Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Question about electrochemical reaction in fuel cell The electrochemical reaction on the anode side of a fuel cell can be expressed as hydrogen split into hydrogen protons and electrons. I'm doing some research about the electrochemical reaction in a fuel cell. Is the anode side electrochemical reaction a heat release process or a heat absorption process?
| The answer lies in your question itself. The left hand side of the anode half-cell equation contains H-H, whereas the right hand side contains isolated H⁺ ions.
What do you have to do in order to break the H-H bond to isolate them? Obviously you must supply them with heat energy equal to or greater than the bond enthalpy of the H-H bond. Hence, the anode process is a heat absorption process.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why did recombination make the universe transparent? It is commonly said that after the universe cooled enough for ionized Hydrogen to settle down into neutral Hydrogen, i.e. recombination, the universe became transparent. A reason I have heard for this is that most photons don’t have the right energy to be absorbed by H atoms.
But the free electrons before recombination weren’t absorbing the photons either, they were scattering them. Doesn’t light still scatter off bound electrons? For instance, my understanding is that Compton’s original experiment on Compton scattering used graphite as the source of electrons. Certainly then, photons were scattering off the electrons bound in carbon atoms?
I suspect the answer has something to do with the scattering cross section of bound electrons in neutral Hydrogen being much less than that of free electrons, but then why is that the case?
| The scattering cross-section for free electrons is known as the Thomson scattering cross-section $\sigma_T$.
It is relatively easy to show that once the electrons are bound, the cross-section can be treated like that of a damped harmonic oscillator. If the oscillator is driven at frequencies below the resonant frequency $\omega_r$, then the (Rayleigh) scattering cross-section goes as $\sigma_T (\omega/\omega_r)^4$.
After recombination most of the hydrogen atoms are in the ground state, so the lowest energy resonance corresponds to Lyman alpha absorption in the UV. But most photons have frequencies in the infrared, much lower than the resonant frequency, so $\omega \ll \omega_r$. Hence the scattering cross-section is orders of magnitude lower than $\sigma_T$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why do electromagnetic waves have the magnetic and electric field intensities in the same phase? My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.
If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $\frac{d(\sin x)}{dx}=\cos x$.
| The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
\begin{align}
\nabla \times \mathbf E & = -\frac{\partial\mathbf B}{\partial t} \\
\nabla \times \mathbf B & = \frac{1}{c^2} \frac{\partial\mathbf E}{\partial t},
\end{align}
where the notation $\nabla \times{\cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $\cos(kx-\omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
} |
Why is blowing so different than sucking? Why is it so easy to blow out a candle from a significant distance, but nearly impossible to suck enough air to do the same?
Even without focusing the airflow through a nozzle or something, this affect seems to be present. For example, it's easy to feel the air coming out of a box fan, but very hard to feel the air going into one.
| Your lungs and mouth are designed to generate overpressure, not underpressure. You can barely suck the liquid out of a straw, while you can expel air with way more pressure.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Why are symmetries in phase space generated by functions that leave the Hamiltonian invariant? Hamilton's equation reads
$$ \frac{d}{dt} F = \{ F,H\} \, .$$
In words this means that $H$ acts on $T$ via the natural phase space product (the Poisson bracket) and the result is the correct time evolution of $F$. In other words $H$ generates temportal shifts $t \to t +dt$.
The function $F$ over phase space describes a conserved quantitiy if
$$ \frac{d}{dt} F = \{ F,H\} =0 \, .$$
Nother's theorem now exploits that the Poisson bracket is antisymmetric
$$ \{ A,B\} = - \{ B,A\} .$$
Therefore we can reverse the role of the two functions in the Poisson bracket above
$$ \{ F,H\} =0 \quad \leftrightarrow \quad \{ H,F\} =0 \,. $$
In words, this second equation tells us that for any conserved quantity $F$, its action on the Hamiltonian $H$ is zero. In other words, $F$ generates as symmetry. This is exactly Noether's theorem.
But usually, we argue that only the Lagrangian has to be invariant. The Hamiltonian can change under symmetries like boosts which increase the potential energy. (While the Lagrangian is a scalar, the Hamiltonian is only one component of the energy-momentum vector and therefore, there is no reason why it should be invariant.)
So why exactly do we find in the Hamiltonian version of Noether's theorem that the Hamiltonian remains invariant under symmetry transformmations?
| This is essentially statement 3 in my Phys.SE answer here, which also provides a proof and some related statements.
Statement 3: "A constant of motion generates a symmetry and is its own Noether charge."
In more detail:
*
*An off-shell constant of motion$^1$ $Q$ satisfies by definition
$$\{Q,H\} +\frac{\partial Q}{\partial t}~=~0,\tag{1}$$
cf. my Phys.SE answer here.
*It generates a quasisymmetry of the Hamiltonian Lagrangian $$L_H~:=~\sum_{i=1}^np_i\dot{q}^i-H.\tag{2}$$
*The corresponding Noether charge is precisely $Q$. $\Box$
--
$^1$ If the quantity $Q$ has no explicit time dependence, then by definition
$$ Q \text{ off-shell constant of motion} \quad\Leftrightarrow\quad Q \text{ and } H \text{ Poisson commute} $$ $$\quad\Leftrightarrow\quad Q \text{ generates a symmetry of the Hamiltonian } H.\tag{3}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Electromagnetic field tensor and antisymmetry Why does the inner product between the four force (caused by the electromagnetic field tensor) and the four velocity equaling zero imply that the electromagnetic field tensor is antisymmetric?
This image is from the textbook General Relativity: An Introduction for Physicists by Hobson, Efstathiou and Lasenby
| The premise is that $F_{ab}u^a u^b=0$ for all timelike $u$. I'll abbreviate this as $uFu=0$.
Given any two timelike vectors $x$ and $y$, we can multiply one of them by a non-zero scale factor so that $x+y$ is also timelike. Then the premise implies
$$
xFx=0
\hskip2cm
yFy=0
\hskip2cm
(x+y)F(x+y)=0.
\tag{1}
$$
This combination of equations implies
$$
xFy+yFx=0.
\tag{2}
$$
Since $x$ and $y$ were arbitrary timelike vectors except for the relative scale factor, and since the relative scale vector doesn't affect equation (2), equation (2) holds for any two timelike vectors.
Any spacelike or lightlike vector $x'$ can be written as the difference of two non-zero timelike vectors, and likewise for any spacelike or lightlike vector $y'$, so equation (2) implies $x'Fy'+y'Fx'=0$ for all vectors $x'$ and $y'$. Since this is true in particular for all of the vectors in some orthogonal basis, this immediately implies that $F$ is antisymmetric.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Unification of gravity and electromagnetism Have there been any attempts at unifying gravity and electromagnetism at least at classical level since Hermann Weyl's idea of gauge principle (1918)? We now have Standard Model which is very successful and many other theories. But gravity and electromagnetism are long range in nature and classical as well. Can these two be unified independent of weak and strong forces?
| Yes, classically, we can unify gravity with electromagnetism. The theories that do so are the famous Kaluza-Klein theories. They are theories of pure gravity in $4+1$ dimensions rather than our usual $3+1$ dimensions. When such theories are viewed from a $3+1$ dimensional perspective, the effects of gravity in the fourth unseen dimension appear in the lowered $3+1$ universe as electromagnetism! This is really amazing (the Professor who taught me GR titled the topic on the Kaluza-Klein theory as The Kaluza-Klein Miracle in his notes!). Now, the lesson physicists learned from the KK miracle is that what makes this miracle possible is the fact that pure gravity in additional dimensions appears as gauge-fields in lower dimensional universes. The more evolved version of the spirit of the KK miracle is survived in string theoretic theories which naturally (and inevitably) unite electromagnetism (and all other forces) with gravity--in addition, these theories are inherently quantum mechanical too.
Now, of course, we don't yet have explicit experimental evidence for any of the string theoretic theories (we also don't know which string theoretic theory reproduces the standard model). But yes, theoretically, we can construct consistent theories that unite electromagnetism with gravity.
Edit
Of course, KK theories are not a part of the "done-deal" physics, among various reasons that I am not well-educated in, the major reason is that we don't have experimental evidence for the extra dimensions. Nonetheless, KK theories are a part of textbook physics and for very good reasons. They do provide an internally consistent classical framework to unite electromagnetism and gravity and teach us how pure gravity in a higher dimension can give rise to gauge-fields in lower dimensions as I already mentioned. And this insight is now a part of the daily toolkit for theoretical physicists.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Why can all solutions to the simple harmonic motion equation be written in terms of sines and cosines? The defining property of SHM (simple harmonic motion) is that the force experienced at any value of displacement from the mean position is directly proportional to it and is directed towards the mean position, i.e. $F=-k(x)$.
From this,
$$m\left(\frac{d^2x}{dt^2}\right) +kx=0.$$
Then I read from this site
Let us interpret this equation. The second derivative of a function of x plus the function itself (times a constant) is equal to zero. Thus the second derivative of our function must have the same form as the function itself. What readily comes to mind is the sine and cosine function.
How can we assume so plainly that it should be sin or cosine only? They do satisfy the equation, but why are they brought into the picture so directly? What I want to ask is: why can the SHM displacement, velocity etc. be expressed in terms of sin and cosine? I know the "SHM is the projection of uniform circular motion" proof, but an algebraic proof would be appreciated.
| One way of deriving it is using Taylor series (although to be fully rigorous, this requires further justification for restricting to analytic functions). We have that $f(x) = \sum a_n x^t$, so $f''(t)=\sum (n+1)(n+2)a_{n+2}t^n$. If $f''(t)=-\frac k m f(t)$, then $(n+1)(n+2)a_{n+2} = -\frac k m a_n$, so $a_{n+2} = \frac {-k a_n}{(n+1)(n+2)m}$. So $a_0$ determines $a_2$, which determines $a_4$, etc., and $a_1$ determines $a_3$ which determines $a_5$, and so on. Since all of the even terms vary linearly based on the 0th term, and don't depend on the odd terms, and vice versa for the odd terms, we can find a solution $f_0$ by taking $a_0=1, a_1=0$ and $f_1$ from $a_0=0, a_1=1$, and all solutions will be a linear combination of $f_0$ and $f_1$. And if you work out what those solutions are, they correspond to cosine and sine, respectively.
You can also get to sine and cosine by taking $f(t)=e^{at}$. Then $f''(t)=a^2e^{at}$, so if $f'' = -\frac km f$, then $a^2= -\frac km$. This leads to the two solutions $a = \pm i\sqrt{\frac km}$. Neither root gives a real solution, but $f(x) = \frac 12 (e^{ct} + e^{-ct})$ and $f(x) = \frac 1{2i} (e^{ct} - e^{-ct})$, where $c = i\sqrt{\frac km}$, do. Those correspond to cosine and sine.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Why does the $\phi$-cubed theory have no ground state? In the book of Sredinicki's, he claimed that the $\phi^3$ theory has no ground state, hence this is not a physical theory.
My question is that I can't see why this system has no ground state. And I don't understand either the explaination he gave. For example, what does "roll down the hill" really mean? What's the case for a harmonic oscillator pertuibed by a $q^3$ term? Maybe it's better if someone can explain it using the quantum harmonic case.
| In quantum theory we usually require that the Hamiltonian $H$ is bounded from below and that the system has a ground state. This is intimately related to unitarity. The $\phi^3$ theory violates this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Beta function in $\lambda_0\phi^4$ theory For a real scalar field $\phi$ after performing all the 1-loop renormalization
for dimensional regulator $d = 4 - \epsilon,\ \epsilon \rightarrow 0^+$, I have found that the renormalized coupling $\lambda$ can be related to the bare one by
$$\lambda\Bigg(1 + \frac{3}{(4\pi)^2\epsilon}\lambda_p\Bigg) = \lambda_0 \tag1$$
I'm stuck trying to get the beta function from that equation. We call beta function to
$$\beta(\lambda) = \mu\frac{\partial\lambda}{\partial\mu}$$
Taking into account that
$$\lambda = \lambda_p\mu^\epsilon,\qquad [\lambda_p] = 0, \qquad [\mu] = 1$$
My problem is that any way I use to get $\beta(\lambda)$ from Eq. (1) gives a dependence on $\epsilon$, but in Peskin it is used a different way to solve this and the solution is
$$\beta(\lambda) = \frac{3\lambda^2}{(4\pi)^2}$$
How can I get the beta function via Eq. (1)?
| The idea is that the bare quantities explicitly do not depend on $\mu$, thus one has the equation
$$
0 = \mu\frac{d \lambda_0}{d\mu} = \left(\frac{\partial}{\partial\mu}+ \beta(\lambda_p)\frac{\partial}{\partial \lambda_p}\right)\left(\mu^\epsilon\lambda_p \,Z_\lambda\right)= \epsilon \mu^\epsilon\lambda_p Z_\lambda+\mu^\epsilon \beta(\lambda_p)\frac{d(\lambda_p Z_\lambda)}{d\lambda_p}\,.
$$
Where $$Z_\lambda = 1 + \frac{3\lambda_p}{(4\pi)^2\epsilon}\,.$$
This determines the $\beta$ function as
$$
\beta(\lambda_p) = - \frac{\epsilon \lambda_p Z_\lambda}{Z_\lambda + \lambda_p\frac{dZ_\lambda}{d\lambda_p}}\,.
$$
Obviously you need to do this perturbatively in $\lambda$, which means that even if $\lambda$ multiplies a pole in $\epsilon$ (which is divergent), you still series expand it as if it was small. If $Z_\lambda \equiv 1 + \lambda_p^2\,z/\epsilon$ one has
$$
\beta(\lambda_p) = - \frac{\epsilon \lambda_p + z\lambda_p^2}{1+2z\lambda_p/\epsilon} =z \lambda_p^2 +O(\epsilon)= \frac{3\lambda_p^2}{(4\pi)^2}\,.
$$
A consistency check is that the result should never be divergent as $\epsilon \to 0$, this can be proved with the Callan-Symanzik equations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sun's energy and grand unification According to Wikipedia,
The approximate grand unification energy value is equal to $1×10^{25}$ eV or $10^{16}$ GeV
This is equal to $1602176$ J.
Now,
The Earth receives 174 petawatts (PW) of incoming solar radiation (insolation) at the upper atmosphere
This is solar radiation which is equal to $1.74×10^{17}$ J
My question is, Since solar radiation is way above the unification scale, near sun, Do forces in nature unify?
PS: Apologies if this was a stupid question, I'd appreciate if someone pinpoints and explains where I went wrong (If I did).
Links for reference:
https://en.wikipedia.org/wiki/Solar_energy
https://en.wikipedia.org/wiki/Grand_unification_energy
| You've got a misunderstanding.
The grand unification energy of $10^{16}$ GeV is meant as an energy of a single particle.
But the solar radiation ($1.74 \cdot 10^{17}$ J/s) is the energy of many photons.
A single solar photon has an energy of only a few eV.
By the way: Even the particle energies achieved with our most powerful accelerators (like the LHC) are "only" in the TeV range.
So, we are far off from grand unification.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Do orbiting planets have infinite energy? I know that planets can't have infinite energy, due to the law of conservation of energy.
However, I'm confused because I see a contradiction and it would be great if someone could explain it.
Energy is defined as the capacity to do work. Work is defined as Force x Distance. Force is defined as Mass x Acceleration. Thus, if we accelerate a mass for some distance by using some force, we are doing work, and we must have had energy in order to do that work.
In orbit, planets change direction, which is a change in velocity, which is an acceleration. Planets have mass, and they are moving over a particular distance. Thus, work is being done to move the planets.
In an ideal world, planets continue to orbit forever. Thus, infinite work will be done on the planets as they orbit.
How can infinite work be done (or finite work over an infinite time period, if you'd like to think of it that way) with a finite amount of energy?
Where is the flaw in this argument?
| Your definitions are incorrect. Force is rate of change of momentum and is a vector. More importantly, the work done by a force is not force x distance, it is the force resolved in the direction of the displacement x the magnitude of the displacement. This is more formally known as the scalar product of force and displacement.
In the case of a (circular orbit), the centripetal force supplied by gravity is at right angles to the displacement, so no work is done.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Quantum logic gates (2-inputs 1-output) Is there a quantum logic gate (or any sequence of QLGs) that takes as input two Qubits and outputs one Qubit?
*
*If yes, could you please write it down in the simple example?
*If no, could you explain why?
PS: Sorry for stupid question
| Quantum gates are reversible and hence must have a one to one correspondence between the input and output bits. Another way of thinking about reversibility is that there is a way to run the circuit backwards from the set of outputs and get back all possible inputs. This means that you have the same number of input and output bits.
However, do note that the way circuits are implemented, you will not necessarily use all the output bits. Say, you're adding two bits. Your "answer" is a single bit and the remaining output bits are irrelevant.
All of the above holds for classical reversible computation as well!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/462912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Electric charge of the Higgs field The Higgs field is
\begin{equation}
\Phi = \frac{1}{\sqrt{2}} \left( \begin{array}{cc} \phi_{1} + i\phi_{2} \\
\phi_{3} + i\phi_{4} \end{array} \right) \tag{1}
\end{equation}
with $\phi_{1}$ and $\phi_{2}$ carrying electric charge $+1$ respectively, while $\phi_{3}$ and $\phi_{4}$ are electrically neutral.
Under the entry "Higgs Boson" in Wikipedia, it states:
It (the Higgs field) consists of four components: two neutral ones and two charged component fields. Both of the charged components and one of the neutral fields are Goldstone bosons, which act as the longitudinal third-polarisation components of the massive $W^+$, $W^−$, and $Z$ bosons.
Before interaction between the Higgs field and the gauge bosons, the total electric charge is the total charge of $\phi_{1}$ and $\phi_{2}$, which is $+2$. After the interaction, however, the total electric charge is the sum of the charges of $W^{+}$ and $W^{-}$, which is $0$; the electric charge is not conserved. What is wrong?
Besides, if two components of the Higgs field carry positive electric charge, the whole space (even the whole universe) is electrically positive since the Higgs field permeates the whole space. This is very doubtful and seems not reasonable to me. Is this case true?
|
Besides, if two components of the Higgs field carry positive electric charge, the whole space (even the whole universe) is electrically positive since the Higgs field permeates the whole space. This is very doubtful and seems not reasonable to me. Is this case true?
You are confusing the fields of quantum field theory, with the particles generated by creation operators on these fields.
After all the whole space in QFT is covered by the electron field and the quark fields etc . these fields have no charges.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Symmetry operations on an infinite uniform sheet of charge My book has a section on symmetry operations.
It says, (if the plane of charge is the yz plane) translation symmetry along the y-axis and z-axis implies that the electric field is constant if one translates along the y and z axes respectively. Also, due to rotational symmetry, the field is is perpendicular to the yz plane. I understand this much.
Further, it says, another symmetry can be invoked to show that the field is independent of the x co-ordinate as well (without mentioning the symmetry).
I thought about translating the plane along the x-axis but it would change the charge distribution in space and hence, is not a symmetry operation. What is the symmetry the book mentions?
| Scale symmetry. An infinite plane looks the same no matter how far away from it you are.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I determine the components of a cinematic jump, for vertical and horizontal velocity? I have been tasked with determining the feasibility of The Rock's jump in the movie 'Skyscraper' I am using projectile motion equations to determine it, but have gotten stuck whilst calculating my horizontal and vertical velocity. However, the values I have been getting have not been close to satisfying a vector sum for the total velocity.
My calculations are listed as below:
Distance Run= 26 metres
Time Taken= 3.03 seconds
Horizontal Velocity = distance/time = 36/3.03 = 8.58m/s
I calculated the angle of the jump to be 70 degrees, from a snippet of the movie, and the maximum height of the jump to occur after 0.73 seconds
Velocity= vsinangle + gravity* time
0=v*sin70 -9.8*0.73
7.15=v*sin*70
v= 7.61m/s
The sum of the vectors is 16.19m/s, which does not satisfy a Pythagorean triangle.
I am wondering where I have gone wrong, and what I have to do to solve for the initial velocity.
Thankyou.
| The initial vertical velocity $v_{y0}$ can be found by using $$0 = v = v_{y0} - gt.$$ The initial horizontal velocity is going to be the sum of the velocity due to the jump and the velocity due to the run. You correctly solved for the horizontal velocity due to the run: $v'_{x0} = 8.58 \, \textrm{m/s}$. The horizontal velocity due to the jump $v''_{x0}$ is given by $$\tan\theta = \frac{v_{y0}}{v''_{x0}}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does theoretical physics suggest that gravity is the exchange of gravitons or deformation/bending of spacetime? Throughout my life, I have always been taught that gravity is a simple force, however now I struggle to see that being strictly true.
Hence I wanted to ask what modern theoretical physics suggests about this: is gravity the exchange of the theoretical particle graviton or rather a 'bend' in space due to the presence of matter?
I don't need a concrete answer, but rather which side the modern physics and research is leaning to.
| Both.
General relativity describes gravity as curvature of spacetime, and general relativity is an extremely successful theory. Its correct predictions about gravitational waves, as verified directly by LIGO, are especially severe tests.
Gravity also has to be quantum-mechanical, because all the other forces of nature are quantum-mechanical, and when you try to couple a classical (i.e., non-quantum-mechanical) system to a quantum-mechanical one, it doesn't work. See Carlip and Adelman for a discussion of this.
So we know that gravity has to be described both as curvature of spacetime and as the exchange of gravitons. That's not inherently a contradiction. We do similar things with the other forces. We just haven't been able to make it work for gravity.
Carlip, "Is Quantum Gravity Necessary?," http://arxiv.org/abs/0803.3456
Adelman, "The Necessity of Quantizing Gravity," http://arxiv.org/abs/1510.07195
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 3,
"answer_id": 0
} |
Does a colored filter reflect their color of light? At the moment I'm somewhat confused by the concept of colored filters; common sense states that they allow only their color of light to pass through(i.e. red filter lets red light through), but, if they appear to be a specific color, wouldn't that indicate that they reflect that color?
| You're right, an ideal filter can either absorb or reflect the spectrum it should not pass, and thus for that ideal filter we should not see that pass-band color reflected from the filter. Unfortunately, practical materials are less efficient and typically reflect a little of the spectrum they are supposed to pass (the degree of inefficiency depends on the type and spectrum of filter). The reason you often see a filter that has a reflection color the same as the color it is suppose to pass, is that it absorbs all other colors (so even the little that gets reflected from some internal structures quickly gets absorbed before it makes it out of the material) but the little pass-band color that does reflect from internal structures does make it out of the material for us to see, because that color isn't absorbed much. On top of that, the eye-brain system causes us to consider something clearly colored (we say its "red") even though it may barely have more red than other colors.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How do you measure the chemical potential? It is clear how to measure thermodynamics quantities such as temperature, pressure, energy, particle number and volume. But I have no idea how to measure chemical potential.
Could someone please provide some examples of how one could measure the chemical potential?
| We can't measure chemical potentials but that's OK because the actual value of the potential is not important, what matters is its difference from some other state. So, chemical potentials are compared. For example, in vapor-liquid equilibrium the chemical potential of the liquid component is equal to the chemical potential of the vapor component. If the vapor phase can be treated as an ideal gas we can calculate (not measure) its chemical potential. If it is not an ideal gas we need some other equation of state along with suitable assumptions about the interaction of components in order to do the calculation. It is also possible to calculate chemical potentials by computer simulation.
The general procedure to measure difference in the chemical potential is to compare it to some standard reference. Two standard references are in common use: ideal-gas state and ideal solution. In both cases the chemical potential of the reference state is
$$
\mu_i = \mu_i^0 + RT \ln x_i
$$
where $x_i$ is the mol fraction of the component and $\mu_i^0$ is the chemical potential of the pure component at the same temperature and pressure. One then calculates the departure of the chemical potential from the reference state using auxiliary properties such as activity coefficients, fugacity coefficients, etc.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Can an atom be split and put back together? I had recent came across this question when unintentionally tearing a piece of my journal paper. These atoms' bonds are pulled apart when the paper is torn, but is there a way to put them back together?
Now i know that, depending on the object, the method can vary. According to an article i read Franken-Physics: Atoms split then put back together this can be done. However i wasnt given enough info on what atom they split, as they can vary per object.
Though i am not well educated in the field of quantum physics i understand in some instances this will work. Can anyone clear this up for me regarding what types of materials can undergo this.
Edit: Forgive my ignorance when i said they are obviously split. i do realize the bonds were pulled apart. i just didn't word it as so
| When you tear a piece of paper, I don't think the atoms are being splitted along the break line because it's far more easy and consumes far less energy separating the atoms into the two pieces that having to cut through every individual atom along the break line. If by splitting you mean something like disintegrating an helium atom into two hydrogen atoms, that happens very well in a fission reactor. It doesn't take just some shear force to split an atom, it requires more of a pinpoint precise energy bombardment to get atoms to give in and split.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why the entropy change is not zero in the irreversible adiabatic process? Why the entropy change is not zero in the irreversible adiabatic process?
...while it is defined as the integral of the heat added to the system over its temperature.
| If you have an irreversible adiabatic process between two thermodynamic equilibrium end states of a system, there exists no possible reversible adiabatic process between these same two end states. So to get the entropy change for the irreversible adiabatic process, you need to devise an alternative reversible path between the same two end states, and this reversible path will not be adiabatic. On the reversible path, you will have to add heat to the system in order to transition between the same two end states. So the reversible path will give a positive change in entropy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/463988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
When the voltage is increased does the speed of electrons increase or does the electron density increase? I am just a high school student trying to self study, please excuse me if this question sounds silly to you.
I know that current is a product of the speed of electrons and the electron density.When current is increased it either means that the speed of electrons has increased or it means that the number density of the flowing electrons has increased.
I also know that voltage is directly proportional to current and when voltage increases(without no change in the resistance) the current will also increase.
But my question is, when voltage increases does an increase in the speed of electrons contribute for an increase in current or does an increase in electron density contribute for it.
If it isn't that black and white, then in what proportion will each of the two components increase? Does it randomly increase?
Related question:Say the electron density of a circuit that lights a light bulb increases.When this happens what change will we see in the brightness of the light bulb?I know that when the speed of electrons increase the brightness increases but what will happen when the electron density increases?
| In a conductive material such as a metal, for all practical purposes, current depends only on the speed of the electrons. The electron density does not change because each metal atom has already given up all of its valence electrons; releasing further electrons would require a very large energy input.
In an insulator or semiconductor, the density of charge carriers may increase during electrical breakdown. This occurs in avalanche diodes, neon lights, lightning bolts, and elsewhere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
What is the Kitaev Model and why has it become so popular? I am seeing Kitaev Model everywhere. It feels like the spin-glass model of our time. How the Kitaev model differ from spin-glass and why it can be used everywhere? Looking at equation 1 here suggests it's basically a spin-glass model.
| The paper linked in the original post already answers some of the post's questions.
*
*What is the Kitaev Model?
It's a lattice model where
nearest neighbor spin degrees of freedom interact via a strongly anisotropic nearest-neighbor Ising exchange [...] The Kitaev interactions along neighboring bonds cannot be satisfied simultaneously, giving rise to ‘exchange frustration’ and driving the system into a Quantum spin liquid (QSL) phase.
*
*Why it became so popular?
The Kitaev honeycomb model is arguably the paradigmatic example of QSL, because of its unique combination of being experimentally relevant, exactly solvable and hosting a variety of different interesting gapped and gapless QSL phases, not the least a chiral QSL that harbors nonabelian Ising anyons
*
*How the Kitaev model differ from spin-glass?
Spin glasses exhibit static, frozen ground states and meta-stable states, while in quantum spin liquids
the spins fluctuate strongly even at zero temperature.
Or, as more explicitly put by Perreault in his PhD. dissertation,
These two states are particularly distinguished by
their dynamics since a spin glass has a very long single-spin autocorrelation time while a spin-liquid lacks such frozen spin configurations.
One could then ask which characteristic of the model (or material) allows for QSL. According to Balents' paper (e-print) a possibility is systems where
quantum effects are strong, and there are no obvious energy barriers [between low-temperature configurations].
*
*Why it can be used everywhere?
It probably can't, but it's analytically solvable, which is a rare beast, so people are gonna apply it whenever possible and try to squeeze as much as possible from it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why is air pressure higher in winter than in summer? At the top of a mountain, say Mt Everest, atmospheric pressure is low.
So shouldn't the same thing be true for winter season.
I.e air pressure in winters should be lesser than that in summers.
But it's the opposite.
Can someone please explain why ?
| One of the reasons the air pressure is higher in winter is because the air is colder so the molecules stick closer together building more pressure. Also Mt. Everest is the highest elevation above sea level so it would seem like it would be cold, BUT there are many things that factor into it that make that statement wrong. One of them is that warm air rises because it is less dense the hotter it gets. Hope this helps! This website has helped me a lot on this subject.
Here's the link.
http://www.wxdude.com/singalongcompanion/windandairpressure/index.html
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Cavity optomechanics Hamiltonian In cavity optomechanics the radiation pressure exerted by light moves a mirror in a cavity. Because of that the resonance frequency of the cavity changes due to change in length of the cavity (cavity frequency, $\omega_{cav} = n\pi c/L$, $L$ is the length of the cavity). The Hamiltonian of the system is given by two harmonic oscillators i.e. the cavity mode and the mechanical mode coupled by the optomechanical Hamiltonian [as discussed in this review article, https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.86.1391, given by Eqs. (18)-(20)]:
$H = \hbar \omega_{cav}a^\dagger a + \hbar \Omega_{m}b^\dagger b - \hbar g_0 a^\dagger a (b + b^\dagger)$.
What I don't understand is that, since the cavity length $L$ is changed due to the radiation pressure, the cavity modes are now changed. So, should the modes not be represented by different creation and annihilation operators because the cavity modes are changing dynamically? How can we use the same annihilation (creation) operator '$a$' ('$a^\dagger$') for the optical mode in the Hamiltonian?
| This is beacause, it is based on the assumption that only one optical and mechanical mode interact. Each optical cavity supports in principle an infinite number of modes and mechanical oscillators have more than a single oscillation/vibration mode. The validity of this approach relies on the possibility to tune the laser in such a way, that it populates a single optical mode only Furthermore, scattering of photons to other modes is supposed to be negligible, which holds if the mechanical (motional) sidebands of the driven mode do not overlap with other cavity modes, i.e. if the mechanical mode frequency is smaller than the typical separation of the optical modes.
I hope this will somehow clear your doubt.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is every open circuit a capacitor? I think that even open-ended wires can let AC current flow through them, just with a low capacitance. I also think an antenna could be a capacitor and open ended. Am I thinking correctly?
| Yes. If two non-insulators (in the Universe) are connected to any emf source, current will flow between them regardless of the source frequency (eg. 60Hz or 60GHz). At DC, the leakage current flow depends on the circuit conductance. At AC, the leakage current flow depends on the circuit admittance. Of course, the current may be insignificant (especially if one non-insulator is on Mars and the other is on Earth) and/or irrelevant.
An antenna is designed to radiate with an AC emf. It depends how it is connected, as to whether it acts as a capacitance or inductance at some frequency, if it is not resonating at its fundamental or a harmonic frequency.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 10,
"answer_id": 8
} |
Demonstration of the completness of an orthonormal set of functions I find this concept of completness a little bit dense when it comes to prove this property of some set of orthonormal functions. In one of my classes, my professor proved this for the orthonormal set of functions $\left\{ \sqrt{2/L} \sin( n \pi x/L) \right\}$, but it did not convince me, even though I can't tell if there is something wrong mathematically speaking. He parted from the very condition of completness, i.e.,
$$\sum_n \frac{2}{L}\sin(\frac{n\pi}{L}x')\sin(\frac{n\pi}{L}x)=\delta(x-x')\;\;\;\;\;\;\;\;\;\;\;(1)$$
and he supposed that, being the set a complete one, then one can describe any funcion in terms of such set. He then wrote that
$$\delta(x-x')=\sum_nC_n\frac{2}{L}\sin(\frac{n\pi}{L}x)\;\;\;\;\;\;\;\;\;\;\;(2)$$
Then, taking advantage of the orthogonality of the set, on the interval $0\leq x\leq L$, from the equation (2)
$$\int_{0}^{L}\delta(x-x')\frac{2}{L}\sin\left(\frac{m\pi}{L}x\right)\mathrm{d}x=\sum_nC_m \delta_{m,n}=C_m$$
$$\therefore \frac{2}{L}\sin\left(\frac{m\pi}{L}x'\right)=C_m\;\;\;\;\;\;\;\;\;\;\;(3)$$
and replacing (3) into (2), then one gets the condition for completeness in (1). Even if this is correct, I can't tell why. Also, I would like to know how the proof for completness would be carried out taking the same condition but in Dirac's notation, that is, $\sum_n |\phi_n><\phi_n|=1$, but I have no idea how to proceed.
| Sorry, I really don't under what the question is in the first part.
For the second part, let
$$|v>=\sum_n c_n |\phi_n> \tag{1} $$
where $|v>$ is a vector (ket) in a finite dimensional space with an orthonormal basis $|\phi_n>$ and
$$<v|=\sum_n c^*_n <\phi_n|$$
is an adjoint or dual vector (bra).
Orthonormality of the basis for the kets and bras gives $<\phi_m|\phi_n>=\delta_{m,n}$.
From the decomposition of $v$ in equation $(1)$,
$$c_n=<\phi_n|v>$$.
Noting that $c_n$ is a scalar, substitute the value of $c_n$ back into equation $(1)$ and re-arrange
$$|v>=\sum_n c_n |\phi_n>=\sum_n <\phi_n|v>|\phi_n>=\sum_n |\phi_n><\phi_n|v>.$$
Breaking apart $<\phi_n|v>=<\phi_n|1|v>$ (you can always insert a $1$ between a bra and a ket) yields
$$|v>=\sum_n {|\phi_n><\phi_n|}1|v>=\sum_n {|\phi_n><\phi_n}|v>$$
which implies
$$\sum_n |\phi_n><\phi_n|=1.$$
I'm using the $1$ (or the identity) in $1|v>$ for clarity - it's typically not necessary.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/464948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
plane vs diverging waves michelson itnerferometer I'm wondering if there's a difference in the interference pattern achieved in the michelson interferometer if the source of light is sending plane waves vs sending diverging waves (for example by putting a diverging lens between the source and the first plate, the one which divides the light).
| Plane waves are a theoretical construct as every aperture produces divergence, so you need an infinite amount of empty space to keep a plane wave plane.
As such every Michelson interferometer deals with diverging waves. These interferometers will take a divergent beam and collimate it before reciprocating the process so that sharp interference can still be seen.
A near exception is a fibre michelson interferometer where the waveguiding properties of the fibre act as collimation, though lensing is still needed to couple to the fibre.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does a car's steering wheel get lighter with increasing speed I've noticed it is difficult to turn the wheels of a car when the car is stationary, especially cars without power steering, which is why the power steering was invented. However, I've noticed it becomes feather light when traveling at speed (some models even stiffen the steering wheel electronically at speed). So, why does a car's steering wheel get lighter with increasing speed?
| The work to turn the wheel is roughly proportional to how much you turn the wheel, and inversely proportional to the distance the car traveled. You feel less resistance at higher speed because the car moved farther for the same amount of steering wheel turn. This is because when the wheels on the ground are rolling, the distortion of the rubber, and the friction, that occurs when you turn the steering wheel is less than it is when those wheels are stationary to the ground.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 2
} |
Does total $\hat{S}^2$ always commute with total $\hat{S}_z$ even for interacting spins? I was given the following operator $\hat{f}$ describing the interaction of two spin-$\frac12$ particles:
$$\hat{f}=a+b{\hat{\bf S}_1}\cdot{\hat{\bf S}_2}.$$
I was told that I can prove that $\hat{f}$ does commute with the total spin operators $\hat{S}^2$ and $\hat{S}_z$ because of the commutation relation $[\hat{S}^2,\hat{S}_z]=0$. Why is this true, and is it necessarily true regardless of the interaction between spins? From what I have learnt about addition of angular momentum, $[\hat{J}^2,\hat{J}_z]=0$ is true for non-interacting particles, but I am concerned about the spin interactions and possible coupling between the two particles.
Additionally, does ${\hat{\bf S}_1}$ necessarily commute with ${\hat{\bf S}_2}$ (for calculating $\hat{S}^2$ as a dot product)? I know that for non-interacting particles, the two particles have different spinor spaces, but once again I am uncertain for particles with interacting spins.
| The form of the intersction is invariant under spin rotation, so we do expect commutativity with total spin operator.
Algebraically, $S_1 \cdot S_2 = \frac{1}{2} \left((S_1+S_2)^2 - S_1^2 - S_2^2\right)$, so we see that $S^2$ and $S_z$ commutes with all the terms. We use the fact that $[S^2, S_z] = [S^2, S_y] = [S^2, S_z] = 0$ as you said.
You can also view the interaction as $S_{x,1}S_{x,2} + S_{y,1}S_{y,2} + S_{z,1}S_{z,2} = \frac{1}{2}\left(S_{+,1}S_{-,2} + S_{-,1}S_{+,2}\right) + S_{z,1}S_{z,2}$ where $S_{+,n} = S_{x,n} + iS_{y,n}$ is the spin ladder operator and so on. We see that the interaction operator preserves the total spin and raises or lowers the individual spin polarization by one unit, but the total spin polarization $S_z$ is preserved.
The commutativity of spin operators is determined by the angular momentum algebra. This is independent of the presence of interaction. So to answer your last question, yes, $S_1$ commutes with $S_2$. The effect of interaction on the system is that it changes which quantum numbers are good (the set of conserved quantities changes). For example, in the isotropic Heisenberg interaction above, we saw that total spin and its polarization are conserved, but for dipolar interaction, this is not true.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Virtual images - Work of a brain or work of a lens? I am just a high school student trying to self study, so please excuse me if this question sounds silly to you.
Many people tell me that virtual images are formed when two rays that are diverging appear to come from a point, therefore our brain thinks that it is coming from an object even though there is no object there.
However I think that a virtual image is formed because two diverging rays converge at a point when made to go through the crystalline lens in our eye to form a real image on our retina.I think that our brain has nothing to do with the formation of virtual images.
What is actually going on here?
| Our brain just gets signals from the retina. The visual cortex interprets these signals, I am guessing, using an image recognition and matching. If a match is found then the owner of cortex believes that that is the reality behind the original signals. All reality is virtual...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Work when there is more than 1 force I know that for an object with an applied force, the work done is
$$W = Fd \cos \theta.$$
I was wondering what would happen when there is another force (e.g. friction)? Is it better to say that the work done for a general case is
$$W = F_{net} d \cos\theta.$$
| First, you must recognize that your formula only works for constant forces and motion in one dimension.
So, if all of those conditions are met, then just add up the work done by each force.
$$W=F_1d\cos\theta_1+F_2d\cos\theta_2$$
If both forces act in the same direction ($\theta_1=\theta_2=\theta$), then you can combine the terms
$$W=(F_1+F_2)d\cos\theta$$
In general, work by a force is given by
$$W=\int\mathbf F\cdot\text d\mathbf x$$
so for multiple forces we just do
$$W=\sum_i\int\mathbf F_i\cdot\text d\mathbf x=\int\left (\sum_i\mathbf F_i\right)\cdot\text d\mathbf x$$
I am not going to say $\sum_i\mathbf F_i$ is the net force, because we can calculate the work done by a subset of the forces. It doesn't just have to be one or all.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Variance of a hermitian operator Take an hermitian operator $O$ such that $O|\psi\rangle = x|\psi\rangle$. The variance of an operator $O$ is defined as
$$ (\Delta O)^2 = \langle{O^2}\rangle - \langle{O}\rangle^2.$$
Let's consider the first term, I would write it as
$$ \langle{O^2}\rangle = \langle\psi|O O|\psi\rangle = x\langle\psi|O|\psi\rangle = x^2\langle\psi|\psi\rangle = x^2.$$
But then for the second term I get the same result
$$\langle{O}\rangle^2 = \langle\psi| O|\psi\rangle^2 = (x\langle\psi|\psi\rangle)^2 = x^2.$$
Therefore I end up with $(\Delta O)^2 = 0$ which is obviously wrong. What's the problem here?
The only idea I have is that $O^2|\psi\rangle \neq O(O|\psi\rangle)$, but i cannot understand why...
What am I doing wrong?
| As you have written things, the variance is indeed $0$ because $\vert\psi\rangle$ is an eigenstate of $O$: thankfully this is so as it means the outcome with eigenvalue $x$ is not uncertain and we can use the eigenvalue $x$ to label the state.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/465923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why do prism split light at angle instead of curving it? I assume that when light goes through matter, it doesnt really slow down, but the waveform is pushed back due to some resonance with the atoms.
EDIT: Interference is probably a better word than resonance here
I also assume that the above effect is responsible for the refraction index of materials.
But according to these assumptions, light rays should curve more as they go deeper through matter shouldnt they ? In other words that effect should be cumulative with the thickness of matter the light does through?
However light doesnt bend at different angles if it goes through thicker glass. So where did I go wrong?
|
I also assume that the above effect is responsible for the refraction index of materials.
Yes. The index of refraction is a quantification of how the speed of light changes due to the medium it is propagating through. So whatever mechanism slows the light down is what is responsible for the index of refraction.
But according to these assumptions, light rays should curve more as they go deeper through matter shouldnt they ? In other words that effect should be cumulative with the thickness of matter the light does through ?
No. There is nothing that would break the symmetry to cause the light to be bending one way or the other once it was in the uniform second medium.
why do prism split light at angle instead of curving it?
The index of refraction has a wavelength dependency, so if we send in something like white light that is not monochromatic, the superposition gets bent at different angles as it crosses the interfaces of the different media. Within each medium the light travels in a straight line.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Do gluons interact with each other by the strong foce? I learned that strong force between quarks are mediated by gluons. What does this say about interactions between gluons? Do they interact with each other by the strong force?
| If you take a look at the QCD Lagrangian describing the strong force
\begin{align}
\mathcal{L}_{QCD}= \overline{q}(i D_\mu \gamma^\mu - m) q - \frac{1}{4} G_{\mu \nu}^a G_a^{\mu \nu}\, ,
\end{align}
you can see in the term $\frac{1}{4} G_{\mu \nu}^a G_a^{\mu \nu}$ we have a contraction of the gluon field strength tensor
\begin{align}
G^a_{\mu \nu} = \partial_\mu A^a_\nu - \partial_\nu A_\mu^a + g f^{abc}A^b_\mu A^c_\nu\, .
\end{align}
For the non-abelian symmetry of QCD, $SU(3)_\text{color}$, the structure constants $f^{abc}$ are non-zero, other than in electromagnetism, where the symmetry group is an abelian $U(1)_\text{em}$.
This means that if we carry out the contraction, we end up with terms containing 3 or even 4 gluon fields. These terms lead to Feynman diagrams where 3 or 4 gluons meet at one vertex, i.e. they interact with each other. So the short answer to this is: yes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Seebeck coefficient of metals In some metals such as Platinum Seebeck coefficient is taking for 0 conditionally.
https://en.wikipedia.org/wiki/Seebeck_coefficient#Seebeck_coefficients_for_some_common_materials
Or graphite.
How it is possible to explain it physically? Why exactly it is "0"?
The Seebeck coefficient of pristine graphite is nearly zero, because the
number of electrons and the number of holes are almost equal.
https://www.jim.or.jp/journal/e/pdf3/50/07/1607.pdf
I wanted to receive an explanation in terms of a charge carriers behavior. Some sources claim that in materials with zero Seebeck coefficient both electrons and holes are equally mobile and when you heat one end of the metal rod an equal amount of electrons and holes flow to the cold end. What I wish to know whether electrons and holes recombine at the cold end. If yes, does it mean such thing as carriers recombination can happen in metals?
| This is explained in the very same article that you linked. The absolute Seebeck coefficient is simply difficult to measure because the voltage probes with which you measure also experience a temperature gradient. Hence, the measured voltage is influenced by both the Seebeck coefficient of the sample material and the Seebeck coefficient of the electrode material. By defining one reference material, platinum in your case, the difference of the Seebeck coefficients is used as an easy-to-measure quantity. You just have to use the same setup to measure platinum and all the materials that you are interested in. If you want to obtain the absolute Seebeck coefficients you just need to determine the Seebeck coefficient of platinum, and not for all the materials that you measured.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are Electrons in a Circuit Subject to Newton's Third Law? Consider a simple electrical circuit made up of a battery, an incandescent bulb, and wire. The battery and bulb are equal in mass and are on opposite sides of a circle made up by the wire. Lastly, the circuit is operating and floating freely in microgravity.
Since an electromotive force propels objects with mass (electrons) around the circuit, can we expect the circuit, given enough battery life, to eventually rotate in the opposite direction of the electrons due to Newton’s third law of motion?
| Yes, conservation of angular momentum applies and the system will not rotate. I assume that the bulb radiates isotropically and also that any other radiation effects are isotropic. The electrons cannot be set into motion without a reaction force on the battery.
However the magnetic field will polarise electron spins, free or located in ferromagnetic atoms. This will align their spin angular momentum possibly causing an undoubtedly tiny opposite angular momentum of the setup.
More importantly any orbital angular momentum of the electrons will be transferred (back) to the lattice. There should be a transient effect on the angular orientation when the current is changing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Peskin & Schroeder eq. 9.26 and functional methods I have been reading chapter 9 in Peskin & Schroeder's QFT book and has been stuck in transition from equation 9.26 to 9.27. Equation 9.26 reads:
$$\frac{1}{V^2} \Sigma_{m,l} \exp{[-i(k_m.x_1+k_l .x_2)]}(\prod_{k_{n}>0}\int{d\ Re\phi_{n}\ d\ Im\ \phi_{n}}) * (Re\ \phi_{m}+iIm\ \phi_{m})(Re \ \phi_{l}+iIm\ \phi_{l}) * \exp[\frac{-i}{V}\Sigma_{k^{0}_{n}>0}(m^2-k^2_{n})[(Re\ \phi_{n})^2+(Im\ \phi_{n})^2]] \tag{9.26}$$
and with $k_l=+k_m$, equation vanishes while with $k_l = -k_m$ the equation becomes,
$$\frac{1}{V^2}\Sigma_{m} \exp[-ik_m\ .(x_1-x_2)](\prod_{k^{0}_{n>0}}\frac{-i\pi V}{m^2-k^{2}_n})\ \frac{-iV}{m^2-k^2_{m}-i\epsilon}$$
I can work out the term in parenthesis using Gaussian integrals but I can't seem to wrap my head around how we get the term
$$\frac{-iV}{m^2-k^2_{m}-i\epsilon}$$
Can anybody point me in the right direction please? Apparently $(Re\ \phi_{m})^2$ adds up to the $(Im\ \phi_m)^2$ when $k_l = - k_m$ and we get this contribution. I don't understand how.
| Consider the case when $k_l=-k_m$. Peskin and Shroeder equation 9.26 would then read
$$ \frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0}\int d\mathrm{Re}\phi_n\ d\mathrm{Im}\phi_n \right)\times (\mathrm{Re}\phi_m\ \mathrm{Re}\phi_{-m} +i\ \mathrm{Re}\phi_m\ \mathrm{Im}\phi_{-m}+i\ \mathrm{Im}\phi_m\ \mathrm{Re}\phi_{-m}-\mathrm{Im}\phi_m\ \mathrm{Im}\phi_{-m}) \times \exp\left[-\frac{i}{V}\sum_{k_n^0>0}(m^2-k_n^2)[(\mathrm{Re}\phi_n)^2+(\mathrm{Im}\phi_n)^2] \right] .$$
In the first term, using $\mathrm{Re}\phi_{-m}=\mathrm{Re}\phi_m^* = \mathrm{Re}\phi_m$, we get
\begin{align}
&\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0}\int d\mathrm{Re}\phi_n\ d\mathrm{Im}\phi_n \right)\times (\mathrm{Re}\phi_m)^2 \times e^{-\frac{i}{V}\sum_{k_n^0>0}(m^2-k_n^2)[(\mathrm{Re}\phi_n)^2+(\mathrm{Im}\phi_n)^2]}\\
&=\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0,n\neq m}\int d\mathrm{Re}\phi_n\ e^{-\frac{i}{V}(m^2-k_n^2)(\mathrm{Re}\phi_n)^2} \right) \times\left(\int_{k_m^0>0} d\mathrm{Re}\phi_m\ (\mathrm{Re}\phi_m)^2\ e^{-\frac{i}{V}(m^2-k_m^2)(\mathrm{Re}\phi_m)^2}\right) \left( \prod_{k_n^0>0}\int d\mathrm{Im}\phi_n\ e^{-\frac{i}{V}(m^2-k_n^2)(\mathrm{Im}\phi_n)^2} \right).
\end{align}
Using the identities $ \int d\xi\ e^{-b\xi^2} = \sqrt{\frac{\pi}{b}} $ and $ \int d\xi\ \xi^2e^{-b\xi^2} = \frac{1}{2b}\sqrt{\frac{\pi}{b}} $, the equation simplifies as
\begin{align}
&\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0,n\neq m}\sqrt{\frac{-i\pi V}{m^2-k_n^2}} \right) \left(\frac{1}{2}\frac{-iV}{m^2-k_m^2-i\epsilon}\sqrt{\frac{-i\pi V}{m^2-k_m^2}}\right) \left( \prod_{k_n^0>0}\sqrt{\frac{-i\pi V}{m^2-k_n^2}} \right)\\
&=\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0}\frac{-i\pi V}{m^2-k_n^2} \right)\frac{1}{2}\frac{-iV}{m^2-k_m^2-i\epsilon} .
\end{align}
The second and the third terms are of the form $\int d\xi\ \xi\ e^{-b\xi^2} =0$. Finally for the fourth term we use $\mathrm{Im}\phi_{-m} = \mathrm{Im}\phi_m^* = -\mathrm{Im}\phi_m $ and then it gives the same contribution as the first term which cancels the factor of $1/2$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is 1 liter always equal to 1 cubic decimeter, independently of temperature, pressure, etc? I recently found this conversion table for the unit conversion $\rm mmol/m^3 \ \leftrightarrow\ \rm mmol/L$ (millimoles per cubic meter to millimoles per liter)
My physics is very rusty, but just to be sure, is it true that a liter of liquid always corresponds to a particular volume? (i.e.: Doesn't change with regards to temperature, pressure, etc?)
|
is it true that a liter of liquid always corresponds to a particular volume?
Yes, this is correct. The relationship
$$
1\:\mathrm{L} = 10^{-3} \:\mathrm{m}^3
\tag 1
$$
(i.e. one cubic meter is a thousand liters) is universal and it does not depend on anything. The liter is a unit of volume - by definition, it's a cubic decimeter.
To convert between $\rm mmol/L$ and $\rm mmol/m^3$, simply insert the relationship above:
\begin{align}
\rm
1\: mmol/L
& =
\rm 1\:mmol / (10^{-3}\:m^3)
\\ & =
\rm 10^{3}\:mmol / m^3,
\end{align}
and vice versa.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Metric for a Collapsing Disk and FLRW I have to obtain a time-dependent metric for a disk which satisfies a simple differential equation but I am stuck with making sure that the physics is correct.
To be explicit, I'll first describe a shrinking disk:
Suppose that we are working in a $4D$ Minkowski spacetime $(\mathcal{M},\eta_{ab})$ where the metric has signature $\text{sign}(\eta)=\text{diag}(-,+,+,+)$ and $\mathcal{M}=\mathbb{R}_t\times\mathbb{R}_{\Sigma}^3$. Consider embedding a compact disk $\Omega_t\hookrightarrow\mathbb{R}^3_{\Sigma}$ which is defined at each time slice by $$\Omega_t=\left\{(x,y)\in\mathbb{R}^2: x(t)^2+y(t)^2\le R(t)^2 \right\}$$ where the radius of the disk solves the differential equation $$\frac{\text{d}R(t)}{\text{d}t} = -\gamma R(t)$$
with the initial and final conditions $R(t_i)=R_i$, $R(t_f)=R_f$, and $R_f<R_i$. The disk starts at some initial radius $(t_i,R_i)$ and shrinks until it reaches the smaller radius $(t_f,R_f)$. The spacetime diagram looks like a cone.
I can solve the differential equation to find that the radius has the explicit form $$R(t) = R_i \exp(-\gamma (t-t_i)) = R_i\exp\left(\frac{t\,- t_i}{t_f-t_i}\log\left[\frac{R_f}{R_i}\right]\right).$$
I'm now stuck trying to construct a metric for the disk. My initial guess was that it should look like FLRW since it looks like a shrinking universe; $$\text{d}s^2 \overset{?}{=} -\text{d}t^2 + \exp(-2\gamma(t-t_i))\left(\text{d}r^2+ R_i^2\text{d}\theta^2\right)$$
but I am unsure how to derive this metric, assuming that it is in the correct form. Could I get some help with deriving a metric for this disk model? Where is a good starting point?
| The metric
$$ds^2=-dt^2+dx^2+dy^2$$
with:
$$x^2+y^2=R^2\,\quad \Rightarrow \quad y=\sqrt{R(t)^2-x^2}$$
$$dy=-{\frac {x{\it dx}}{\sqrt { \left( R \left( t \right) \right) ^{2}-{x
}^{2}}}}+{\frac {R \left( t \right) \left( {\frac {d}{dt}}R \left( t
\right) \right) {\it dt}}{\sqrt { \left( R \left( t \right)
\right) ^{2}-{x}^{2}}}}
$$
Goto
$$ds^2= \left( -1+{\frac { \left( R \left( t \right) \right) ^{2} \left( {
\frac {d}{dt}}R \left( t \right) \right) ^{2}}{ \left( R \left( t
\right) \right) ^{2}-{x}^{2}}} \right) {{\it dt}}^{2}-2\,{\frac {{
\it dx}\,R \left( t \right) \left( {\frac {d}{dt}}R \left( t \right)
\right) x{\it dt}}{ \left( R \left( t \right) \right) ^{2}-{x}^{2}}}
+{{\it dx}}^{2} \left( 1+{\frac {{x}^{2}}{ \left( R \left( t \right)
\right) ^{2}-{x}^{2}}} \right)
$$
$$g=
\left[ \begin {array}{cc} -1+{\frac { \left( R \left( t \right)
\right) ^{2} \left( {\frac {d}{dt}}R \left( t \right) \right) ^{2}}{
\left( R \left( t \right) \right) ^{2}-{x}^{2}}}&-{\frac {R \left( t
\right) \left( {\frac {d}{dt}}R \left( t \right) \right) x}{
\left( R \left( t \right) \right) ^{2}-{x}^{2}}}
\\ -{\frac {R \left( t \right) \left( {\frac {d}{dt
}}R \left( t \right) \right) x}{ \left( R \left( t \right) \right) ^
{2}-{x}^{2}}}&1+{\frac {{x}^{2}}{ \left( R \left( t \right) \right) ^
{2}-{x}^{2}}}\end {array} \right]$$
$$\det(g)={\frac { \left( R \left( t \right) \right) ^{2} \left( -1+ \left( {
\frac {d}{dt}}R \left( t \right) \right) ^{2} \right) }{ \left( R
\left( t \right) \right) ^{2}-{x}^{2}}}
$$
Edit
Metrix with polar coordinate
with $x=R(t)\,\cos(\phi)\quad, y=R(t)\,\sin(\phi)$ this satisfy the constraint equation $x^2+y^2=R(t)^2$
you get the metric:
$$ds^2=-dt^2+dx^2+dy^2=-dt^2+\left( {\frac {d}{dt}}R \left( t \right) \right) ^{2}{{\it dt}}^{2}+
\left( R \left( t \right) \right) ^{2}{d\phi }^{2}=\left( -1+ \left( {\frac {d}{dt}}R \left( t \right) \right) ^{2}
\right) {{\it dt}}^{2}+ \left( R \left( t \right) \right) ^{2}{d
\phi }^{2}
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/466887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Maxwell Tensor Identity In Schawrtz, Page 116, formula 8.23, he seems to suggest that the square of the Maxwell tensor can be expanded out as follows:
$$-\frac{1}{4}F_{\mu \nu}^{2}=\frac{1}{2}A_{\mu}\square A_{\mu}-\frac{1}{2}A_{\mu}\partial_{\mu}\partial_{\nu}A_{\nu}$$
where:
$$F_{\mu\nu}=\partial_{\mu} A_{\nu} - \partial_{\nu}A_{\mu}$$
For the life of me, I can't seem to derive this. I get close, but always with an extra unwanted term, or two.
Anyone have a hint on the best way to proceed?
| Hint: Try introducing an integral to the expression so it becomes $$-\frac{1}{4}\int F_{\mu\nu}F^{\mu\nu}\text{d}^d x$$
and take the total derivative terms to vanish at infinity. A much more careful argument can be made here in the presence of boundaries, but this should get you started.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/467007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why parity required symmetry? I'm studying parity for the first time but there is something I don't understand.
I read that a system conserves parity if every experiment is the same in a mirror that is also $180^{\circ}$ flipped.
When I look myself doing something in a mirror everything is the same and so I'm a system that conserves parity.
But I also read that to conserve a parity a system must be symmetric (or antisymmetric) and I'm not, so how it's possible?
| rubbish You are not a system that conserves parity. Maybe you need a more rigorous definition of a parity transform: https://en.wikipedia.org/wiki/Parity_(physics) /rubbish
Parity transform means that you mirror the system in three perpendicular planes. So lets see. First we mirroring: your left and right hands swap places, same goes for eyes, ears etc. Second mirroring: your nose goes to the back of your head etc. Third mirroring: your head goes to the follow, whilst your feet go up into the air. Now at that point you clearly do not look like you did initially, so you are odd under parity transform.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/467273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Poisson's ratio in analytical beam deflection equations After looking at some of the analytical expressions for analyzing beams, I noticed that none of the equations depend on the material's Poisson ratio. Some analytical expressions can be found in https://www.linsgroup.com/MECHANICAL_DESIGN/Beam/beam_formula.htm
I believe these equations are derived from the Euler Bernoulli equations and not from 3-D Linear elasticity, and Poisson's ratio isn't included in Euler Bernoulli. What underlying assumption in Euler Bernoulli allows for the exclusion of the Poisson's ratio?
If I want to model a beam under the Euler Bernoulli assumptions in a 3-D linear elasticity code, where the Poisson's ratio is required as an input, how can this be done? Essentially, I am trying to perform a one to one comparison of the linear elasticity code I am using with the analytical beam expressions. To do this, I need to figure out what value to input for the Poisson's ratio in the code. When I model the beam with the code and vary the Poisson's ratio, the deflection changes as a function of the poisson's ratio. However, the analytical beam expressions do not depend on the Poisson's ratio, so it is not clear to me how this comparison can be performed.
| One of the assumptions of Euler-Bernouli theory is that the cross sectional plane remains rigid.
Consider a cantilever beam (slender) with rectangular cross section. The above assumption implies that breadth and height can NOT change even though there is extension in the longitudinal direction (unrealistic but that is the implication of the assumption). Thus poison ratio does not come into the picture. Section5 5.2 of the text book Structural analysis by O.A.Bauchau and J.I.Craig has good illustration of the implications of Euler-Bernouli assumptions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/467394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How does the green function for the wave equation in three dimensions preserve the ordering of noises between a speaker and a listener I was provided with the following equation in class for the Green's function of a three dimensional wave equation:
However, I am confused as to how this form of the Greens function preserves the ordering of noises between a speaker and a listener. Any explanation would be much appreciated.
| This Green’s function is for a single sound pulse emitted from $r=0$ at $t=0$. It doesn’t explicitly describe a sequence of sounds.
However, from it you can see that the sound arrives at distance $r$ at a time $r/c$ after it is emitted. So noises emitted later are going to arrive later.
For a better understanding of this, consider the more general Green’s function $G(\mathbf{r},t;\mathbf{r}’,t’)$ for a pulse emitted from $\mathbf{r}’$ at $t’$, and think about convolving it with a sound source that emits sound over time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/467643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Principle of friction force When two equal and opposite forces act on a body at rest or in motion we say it is rest or in its state of motion. But when a body kept on a table starts moving uniformly due to a force applied, it overcomes force of friction.Does it really overcome the force of friction or is just equal to it. ie for force F = ū.N if it's so then why would it start to move,why not stationary as the two forces balanced?
| It's just that the scenario "when a body kept on a table starts moving uniformly due to a force applied, it overcomes force of friction" is a bit nastier than the simple sentence would lead us to believe. It makes us think of an object which (i) immediately starts moving, and (ii) keeps a constant speed. Well, it would have to accelerate first. So the event that most closely resembles the scenario expressed in the sentence, but yet physically sound, is the one @harshit54 alluded to: The force would have to be somewhat larger than the maximum static friction to initiate motion. Then the object will accelerate to the speed in question (during which time the force may be reduced, since the force of friction will reduce due to typical kinetic friction coefficients being smaller, but that's not crucial, I think). Then once the speed is at that value, the force would need to equal the force of kinetic friction from then on.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Propagation of light Suppose we are able to make any place free from the magnetic and electric field; then we turn on a light source at any place in that region (where there is no electric and magnetic field). Does light propagate in that region?
| There is the scientific method of limit value analysis. Let’s see what happens to the light in extreme situations.
The local speed of light is influenced by the local gravitational potential. About an influence of electric or magnetic fields on the speed of electromagnetic radiation is nothing published. So even the theoretical absence of any external electric or magnetic fields shouldn’t change the propagation of light in this region.
Nearby strong magnetic fields no deflection of light was observed. Means, that, if on switch on and off of strong magnets, no change in the image behind the magnet is observed.
Now the question arises since strong electric or magnetic fields neither influence the speed of light nor the deflection of light, at which lower limit of these fields does the propagation of light stop? For me, an abrupt change of the behavior of electromagnetic radiation (a discontinuity of the value of the speed of light) doesn’t make any sense.
The conclusion is, that even in the theoretical absence of any electric or magnetic fields except the EM fields of the propagation light, the light will propagate through such a region.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
In a pilot-wave model, is knowing the position of the particle sufficient for predicting its behavior? Suppose that we somehow exactly know the position of an electron before hitting the double-slit structure (for example we know it's 20cm away from the structure and it's closer to the left slit). In fact we are no longer ignorant of the electron's position.
Now, is it possible to predict which slit the electron will go through, or we also need to know the wavefunction (pilot wave) to do so? In other words, given only the position and knowing that the electron is closer to the left slit, are there still possible guiding waves that will lead it to the right slit, or we can be sure it goes throw the left one?
| Yes.
To expand: (I am going to use Bohm's pilot wave theory for concreteness).
In normal Newtonian physics, predicting the path of a billiard ball in say a gravitational field requires knowing its position and velocity.
In Bohm's pilot wave theory, the velocity of the particle is defined by the Bohmian velocity formula, which is a function of the usual wave function and is dependent only on the particle position and is completely deterministic. So we only need to know the initial position of the particle to determine its future behavior.
The trick is we can't know the particle's position, it's hidden. All we know is the distribution of positions at the start of an experiment, individual runs only have one particle in a well-defined position.
One consequence of this is that in a two slit experiment, an electron starting on the left stays on the left, and cannot cross to the right, (which is a general result for Bohmian mechanics) as in this image:
The lines in the image are particle tracks if you did the experiment lots of times.
Also see the accepted answer here:
How does the de Broglie-Bohm picture explain the double-slit experiment with single particles?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Flux received by a negative charge Consider two charges $+q$ and $-Q$ placed at a distance, note charge $q$ and $Q$ are different in terms of magnitude.
My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?
As according to gauss law
Source of image: Britannica
The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.
| The number of flux lines of each charge is proportional to its own charge.
The other charge has nothing to do with that.
See this image with two unequal charges. The right negative charge ($-3Q$)
has three times the size of the left positive charge ($+Q$):
(image from Chegg Study: physics questions and answers)
This is in accordance with Gauss's law for the electric field:
*
*Draw a closed surface around the left charge ($+Q$) only.
There are 6 field lines coming out of this surface
*Draw a closed surface around the right charge ($-3Q$) only.
There are 18 field lines going into this surface.
*Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).
There are 12 field lines going into this big surface.
*Draw a closed surface which does not enclose any of the charges.
There are $n$ field lines going into and the same $n$ field lines
coming out of this surface, thus giving a sum of zero.
In all cases the number of field lines (i.e. the electric flux)
through the closed surface is proportional
to the charge inside the surface.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Self-interaction in spin-orbit coupling? In spin orbit coupling, in an atom motion of electrons about nucleus generates magnetic field and we consider this field to interact with magnetic moment of electron. It sound strange as in electrostatics a field is generate by a charge particle but this field does not interact with its origin charge. Doesn't this contradict ?
| It's in the name $\textbf{spin-orbit}$ coupling. The source of the magnetic field is the orbital angular momentum, whose origin was discussed by Andrew's answer. And the dipole that is interacting with it is due to the spin. So the field does not interact with the origin of the dipole as the source of the two are different.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Difference between voltage, electrical potential and potential difference I am having hard time to visualize these two concepts in my mind seriously.
First of this confusion came from two parallel plates that was connected to a power supply, charged then disconnected from power supply and then separated from each other, strangely potential difference increased but why? I have learn that the electrical potential is$$ V= k \frac{q}{d} $$and when the distance increases the potential of a point must drop but why when we are talking for potential difference it is increasing, doesn't every point between these plates feel less stress as the plates moves apart and doesn't this mean potential is dropping and so the potential difference as well?
| In a parallel plate capacitor, the electric field is homogeneous between the plates (we neglect boundary effects here) and given by
$$ E = \frac{\sigma}{\epsilon\epsilon_0},$$
where $\epsilon$ is the relative dielectric constant of the medium between the plates, and $\sigma$ is the charge density on one plate. The direction of the electric field vector is normal to the surface of the plates. The formula follows from a direct application of Gauss' law of electrostatics.
The potential difference between the plates is the line integral of $E$ along a line connecting the two plates, giving
$$ \Delta V = \frac{\sigma}{\epsilon\epsilon_0}\times d.\quad (1)$$
Here, $d$ is the separation of the two plates.
If you charge the plates by connecting them to a voltage source supplying the voltage $\Delta V$, then you get the surface charge density of magnitude
$$ \sigma = \frac{\epsilon\epsilon_0}{d}\Delta V$$
on each plate (the sign of the charge density will be different on the two plates).
Disconnecting the plates from the voltage source leaves the charges on the plates unchanged. Likewise, the charge density $\sigma$ does not change, if you further increase the separation $d$ of the plates. The voltage will increase with $d$ according to eq. (1).
In fact, the quantity $\epsilon\epsilon_0/d$ is called the capacitance per unit area of the parallel plate system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/468938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Does a vacuum solution to the Einstein equation imply flat spacetime? I have read that a solution to the vacuum Einstein equation has a vanishing Einstein tensor, and therefore a vanishing stress-energy tensor. This means that there is no matter to generate spacetime curvature. If there is no matter to generate spacetime curvature, I assumed (apparently naively) that spacetime is flat.
But the Schwarzchild metric is a solution to the vacuum Einstein equation and is obviously not flat. For instance, one can calculate the deflection of light and Shapiro time delay which are obviously curvature effects. Moreover, the Schwarzchild metric describes spacetime outside a spherically symmetric mass $M$ and so there is necessarily some mass present.
How does one reconcile the fact that the Schwarzchild metric depends explicitly on a mass $M$ yet is also a solution to the vacuum equations? More generally, how can solutions to the vacuum Einstein equations have curvature if "mass tells spacetime how to curve"?
| No, a vacuum solution does not imply flat spacetime.
It is possible, as in a Schwarzschild metric, to have a zero Einstein tensor but a nonzero Riemann tensor. The Riemann tensor is the most detailed indicator of curvature, with 20 independent components (out of 256 nominal components) at each point in spacetime. The Einstein tensor is more like a curvature average, because each of its components is a sum over multiple components of the Riemann tensor. It has only 10 independent components (out of 16 nominal components) at each point.
“Mass tells spacetime how to curve” is oversimplified. “The density and flow of energy and momentum tells spacetime how to curve on average” is more accurate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Lattice gauge theory under a Lorentz transformation Taking a grid of 'evenly-spaced' space-time points. e.g. at integer values of (x,y,z,t). Now do a Lorentz boost on this grid. We end up with a grid of points which are much closer together.
It is bothering me that the grid of points only looks evenly distributed in certain frames (related by spacial rotations) but looks different under Lorentz boosts. The points that looked like neighbours no longer look like neighbours.
When working out results in lattice QCD, does this mean that the results you get don't exhibit Lorentz symmetry. And how can we be sure in the limit the Lorentz symmetry is conserved?
Edit: I just came across this animation from John Baez on time crystals.
Which seems to suggest that for certain lattices like a triangular lattice, the Lorentz transformation can preserve the Lattice form. Although looking at the middle point, it's neighbours are continually changing. And Lattice QCD is based on calculations of nearest neighbours (discrete derivative). I guess one could reformulate the calculations to all neighbours a unit Minkowski-distance away. But this would increase the number of neighbours a lot for large volumes of space-time. (and does this obey the concept of `locality'??)
| Lattice theories are not invariant under Lorentz transformations. They are usually defined in Euclidean signature (imaginary time) so the symmetry woud be ${\rm SO}(N)$ rather than ${\rm SO}(N-1,1)$ --- but apart from this the lattice breaks the rotational symmetry down to a discrete subgroup of 90 degree rotations. It's only the in the continuum limit (i.e close to a critical point) that full rotational invariance is restored.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Weak Energy conditions for Perfect fluid derivation
I want to derive the restraints $$\rho + p \geq 0\qquad\text{and}\qquad \rho \geq 0$$ which correspond to applying the weak energy condition ($T^{\mu \nu} v_\mu v_\nu \geq 0$ for $v_\mu$ timelike) on the stress energy tensor $T_{\mu \nu}$ of a perfect fluid given by
$$
T_{ab} = (\rho + p ) u_a u_b + p g_{ab}.
$$
Applying the definition:
\begin{align*}
0 & \leq T^{\mu \nu} v_\mu v_\nu
\\
& = ( (\rho + p)u^\mu u^\nu + p g^{\mu \nu}) v_\mu v_\nu
\\
& = (\rho + p) u^\mu u^\nu v_\mu v_\nu + p g^{\mu \nu} v_\mu v_\nu
\\
& = (\rho + p) u^\mu v_\mu u^\nu v_\nu + p v^\nu v_\nu
\end{align*}
I'm not sure how to evaluate $u^a v_a$. I know that $u_a$ is the four velocity so we have $u^a u_a = -1$ in natural units and also that $v_a v^a < 0$ since $v$ is timelike but I dont see how I can simplify this expression to get the constraints above.
| The statement of the weak energy condition is that
\begin{equation}
T^{a b} v_a v_b \geq 0
\end{equation}
holds for any timelike vector $v$.
The first important step is to prove that the weak energy condition implies the null energy condition, if the previous inequality holds for any timelike vector $v$, it will also hold for any null vector $k$.
I will leave the mathematical details aside, but the rough idea in proving this statement is using the fact that null vectors constitute the boundary of the light cone defined by the timelike vectors, so we can always take a sequence of timelike vectors that converges to any null vector. (i.e $\forall$ null $k$ $\exists$ a sequence of timelike vectors $\{v_n\}$ such that $\lim \, v_n = k$).
This sequence leaves the first inequality invariant, proving the weak energy condition. Now we have to add to the first constraint that
\begin{equation}
T^{ab} k_a k_b \geq 0
\end{equation}
holds for any null k.
Henceforth I will use the notation $a_ab^a \equiv a\cdot b$ and $a_a a^a \equiv a^2$.
Focusing first on the statement of the null energy condition
\begin{align*}
0 & \leq T^{a b} k_a k_b
\\
& = ( (\rho + p)u^a u^b + p g^{a b}) k_a k_b
\\
& = (\rho + p) (u\cdot k) ^2 + p k^2
\\
& = (\rho + p) (u\cdot k) ^2
\end{align*}
where, in the last step, we used the fact that k is null. Using $(u\cdot k) ^2 \geq 0$ we have
$$(\rho + p) \geq 0 $$
as required.
Now, looking at the statement of the weak energy condition, we end up with
$$0 \leq (\rho + p) (u\cdot v) ^2 + p v^2 $$
for $v$ timelike.
Writing $v = \|v\| n$, where $\|v\| = \sqrt{-v^2}$ and $n^2 = -1$, we get that
$$\frac{p}{(u\cdot n)^2} \leq \rho + p$$
Now we need just the final bit of mathematics. Due to the Lorentzian nature of geometry in GR, the Cauchy - Schwartz inequality we all know and love from Riemannian geometry, is changed. Specifically, we know that if $x$ and $y$ are null or timelike
$$ |x\cdot y| \ge \|x\| \|y\|.$$
This means that $(u\cdot n)^2 \ge 1$ with equality iff $n = u$
Hence:
$$\frac{p}{(u\cdot n)^2} \leq p \leq \rho + p \iff \rho \geq 0$$
Q.E.D
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What are equations of state in thermodynamics? So I am having real trouble understanding what equations of state are and how we form them. My issue stems from reading multiple sources. So I understand that an equation of state is used to build a relationship between variables to describe a state of a system.
For example $P=P(V,T)$, where $P=NKT/V$, is a function of state, but then I started reading about Gibbs, Helmholtz, enthalpy, etc and suddenly am very confused.
So looking at Gibbs, for example, we have two equations
$$G=E+PV-TS$$
and
$$dG=-SdT+VdP$$
but both describe the state of the system. I did not think an equation of state could be a differential equation but according to one book I have read, $dE$ is related to equation of state and this has caused me great confusion.
| An equation of state is simply an equation that shows the relationship between the properties of a system when the system is in equilibrium, that is, when the properties (Temperature, Pressure, Volume, etc.) are not changing in time. For example, the equation of state of an ideal gas is given by
$$PV=NRT$$
Where $P$ is the absolute pressure, $V$ is the volume, $N$ is the number of moles of gas, $T$ is the absolute temperature in deg Kelvin, and R is the universal gas constant.
As has been pointed out by Ole Krarup, Helmlholtz free energy, Gibbs free energy, Enthalpy are not equations of state. They are called thermodynamic potentials. To get more info on these, check out the Hyperphysics web site.
Hope this helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How can a day be exactly 24 hours long? The longest solar day of year is approximately 24 hours 0 min 30 seconds (occurs at mid to late December) while the shortest solar day of year is approximately 23 hour 59min 38 seconds. If I average out both of these I come up with average solar day of 24 hour +4seconds. Why then it is said it is 24 hours 0min 0 seconds exactly??
Wouldn't using a 24 hour solar day as a definition of day cause the offset of 4 seconds each solar year
| The original definition of an hour was 1/24th of a day, no matter how long the day was at the time. The durations you're quoting are only possible with the help of extremely modern definitions made possible by redefining the second (keeping the minute and hour fixed as 60 seconds, and 3600 seconds, respectively).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 3
} |
How can I calculate the speed at which a gas flows from one volume to another? Suppose that there are two rigid Volumes A and B. Each of these contains a gas with know properties (pressure, temperature, number of moles, volume, composition). Suppose also that there is a pipe with a valve that connects the two Volumes.
How can I calculate the speed at which gas goes from one volume to the other. Using the ideal gas laws, I can calculate the final properties of the gases in each volume, but that isn't what I'm looking for.
I'd imagine that the rate of flow is dependent on the difference of pressure in the two volumes or the ratio of pressure; in real life, when a high pressure volume is punctured, gases flow out of it quickly, and when a volume at room temperature is punctured, the gases flow out of it more slowly.
| The ideal gas equation of state that you've mentioned or the equation of heat transfer provided in the other 2(!) self-advertising answers don't help in this case. Heat transfer only talks about.. well the heat of the fluid, not its bulk movement.
For your problem, you need a variant of the Euler equations, which control the acceleration of fluid or gas flow under the action of forces, like pressure gradients.
Those equations might look scary, but fortunately for your problem, there is a beautiful simplification of the Euler equations, into just one equation, which is the Bernoulli equation. You're interested in the part about compressible flows.
If you're only interested in plugging in some numbers, then you're already there. Otherwise if you're interested in further reading, I can answer some questions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/469999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Can someone provide to me an intuitive explanation of the second integral of position with respect to time? I am aware of what the first integral of position, absement means (at least to a very superficial level).
However, I can find nothing regarding the physical intuitive meaning of absity, the second integral of position.
If anyone is able to explain to me, that would be great.
| You can say that the position is the second derivative of it, so that the relation between this quantity and the position is the same between the position and the acceleration.
I've never seen this quantity be used for a physical purpose, it carries indeed very redundant information.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Longitudinal magnification I want to prove that if an object is small in length and lying along the principal axis then
$$M = -\frac{dv}{du} = -\left(\frac{v}{u}\right)^2$$
Where, $M$ is the longitudinal magnification.
| In geometrical optics the following relation between the longitudinal positions of object and image (respectively $u$ and $v$) together with the focal length $f$ is valid:
$$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$$
If the object is small and it has one of its ends at $u_1$, with the corresponding image at $v_1$, we can calculate the position of the image of the other end, $v_2$, in an approximate way using derivatives:
$$v_2 \approx v_1 + \frac{dv}{du} (u_2 - u_1)$$.
where $\frac{dv}{du}$ is the derivative calculated at $u_1$.
The longitudinal magnification is the ratio between the length of the image and the length of the object:
$$ M = \left|\frac{v_2 - v_1}{u_2 - u_1}\right|, $$
and using the approximate equation for the position in terms of the derivatives that we wrote above we see that
$$M \approx \left| \frac{dv}{du} \right|.$$
An easy way to calculate the derivative is considering that the variation of the quantity $\frac{1}{u} + \frac{1}{v}$ is zero for any variation of the position of the object $u$ and the corresponding variation of the position of the image $v$. So:
$$d \left (\frac{1}{u} + \frac{1}{v}\right) = 0$$.
We can express the variation using the variations of $u$ ($du$) and $v$ ($dv$) as
$$d \left (\frac{1}{u} + \frac{1}{v}\right) = -\frac{1}{u^2}\,du -\frac{1}{v^2}\,dv$$
still equal to zero. From $-\frac{1}{u^2}\,du -\frac{1}{v^2}\,dv = 0$ we obtain the expression for $\frac{dv}{du}$:
$$\frac{dv}{du} = - \frac{v^2}{u^2}$$
The longitudinal magnification is then
$$M_{\mathrm{long.}} = \frac{v^2}{u^2}$$
Since the transverse magnification is
$$M_{\mathrm{transv.}} = \frac{v}{u}$$
then
$$M_{\mathrm{long.}} = M_{\mathrm{transv.}}^2$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why the time period of pendulum with infinite length is $84.6$ minutes? In a book I was reading about SHM it stated:
If the length of a simple pendulum is increased to such an extent that $\ell\to\infty$, then its time period is given by,
$$T=2\pi\sqrt\frac{R}{g}\approx84.6\text{ min}$$
Now I have many confusions like:
*
*Doesn't the gravity change with such a large length and amplitude?
*How can an infinitely long pendulum have a confined time period?
*Some parts near the end may have a velocity greater than light.
| $\def\PD#1#2{{\partial#1 \over \partial#2}}$
I would suggest a different approach. If $l$ is very large the circumference of radius $l$ becomes a straight line, tangent to a circumference centred in Earth's centre C and radius $R$. If $x$ is bob's displacement along this line, bob's distance from C is $D=\sqrt{R^2+x^2}$ and gravitational potential energy is
$$V(x) = -{G\,M\,m \over D}.$$
The component of gravitational force along bob's trajectory is
$$F_x = -\PD Vx = -{G\,M\,m \over R^3}\,x =
-{m\,g \over R}\,x$$
and the equation of the motion is
$$\ddot x = -{g \over R}\,x.$$
Then
$$\omega = \sqrt{g \over R} \qquad
T = 2\,\pi\,\sqrt{R \over g}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If a downed power line hits your car, why should you shuffle away from it? A bit of fun! My colleague showed me a video at work about what to do if a downed power line lands on your car.
https://www.youtube.com/watch?v=Psv3ySUoF3Q&feature=youtu.be&t=54
In the video, the narrator says this about how to get out of the car:
Remove all loose items or clothing and jump clear of the vehicle. Avoid
touching the car and the ground at the same time. Land with both feet together, keep your feet as close together as possible, and shuffle away from the car.
My question is, what is the point of landing on both feet, keeping them close together, and shuffling away? How would this be more safe than simply sprinting away from the vehicle?
Thanks!
| A downed power line can create a voltage gradient across the ground. If your two feet touch areas that have a voltage difference, a current can flow up one leg and down the other, electrocuting you. By keeping your feet close together, you are minimizing the potential for this to happen, as nearby points on the ground will in general exhibit a lower voltage difference than distant points.
The better option is to just stay in the car, though, so long as there's no imminent danger from fire or explosion. Evacuating a car with a live wire on it is dangerous; it's better to stay in the car until the line has been de-energized, if at all possible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground? When we drop a ball, it bounces back to the spot where we dropped it, due to the reaction forces exerted on it by the ground. However, if a person falls down (say, if we push them), why don't they come back to their initial position where they started their fall?
According to Newton's 3rd law of motion, to every action there is always an equal but opposite reaction. If we take the example of ball then it comes back with the same force as it falls down. But in the case of a human body, this law seems not to be applicable. Why?
| Energy is not lost, but it is expended in different ways when different objects collide.
A more visceral statement of these principles:
Drop a rubber ball on the floor. As it hits the floor it stores up spring energy inside itself, and then bounces back almost as high as the start point, losing only a little energy to friction.
Drop a hard billiard ball on the floor. It bounces a bit, but much of the energy gets spent making a dent in the floor.
Drop a sphere of soft clay on the floor. It goes "splat" - most of the energy is spent pushing the clay out sideways. (There may be a small crown on the blob of clay on the floor where some of it did bounce back a bit.)
Drop a human being on the floor. Where does the energy go? It goes into bruising skin and muscles, bouncing internal organs back and forth, forcing air out of the person's lungs, swear words, etc. It takes quite a forceful impact to have enough leftover energy yo make a human being bounce enough to see.
The concept of a spherical cow is often used to describe the way real-world objects are simplified to make the underlying physics easier to describe; in this case you seem to be thinking of a spherical person.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 5
} |
If atmospheric pressure is 76 cm of $\text{Hg}$ , why won't 76 cm of mercury stay in an open tube when suspended in air? If we keep an hold a tube in air with the closed end up and open end downwards, containing mercury upto a length of 76 cm, why does the mercury not stay in place? Shouldn't atmospheric pressure exert a force equal and opposite to its weight and balance it?
| You probably mean this:
Let we fill the tube with closed end at the bottom with 76 cm of mercury. Then we will turn the tube upside down. The mercury will flow out from it.
It happens because during the rotation of the tube some amount of mercury flowed off, and is instantly replaced with air, which moves up to the closed end of the tube.
So after the turn of the tube, there is an air pressure not only from the bottom, but from the top of the mercury column, too.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Minkowski space
*
*In Minkowski space, coordinates which satisfy
$x^2 = t^2 - X^2 > 0$
are in the region of spacetime that is time-like.
*If it's
$x^2 = t^2 - X^2 < 0$
the region is space-like.
*But if
$x^2 = t^2 - X^2 > 0$
then its "trajectory of light-like particles".
I have understood the first two points about time- and space-like regions but I could not get the third one about "light-like particles".
My confusion is - why just light-like particles? There are many other particles at quantum level.
| Only particles with zero mass can travel between two events which are separated by a light-like distance. The trajectory is called light-like because photons (light) are massless, and historically the first example of a massless particle, as well as the only example in the 1910's. There are other massless particles, like gluons which would also be able to travel between two events separated by a light-like distance.
The reason why only massless particles are able to travel between two events separated by a light-like distance is that it requires you to travel at exactly the speed of light. You can see this by considering the equation $t^2-x^2=0$, this means that $x=\pm t$. These equations are with the units such that the speed of light $c=1$. Thus the particle taking this trajectory is travelling at the speed of light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/470979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Geometrical Optics: Infinite Rays Normally in ray optics, we draw a parallel line from the top of the image to the lens and stop when this line intersects an angled line (drawn from the height of the real object) and intersects. However, why do we stop? We can draw infinite rays from this object and they should be able to go out for infinitely far distances. Thus, we should be able to get a huge image (albeit a bit darker).
| I have added some more rays onto your diagram and omitted the arrowhead to make the diagram clearer. I have also ignored the effects of spherical aberration.
Every point on the object produces light which travels along rays and after reflection from the mirror meet at a corresponding point on the image - neighbouring points on the object are neighbouring points on the image.
All the light travelling along the red rays which originate from point $A$ on the object and after reflection from the mirror arrive at point $B$ on the image having taken the same time to travel from $A$ to $B$.
This means that the path lengths of all rays which travel from $A$ to $B$ is the same.
Waves leaving point $A$ in phase arrive at point $B$ in phase.
All the light travelling along the blue rays which originate from point $C$ on the object and after reflection from the mirror arrive at point $D$ on the image having taken the same time to travel from $C$ to $D$.
This means that the path lengths of all rays which travel from $C$ to $D$ is the same.
Waves leaving point $C$ in phase arrive at point $D$ in phase.
Light which leaves every point on the object after reflection meets at a corresponding point on the image.
The path length from point $A$ to point $D$ after "reflection" from the mirror is not constant and so all the light which originates from point $A$ would not arrive in phase at point $D$ so no "image" of point "A" is formed at point $D$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/471223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why is normal force at the bottom dependent on normal force on top?
Why does the normal force on bottom of the track have anything to do with the normal force on top of the track? Why isn't the normal force at the bottom simply $mg$?
| You need to know that a motion can be accelerated in two independent ways or any combination of these; either when there is a change in velocity, or when there is a change in direction of motion.
The above problem has a constant velocity but the direction of car at every point of the loop changes, which mean the motion is accelerated.
Newton's laws tell us that for a motion to be accelerated, a net force must act on the body.
Enters centripetal force:
If $mg$ was the only normal, it would mean there is no force to keep the car in circular motion and it would continue in a straight line which is not the case, so there must exist some mysterious force that is keeping the car from going straight, which is indeed, the centripetal force.
Imagine this: from the car's frame of reference, it would be exerting more than just its weight on the loop; if you were inside, you'd be pushed to the floor of the car.
This is why you have another force in play. I hope you can take over from here.
P.S. you intuition isn't wrong but it would apply when the car has not to move in a loop in the next lap, meaning, if it leaves the circular motion at that very precise moment when it reaches the bottom. Then, there would be no force required to keep it in circular motion and thus only mg would act in it in vertical direction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/471561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Why does $\sqrt{\frac km}$ represent angular velocity and not frequency? When I break down $\omega = \sqrt{\frac km}$ (angular velocity for a simple harmonic oscillator) into its units, I get:
$$\omega = \sqrt{\frac{kg * \frac {m}{s^2}}{kg *m}}$$
which simplifies to:
$$\omega = \frac 1s$$
If I'm not mistaken, that is the unit for frequency (hz). I know that radians are considered "unitless", but I do not understand intuitively how it translates to angular velocity ($\frac{rad}{s}$) or how radians are involved.
| Looking into the equation of motion of a mass attached with a spring of spring constant, we get
$$\frac{d^2x}{dt^2}= - \left( \sqrt{\frac{k}{m}} \right)^2 x \;,$$
which is analogues to the equation
$$\frac{d^2x}{dt^2}=-(\omega)^2 x \;.$$
(Obtained by double differentiating $x=ACos(\omega t+\phi)$). Here $\omega$ is the angular velocity not frequency.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/471641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Does the Central Limit Theorem hold for position measurements? A friend asked me recently if the Central Limit Theorem holds for quantum systems: i.e., if the distribution of measurements (e.g., of position) for any wavefunction would prove approximately normal, given enough samples.
My gut response was no, because I've encountered plenty of position wavefunctions in introductory QM that wouldn't lead to a normal distribution via Born's rule. But the CLT's requirements seem fairly permissive. What am I missing?
| Let's be clear here: You're asking whether the probability distribution of the normalized sum of many individual quantum measurements of the same quantity necessarily tends to a normal distribution, not whether the probability distribution of the possible outcomes of any single quantum measurement is necessarily a normal distribution. The answer to the latter question is clearly "no". As for the first question, I would say that the answer is "yes". The central limit theorem can be stated as
In probability theory, the central limit theorem (CLT) establishes that, in some situations, when independent random variables are added, their properly normalized sum tends toward a normal distribution (informally a "bell curve") even if the original variables themselves are not normally distributed. Wikipedia: Central Limit Theorem
So whether the individual variables you're averaging result from classical measurements (e.g., the roll of a die) or from quantum measurements (e.g., spin up=1 or spin down=0) and what their probability distributions are is irrelevant to the central limit theorem. If you sum up a lot of individual measurements (either classical or quantum) and normalize them, the outcome will tend to a normal distribution.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/471789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why the source function is equal to Planck function when we have a local thermodynamic equilibrium? I understand that the source function $ S_λ $ for the special case of blackbody radiation is equal to the Planck function $B_λ $.
However, in the broader case of a local thermodynamic equilibrium (and not the special case of a blackbody) I would expect that
$$ S_λ=εB_λ $$ where $ε$ the emissivity
and the equation of radiative transfer to be:
$$ \frac{dI_λ}{k_λρds}=-I_λ+εB_λ $$
and not
$$ \frac{dI_λ}{k_λρds}=-I_λ+B_λ $$
Where do I make a mistake?
| The source function is a property of the body itself and not the radiation field it is immersed in. If the object is in local thermodynamic equilibrium then the source function only depends on temperature.
In principle you could put the body in a cavity filled with black body radiation at a given temperature and it would come into equilibrium at that same temperature. Since a blackbody radiation field is isotropic then there can be no gradient in the specific intensity and therefore the source function at temperature $T$ must equal the Planck function at the same temperature.
In your prescription the radiation field could never be isotropic since there would be a gradient, $dI_\lambda/ds \neq 0$, in the specific intensity, but this is inconsistent with a blackbody radiation field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/472030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What does 'coupling' mean? (Laser) I assume this question is so basic that no YT-video or paper I found answers it. But in order to understand I have to know what it refers to.
I'm talking about laser coupling to be exact. I have a 'coupler' device. What does it do with the laser?
| Coupling is the term used to describe the coupling of laser equipment together in the same way as a mechanical coupling.
One such laser coupler is the laser to fiber-coupler as can be seen in this youtube video.. The goal in this device is to get the laser output beam coupled as good as possible into the fiber cable.
What it does with the laser beam depends on what you are going to couple it from an to. In the example of the fiber-laser coupler, the beam is adjusted to match the acceptance angle of the fiber, the beam diameter, and to remove internal reflections from the coupling.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/472517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Derivation of length contraction in Einstein, Relativity: The special and general theory I was reading the book, "Relativity: The special and general theory" by Einstein. At some point it discusses the awkwardness of "relativity of simultaneity" with using universal time axis for all inertial frames. And to fix it uses Lorentz's transformation(the attached image shows two inertial frames(page 22 of the book).
To explain length dilation after this, it uses below example,
I am not able to derive the last line.
| From the first equation, at time t , we can write $$x= x' \sqrt{1-\frac{v^2}{c^2}}+vt.$$ Substitute for $x'=0$ and $x'=1$, and do $x_{(x'=1)}-x_{(x'=0)}$. You will get the answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/472810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Light in classical electrodynamics I am starting to learn elementary electrodynamics with Griffiths.
In the book, he has shown the natural correspondence between light and electromagnetic plane waves.
The problem that has agitated me is that plane waves are "global solutions", i.e. they have non-trivial EM field almost everywhere; while light seems to be localized phenomenon, when considered as stuff generated in region A and traveled to region B.
How is this paradox resolved classically(without QM)?
From Fourier Theory, it seems viable to create localized solutions by adding plane waves of different frequencies. However, this does not solve the paradox when monochromatic light is considered.
| The reason is that light in real life is better approximated as "spherical", although when you zoom in, it looks like planes.
Spherical waves expand as they propagate, and therefore decrease in intensity due to conservation of energy. So if a wave carries a certain amount of energy power P (Joules per second), the intensity, which is P/A where A is area decreases, since A increases as the wave propagates. This creates the effect where light is "stronger" near the source and therefore "localized".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Infinite Electricity from Photon Inside Closed Box with Magnet Attached in a Coil? (Thought-experiment) Let's say I have a very large box with a magnet attached to the outside. Inside the box there is a perfect mirror on each side.
I open my box and quickly shoot in one photon before shutting it. After some time the photon reaches a side of the box and is reflected. From the momentum of the photon the box receives an impulse causing it to move however the box is gradually slowed down by the magnet attached to it which is inside a coil. This is also how the electricity is produced. The box has now been slowed to a halt and the photon is yet to reach the other side.
Once the photon does hit the other side of the now stationary box it'll create another impulse and hence the cycle repeats creating electricity forever. Brilliant!
So what idea have I overlooked that makes my design flawed and me silly?
| Effectively, what you're doing is Compton scattering with a very massive object rather than a subatomic particle. In particular, this means that when the photon collides with the walls of the box, its energy will decrease slightly. This means that the photon will slowly lose energy to the box; and as time goes on, its collisions with the interior of the box will impart less and less energy (and momentum) to the box. The total amount of energy that is imparted to the box will necessarily be less than the original energy of the photon.
If you imagine replacing the photon in your contraption with a marble bounding back & forth, you couldn't extract infinite energy from it either, for exactly the same reason.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does light 'choose' between wave and particle behaviour? Light exhibits wave behaviour in phenomenon such as interference but particle behaviour in the photoelectric effect. How does light 'choose' where to be a wave and where to be a particle?
|
How does light 'choose' where to be a wave and where to be a particle?
It doesn't. It always behaves as a wave (obeying the principle of superposition), and it always behaves as a particle (particle number being quantized).
It sounds like you may have been influenced by someone who told you that light behaves like a particle in some experiments, and like a wave in others. That's false.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
"answer_count": 11,
"answer_id": 0
} |
Why do stars appear so close in universe photos? I saw an extremely sharp picture (from NASA/ESA) of the Andromeda Galaxy recently, and it made me wonder why the stars appear so close together, when I know in fact that they are not.
Is it simply because there are many stars behind stars, so that even if two stars are far away in terms of "depth" from our line of sight, one star may be offset from this line of sight by a little bit, which makes two stars appear "side-by-side" and thus close? Or are there multiple factors at play?
Also, if the above explanation about the line of sight is correct, why do we see distinct points as stars and distinct dark spaces between them? Shouldn't stars that are much further away still emit a faint light, thus making the background darkness of space brighter (or create gradients of light, rather than being completely black)?
Sorry if this is a bad question; I wasn't able to find very much information online.
| Your guess is correct; the stars appear close together because they are far away.
Relatively nearby stars are resolvable as distinct points, but in very distant galaxies it is often not possible to resolve individual stars.
Dark spaces between individual stars are actually full of stars and galaxies full of stars. Those spaces appear dark either just in contrast with the relatively nearby stars, or because of dust clouds between us and the stars/galaxies in those spaces.
One other factor contributes: Light from extremely distant galaxies is red-shifted due to cosmic expansion, such that the farther away a galaxy is, the more red-shifted its light is. If the light spectrum is shifted far enough to be below the visible range, we cannot see it. As a result, the universe can look very different when imaged using a radio telescope vs a visible light telescope.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Uniqueness Theorem and the 1D Infinite Square Well Consider the 1D infinite square well problem: $$\frac{d^2\psi (x)}{dx^2} = -k^2\psi (x)\tag{1}$$ along with the boundary conditions $\psi (0) = \psi (L) = 0$. This seems to be a well posed problem with enough information to have a unique solution but the solution $\psi (x) = A\sin(kx)$ has an ambiguity in $A$. Only with the normalization condition is the problem uniquely solved. On top of all this, I could not have specified any value of $k$ at the beginning as that would be an over-posed problem with potentially no solution (Eg: in retrospect, if I had specified a value of $k$ corresponding to an energy that is not allowed, there would not exist any solution).
This is in sharp contrast to what happens in the case of a simple harmonic oscillator from classical mechanics:
$$\frac{d^2\psi (t)}{dt^2} = -\omega^2\psi (t)\tag{2}$$
along with the initial conditions $\psi (t = 0) = 0, \psi^{\prime} (t = 0) = 1$. This has a unique solution $\psi (t) = \frac{1}{\omega}\sin(\omega t)$. Not only is some normalization condition not required to fix the value of the amplitude but this solution is valid for any value of $\omega$ that could have been specified at the start of the problem.
I can guess that this difference in behavior comes from the difference in boundary conditions but I want to know what exactly it is that makes the first one different? It seems like you need more information to pin down a unique solution for the first problem. Does the uniqueness theorem not hold due to some technicality?
| The 1D square well is quite different from the classic harmonic oscillator. You need to remember that the wave function in the square well is actually a function of $x$ and $t$. While it is normal to compute the amplitude from normalization, we could compute it from specifying initial conditions.
If you want to compare it to a classical system, the 1D heat conduction equation is closer. For instance, finding the temperature in a 1D bar as a function of $x$ and $t$ while specifying the temperature at the end points as well as some initial condition. Separating variables in the heat conduction equation will give you the same form in the $x$ equation as you get in the square well problem. You will similarly get a discreet set of eigenvalues.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If a satellite speeds up, does that make it move farther away or closer? If a satellite is in a stable circular orbit and goes about 41% faster (escape velocity) then it leaves its host forever. I get that. However, what if it speeds up by less than 41%?
Intuitively, it would seem to make the satellite move farther away from the host and thus enter a higher (more distant) orbit.
However, according to my understanding, a stable orbit requires the satellite to move more slowly the farther away it is from the host. For example, the earth moves more slowly around than the sun than Venus because it is farther away from the sun than Venus.
So, if a satellite speeds up then the stable orbit would be closer to the host, not farther away. What am I missing here?
| What you are missing is that as the satellite moves farther away from the Earth, it slows down because of work done by the gravitational field.
Since you have given the satellite more total energy (kinetic plus potential) and because the total energy of a bound orbit is $-GMm/2a$, where $a$ is the semi-major axis, then to increase the energy (make it less negative), $a$ must increase.
For a tangential $\Delta v$, the resulting orbit is an ellipse with a perigee at the point where the velocity was added to the satellite. This is because both energy and angular momentum are conserved - when the satellite returns to the same radius it must have the same tangential velocity (to conserve angular momentum), but then cannot have an extra radial velocity component because that would change the total energy. This is a Hohmann transfer orbit.
On the other hand, a radial impulse adds energy but does not change the angular momentum. The result is an elliptical orbit with a larger $a$, but with a perigee closer to the Earth. The satellite moves outwards first but then falls back. When the satellite returns to the same radius then the tangential part of its velocity must be the same as before the impulse, but to conserve energy there must be an inward radial velocity taking it closer to the Earth. i.e. a more eccentric ellipse is produced than if the same impulse is given tangentially.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/473934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
How did the Intersecting Storage Rings Actually Perform Tests I am reading the book written in 1982 titled "Quarks, The Stuff of Matter" by Harald Fritzsch. In it he writes about the ISR at Cern, that slammed protons together and in some of those collisions, 2 quarks bounced off each other at right angles to the protons.
I understand this part of it. What I don't understand is how they detected anything. None of the images of the ISR Interaction Areas show any detectors.
https://www.researchgate.net/figure/The-Intersecting-Storage-Rings-ISR-the-worlds-first-hadron-collider-photo-CERN_fig1_44218180
In this photo at almost the exact center is where the protons collided. I do not see a large detector. What am I missing?
Thanks
| The ISR was, in large part, built as an experiment in accelerator technology. When it succeeded, CERN considered it a "facility for experiments". Unlike modern colliders, which are built along with detectors aimed at specific physics, the ISR was well along on construction while the physics was still being discussed.
Eventually, a number of detectors were built and given beam time. Some sample pictures (from this retrospective):
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/474039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Location of free charge in insulators I'm going through the introductory section to Electrostatics in Materials in Griffiths, and I have a question that I can't seem to find a satisfactory answer to.
If I have an insulator with free charge, is it necessarily confined to the surface?
In the case of a conductor, Gauss' law immediately gives a "yes", because no electric field can exist in a conductor, leading us to conclude that there is no free charge inside the surface. But insulators can have electric fields inside them. Does this mean that free charge can exist inside the volume? Or does the free charge still move to the surface of the insultator?
| The question is not entirely clear; I wonder if you are really asking about bound charge (polarization charge).
Free charge is charge that is free to move macroscopically; bound charge is charge that can only move on a microscopic or submicroscopic scale. Free charge ends up on the surface of a dielectric medium under static conditions.
In a uniform dielectric slab subjected to a uniform electric field, bound charge is displaced in proportion to the strength of the electric field, but there is no net bound charge density except at the surfaces of the slab.
However net density of bound charge is confined to the surface of a dielectric only if the dielectric and the electric field is uniform. See this Feynman lecture. If, for example, there are two slabs of different dielectric media laminated together and an electric field is imposed across the combined slab, free charge will exist at the interface between the two. Or, if a slab of material were prepared containing a gradient of ratio of two substances with different dielectric constant - so that the net dielectric constant in the material has a gradient - and an electric field is applied across the slab, bound charge will be distributed throughout the volume of the slab.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/474160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating percentage error with significant figures I have a problem with significant figures. In an assignment, the paper gives us data to $3$ significant figures to calculate some values but afterwards requires us to calculate the percentage difference given an accepted value that is to $4$ significant figures. If we round the accepted value to $3$ significant figures as well, it yields $0 \%$ error. How should I deal with this?
| That's the point, you weren't meant to round it up. Approximation of numbers comes with error because you are, for instance, adding .005 to 2.545 to approximate it to 2.55. The .005 has a value and adding it when it never existed generates error. Now the question is requesting you to calculate the percentage error (.005s) which I believe should be $$\frac {\sum errors}{\sum accurate} × 100$$
So approximation causes error.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/474271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lorentz Transformation: Message sent before finish line As she wins an interstellar race, Mavis has a “hooray” message sent from the back of her 300m long ship as she crosses the finish line at v=0.6c. Stanley is at the finish line and at rest relative to it. He claims the message was sent before she crossed the line.
I understand how to get the answer using the Lorentz transformation. However, I am having trouble conceptually understanding why he observes the message before she crosses the finish line.
| Here's a spacetime diagram drawn on rotated graph paper (so that one can more easily measure displacements in time and space along segments and so that one can visualize the orthogonality between an observer's time and space axes).
It encodes the calculation (without explicit use of the Lorentz Transformation formulas), as well as the statements made in the earlier comments and answers.
As @PM 2Ring says, events ev1 and ev2 are spacelike-related, and are simultaneous according to Mavis. (So, the temporal ordering of these events in frame-dependent.)
As @Ollie113 says, starting from event ev2, the light-ray wins the race to the finish line, arriving at event A, compared to Mavis's arrival at event B. (Events A and B are timelike-related. So, the temporal ordering of these events in frame-independent: B happens after A.)
In fact, no time dilation or length contraction formulas are explicitly needed.
The key relationships are that
the area of all "light-clock diamonds" are equal and
that the diagonals of a diamond are orthogonal...
one diagonal represents
an observer's tick along her time-axis and
an observer's stick (= 1 "light-tick") along her space axis.
(For $v=(3/5)c$, the Mavis's diamond is stretched by a factor $k$ in the future-forward direction and shrunk by a factor of k in the future-backward direction, where $k=\sqrt{\frac{1+(v/c)}{1-(v/c)}}=2$ is the Doppler-(Bondi k)-factor.
This construction is essentially the Lorentz Transformation in light-cone coordinates.)
One can transcribe the diamonds into Mavis' frame:
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/474384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What would happen if we use two of Schroedingers cats? I am worrying about Schroedinger's cat's experiment and I am wondering about the phenomena. First of all, I have to say that I am not a physicist and I have no clue about more than advanced high-school physics.
As far as I understood the experiment there is a chance that an atom which sends out gamma - radiation in case of decay decays. This chance exists only as far as we do not detect the particle. Now we wait some time and then look up the cat. The cat was before dead and alive at the same time.
Now it is either dead or alive.
Could it be that if we would have two cats, one cat would be dead and one would be alive, or is this impossible?
Thanks
| If the experiment is set up in such a way that the quantum mechanical event that kills a cat is guaranteed to kill both cats (e.g., it releases poison gas into a chamber containing both cats), then the experiment entangles the cats' alive/dead states in such a way that either both cats live or both cats die. That's the whole point of the Schroedinger's Cat story: quantum mechanical events can have macroscopic consequences. Contemplation of the story leads directly to the Many Worlds Interpretation of quantum mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/475921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Variation of Refractive index We know that refractive index, for any medium,
$$n=1/\sqrt{\epsilon\mu}.$$
Also, according to Cauchy's relation
$$n=A+B/\lambda^2,$$
where $A$ and $B$ are constants related to the medium.
According to the first relation, refractive index isn't in any way related to the wavelength of the light, it is only related to the permitivitty and permeability of the medium. According to the second relation, however, it depends on the wavelength too. Which formula is right? Why does the discrepancy arise?
| There is no discrepancy. Generally speaking, the permittivity of the medium depends on the wavelength, which means that $n=1/\sqrt{\mu\epsilon}$ also depends on the wavelength. (The magnetic permeability could also depend on the wavelength, but this is much less frequent.)
It's also important to note that Cauchy's relation is only ever an approximation which holds for limited ranges in wavelength, and only ever to a finite precision. As one example, many materials can be modelled very well with the Drude model, which predicts $\epsilon = \epsilon(\omega) = 1-\omega_p^2/\omega^2$, where $\omega_p$ is a parameter that describes the material. For certain ranges of $\omega$ in relationship with $\omega_p$, the Cauchy relation can be a good fit - but even then it is still a phenomenological model that's only a vague description of the rough behaviour of the underlying physics over a limited range of the controlling parameters.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/476165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do we observe the nuclear force only in scatterings and decays? Why, at first glance, are the only forces we perceive to be gravity without quantizing, electromagnetism and nuclear forces only in disintegrations?
| Gravitational and electromagnetic potentials of a point source both fall of with the distance $r$ as $1/r$. This is a "long-range" characteristic compared to the effective nuclear forces that are mediated by massive particles (pions) and have a Yukawa potential, where the $1/r$ Coulomb behaviour is screened by an exponential $\mathrm{e}^{-mr}$. This makes nuclear forces effectively short-range.
As for "why" this is so, I discuss the origin of the Coulomb and Yukawa potentials from the underlying quantum field theories also in this answer, and in more mathematical detail in this answer for the electromagnetic form and here for forces mediated by massive particles.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/476341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How is the relative force of the fundamental forces measured? My physics textbook includes the following table:
My question is about the fourth row, where it compares the relative strengths of the fundamental interactions. How are these determined? Is the ratio of electromagnetic and gravitational simply the ratio of the force between 2 1kg point masses separated by 1m, and the force between 2 1C point charges separated by 1m? (that was the explanation my teacher gave me) If so, how can this be justified, since the C and kg are just arbitrary units?
| Here is another table of fundamntal forces as used in particle physics.
Fundamental means the basic framework, and the basic framework is quantum mechanics, from which macroscopic forces emerge which can be demonstrated mathematically. These forces exist at the particle level, and their strength relative to each other enter in the Feynman diagram description of the complicated integrals that have to be computed in order to describe data at the particle level.
The column called "strength" gives the coupling constants that have to enter in a multiplicative manner whenever a vertex in a feynman diagram involves the corresponding force. The coupling constants are measured from data. Example here for weak and here for electromagnetic. Mesurements are fitted by specific calculations, and thus the valuce of the coupling constants are defined.
The relative strength is gotten from data that need two different couplings, so the Feynman diagram calculations are fitted to the data, and the relative strength is determined.
In a hand waving way, the diagrams with the weak constant gives small crossections with respect to the ones with electromagnetic vertices, that is why the were called weak. The calculations quantify this.
Here is a link for measuring the strong force coupling.
Gravitation is still not definitively quantized, but the coupling constant is used in effective quantizations. Here is a link on how it is defined.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/476564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Will sound be generated by the capture of air from a vacume? I have a thought experiment. If I had a box sized vacuum sitting on the floor in front of me and I were to instantaneously remove all the walls of my box. Would the capturing of air from that vacuum make a noticeable sound?
| Yes. Once the walls are magically removed, the air surrounding the box will suddenly rush inwards toward the center of the box volume- thereby propagating a rarefaction wave outwards at the speed of sound- which you will hear. Very soon thereafter, all the inrushing air will meet at the center of the box volume and rebound, propagating a compression wave outwards.
You would therefore hear something like a "KA-POP" noise.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/476822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is the neutron magnetic moment negative? I understand that the magnetic moment is due to the quarks, but specifically why is it negative? Is it due to the two down quarks or something?
| Because the "large" pieces in the baryon octet wavefunctions usually win.
Bég,Lee,&Pais, PRL 13 (1964) 514, a historically significant demonstration that the naive constituent quark model with a symmetric wavefunction was there to stay, one way or another-- and thus led to the inference of color to antisymmetrize it consistently with Fermi statistics (but that is another story...).
Pick your spin axis in the z direction. Take your proton and the neutron as bland symmetrized assemblies of just three inert constituent quarks, and their magnetic moments as the naive sum of the magnetic moments of each of their quark constituents. So, you only simply read off
$$
\mu_p=\langle p\uparrow|\tfrac{e}{2m_q}\sum_i Q_i \sigma_i^3 |p\uparrow \rangle ,
$$
and likewise for the neutron, where Q indicates the (fractional) value of the quark's charge in terms of the elementary (proton) charge, and $m_q$ its common mass, but we'll only consider baryon moment ratios, so it washes off.
But since the sum over the three quark constituents is symmetric among quarks, we need not write the full messy wavefunctions of the baryons: we'll simply imply symmetrization (and color antisymmetrization, today).
The calculation is then trivial, running on simplified collapsed wavefunctions,
$$
|p\uparrow \rangle \sim \frac{1}{\sqrt 6} (2 u\uparrow u\uparrow d\downarrow -u\uparrow d\uparrow u\downarrow - d\uparrow u\uparrow u\downarrow ), \\
|n\uparrow \rangle \sim \frac{-1}{\sqrt 6} (2 d\uparrow d\uparrow u\downarrow -d\uparrow u\uparrow d\downarrow - u\uparrow d\uparrow d\downarrow ).
$$
It is then evident by inspection that
$$
\frac{\mu_p}{\mu_n}= \frac{\langle p\uparrow| \sum_i 3Q_i \sigma_i^3 |p\uparrow \rangle }{\langle n\uparrow| \sum_i 3Q_i \sigma_i^3 |n\uparrow \rangle}\\ =\frac{4(2+2+1)+(2-1-2)+(-1+2-2)}{4(-1-1-2)+(-1+2+1)+(2-1+1)}\\ =\frac{18}{-12}=-3/2.
$$
This is the classic "cannot be a coincidence" moment. The experimental value is -1.45989806(34), from Wikipedia. (You can all but hear the palms slapping on foreheads all over the planet, at the time...).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why are there rings (halos) around street lights? Especially when it's foggy I was in a car that was turned off last night for some time and the windows became foggy via condensation (moisture droplets building up on one side of window). Looking outside, I could see that street lights which were near me had a halo or a ring around them. They would disappear if I wiped the moisture from the window. Why do I see halos around light through a foggy glass?
Also, after stepping outside I DID notice a ring around the street light, but it had a larger diameter and it was VERY faint. Why does this light effect occur even without a foggy window? What is going on?
Thank you.
| A couple of years ago a little piece of debris met my left eye. Later that day at night I started to notice a halo around streetlights as seen by my left eye. After an eye doctor visit and some treatment and time, they were gone for good. That made me ask myself the same question.
The rings are not around the street lights but in our views of them. Nearby rays coming out from the lamp are focused by your eye lenses to form a point in your retina. The halo you see is formed by rays that otherwise weren't aimed at your retina, but stuff refracted them midways and redirected them to your retina.
This could be by the droplets of water in the moisture of your glass (that's why cleaning it removed the halo) or by any other thing in the path of the rays. It is axially symmetric because only the rays that make a specific angle with the central ray get refracted into your eyeball, the rays at other angles get refracted elsewhere. And the refraction index marks this angle, which together with distance will give the perceived radius of the rings or halo.
As an experiment next time you can try to occlude with a finger the lamp and see if the halos are still there. Do you expect them to still be there for a) you inside your car, foggy glass? b) you outside your car?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the definition of beam energy in particle physics? For example, the proton beams in the LHC collider have 7 TeV energy. Does this mean that the individual protons in the beam have 7 TeV energy or that the energy of all the protons in the beam add up to 7 TeV?
| In the LHC, each individual collision has a center-of-mass energy of roughly 14 TeV. Since the collisions are symmetric (two protons with equal energy, moving in opposite directions, collide), we can say that each individual proton has roughly* 7 TeV of energy.
As you can probably tell, adding up the energy of all of the protons in the ring at any given time gives you a pretty colossal amount of energy.
*In reality, there's a distribution of energies that the protons in the beampipe can have, due to the fact that accelerators can only be so precise.The distribution is centered around 7 TeV, and is sharp enough that we can safely talk about collisions having a pretty uniform center-of-mass energy in most cases.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is throwing dice a stochastic or a deterministic process? As far as I understand it a stochastic process is a mathematically defined concept as a collection of random variables which describe outcomes of repeated events while a deterministic process is something which can be described by a set of deterministic laws. Is then playing (classical, not quantum) dices a stochastic or deterministic process? It needs random variables to be described, but it is also inherently governed by classical deterministic laws. Or can we say that throwing dices is a deterministic process which becomes a stochastic process once we use random variables to predict their outcome? It seems to me only a descriptive switch, not an ontological one. Can someone tell me how to discriminate better between the two notions?
| Physics models rarely hint at ontological level. Throwing dice can be modelled as deterministic process, using initial conditions and equations of motion. Or it can be modelled as stochastic process, using assumptions about probability. Both are appropriate in different contexts. There is no proof of "the real" model.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/477910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 5,
"answer_id": 0
} |
Does a particle exert force on itself? We all have elaborative discussion in physics about classical mechanics as well as interaction of particles through forces and certain laws which all particles obey.
I want to ask,Does a particle exert a force on itself?
EDIT
Thanks for the respectful answers and comments.I edited this question in order to make it more elaborated.
I just want to convey that I assumed the particle to be a standard model of point mass in classical mechanics. As I don't know why there is a minimum requirement of two particles to interact with fundamental forces of nature,in the similar manner I wanted to ask does a particle exerts a force on itself?
| What even is a particle in classical mechanics?
Particles do exist in the real world, but their discovery pretty much made the invention of quantum mechanics necessary.
So to answer this question, you have to set up some straw man of a "classical mechanics particle" and then destroy that. For instance, we may pretend that atoms have the exact same properties as the bulk material, they're just for inexplicable reasons indivisible.
At this point, we cannot say any more whether particles do or do not exert forces on themselves. The particle might exert a gravitational force on itself, compressing it every so slightly. We could not detect this force, because it would always be there and it would linearly add up with other forces. Instead, this force would show up as part of the physical properties of the material, in particular its density. And in classical mechanics, those properties are mostly treated as constants of nature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/478060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "127",
"answer_count": 8,
"answer_id": 3
} |
Can you change the wavelength of light keeping frequency constant and can you do the opposite as well? Can you change the wavelength of light keeping frequency constant and can you do the opposite as well?
I understood the basics but please don't hesitate to go deeper into the concept. Also, If you happened to have an elegant explanation please drop it here if you can.
| Frequency: acoustic optics can change the frequency.
Wavelength: the speed of light $=3\cdot 10^8 m\cdot s^{-1}$ in vacuum, change a material with different reflective index change the wavelength.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/478686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Displacement currents, aren't really currents? I'm confused with this definition of displacement currents within capacitors via Wikipedia:
However it is not an electric current of moving charges, but a
time-varying electric field.
It's a time-varying electric field, but there isn't any actual flow of current, while at the same time... it produces a magnetic field. It's counter-intuitive.
If air was in the center between the plates, it's possible to have some "current flow" due to atoms within the gas molecules, likewise, for a dielectric.
Yet, displacement currents aren't considered "real/free" current.
Could someone clarify this confusion for me.
How significantly different are displacement currents from normal/free currents?
| In the absence of current the Maxwell equation reads $\partial \vec E = \vec \nabla \times \vec B /c^2$. This equation means that both sides denote the very same thing. It does not mean that E induces B or vice versa.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/478813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Why hydrogen lines are less visible in the Sun spectrum than in supernovae clouds? Supernovae clouds are very colorful, and if I trust documentaries I watched, the colors are due to excitation of elements, as in fireworks. Since the Sun is mostly made of hydrogen, I suppose those lines should be very apparent but they are not so much, its light looking like a blackbody radiation. What contributes to the rest of the spectrum up to the point it masks hydrogen lines?
|
What contributes to the rest of the spectrum up to the point it masks
hydrogen lines?
It's basically due to the optics.
In the case of the sun, you're seeing the energy of the gas in the outer shell. This is highly random, moving in all directions and over a huge distribution of speeds. So everything is being blurred out in frequency by Doppler shifting, which is why...
its light looking like a blackbody radiation
In the case of a supernova remnant, you're looking at a bubble of hot gas caused by the outward moving particles shock-heating the interstellar medium when it hits it. So, in this case, in any one area of the cloud, you're seeing particles that are moving at a much more limited range of speeds. This is also why you see a sort of color shift, notice that the crab nebula looks redder at the edges than the center.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/479047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Resource recommendation for spintronics What are some good beginner level books to understand spintronics. I am looking for some sources in which basic mathematics of spintronics is properly explained.
| You should start with books on solid state physics, obviously. Beginner level material is "Kittel" or "Ashcroft/Mermin". You may continue with mathematically more involved books as, for instance, Bruus/Flensberg to understand the modern solid state physics notation. There some basic spintronics phenomena are already mentioned (Kondo effect and Luttinger Liquid). At some point you should be able to understand the following:
*
*I. Zutic, J. Fabian, S. D. Sarma, "Spintronics: Fundamentals and applications", arXiv:cond-mat/0405528, Rev. Mod. Phys. 76, 323-410 (2004).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/479159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Does kinetic energy warp spacetime? My interpretation of GR leads me to think that energy (namely kinetic) also adds to the curvature of space-time. Which, has raised a thought experiment. If a $10000$ kg ship closely passed a $1$ kg glass ball at $0.8c$ relative to the glass ball, would the glass ball be moved in the direction of the ship for the tiny fraction of a second that its passing by, more so than if the ship popped in and out of existence at rest for the same time period relative the glass ball?
| The answer is yes, and this is actually an important point when considering extremely high-energy physics. As a result of this effect, gravity is the dominant interaction at sufficiently high energies, at least if we can trust the most straightforward extrapolation of the current foundations of physics. This is highlighted in 't Hooft (1987), "Graviton dominance in ultra-high-energy scattering," Physics Letters B 198: 61-63. The abstract says:
The scattering process of two pointlike particles at CM [center-of-mass] energies in the order of Planck units or beyond, is very well calculable using known laws of physics, because graviton exchange dominates over all other interaction processes. At energies much higher than the Planck mass black hole production sets in, accompanied by coherent emission of real gravitons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/479299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
} |
Why are solar panels kept tilted? I have noticed that, in my country India, most of the solar panels are tilted southward at an angle of ${45}^{\circ} .$ Even on buildings with inverted V-shaped roofs, solar panels are still oriented southward on both the sides of roof.
Research
Many sites suggests that the tilt aids in self-cleaning also another site stated that tilt depends on factor like latitude
My questions:
*
*Why are solar panels tilted southward?
*How is latitude of the location of a solar panel relevant in increasing efficiency?
| *
*Because in northern hemisphere sun is at the south (it moves from east to west - but always on the south side of the sky), by tilting the solar panel towards the sun (to the south) you increase energy generation. Solar panel at 90° angle towards the sun gives you near 0 energy, at 0° (directly looking at the sun) you are getting maximum rated power. Of course without solar tracking which turns solar panel towards the sun - you will not be at 100% efficiency all the time. But this tilt increase your results in average.
*It is directly relevant: At the equator your optimal angle is around 0° as sun rises to zenith, at 45° latitude optimal angle is around 37°. So maximum and average sun elevation angle directly depends on latitude.
You can see more on angle variation here: https://www.solarpaneltilt.com/
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/479515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 4,
"answer_id": 0
} |
Why is it that magnetic fields(or any field)not move in space? When I imagine a magnetic field produced by a magnet, or the electric field produced by a charge, I've learned that the fields are stationary, however, their value(across space) changes.
If I placed the magnet at a point $P$($0,0,0$), and then moved the magnet to $P_2$$(1,1,1)$ Why wouldn't it's associated magnetic field move with it?
| The fields are stationary with respect to a coordinate system fixed to the magnet.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/479627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simple harmonic motion as projection of circular motion Can we consider $\omega$ (angular frequency) in equation of simple harmonic motion (SHM) as the angular velocity of the object in circular motion, when we see simple harmonic motion as projection of circular motion?
|
Yes. The point on the circle with radius $R$ revolves around the centre, with angular velocity:
$$\omega=\dot{\theta}=\frac{\mathbf{d}\theta}{\mathbf{dt}}$$
The projection on the RHS axis, call it $y$, is:
$$y(t)=R\sin(\omega t+\phi)$$
where $\phi$ is the angle $\theta$ at $t=0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/479856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Grassmann-even action I am currently studying supersymmetric quantum mechanics with the help of the book Mirror Symmetry by Kentaro Hori (and others). On page 155 where they introduce Grassmann variables they say that the action is Grassmann-even without an explanation. But i do not quite understand why this is the case and a Grassmann-odd action is not allowed?
| *
*At the classical level, the action $S$ can have any Grassmann-parity.
*At the quantum mechanical level in a path integral/partition function, it would be rather strange/exotic$^1$ to consider a Boltzmann factor $$\exp(\frac{i}{\hbar}S)~=~1 +\frac{i}{\hbar}S$$ for a Grassmann-odd action $S$. In particular, the Boltzmann factor would no longer have definite Grassmann parity.
--
$^1$ For an example of a Grassmann-odd action in the Literature, see e.g. Ref. 1.
References:
*
*H. Hata & B. Zwiebach, arXiv:hep-th/9301097; Section 3.2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/479979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why aren't satellites disintegrated even though they orbit earth within earth's Roche Limits? I was wondering about the Roche limit and its effects on satellites.
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
I think it's due to the satellites being stronger than rocks?
Is this true?
Also, is the Roche limit just a line (very narrow band) around the planet or is it a range (broad cross sectional area) of distance around the planet?
| The Roche limit is a limit on objects being held together by their own gravity. Satellites are held together by much stronger forces. Different parts of the satellite are ultimately connected by chemical bonds, which are electromagnetic.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/480190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 6,
"answer_id": 3
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.