_{system}(t_i) = (m_{skater}+m_{sandbag})v_i \\
p_{system}(t_f) = (m_{skater}+m_{sandbag})v_f$
Since the mass of the sandbag is smaller, the $v_f$ must increase, no?
| The problem is that you have to consider the full $p_{system}$ when you look at $p_{system} (t_f)$.
The full momentum of the system has to consider every part of the system for it to remain conserved, and that includes the falling sand. So really you have:
$$p_{system} (t_f) = (m_{skater} + m_{sandbag} + m_{sand}) v_f$$
We can assume the falling sand has the same velocity $v_f$ because it should retain the velocity from when it fell (neglecting slight air drag), and since the surface is frictionless it should keep sliding along. If the sand were slowed by friction or air drag, that momentum from the sand would go into the ground or the air respectively; but wouldn't take momentum away from the skater or sandbag, or add any momentum to them.
If the sand were being pushed backwards relative to the skater when it fell out (not like here), then the velocity of the skater would increase; and there would be a force due to the sand pushing away from the skater.
| {
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Does the oscillating electric and magnetic field of a photon generate gravitational waves? From my understanding, little as it may be, because photons have energy they warp spacetime. The energy is expressed as an oscillating electric and magnetic field. Would this mean that the energy is also oscillating and would generate a gravitational wave?
| You are correct, photons do have stress-energy and they do create their own gravitational effects, like bending spacetime.
So the idea that photons bend spacetime is part of mainstream cosmology, such as the standard Lambda-CDM model.
Do photons bend spacetime or not?
Now you are saying that the oscillating EM field would create gravitational waves.
In reality, to create GWs, you need objects whose motion involves acceleration and its change, provided that the motion is not spherically or rotationally symmetric.
A simple example of this principle is a spinning dumbbell. If the dumbbell spins around its axis of symmetry, it will not radiate gravitational waves; if it tumbles end over end, as in the case of two planets orbiting each other, it will radiate gravitational waves.
https://en.wikipedia.org/wiki/Gravitational_wave
More technically, the second time derivative of the quadrupole moment of an isolated system's stress-energy tensor must be non-zero to create gravitational waves. Since the photon alone (as an isolated system) in your case does not satisfy this, it would not emit gravitational waves, but for example a system of a photon orbiting (in the photon sphere) a black hole could emit gravitational waves (though to have detectable GWs at our current technological level, you need two actual black holes merging).
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Perpetual motion: Conditions for rolling a wheel sliding on a bar This is a basic question about energy conservation and classical mechanics:
Question: Under what situations can this motion be perpetual?
*
*Without gravity and without frictions.
*Without gravity and with frictions.
*With gravity and without frictions.
*With gravity and with frictions.
*Others setup (please state the setup)
*Impossible to be perpetual
Gravity (say) is along the vertical $y$ direction, with a constant gravitational force and a linear gravitational potential $V(y)=mgy$.
| Look , the object in question is undergoing accelerated motion so it will create ripple in spacetime fabric . The energy will be released in the form of gravitational waves which will ultimately make it slow down .
Had there been friction the ceasing of motion would be clearly apparent. But in case of energy released due to G - waves it may take a billion years to observe significant change in its motion.
G - waves are one of the reasons black holes which were initially a member of binary system rotating about their common centre of mass cease to rotate and merge together.
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Is there any physical interpretation of the constant which is seen in the constraint curve of an adiabatic process? What is the $C$ in $ PV^{\gamma} = C$? I always saw it as a result out of the mathematical calculations that we do but I recently saw this video which made me think that the constant may have more meaning that meets the eye.
See this video at 4:11
He writes $ S = PV^{\gamma}$ .. but where exactly is this equation from? I don't think I've seen it anywhere else.
| An adiabatic reversible process keeps the entropy the same. So adiabats are lines of constant entropy, i.e. the entropy depends on $P$ and $V$ only through the combination $PV^\gamma$.
Of course, that doesn't mean the entropy is literally $PV^\gamma$. It's a more complicated function given by the Sackur-Tetrode equation, which states that
$$S = n C_V \log(PV^\gamma / n^\gamma) + \text{constant}.$$
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If objects in motion experience time differently, how does my body stay synced when I move my legs or arms? If I move my swing my arm really fast, the matter in my arm should experience time slower than the matter in my body.
So how does my body still sync with each other?
And a more general question that derives from this: A lot of matter move at different speeds inside our body, how does anything ever stay synced?
| Forgive me for answering a question with a question, but what makes you think that your limbs do stay "in sync" as you put it?
Let's examine one particular molecule in your wrist. As you swing your arm, the atom of this molecule that's closer to your arm probably experiences one or two more Planck Times, and thus has a tiny fraction of a femtogram less mass, than the atom that's closer to the hand. Molecules experience trillions of times more force than that just existing in a world with other molecules to bounce off of.
The fact that one atom of this molecule is now "younger" than another doesn't matter in the slightest. In fact, the concept of "age" has no meaning for an atom at all. It will continue to participate in every chemical reaction it would have otherwise, exactly the same.
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Density of States of 1D (ideal) Fermi Gas discrepency - Missing factor? I wanted to find the density of states of a 1D ideal, noninteracting Fermi gas. My workings are below:
$$D(\epsilon) = \frac{1}{2\pi}\int_{0}^{\infty}\delta(\epsilon-\epsilon_k)dk \times2$$
$$\epsilon_k=\frac{\hbar^2k^2}{2m} \rightarrow dk=\frac{\sqrt{2m}}{\hbar}\cdot\frac{1}{2\sqrt{\epsilon_k}}d\epsilon_k$$
Which means that:
$$D(\epsilon)=\frac{1}{\pi}\frac{\sqrt{2m}}{2\hbar}\int_0^{\infty}\frac{\delta(\epsilon-\epsilon_k)}{\sqrt{\epsilon_k}}d\epsilon_k=\frac{1}{2\pi\hbar}\sqrt{\frac{2m}{\epsilon}}$$
The irritating thing is, on sources I've checked, the answer "should" be:
$$D(\epsilon)=\frac{1}{\pi\hbar}\sqrt{\frac{2m}{\epsilon}}$$
I'm not too sure as to where the additional factor of 2 may be coming from, as to me my logic (including the spin degeneracy of 2, etc) seems to make sense.
As an additional point to this, I had to do a similar procedure for 2D, and I got the correct answer that I anticipated I should have.
Am I missing something that would yield this factor of 2?
| It looks to me like you're only integrating over positive values of $k$. If you want to take account of electrons going the other way, you'll need an additional factor of $2$ (which is taken care of by the polar coordinate angular integral in $>1$ dimensions).
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Symmetry for dipole conservation in field theory In article The Fracton Gauge Principle complex scalar field is considered.
There's statement, that for conservation of charge one needs usual U(1) global symmetry:
$$
\phi \to e^{i\alpha}\phi \Rightarrow Q =\int d^Dx \rho
$$
For conservation of dipole moment:
$$
\phi \to e^{i\vec x \cdot \vec \lambda}\phi \Rightarrow Q^i =\int d^Dx x^i\rho
$$
Why for dipole moment we need such symmetry?
| For a symmetry
$$ \phi \rightarrow \mathrm{e}^{\mathrm{i}\lambda(x)}\phi, $$
Noether's theorem gives you a conserved four-current:
$$ J^\mu = \mathrm{\lambda}\frac{\partial \mathcal L}{\partial\frac{\partial \phi}{\partial x_\mu}}\phi $$
which satisfies the continuity equation:
$$ \partial_\mu J^\mu = 0 \Leftrightarrow \frac{\partial J^0}{\partial t} + \nabla \cdot\mathbf{J} = 0. $$
If there are no current sources or sinks, then $\nabla \cdot\mathbf{J} = 0$ and the quantity $J^0$ is conserved:
$$ \frac{\partial J^0}{\partial t} = 0.$$
Let's take the Dirac Lagrangian as an example, which gives you
$$ J^\mu = \lambda \bar\psi \gamma^\mu\psi.$$
*
*With $\lambda = \alpha$, then the conserved quantity $J^0$ is:
$$ Q = \int\mathrm{d}^3\mathbf{r} \, J^0 = \int\mathrm{d}^3\mathbf{r} \, \alpha |\psi|^2,$$
where $\bar\psi = \psi^\dagger$ allowing to take the absolute mod squared, $|\psi|^2$ being the (normalised) probability density. $\alpha$ is then the electrical charge of the particle under consideration.
*With $\lambda(x)= x^i \lambda_i$, then the conserved quantity $J^0$ is:
$$ Q = \int\mathrm{d}^3\mathbf{r} \, J^0 = \int\mathrm{d}^3\mathbf{r} \, x^i \lambda_i |\psi|^2 = \lambda_i \int\mathrm{d}^3\mathbf{r} \, x^i |\psi|^2 ,$$
where $\int\mathrm{d}^3\mathbf{r} \, x^i |\psi|^2$ is the definition of the $p^i$ dipole moment, $|\psi|^2$ being the charge distribution.
Now, the conserved quantity is actually $\boldsymbol{\lambda}\cdot \mathbf{p}$, but I guess you can take $\boldsymbol{\lambda}$ to point along $x$, or $y$, or $z$ and each of these would give you a conserved $p_x, p_y$, or $p_z$. Because $\boldsymbol{\lambda}$ is arbitrary?
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is kinetic friction present when two objects are slipping across each other my mechanics sir was telling us the other day that when the relative velocity is zero between the surface and the object then the object starts sliding, eg in rolling motion without slipping, the v rel is zero at the bottom most point .how does kinetic friction come into play here. I understand that since the object has started moving then kinetic friction will come into play. Is this the one and only reason?
he, secondly, mentioned that slipping and sliding are essentially the same phenomena. I don't quite agree with this statement. can someone pls explain this as well?
Also, this is not really a question but a few thoughts that came to my mind that I wanted to discuss.
| When a wheel is rolling without slipping or sliding, it has static friction with the ground not kinetic friction. When a wheel is slipping or sliding, such as on ice or mud, it has kinetic friction not static friction. The words slipping and sliding are fairly synonymous, but we might be more inclined to say that when a wheel has lost static friction and has kinetic friction it is slipping, and a block moving with kinetic friction on a surface is sliding. However either word could work in either situation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proof Rényi entropy is non negative The Rényi entropy is defined as:
\begin{equation}
S_\alpha = \dfrac{1}{1-\alpha}\log(\text{Tr}(\rho^\alpha))
\end{equation}
for $\alpha \geq 0$. This can be rewrited in terms of $\rho$ eigenvalues, $\rho_k$, which verify $0 \leq \rho_k \leq 1$, as:
\begin{equation}
S_\alpha = \dfrac{1}{1-\alpha}\log(\sum_k \rho_k^\alpha)
\end{equation}
How can one proof rigurously that $S_\alpha \geq 0$? I am having trouble with this proof eventhough it seems pretty easy.
| Consider 3 cases : ($\alpha > 1$, $\alpha < 1$, $\alpha = 1$)
*
*$\alpha > 1$
$$
\sum p_k^{\alpha} \leq \sum p_k = 1 \Rightarrow \log (\text{Tr}(\rho^\alpha)) \leq 0\Rightarrow S_\alpha \geq 0
$$
*$\alpha < 1$
$$
\sum p_k^{\alpha} \geq \sum p_k = 1 \Rightarrow \log (\text{Tr}(\rho^\alpha)) \geq 0\Rightarrow S_\alpha \geq 0
$$
*$\alpha = 1$
$$
S_\alpha =\frac{1}{1 - \alpha} \log (\text{Tr}(\rho e^{(\alpha - 1)\log \rho})) = - \text{Tr}(\rho \log \rho) = - \sum p_k \log p_k \geq 0
$$
And in the last sum we used, that $x \log x \leq 0$ for $x \in [0, 1]$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why do the Dirac-Maxwell Lagrangian and the QED Lagrangian look the same? I know that QED is some kind of second quantized version of the Maxwell-Dirac theory. But why is it that this modification to a second quantized version is just to replace the scalar function $\Psi$ by a field operator $\hat{\Psi}$?
| When going to second quantisation, the fields once denoted by $\phi$ and $\psi$ are replaced by field operators $\hat \Psi$ and $\hat \Phi$ which are essentially operators which create/annihilate particles at points in space represented by the fields.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Can the universe ever contract? I am currently going through this answer related to the Big Bang theory and from there a question arose in my mind:
*
*Can the universe ever contract?
*Can it ever contract to singularity?
I wonder, if it is possible, how it would happen? Is there any chance at all?
I am unaware about the physical reality of this question.
| Nobody knows for sure how the universe will evolve. It's accepted by almost all cosmologists that the universe is expanding at an increasingly accelerated rate.
Almost all. There are indeed convincing experimental facts (type IA supernovae, the CMBR), but there is also counter-evidence. I think this is not taken too seriously because there were already Nobel prizes awarded for the discovery that the universe is expanding at an accelerated rate.
See for example this paper, containing much math. The link to this paper is made in this article (many more links are included here).
So maybe dark energy (the Nature of which is completely unknown; it's supposed to become non-diluted by the expansion of space) is not necessary after all. The Dutchman Erik Verlinde (who received the Spinoza prize in 2011 for his in the Netherlands so-called revolutionary new theory of gravity, including a premium of 2 million euros...) proposes that dark energy, as well as dark matter, are emergent properties, though I think observations on the Bullet Cluster defuse his theory.
So the situation is far from settled. And who knows what the future in cosmology holds in store?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why doesn't planet Earth expand if I accelerate upwards when standing on its surface? According to General Relativity I am being accelerated upwards by planet earth while writing this question. But a curious person on the the other side of the planet relative to me would have the same experience. That means we are accelerated in opposite directions, although earths diameter do not seem to increase. How can this be?
| What is asserted is that gravity isn't a force at all; rather, it's acceleration such that if I sit on a chair here in California, the acceleration is upward by Earth against me and my chair. What about my friend in China sitting on his chair? What about everyone else, Earth-wide, sitting on their chairs. I don't know if gravity is force or not; I do know that gravity is still unexplained. I took a lot of physics in college but I'm no scientist. Admittedly, I don't completely get space-time which is probably where gravity's explanation lies. But I'm pretty sure it's not fabric.
Einstein was brilliant for figuring it out. But he's dead and now we need another genius to explain.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is there a need for the concept of energy if we have the concept of momentum? We have two concepts that are energy and momentum. To me, momentum is more fundamental than energy and I think that momentum was the thing which we wanted to discover as energy.
Now momentum can describe several things that that energy does and if it is not able to describe it then it can be somehow extended to describe that thing.
For example, momentum can not describe the potential energy let's say due to the gravitational field of Earth on an object but it can be easily twisted to be able to describe it.
Momentum can describe quantum stuff as well. Also, we know that momentum is conserved just as energy.
In short, I want to know the physical difference b/w momentum and energy.
| Your impression that "momentum can describe several things that that energy does and if it is not able to describe it then it can be somehow extended to describe that thing" is a reflection of something real and deep.
When you move to the framework of general relativity, in which space and time are combined to form the 4-dimensional manifold "spacetime", then momentum and energy also combine into a 4-dimensional vector, given the name four-momentum:
$$
p^\mu=(E,p_x,p_y,p_z)
$$
So, in a sense, energy "really is" momentum in the time dimension.
However, working in this framework is a waste of effort in any context where Newtonian physics are "good enough"—you just wind up doing extra math to get the same result. And one of the really useful ways Newtonian mechanics is simpler than relativistic mechanics is that you can treat energy and momentum as independent quantities. Don't get out the tensor calculus if you don't need it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/585384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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What does it mean for particles to "be" the irreducible unitary representations of the Poincare group? I am studying QFT. My question is as the title says. I have read Weinberg and Schwartz about this topic and I am still confused. I do understand the meanings of the words "Poincaré group", "representation", "unitary", and "irreducible", individually. But I am confused about what it means for it to "be" a particle. I'm sorry I'm not sure how to make this question less open-ended, because I don't even know where my lack of understanding lies.
| This is really a deep question.
I am still learning, so any feedback is more than welcome
My key takeaways and interpretations are:
*
*Particles are interpreted as field excitations
*The complexified (thanks ZeroTheHero, for clarifying) full Poincaré ISO(3,1) when studied, e.g. through the Little Group (Wigner method), turns out to have an algebra that is isomorphic to $su (2) \oplus su(2)$
*That decomposition gives us the allowed subspaces (incarnated through Weyl spinors, the Electromagnetic Tensor and such) for physical theories
*From that, irreps are found
*Irreps are the foundational blocks to represent any group in a physical theory
*Particles, being field excitations, have its quantum numbers (spin for instance)
*The irreps give natural quantum numbers, which can be discriminated through the Spin-Statistics Theorem as bosons or fermions (anyons if we're working with different dimensions)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/585475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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} |
Computing a quality factor of multiple measurements Suppose I measure the same quantity twice with two methods, first I get 0 with 0.001 uncertainty, then I get 1 with 0.000001 uncertainty. We can see from this most likely there is something wrong with the uncertainties or measurements. I am faced with such a problem, I have multiple measurements of the same quantity with prescribed uncertainties, and I would like to calculate how compatible these measurements are, or if they are correct, how unlikely it is to get those values. A quality factor or something is what I'm looking for.
What is the most sensible way to do this?
The goal is to know if there is something wrong with the uncertainties or not.
| If you assume the uncertainties are Gaussian distributed, then 1 is 1000 sigma away from 0 +/- 0.001. To a first approximation, you can ignore the much smaller uncertainty on 1.
You can use the complementary error function to determine the probability of finding a value at $n$ sigma away from the mean value (for Gaussian uncertainties).
For example,
\begin{equation}
{\rm 1 \sigma:} \ \ {\rm erfc}\left(\frac{1}{\sqrt{2}}\right) = 0.3173105...\\
{\rm 2 \sigma:} \ \ {\rm erfc}\left(\frac{2}{\sqrt{2}}\right) = 0.0455003...
\end{equation}
By the way, the $2\sigma$ result is very close to the $p$-value of 5% used in some sciences as a threshold for statistical significance.
In your example, we want to know the probability of getting a result at 1000$\sigma$ or higher. I evaluated this on WolframAlpha and got
\begin{equation}
{\rm 1000 \sigma:} \ \ {\rm erfc}\left(\frac{1000}{\sqrt{2}}\right) = 4.58 \times 10^{-217151}\\
\end{equation}
Suffice it to say, it is very unlikely your two measurements differ only by a statistical fluctuation. You should be looking for a systematic error affecting one or both measurements.
| {
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"url": "https://physics.stackexchange.com/questions/585565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why two orbitals having same phase is not a random phenomenon? I have been reading about Molecular Orbital Theory for Chemistry.
*
*I tend to believe that when two Hydrogen atoms approach each other, whether the $1s$ orbitals are in-phase or out-phase is a random phenomenon. However I know that this is not so. Please provide some arguments to counter it.
Please try not to indulge in complex Mathematics as I am relatively new to the subject. My mathematical understanding is "negligible".
| The total wavefunction is a function of both electron positions. There is no phase between them. The wave function is also a function of the electron spins. The total wave function (both spin and spacial part) must be anti-symmetric. Although a clever experimenter could prepare a pair of hydrogen atoms which had specified spins, in most cases the total (spin) angular momentum will be random, and if the spins are the same then the spacial part of the wavefunction will be antisymmetric and one of the electrons will be forced into an anti-bonding orbital. In a spin 0 case, the spin wavefunction is ant-symmetric, so to spacial wave function is symmetric, and both electrons can fit into a bonding orbital. The attractive force between the atoms will depend on the spins.
You may want to look up the Woodward-Hoffmann rules of orbital symmetry, which predict the stereochemistry of reactions based on simple physical descriptions of the electron orbitals.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/585648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is spin necessary for electromagnetism? I know that spin is needed for defining the magnetic moment of any particle, and I have also read that the spin actually is the reason why some materials are magnetic. What I want to know is whether spin is necessary for the some interactions in the electromagnetic field.
Let me expound a bit: in classical electromagnetic field theory, the electric and the magnetic fields could be considered as some combinations of partial derivatives of the vector potential ($A_\mu$). Any particle couples with the field and interacts with other particles through it.
Moving on, if we consider the quantum field theory version, we have two particles coupled with the electromagnetic field, which then interact with the exchange of bosons (photons). My question is: how big of a role does spin play in the interactions which happen through the electromagnetic field? Are there some interactions which spinless particles cannot have, but those with spin can?
| The charged pions, for example, $\pi^+$ and $\pi^-$, have zero spin but interact with a magnetic field, as can be seen from the curved tracks they leave in a bubble chamber with a magnetic field. So, to answer your question, spin is not theoretically necessary for electromagnetism. You could have a perfectly good and complete EM theory with spinless particles. But the real world doesn't work like that.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/586741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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