Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
What happens to circularly polarized light when it hits a linear polarizer? I have seen a lot of examples of what happens when circularly polarized light passes through a circular polarizer composed of a quarter-wave plate and a linear polarizer, but what would happen to the circularly polarized light if it passed through only the linear polarizer without a quarter-wave plate?
The reason I’m asking is because I’ve heard that in photography linear polarizers can cut through smog but not through fog, which generates circularly polarized light. This seems strange, because one would assume that the linear polarizer would absorb all of the circularly polarized light and would thus cut through the fog. Why is this not so?
| Circularly polarised light can be decomposed into two electromagnetic waves, with their respective electric fields linearly polarised at right angles to each other, of equal amplitude but 90 degrees out of phase. One of the polarisations can be chosen to line up with the polariser and the other at right angles to it. e.g.
$${\bf E} = E_0 \sin(\omega t - kz)\ {\bf i} + E_o \sin(\omega t -kz -\pi/2)\ {\bf j}\ .$$
Note that you can choose any pair of orthogonal unit vectors that are in a plane at right angles to the wave motion. So you can choose to have one of those unit vectors be along the axis of the linear polariser.
As a result, only one of the waves makes it through the polariser and the transmitted light will be linearly polarised and of half the intensity of the original beam. This will be true, irrespective of how you rotate the polariser.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/742254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The "speed of length contraction" in switching between inertial frames Suppose I want to measure the length of an object in front of me, along the axis separating us (for instance looking headlong at a bus, we could talk about someone standing inside the middle of the bus with a light signal and mirror, but you get the idea).
Let's say in the rest frame of the bus I measure a length $\mathcal{L}_{0}$. Now, suppose I boost into an accelerated frame with relative velocity v, the length is now observed to be:
$$\mathcal{L}=\mathcal{L}_{0}\sqrt{1-\left(v/c\right)^{2}}$$ I could imagine repeating the boosting several times, while only going into an inertial frame momentarily to measure the length. It would appear as though the “bus” is shortening with a measurable speed:
$$\frac{d\mathcal{L}}{dt}=\frac{d}{dt}\mathcal{L}_{0}\sqrt{1-\left(v/c\right)^{2}}$$
$$=\frac{\mathcal{L}_{0}}{c^{2}}\frac{\left(-v\right)}{\sqrt{1-\left(v/c\right)^{2}}}\frac{dv}{dt}$$
I understand special relativisticaly how this occurs, but now we have acceleration and noninertial reference frames. This is more the purview of General relativity. How would I go about calculating this observed “speed of length contraction” properly for some constant acceleration and relative velocity? I'm guessing there's more to the story? I'm very familiar with GR, but for some reason this is stumping me.
Also, an accelerating observer should have a time dilation due to acceleration and not just velocity, I'm not sure where that fits in
| If $\rm \ L(t)=L_0/\gamma(t)=L_0 \sqrt{1-v(t)^2/c^2}$
then $\rm \ L'(t)=dL/dt=-L_0 v'(t) \ v(t) \ \gamma(t)/c^2$
see here and here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/742455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to explain radiation of an accelerated charged particle via the Unruh effect? From the point view of a noninertial observer, if a charged particle accelerates then it can catch the Unruh particle and excite. After that, the charged particle emits it and falls back to its ground state.
If we accept radiation of an accelerated charged particle like that then we can say the same thing for an accelerated normal particle (without charge) and conclude it radiates too.
How to solve this paradox?
| Indeed, electromagnetic radiation of accelerated charges can be understood in the accelerated reference frame by means of the Unruh effect, as shown by Higuchi, Matsas, and Sudarsky in Phys. Rev. D 45, R3308(R) (1992) and Phys. Rev. D 46, 3450 (1992). The emission of a photon in the inertial frame corresponds to either the emission or the absorption of a photon in the accelerated frame.
An important remark though is that an accelerated particle will not absorb or emit a photon in either reference frame unless it is charged. This is because this absorption or emission will only happen if there is an interaction between the particle and the photon field, which only happens if the particle is charged. Hence, neutral particles won't radiate.
In other words, neutral particles don't radiate because they can't "catch" (or release) Unruh particles. To do so, they would need to interact with the photon field, but they don't.
Remark: I'm considering only quantum electrodynamics in this answer, meaning I'm ignoring more complicated interactions that could happen at higher loop levels in the Standard Model. After all, we are trying to understand a prediction of classical electrodynamics (radiation by accelerated charges).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/742608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why do we choose the Dirac delta function as the eigenstate of position operator? When we try to find the eigenstates of the position operator, we get that the product of (x-y) and the eigenstate must be zero. It is obvious then that for x different than y, the eigenstate must be zero.
Now for x equal to y, how do we know that the eigenstate is infinite so that we get the Dirac delta function? What if we choose any other form of the eigenstate for x=y - it would still be zero and satisfy our math, right?
| This is because the position operator is $x\delta (x-x') $. This may sound like I'm just kicking the can down the road but I'm not.
Naively, you would expect the position operator to be $x \delta _x ^{x'}$, where the $\delta$ is the Kronecker delta. But keep in mind that $x$ is a continuous index. So the action of the operator on a state cannot be defined to be a sum:
$$\sum _ {x'}A(x, x') f(x') $$
It's instead defined to be an integral:
$$\int A(x, x') f(x') dx'$$
You can see that if we choose $A=x\delta (x-x') $, this yields the correct action of the position operator:
$$X (f(x))=xf(x) $$
The action of the operator, defined using the integral, will be zero on any vector if you chose the Kronecker delta as the position operator.
You can try using the eigenvector as $f(x)=1$ at $x=a$ and $=0$, everywhere else and the Kronecker delta as the position operator. The action of this operator will yield zero because the integral will be zero.
This still leaves the option of having the operator as $x\delta (x-x') $ but your proposed functions as eigenvectors. The problem with this is that you cannot write an arbitrary vector $f(x)$ as a linear combination of your proposed eigenvectors:
$$f(x)=\int f(x_0) E(x, x_0) dx_0$$
$f(x_0) $ is the component of the vector projected along the eigenvector $E(x, x_0) $. You can see that $E(x, x_0) $ must be the set of delta functions to satisfy the linear combination equation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/742750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Open-close string amplitude My question is regarding Polchinski question number 6.12 in which we have to find the amplitude for two open and one close tachyon strings.
I found this solution by Matthew Headrick ("A solution manual for Polchinski's "String Theory""):
Why does the calculation is done in the disk $D_2$ (as usually done for open strings) and not on the sphere (as usually done for closed strings)?
Edit:
I think that most of my confusion is that it I don't understand how we can compare open with close states. Don't they have different vacuum state and hence totally different Hilbert space?
| The calculation done on the disk because in order to be able to have open states we need that our surface will have a boundary.
On the disk we have the boundary that allow us to represent open strings and we also have the interior (bulk) that allow us to represent close strings.
Regarding the vacuum, they do have the same vacuum states and the difference in the excited states is given by the difference between acting with vertex operator on the boundary vs vertex operator in the interior.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/743213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why $F = m(v_f - v_0)/2$? Force is directly proportional to mass and velocity and inversely proportional to time so why don't we write $F=1/t+m+v-v_0$ where $m$ is mass, $v$ is final velocity, and $v_0$ is initial velocity?
|
Force is directly proportional to mass and velocity and inversely proportional to time so why don't we write $F=1/t+m+v-v_0$
As others mentioned, the units don’t work. However, suppose we modify it to $$F=k_t/t+k_m m+k_v(v-v_0)$$ where the various $k$ are constants with appropriate dimensions that make each term a force.
Now, you have an equation that is dimensionally consistent. However, it is not directly proportional to mass and velocity and inversely proportional to time. If you double $m$ then according your formula you do not double $F$. Instead you get $$k_t/t+k_m 2m+k_v(v-v_0) \ne 2 F$$
For $F$ to be proportional to $m$ means $F=km$, and similarly with the other factors.
Also, one nitpick. Force is not proportional to velocity but the average force is proportional to the change in velocity. Those are slightly different statements.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/743353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How does $\sum_s\left[u_a^s(p)\bar u_b^s(p) + v_a^s(-p)\bar v_b^s(-p)\right] = 2E_p\gamma^0_{ab}$? For Dirac spinors, we have the spin sums for particles,
$$
\sum_{s=1,2}u_a^s(p)\bar u_b^s(p) = (\not\!p+m)_{ab},
$$
and for antiparticles:
$$
\sum_{s=1,2}v_a^s(p)\bar v_b^s(p) = (\not\!p-m)_{ab}.
$$
As $\{u^s(p),v^s(-p)\}$ span the spinor space, and are normalized to $2E_p$, we have
\begin{equation*}
\sum_s\left[u_a^s(p)\bar u_b^s(p) + v_a^s(-p)\bar v_b^s(-p)\right] = 2E_p\gamma^0
\end{equation*}
I'm struggling with verifying this relationship. If we sum over the two spin sums for particles and antiparticles, shouldn't we get $2\!\not\!p_{ab} = 2(\gamma^0p_0-\gamma^i p_i)$? It seems like the third equality contains $2\gamma^0p_0$ only.
| The correct relation involves inverting only the three-momentum of the spinors,
$$\sum_{s}\left[u^{s}(\mathbf{p})\bar{u}^{s}(\mathbf{p}) + v^{s}(-\mathbf{p})\bar{v}^{s}(-\mathbf{p})\right] = 2E_p\gamma^{0}.$$
This follows straightforwardly from the closure relations, which may be written
$$\sum_{s}u^{s}(\mathbf{p})\bar{u}^{s}(\mathbf{p})=E_{\mathbf{p}}\gamma^{0}-\boldsymbol{\gamma}\cdot\mathbf{p}+m \\
\sum_{s}v^{s}(\mathbf{-p})\bar{v}^{s}(\mathbf{-p})=E_{\mathbf{p}}\gamma^{0}+\boldsymbol{\gamma}\cdot\mathbf{p}-m,$$
which obviously sum to the required expression.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/743615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do we maintain polytropic processes? In polytropic processes for an ideal gas,
$$PV^{\alpha}=constant$$ where $\alpha \neq 0,1,\gamma$
And $\gamma$ is adiabatic exponent of gas
So, how these processes are maintained?
What things are done to initiate this process?
| By definition a a polytropic process is one for which $TdS=\mathcal K dT$ and $\mathcal K$ is a constant. Using the $dU=TdS-pdV$ equation it follows that
$$\frac{dp}{p}+\alpha \frac{dV}{V}=0 \tag{1}\label{1}$$
and upon integration you get $$pV^{\alpha}=K_0\tag{2}\label{2}$$
where $\alpha=\frac{C_p-K}{C_V-K}$.
This means that if you change the volume by an amount of, say, $\delta V$ then you have to change the pressure by $\delta p = -p \alpha \frac{dV}{V}$. This can be achieved by absorbing $\delta S=\frac{\mathcal K}{T}\delta T$ entropy from a thermal reservoir at temperature $T+\delta T$ where $pV=RT$ and $\delta T=({V \delta p+p \delta V})/R\alpha=\frac{1-\alpha}{R\alpha}p \delta V$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/743735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why do we Wick rotate before regularizing Feynman diagrams? In Folland's Quantum Field Theory he mentions that we can apply Feynman's formula (Feynman parameterization) to either the Wick rotated integrals or the non-Wick rotated integrals corresponding to Feynman diagrams. However, later in the section he outlines a regularization procedure, in which he says we must first Wick rotating the integrals. Why is this Wick rotation necessary? Is it to make use of certain properties of Euclidean space that are not shared by Minkowski space? If so, which ones?
| Well, as long as the singularities of the propagators in Minkowski signature are regularized with a Feynman $i\epsilon$ prescription, Wick rotation is in principle not necessary. However in practice, the loop momentum integrals of the Feynman diagram are simpler to evaluate in Euclidean signature, e.g. because $SO(d)$ symmetry is simpler to handle than $SO(d-1,1)$ symmetry.
See also this related Phys.SE post.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/744242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Understanding this Lagrangian calculation I was trying to understand this section of a Wikipedia article:
$$0 = \delta \int \sqrt{2T} d\tau =
\int \frac{\delta T}{\sqrt{2T}} d\tau =
\frac{1}{c} \delta \int T d\tau$$
For the life of me, I can't figure out how does one get from $\displaystyle \delta \int \sqrt{2T} d\tau$ to $\displaystyle \int \frac{\delta T}{\sqrt{2T}} d\tau$. What is $\delta$? I assume it is some kind of differential operator but in respect to what variable?
|
I was trying to understand this section of a Wikipedia article:
$$0 = \delta \int \sqrt{2T} d\tau =
\int \frac{\delta T}{\sqrt{2T}} d\tau =
\frac{1}{c} \delta \int T d\tau$$
For the life of me, I can't figure out how does one get from $\displaystyle \delta \int \sqrt{2T} d\tau$ to $\displaystyle \int \frac{\delta T}{\sqrt{2T}} d\tau$.
Let the function $T(\tau)$ be changed to another function that is "close" to $T(\tau)$:
$$
T(\tau) \to T(\tau) + \epsilon \eta(\tau)\;,
$$
where $\eta(\tau)$ is a function and $\epsilon$ is "small" constant in a sense we will describe more below.
We note that the relationship between our notation and the Wikipedia notation is:
$$
\delta T = \epsilon \eta(\tau)\;,
$$
where our notation makes explicit that the variation $\delta T$ of $T$ is an arbitrary function $\eta(\tau)$ times a "small" expansion parameter $\epsilon$.
We literally make the above replacement in the integral of interest:
$$
I = \int \sqrt{2T}d\tau \to \int \sqrt{2T+2\epsilon \eta(\tau)}d\tau\;,
$$
and then we look at only the first order change in the integral with respect to $\epsilon$. We look at the first order change because the integrand of the first order change in $I$ is what we define as the functional derivative of $I$ (in analogy with the derivative of a multivariate function). We are "allowed" to do this because $\epsilon$ is "small" in the sense that we can ignore all the higher order terms in the $\epsilon$ expansion.
Expanding to first order in epsilon can be performed in this case via the known McLauren/Taylor series for the function $\sqrt{1+x}$. (Which is $\sqrt{1+x} \approx 1 + x/2 + \ldots$, where the terms being ignored are of order $x^2$ and higher.)
$$
\int \sqrt{2T+2\epsilon \eta(\tau)}d\tau
=\int \sqrt{2T}\sqrt{1+\frac{\epsilon \eta(\tau)}{T}}d\tau
\approx \int \sqrt{2T}\left({1+\frac{\epsilon \eta(\tau)}{2T}}\right)d\tau
\equiv I + \delta I\;,
$$
where, now we see that:
$$
\delta I = \int \sqrt{2T}\left({\frac{\epsilon \eta(\tau)}{2T}}\right)d\tau
=\int \frac{\epsilon \eta(\tau)}{\sqrt{2T}}d\tau
\equiv\int \frac{\delta T}{\sqrt{2T}}d\tau\;.
$$
We could also write this result as a functional derivative, by definition of the latter, as:
$$
\frac{\delta I}{\delta T(\tau)} = \frac{1}{\sqrt{2T}}
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/744328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why "into screen" and "out of screen" markings in electromagnetism are the way they are? Let's say that you have a bullet going through the metal screen. Into screen. It makes a dot-like hole, like here:
Now, let's say that we have a bullet going out of screen. It makes a jagged, sharp edges, often 3-4 of them, making it often look like an X, just like here:
It doesn't necessarily have to be a bullet. It can be a nail, a dart, a compass needle point or many other everyday items, going through different materials with the similar effect as above, making this result something very intuitive.
However, in physics, mostly electromagnetism, we are presented with completely opposite denomination, i.e. dot meaning out of screen and X meaning into screen, which is reverse of the intuitive logic above:
What is the reason behind it? Is it because of some tradition or historical reasons?
I studied math (and some physics, though I quit on 4th semester) for quite several years and never understood it, and it was counter-intuitive for me. Even more for some people I know that always had a problem with this symbolism.
I memorized it myself by using the very fact of it being reversed (for me, in my personal opinion), as to never mistake it, which is kinda funny, to always have to think of opposite!
| You just need to have the right picture in mind, then it's actually very intuitive and easy to remember. From rfcafe.com:
We anyway draw fields as arrows. Now, imagine someone shot an arrow in the plane and it got stuck, so you either see the pointy end just barely sticking out or the fletching (the feather cross at the tail end).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/744431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do atoms emit an electromagnetic wave (infrared radiation)? From what I understand, when an object has a certain temperature, its atoms vibrate and this atomic vibration accelerates the electrically charged particles and this generates infrared radiation.
To generate infrared radiation, it is therefore necessary to accelerate electrically charged particles, but since atoms are electrically neutral, how can their acceleration generate infrared radiation?
| I would taks H atom as an example. One electron and one proton. There are a number of discrete energy levels (negative energies since the electron is bound to the proton). Normally the electron is in the lowest state of energy called the ground state which is -13.6 eV. If the electron due to collisions with other atoms or some other reason such as temperature goes to a higher energy then it cant stay there for long it comes back to its ground state. Depending on the difference of two levels the E.M. radiation is emmited. This is typically the case of all atoms. IR visibile UV X-rays etc radiations can therefore be emmitted.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/744698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Pauli Exclusion Principle with two electrons in box Suppose that I measure the spin of two electrons in the z direction, and that they have the same spin. Then we put both electrons in a box. After a long time, we know nothing about the spatial distribution of the electrons in the box, so their spatial distributions are equal. But their spins in the z direction are still equal. This seems to contradict the Pauli Exclusion Principle. How can this be resolved?
| Why would we know nothing? If you place the electrons in an antisymmetric spatial state, say
$$
\psi_{12}(x_1,x_2)=\frac1{\sqrt{2}}\left(\psi_1(x_1)\psi_2(x_2)-\psi_2(x_1)\psi_1(x_2)\right)
$$
so the full spatial+spin state is antisymmetric as required by Pauli:
$$
\psi_{12}(x_1,x_2)\vert \uparrow\rangle_1\vert\uparrow\rangle_2
$$
then the state will evolve
$$
\Psi(x_1,x_2,t)=e^{-i t E_{12}/\hbar}
\psi_{12}(x_1,x_2)\vert \uparrow\rangle_1\vert\uparrow\rangle_2
$$
so the distribution is well known.
Even if you use a linear combination of antisymmetric states:
$$
\psi(x_1,x_2)=\sum_{ij}c_{jk} \frac1{\sqrt{2}}\left(\psi_k(x_1)\psi_j(x_2)
-\psi_j(x_1)\psi_k(x_2)\right)
$$
the time evolution will be given by
$$
\sum_{ij}c_{jk} e^{-i E_{jk}t/\hbar}\frac{1}{\sqrt{2}}\left(\psi_k(x_1)\psi_j(x_2)
-\psi_j(x_1)\psi_k(x_2)\right)\vert\uparrow\rangle_1\vert\uparrow\rangle_2
$$
which is still (legitimately) antisymmetric.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/744947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can we say that the center-of-momentum frame is the frame in which the center of mass is at rest? Isn't the center-of-momentum frame is same as the frame in which the center-of-mass is at rest? Since the position of the center-of-mass of a system of particles is defined as $\vec R=\sum_i m_i \vec r_i/M\Rightarrow\frac{d\vec R}{dt}=\sum_i m_i\vec v_i/M=\sum_i\vec p_i/M$. Therefore, the frame in which the center of mass is at rest, $d\vec R/dt=0\Rightarrow \sum_i \vec p_i=0$. Now, by definition, the center-of-momentum frame is one in which the total linear momentum vanishes. So am I right in saying that the "center-of-momentum frame is the same as the frame in which the center-of-mass is at rest."?
| Let's check a simple 1D situation: 2 equal mass particles on a collision course. One (1) at rest at the origin, and another (2) moving in the $+$ direction towards it:
$$ x_1m = 0$$
$$ x_2(t)m = (x_2(0)+\beta t)m$$
Center of mass is clearly:
$$ X_{cm}(t) = \frac 1 2 (x_2(0)+\beta t) $$
Wait. We could do this problem in the center of mass frame where (maybe) both particles have equal and opposite velocities, so let's boost everything by $\beta/2$:
$$ v_1 = \frac{0-\beta/2}{1+\frac 1 2 0\beta}= -\beta/2 $$
$$ v_{cm} = \frac{\beta/2 - \beta/2}{1+\frac 1 4 \beta^2} = 0 $$
Now here comes the rub:
$$ v_2 = \frac{\beta + \beta/2}{1+\frac 1 2\beta^2} = \frac 3 2 \frac{\beta}{1 + \frac 1 2 \beta^2} \ne + \beta/2 = -v_1$$
So there is clear asymmetry.
Another way of looking at this is that the center of mass (in 1D) is the mean of $x$ weighted by $m$, while the center of momentum is weighted by $\gamma\beta m$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/745211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the corresponding energy given the wave function? So I was struggling doing the following question:
Given the wave function $\psi(x) = A\, \mathrm{e}^{-ax^2} $ with potential $V = \frac12 kx^2$, find the corresponding total energy in terms of $k$ and $m$.
I did the calculation for $\left<x^2\right>$ and $\left<p^2\right>$ but it turns out to be expressions including $a$. How can I express $a$ in terms of $k$ and $m$?
| The energy of a state is the eigenvalues of the Hamiltonian. So calculate the expectation value of it for that state.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/745599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
In the Lorentz force equations, can the magnetic field be assumed to be singular at one or more points? Consider the relativistic Lorentz force equation (in a simplified form) given by
\begin{equation}\
\left(\frac{u'}{\sqrt{1-|u'|^2}}\right)'= E(t,u)+u'\times B(t,u).
\end{equation}
Here, $E$ and $B$ denote respectively the electric and magnetic fields and are given by
\begin{equation}
E=-\nabla_u V-\frac{\partial W}{\partial t}, \qquad B=\mbox{curl}_u\, W,
\end{equation}
with $V:[0,T]\times (\mathbb R^3\setminus\{0\}) \to\mathbb R$, $W:[0,T]\times\mathbb R^3\to\mathbb R^3$ two $C^1$-functions.
The models I have seen so far deals with $V$ singular at one (or more) points ($0$ in this case, e.g.). My question is: there exists a relevant physical model in which also $W$ is singular at one or more point?
| A general solution for the magnetostatic case for the magnetic Vectorpotential (what you call $W$) is given by:
\begin{align}
{\displaystyle {\vec {A}}({\vec {r}})={\frac {\mu _{0}}{4\pi }}\int {\frac {{\vec {j}}({\vec {r}}')}{\left|{\vec {r}}-{\vec {r}}'\right|}}\mathrm {d} ^{3}r'\,}
\end{align}
If your $\vec{j}$ becomes singular, then $\vec{A}$ would be singular at that singularity as well. An often used model is the model of an infinitely thin wire that carries an electric current. This infinitely thin wire then carries a current distribution that is zero everywhere, except for points in the wire, where it takes on the values of a delta distribution. For that arrangement, the magnetic vector potential will become singular at all that points.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/745859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Nodal circles on spherical harmonics Homework question:
For the spherical harmonic with $l=2$, $m=0$ , at what angle theta relative to the polar axis is the nodal circle in the northern hemisphere?
My attempted answer:
I know that I need to find the values of $\theta$ for which the function is equal to zero. $Y(2,\theta) = 0$ and solving for $\cos(\theta)$, I got $3\cos^2(\theta) - 1 = 0$, $\cos(\theta) = \pm√(1/3)$ and so $\theta$ = around $109.47$° and $70.52$°. So for the northern hemisphere, I picked $109.47$° as my answer but it is not the answer. What is my mistake here? Thank you.
| Remember the north pole is at $\theta=0°$, the equator is at $\theta=90°$,
and the south pole is at $\theta=180°$.
From this you know which solution to pick for the northern hemisphere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/745963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is so significant about electron spins and can electrons spin any directions? I just want to know what is so significant of with direction electron is spinning. Does it have any effect on the element or on the atom?
Also, does electron must spin up or down or can they also spin sideways or vertically?
| Electron spin is proportional to the magnetic dipole moment produced by an electron that is not traveling. This acts very much like the dipole moment of a current traveling through a circle of wire. The magnitude of the spin vector is set, but it can be in any direction.
When you measure a component, any component, of that magnetic dipole moment, and thus the spin, you will get only one of two results. It will be fully along that axis in one of the two possible opposite directions. If you change to a perpendicular axis, you will still get only one of these two results but along the new axis. To measure something, you must interact with it. At the quantum level, making a measurement forces the object into one of the allowed states of what you are measuring.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/746063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to statically generate lift with the difference in pressure like wings? If I understood it correctly, the shape of the wings and/or propellers generates lift/thrust with the difference in pressure in both sides of the wings/propellers; where the lower side has higher pressure airflow and the uper side has low pressure airflow.
With this in mind, I was wondering if it is possible to generate an area of low pressure around the upper part of the an aircraft without the moving balloons, wings or propellers/rotors.
A "static lift" is the best way I could put it.
So, would such thing be possible? Or lift would only be achieved with the airflow that wings already work around?
| If you have a difference in pressure, then fluid will flow to equalise that pressure unless restricted by a barrier. If you have a barrier then that's a balloon, if you don't have a barrier than you don't have a static system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/746531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Metric compatibility and torsion-free condition of GR In an introduction to general relativity, we see the unique connection of a manifold is described by both the conditions, matric compatibility and torsion-free condition. The metric free condition can be physically understood which allows us the parallel transport of vectors between two tangent spaces. But What is the importance of the torsion free condition? The problem would be no unique connection for a manifold. Is it an issue in GR? I mean is there any way which can relax this condition
| For knowing why the Torsion tensor is important it is important to know what it is.So,
$$[\nabla_\mu ,\nabla_\nu]f=T^\lambda_{\mu \nu}\nabla_\lambda f$$
Where T is the torsion tensor, so if your covariant derivatives commute you have torsion free manifold. Quickly working out the commutator of the covariant derivatives will shows you that
$$T^\lambda_{\mu \nu}=\Gamma^\lambda_{\mu \nu}-\Gamma^\lambda_{\nu \mu}$$
So when we talk about your metric compatibility we are assuming that the metric is diagonal. In that case your Riemann Christoffel Tensor is also symmetric under exchange of μ and ν which leads to a torsion free manifold. You bring out a metric that has a non diagonal and non zero component and suddenly both conditions vanish such as in the case of the Einstein Cartan Theory which is mentioned by@gold
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/746638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Thermodynamic work done on a free-falling object By definition the thermodynamic work is the opposite of the mechanical work (engineers'convention).
Consider then an object of mass $m$ falling at constant velocity above the Earth surface (imagine that someone is pushing in order to balance the gravitational pull) for an height $h$. We know that the potential energy decreases by $-mgh$, hence the internal energy should also decrease by the same amount however, by applying the First Law of Thermodynamics I get that $\Delta U = 0$ because there is no net work done on the object.
I would like to know what I am doing wrong.
| The complete form of the first law is:
$$
\Delta (U + E_K + E_P + \cdots) = Q + W,
$$
where $U$ is internal energy, $E_K$ is kinetic energy and $E_P$ is potential energy. The left-hand side includes all forms of energy in the system; the right-hand side is the net inputs of heat and work. For a stationary system, like a cylinder/piston that we often use in thermodynamics, $\Delta E_K=\Delta E_P=0$, but in the case of the example that you give we have
$$\Delta E_K +\Delta E_K = Q = W = 0$$
The net result is that potential energy is converted to kinetic. $\Delta U$ is zero in this case. Internal energy will increase when the mass impacts on the ground. Unless the collision with the ground is perfectly elastic some energy will be converted into internal but the body will rebound with less kinetic energy than when it hit the ground.
(I am ignoring in this analysis the drag due to air, which would convert some of the kinetic energy into internal of the air and of the falling body).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/746969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does rate of acceleration change as object gets closer or further to center of a mass? I learnt that newton's law of universal gravitation F = G(m1m2)/R^2, and thought if the R is distance and determined gravitational strength, why do we use 9.81 as default acceleration of earth's gravity when it is not even constant at different heights?
| We use 9.81 m/s$^2$ only when dealing with objects close enough to the surface of the earth (heights negligible compared to the radius of the Earth) that the acceleration due to gravity can be considered constant.
The value of $g$ near the surface of the Earth can be computed when you substitute the radius of the earth for $R$ and calculate the acceleration $g$ using the universal law of gravitation equation, where $M$ is the mass of the earth,
$$g=\frac{GM}{R^2}$$
Hope this helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/747083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Linear Harmonic motion (simple oscillator) We know that for a simple harmonic linear oscillator, the displacement is given by $x(t)=A\sin(\omega t + \phi)$, where $\phi$ denotes the phase angle. Now as per my understanding this $\phi$ is only significant when considering SHM in form of a sinusoidal wave. Is there any physical meaning in reality. Is there a way to measure the phase angle in reality just by virtue of the particle's (which is oscillating) position with respect to the mean position?
Some Assumptions
*
*I am considering SHM in only one axis.
*I am also considering SHM linearly for ex Spring Block System.
| SHM has this differential equation
$$\ddot x(t)+\omega^2\,x(t)=0\tag 1$$
to solve this differential equation you need two initial conditions.
$$x(0)=x_0~,\dot x(0)=v_0$$
with your Ansatz $~x(t)=A\,\sin(\omega\,t+\phi)~$ in equation (1) you obtain $~0=~0~$ thus $~x(t)~$ is the solution of the SHM. to obtain the constants $~A~,\phi~$ you have two equations
$$x(t=0)=A\,\sin(\phi)=x_0\\
\dot x(t=0)=A\,\omega\cos(\phi)=v_0~,$$
for the unknowns $~A~,\phi~\Rightarrow$
$$A=\frac 1\omega\,\sqrt{\omega^2\,x_0^2+v_0^2}\\
\phi=\arctan\left(\omega\frac{x_0}{v_0}\right)$$
this work also for $~v_0=0~$
$$\phi=\rm arctan2\left( {\frac {x_0\,\omega}{\sqrt {{x_0}^{2}{\omega}^{2}+{v_0}
^{2}}}},{\frac {v_0}{\sqrt {{x_0}^{2}{\omega}^{2}+{v_0}^{2}}}}
\right) $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/747185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How did we get the formula $d U = nCvdT$? Our teacher taught us that for any thermodynamic process, dU=nCvdT where Cv is molar specific heat capacity at constant volume and dU is change in internal energy. How did we get this formula and why is it valid for all processes
| The general relation for all materials is $$dU=C_V\,dT+(\alpha TK-P)\,dV,$$
with internal energy $U$, constant-volume heat capacity $C_V\equiv T\left(\frac{\partial S}{\partial T}\right)_V$, temperature $T$, constant-pressure thermal expansion coefficient $\alpha\equiv\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$, constant-temperature bulk modulus $K\equiv-V\left(\frac{\partial P}{\partial V}\right)_T$, pressure $P$, and volume $V$. We derive this from expanding $U$ in its natural variables $T$ and $V$:
$$dU=\left(\frac{\partial U}{\partial T}\right)_V dT+\left(\frac{\partial U}{\partial V}\right)_T dV,$$
with the rest being just application of the identities above along with the fundamental relation $dU=T\,dS-P\,dV$ for a closed system. (Note that $\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V=-\left(\frac{\partial V}{\partial T}\right)_P\left(\frac{\partial P}{\partial V}\right)_T$ via a Maxwell relation and the triple product rule.)
So $dU=C_V\,dT$ clearly holds for constant-volume processes ($dV=0$). Interestingly, it also always holds for the ideal gas, whose equation of state $PV=nRT$ causes $\alpha TK-P$ to be identically zero (as $\alpha =\frac{1}{T}$ and $K=P$). That is, even if $dV\neq 0$, its coefficient is zero. This has frustrated endless numbers of new thermodynamics practitioners because the resulting equation (which contains a material property with "constant-volume" in its name) applies to all processes, and in general—but only for the ideal gas. It's as if the equation is too simple for its own good. Much more discussion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/747357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Why is the acceleration vector the spatial gradient of the lapse function? If we have a Lorentzian manifold $(M, g)$ with a foliation by spacelike surfaces $\Sigma_t$ with unit-normal vector field $n$, we can define the lapse function $N$ by
$$
\partial_t = N n + X
$$
where $X$ is the shift vector. I have seen several claims that, for any $Y$ tangent to $\Sigma_t$, we have
$$
g(\nabla_n n, Y) = \nabla_Y \ln N = N^{-1} \nabla_Y N = N^{-1} Y(N),
$$
but I cannot find a proof for this. It is often stated as a trivial consequence of the definitions but I cannot derive it myself. Is there something obvious I am missing?
| Write
\begin{align*}
\langle \nabla_n n, Y\rangle &= -\langle n, \nabla_Y n\rangle-\langle n, [n,Y]\rangle
\\
&=-\frac 12 Y \langle n,n\rangle - \langle n, N^{-1} Y(N)\rangle
\\
&=N^{-1} Y(N).
\end{align*}
You can get the formula for $[n,Y]$ by writing $n=\frac 1N(\partial_t-X)$ and using the coordinate expression for the Lie bracket.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/747472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can red and blue light interfere to make fringes in young’s double split experiment? Supposing in young’s double split experiment, I cover one slit with red filter and the other slit with blue filter. The light coming out from the first slit would be red and the second slit would be blue. Would there be any interference fringes? I tried googling this question but all the answers say that two different monochromatic lights cannot interfere and hence no interference pattern. But, we do know that if we did the experiment using white light, there is a pattern(for a few fringes). So what should be the right answer?
| So the concept of double slit "interference" is historical dating back to Young and Fresnel (early 1800s)... and is still the basis for the typical high school/university textbook. When quantum mechanics came about in the early 1900s some of the most famous scientists (Dirac and others) pointed out that "each photon interfered with itself" which was just their way of saying each photon determines its own path. (But it was complex, subject to argument, and textbook writers were happy with what they had written.)
So photons do not cancel each other, that would be a violation of conservation of energy. When an atom/electron is excited it is already generating virtual EM forces .... when the EM forces see 2 slits they cause the pattern ... the real photon is eventually released and follows the pattern.
So when you filter the slits the red "to be" photons/electrons/atoms only see one slit ..... so no "interference".
Unfortunately there may very likely be high school textbook problems that want you to calculate an interference pattern based on the fact that the 2 colours have a different wavelength value .... it becomes more of a math problem than a physics problem!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/747560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Coefficient of restitution bouncing ball I was researching for my physics investigatory project and I have very big confusion.
The ratio between the initial height of the ball (the independent variable) and the height the ball bounces (the dependent variable)
$$\mathrm{COR} = e =\sqrt{\dfrac{\text{final height}}{\text{initial height}}}.$$
As the drop height increases, COR should decrease or will it be equal, regardless of the changes in the size of the ball and the drop height?
Also if the drop height increases will the bounce height also increase?
| The COR is technically only defined for collisions. In this case the ball collides with the floor so we use the COR. The COR is more precisely defined as
$$e=\frac{v_f}{v_i}$$
where $v_i$ is the (relative) velocity before the collision and $v_f$ is the (relative) velocity after the collision. With relative I mean the relative velocity between the two objects. If one ball was moving with $v=5$ to the right and another ball with $v=1$ to the left, the relative velocity would be $5-(-1)=6$: the rate at which the gap between them is closing.
The COR tells you how much energy is lost during a collision. For $e=1$ no energy is lost and for $e=0$ all energy is lost. A collision between two pieces of clay which stick together after the collision is an example where all energy is lost. You can also write the COR using the kinetic energies using $v=\sqrt{2E_k/m}$, which gives
$$e=\sqrt{\frac{E_{k,final}}{E_{k,initial}}}$$
You can then use conservation of energy,
$$E=\tfrac 1 2mv^2+mgh,$$
to derive the equation you mentioned. You have to relate the energy before the drop to the energy just before the collision. This becomes
$$e=\sqrt{\frac{h_f}{h_i}}.$$
Note that this equation is only a valid expression for the COR if no energy is lost during the fall. Generally there will be air friction which can dissipate energy. If no energy is lost to friction, $e$ will be independent of drop height. Now my question for you is: if there is air friction, will the COR decrease or increase if you increase the drop height?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/747702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do non-periodically varying currents produce electromagnetic waves? Electromagnetic radiation is created by the varying/accelerating of a system of charges and currents. Suppose that the time dependence of the charges and currents are $\rho(x,t)$ and $J(x,t)$. Then the subsequent radiation will have the same time dependence. In Jackson it is stated that we can assume $\rho$ and $J$ have harmonic time dependence because we can build up any "arbitrary" function as a superposition of sinusoidal functions via Fourier analysis. My understanding of Fourier series is that we can only do this for periodic functions. We always refer to electromagnetic radiation as a wave because of the harmonic time dependence but the radiation has the same time dependence as $\rho(x,t)$ and $J(x,t)$. So what do we do, if the charges and currents are accelerating but not periodically? Then the electromagnetic radiation would not be a wave I think. So why does Jackson state that we can assume harmonic time dependence without losing any generality?
| Fourier analysis does not require periodicity. This was one of the things about Fourier's work that shocked mathematicians. Essentially, any function you can graph may be decomposed into sinusoidal waves.
Imagine electric field lines converging at a charge in a particular position. Now, move that charge to a new position. The electric field lines will converge at the new position. But the news that you've moved the charge can't travel faster than the speed of light. So, there's a kink in the lines joining the old lines with the new.
It turns out that that kink moves at the speed of light. It's a non-periodic electromagnetic wave.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/748085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Wavefunction of distinguishable spin 1/2 fermions Does the total wave function for distinguishable (i.e. not identical) spin 1/2 fermions need to be anti-symmetric under particle exchange? Or does the Pauli exclusion only hold for indistinguishable fermions?
| No. Particles may be indentical but may or may not be indistinguishable. Indistinguishable particles will be described by a quantum state fully symmetric or fully antisymmetric under permutation.
It is perfectly possible to have partially distinguishable particles: two photons with non-orthogonal polarization made to interact at a beam splitter are an example of system of partially distinguishable particles. If the photons are described by wave packets, then the overlap of the wave packets serves as a measure of distinguishability. The most spectacular example of this is the Hong-Ou-Mandel effect, where certain coincidence rates basically go to 0 only when the photons are fully indistinguishable:
In the figure, the overlap between two Gaussian wave packets is tuned using an ajustable delay. When the two packets overlaps and the photons are fully indistinguishable, there is no probability of the two photons exiting to different ports of the interferometer. As one increases the distinguishability, this probability increases.
If you’re looking for an example with two fermions, imagine two wave packets describing identical spin-1/2 fermions but such that the wave packets are only partially overlapping. In this case the state is described by a linear combination of a fully symmetric state, and a fully antisymmetric state (here we’re talking about symmetric/antisymmetric w/r to the action of the permutation group $S_2$), and the coefficients are such that, as the wave packets fully overlap, only the antisymmetric combination remains. Indeed the HOM experiment has been done with electrons, as reported in
Liu, R. C., et al. "Quantum interference in electron collision." Nature 391.6664 (1998): 263-265
and in
Oliver, William D., et al. "Hanbury Brown and Twiss-type experiment with electrons." Science 284.5412 (1999): 299-301.
Note the similarity between the results of Oliver:
and the results of Hong et al.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/748379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why can you hear loud TV in the next room despite the wall and door? In a house, when two rooms are next to each other, why can you hear the loud TV on the next room, despite the wall between them and despite that their two doors are closed. (I don't know a lot on physics, but isn't there something like sound travelling, does the sound travel through the wall ? It must be 3 to 5 centimetres). The loud TV is especially annoying because I hear all the bass sounds (less of the higher-pitched sounds). And I read about how just bass sound makes you anxious (especially if you are trying to sleep).
| Their are two main transmission mechanism: The first one is air gaps: residential doors are not airtight and even a small gap will transmit a fair bit of sound.
The second is structure borne: The TV or loudspeaker will vibrate quite a bit and mechanically excite the surface they are resting on: the shelf, cabinet, floor, wall etc. That vibration makes it through to the next room and gets radiated by the walls, floor and/or ceiling. To a lesser extent the air-borne sound in the TV room will also vibrate other objects in the TV room (especially if the sound hits the resonance frequency) and that vibration will also end up in walls and floors.
Finally, there is direct transmission through the wall. The wall itself is also a mass/spring system which can be excited by an incident sound wave but for a decently constructed wall there is typically not a lot transmission since the wall is so much heavier than the air.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/748664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to justify this small angle approximation $\dot{\theta}^2=0$? Suppose the equation of motion for some oscillating system takes the following form:
$$\ddot{\theta}+\dot{\theta}^2\sin\theta+k^2\theta\cos\theta=0$$
Applying small angle approximation to $\theta$ gives $\sin\theta\approx\theta$ and $\cos\theta\approx1$,
$$\ddot{\theta}+\dot{\theta}^2\theta+k^2\theta=0$$
But I am wondering if it is alright to set $\dot{\theta}^2=0$ so that the equation simplifies to
$$\ddot{\theta}+k^2\theta=0$$
If so, how can this approximation be justified?
| There is no mathematical justification for setting $\dot \theta^2$ to $0$. If this equation represents some specific physical system and if it is known for physical reasons that $\dot \theta^2 \approx 0$ then it could be justified. But that will require a specific physical argument for the specific system in question, not a generic mathematical justification.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/748837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is relativity of simultaneity an "observer issue"? There are some threads about this, but some answers seem to disagree.
First, this is what Einstein said on this matter:
The light rays emitted by the flashes of lightning A and B would reach him simultaneously", and again: " the observer will see the beam of light emitted from B earlier
and this is the reason why he says:
Events which are simultaneous with reference to him are not simultaneous with respect to the train, and vice versa.
(Source)
He basically says that it is the observation of two event that can either be simultaneous or not depending on the frame of reference.
But do the observers disagree on the simultaneity of the events after the "post-processing" (tracing back the actual events in spacetime)? It is at this point that things start to become difficult: after the "post processing", will the observers disagree on the order of the actual events, or will they agree?. Which is correct and why? And what does a Cauchy Hypersurface have to do with this?
|
do the observers disagree on the simultaneity of the events after the "post-processing" (tracing back the actual events in spacetime)?
Yes. The relativity of simultaneity specifically refers to the disagreement about simultaneity that remains after the observers have accounted for the finite speed of light.
In other words, if I receive light from two novas tonight, one 100 light years distant and one 200 light years distant, I will say that they did not occur simultaneously. Similarly, if I receive the light from one 200 light years distant and read that a 100 light year distant nova was observed here 100 years ago, then I will say those did occur simultaneously.
Someone traveling with respect to me will make the same computations and disagree about both results.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/749235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Does the spin have to be of opposite sign for fermions to form cooper pairs in bcs theory? I recently started exploring superconductivity and here's what I've understood wrt to bcs theory of superconductivity:
Two electrons can have a close association as there is low agitation of the lattice below the critical temperature. Two fermions together can act as a boson thus having an integral spin of 0 or 1 instead of + or - half so there's no need to worry about pauli's exclusion principle and all that. So does the spin of the electrons acting as a single cooper pair have to be opposite? If so, why
| While $^3He$ is superfluid, rather than superconductive, Cooper pairs in $^3He$ have two fermion atoms/quasiparticles with the same spin.
While $^4He$ is a boson, $^3He$ with three rather than four nucleons
is a fermion, and superfluid $^3He$ is formed by condensation of
Cooper pairs of $^3He$ atoms (or more precisely of “quasiparticles” of
atoms each with a surrounding cloud of other atoms) that have internal
degrees of freedom. This is because the two partners form a
spin-triplet in a relative orbital $p$-state. https://www.nobelprize.org/uploads/2018/06/advanced-physicsprize2003-1.pdf
I don't know if a similar situation is possible in superconductivity.
EDIT (Feb 12, 2023): Looks like spin-triplet superconductivity is also possible (Science, 16 Aug 2019, Vol 365, Issue 6454, pp. 684-687, DOI: 10.1126/science.aav8645)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/749808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to move from AdS to dS space? I studied different black holes in different spacetime and I also checked their differences, for example, the difference that exists in dS and AdS spaces. The question that has been created for me is whether it is possible to change the space of black holes from dS to AdS or vice versa with just one sign change in the cosmological constant Λ
. For example, if I change the sign of the last term of equation (2) f(R)
of the (Thermodynamics of Accelerating Black Holes), will it be transformed into dS space? https://arxiv.org/abs/1604.08812
| Simply changing the sign of cosmological constant in an expression for a metric should give a solution of Einstein equations for the new value of cosmological constant (and in particular, the Plebanski–Demianski family, to which accelerating black holes belong, exists for all values of cosmological constant). However, apriori there is no guarantee that such solution represents a black hole in an asymptotically de Sitter space and if it does, what ranges the other parameters of solution can assume. There is also no guarantee that coordinate patch obtained for that new metric expression would be sufficient to do analysis (e.g. of thermodynamic properties) one intends to do.
However, we could hope that a small black hole in AdS (black hole mass is much smaller then (A)dS length scale, $M\ll \ell$) would remain a small black hole in dS through such change of sign, yet even then we can be certain that the global structure of this new spacetime would be completely different.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/750083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Diode $I$-$V$ characteristic for inverse region This is the Shockley equation for the current of the diode:
$$I_D = I_S \left( e^{\frac{V_D}{nV_T}} - 1 \right)$$
It is also valid for $V_D < 0$, when this equation tends to be
$$I_D = - I_S$$
Wikipedia specifies that the equation fits for a "moderate" reverse bias.
But which is the equation describing the diode current in breakdown region? Is it still an exponential current?
| The Shockley equation assumes that the breakdown wasn't produced yet. If you are in the breakdown region things get complicated. You need the ionization rates because the impact ionization generation is what predominates. Read chapter 7.4 from Donald Neamen's - Semiconductor Physics and Devices, but the formula you would use:
$$\frac{d I_n(x)}{dx} = I_n(x) \alpha_n + I_p(x) \alpha_p \approx I \alpha$$
Where I simplified the ionization constants of electrons and holes as equal $\alpha_n \approx \alpha_p = \alpha$ and the total current is $I = I_n(x) + I_p(x)$. Now, for a junction of width W:
$$I_n(W) - I_n(0) = I \int_0^W \alpha(E) dx $$
On avalanche $I_n(W) >> I_n(0)$ and $I_n(W) \approx I$, so, by definition, to find the breakdown voltage you have to find the electric field that produce $\int_0^W \alpha(E) dx = 1$. If you want to find the current at a given point $x_0$ for a given voltage other than the breakdown, first you have to find $\alpha(E)$, then $I_n(0)$ (I guess this last from Shockley equation) and then solve the above equation.
Is it still an exponential current?
From this I found that:
$$\alpha(E) = A \exp\left(-\frac{B}{E}^{\beta}\right)$$
With $E$ the electric field, I assume proportional to the voltage and $A$, $B$, $\beta$ constants. The graph of $I(V)$ is more weird than an exponential, obviously will decrease faster than an inverted exponential.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/750268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Mass-Energy Equivalence and First Law of Thermodynamics Einstein showed mass can be converted into energy and vice versa.
$E=mc^2$
However, in school we are taught that according to the First Law of Thermodynamics, energy can neither be created nor destroyed.
Are they not contradicting each other? I already tried finding it on other sites but was surprised that there was little information regarding this.
| I think the best way to resolve this question is to forget about more colloquial descriptions of $E=mc^2$ and just take a look at special relativity. SR imposes that the famous mass-energy relation (in natural units): $$ p_{\mu} p^{\mu} = E^2 - \vec{p}^2 = m^2$$ holds for all particles in all reference frames since $p_{\mu} p^{\mu}$ is a Lorentz invariant quantity. Furthermore, in SR we also have a conservation of total 4-momentum: $$p_{\text{initial}}^{\mu} = p_{\text{final}}^{\mu}$$ for any closed system.
In the nonrelativistic limit, the mass-energy relation reduces from $E^2 = \vec{p}^2 +m^2$ to $E = mc^2$ since the 3-momentum is small relative to the $mc^2$ term. However, the fundamental invariant quantity in SR is $p_{\mu} p^{\mu}$.
An example to think about mass and energy is found in nuclear physics. The proton and neutron can form a bound state called the deuteron with a mass $m_d = m_p + m_n - 2.2 MeV$. Thus, to break apart the deuteron, you need to put in at least 2.2 MeV of energy and you'll get out two particles with "greater mass than you started with" (seemingly $m_d \rightarrow m_p + m_n $, which apparently violates mass conservation). But to correctly understand what happened you must take both the energy and mass into account and you'll see conservation: $m_d + 2.2 MeV \rightarrow m_p + m_n$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/750429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Experimental searches for magnetic quadrupole moment My professor mentioned that a particle with an intrinsic magnetic quadrupole moment would be CP violating in an analogous manner to how a particle with an electric dipole would be evidence for new sources of CP violation. Are then any active or past experiments which are trying to measure the magnetic quadruple moment of particles like the electron for example?
I know of at least two ongoing experiments to measure the electric dipole moment of the electron and neutron, but I'm wondering if anyone is studying magnetic quadrupole moments in the same way.
| A spin-half particle cannot have an intrinsic quadrupole moment, because the two-state system does not have enough degrees of freedom. See this answer for a hand-waving explanation, which is related to the Wigner-Eckart theorem.
In a comment, you link to this preprint, which discusses magnetic quadrupole moments in a number of odd-$A$ nuclei with spin $3/2$ or higher.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/750515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does the double slit experiment not prove that the wave function is ontological? To me, it seems that the interference pattern is the evidence that the wave function is a physical aspect of reality, but people still seem to be trying to decide whether or not it's ontological or just a mathematical construct.
Why is the double slit experiment not considered proof that the wave function is ontological?
| Even classical physics predicts a wave pattern on the screen when you do the double-slit experiment with light. Not all waves are wave functions. In fact, any wave that you can see isn't a wave function. The wave function encodes the probabilities of various measurement outcomes before measurement happens. To say that it's real amounts to saying that the measurement outcomes that don't happen, and that you don't see, have some sort of reality as well.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/750647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Is the attractive Fermi-Hubbard model solvable by Bethe Ansatz? I know that the one-dimension Fermi-Hubbard model is solvable by using the Bethe Ansatz method. The results I have seen, however, seem only to treat the repulsive case, i.e. $U > 0$, and I have not come across any discussion of the attractive case ($U < 0$). My question is: Can the attractive Fermi-Hubbard model also be solved by Bethe Ansatz?
| In brief, yes. The attractive case is addressed in chapter 16 (and references therein) of the Hubbard chain "bible": Essler, F. H., Frahm, H., Göhmann, F., Klümper, A., & Korepin, V. E. (2005). The one-dimensional Hubbard model (Cambridge). A draft of the book is available here, on Korepin's website. As the chapter describes, it is possible to solve the attractive model independently using Bethe ansatz methods, but it's also possible to construct a unitary transformation that reduces the attractive case to the repulsive one and use the solution of the repulsive model. A key paper on the latter approach is F. Woynarovich, J. Phys. C: Solid State Phys. 16, 6593 (1983).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/751014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can eigenvalues of the density matrix in the Lindblad equation be negative? Can the density matrix in the Lindblad equation for an open mixed quantum system have (real part) negative eigenvalues?
| The density matrix by definition is a positive semi-definite matrix. It therefore has real positive eigenvalues.
Physically, these eigenvalues correspond to the probability $p_i$ of being in one of the pure eigenstates $|\phi_i\rangle$ according to
$$\rho = \sum_i p_i |\phi_i\rangle\langle \phi_i|.$$
Clarification in response to the comments
The above answer only uses the definition of the density matrix. Positive semi-definiteness of $\rho$ is a very basic property which follows from its probability interpretation.
The above also allows for mixed states. As @ZeroTheHero states, a pure state would be the case where one of the $p_i=1$, such that $\rho=|\phi_i\rangle\langle \phi_i|$.
According to the comments, the OP may contain a second question, which is related to the dynamics generated by the Lindblad equation. Indeed, the Lindblad equation manifestly conserves positive semi-definiteness of the density matrix. It is also trace preserving, since another important property of the density matrix is that the probabilities $p_i$ sum to 1.
Note that the Lindblad equation is particularly nice because of this property. While it usually is derived using multiple approximations, it produces a fully consistent density matrix at all times of the dynamics. This is not the case for many other ways to treat open system dynamics. E.g. the Redfield equation contains less approximations, but does not generate positive time evolution.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/751206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does magnetic force only act on moving charges? I don't understand why the magnetic force only acts on moving charges. When I have a permanent magnet and place another magnet inside its field, they clearly act as forces onto one another with them both being stationary. Also, I am clearly misunderstanding something.
| This is basically just a matter of definition. The magnetic force is not something independent or complete on its own. Instead, it is the electromagnetic force that is the underlying concept. The electromagnetic force can be split into a portion that acts on stationary charges and a portion that acts on moving charges. We call the first portion the electric force and the second portion we call the magnetic force. But they are not independent. They exist together and it is just an essentially arbitrary division to split the overall electromagnetic force into an electric force and a magnetic force.
Regarding a permanent magnet: a permanent magnet is typically uncharged, so you would not consider a permanent magnet as though it were an electric charge. It would typically be better to consider a permanent magnet as an uncharged magnetic dipole. Inhomogeneous magnetic fields can exert forces on stationary magnetic dipoles. The formulas are not at all the same for magnetic dipoles as for electric charges.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/751358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
} |
Kepler third law for circular orbits This question may be uber trivial, but it has been stuck in my head for a while.
Kepler's third law states that the period of the orbit $T$ is related to the semi-major axis $a$ though
\begin{equation}
T^2 = 4\pi^2\frac{a^3}{G(m_1+m_2)}.\tag1
\end{equation}
where $m_1$ and $m_2$ are the masses of the two bodies.
Consider a simple system where $m_1 = m_2=m$ and the orbit is circular so $a$ is the radius.
Kepler's Law (1) tells us that
\begin{equation}
T^2 = 4\pi^2\frac{a^3}{2Gm}.\tag2
\end{equation}
If I wanted to derive this equation from the equation of motion of $m_1$, for example, I would write
\begin{equation}
G\frac{m_1 m_2}{(2a)^2} = m_1 a \omega^2.\tag3
\end{equation}
which in the case that $m_1 = m_2 = m$ reduces to
\begin{equation}
\omega^2 = G\frac{m}{4a^3} .\tag4
\end{equation}
Considering that $T = 2\pi / \omega$, I obtain
\begin{equation}
T^2 = 4\pi^2\frac{(2a)^3}{2Gm} .\tag5
\end{equation}
Why do I have here the diameter instead of the radius as in equation (2)?
Two-body problem approach
If I approach the problem instead of looking at the motion of a single mass, but considering the two-body problem, the equation of motion should be
\begin{equation}
G\frac{m_1 m_2}{(2a)^2} = \mu (2a) \omega^2.\tag6
\end{equation}
where $\mu = \frac{m_1 + m_2}{m_1 m_2}$ is the reduced mass. Considering again $m_1 = m_2 = m$, the previous equation simplifies to
\begin{equation}
G\frac{m^2}{(2a)^2} = \frac{m}{2} (2a) \omega^2.\tag7
\end{equation}
which is equivalent to the equation of motion for the single body (3) and should lead to the same result.
| You are trying to apply Kepler's third law in the form
$\displaystyle \omega^2 = \frac {G(m_1+m_2)} {r^3}$
where $r$ is the radius of a planetary orbit, $m_1$ is the mass of the primary and $m_2$ is the mass of the secondary. However, this form only applies when $m_1 >> m_2$. If $m_1$ and $m_2$ have similar values then the appropriate form of Kepler's third law is the binary mass function
$\displaystyle \omega^2 = \frac {Gm_1^3} {(m_1+m_2)^2r_2^3}$
where $r_2$ is the distance of $m_2$ from the centre of mass. In you example, $m_1=m_2=m$ and $r_2=a$ so we have
$\displaystyle \omega^2 = \frac {Gm^3} {(2m)^2a^3} = \frac {Gm} {4a^3}$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/751511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maxwell stress tensor on two steady current carrying wires I am having trouble thinking about the force on a wire via maxwell stress tensor. The problem is as follows; given two parallel infinite wires separated by a distance of 2a with steady currents in the same direction, find the force per unit length on the two wires. I have calculated the maxwell stress tensor given the field of a single wire, in SI units it's something like $T_{ij}=\frac{I^2\mu_0}{4\pi^2r^2} \begin{pmatrix} \sin^2{\phi}-\frac{1}{2} & -\cos{\phi}\sin{\phi} & 0\\
-\cos{\phi}\sin{\phi} & \cos^2{\phi}-\frac{1}{2} & 0 \\
0 & 0 & \frac{-1}{2}
\end{pmatrix}$
Now if I understand correctly, the trick to making use of the tensor here is to enclose one of the currents by a surface where you let some part of it go off to infinity where the field strength is zero so you are only left with integrating over some nice surface (does it have to be compact?). My real question is where do I start with parameterizing this surface? I thought about making a kind of dome enclose one of the wires and having the non zero curvature part of the dome go out to infinity and therefore im left with integrating over the plane in between and orthogonal to both the wires. Is this the right approach? Is my understanding ok? Do you agree with my calculation thus far?
I would really appreciate some insight and I'm sorry if I'm asking to be spoon fed here in advance.
| In a suitably chosen Cartesian coordinate system, the current density (assuming equal currents $I$ in both wires) is given by $$\vec{j}(x,y,z)= I \delta(y)[\delta(x+a)+\delta(x-a)]\vec{e}_z,$$ generating the magnetic field $$\vec{B}(x,y,z)= \frac{\mu_0 I}{2 \pi} \left[\frac{-y \, \vec{e}_x +(x+a) \, \vec{e}_y}{(x+a)^2 +y^2}+\frac{-y \, \vec{e}_x+(x-a) \, \vec{e}_y}{(x-a)^2+y^2} \right].$$
The force $\vec{F}$ acting on the segment $S=\left\{(-a,0,z)|0\le z \le \ell\right\}$ of the left wire is obtained from the Maxwell stress tensor $$T_{ik}= \frac{1}{\mu_0} \left(B_i B_k- \frac{1}{2}\delta_{ik}\, \vec{B}\!\cdot\! \vec{B} \right)$$ by the surface integral $$F_k= \int\limits_{\partial V} \!d\sigma_i \, T_{i k}, $$ where $\partial V$ is the boundary of an arbitrary volume $V$ containing the segment $S$ but neither any other part of the left wire nor any piece of the right one. As already remarked in a comment by Jan Lalinsky, a convenient choice of the domain $V$ is given by $V=\{(x,y,z)|x^2+y^2\le R^2, \, x\le 0, \, 0\le z \le \ell\}$ in the limit $R\to \infty$. In the surface integral, only the contribution from the slice of the $y$-$z$-plane with $0 \le z \le \ell$ survives. Using $d \vec{\sigma} = dy \, dz \, \vec{e}_x$, we obtain $$F_k =\frac{1}{\mu_0} \int\limits_0^\ell \! dz \int\limits_{-\infty}^\infty\! dy \, \left[(B_x(0,y,z) B_k(0,y,z) - \frac{\delta_{x k}}{2} \, \vec{B}(0,y,z) \cdot \vec{B}(0,y,z) \right]. $$ Inserting $B_x(0,y,z)=-\frac{\mu_0 I y}{ \pi (a^2+y^2)}$ and $B_y(0,y,z)=0$, $B_z(x,y,z)=0$, one finds $$F_x= \frac{\mu_0I^2 \ell}{2 \pi^2 a} \int\limits_{-\infty}^\infty \!\! d \eta \, \frac{\eta^2}{(1+\eta^2)^2}= \frac{\mu_0 I^2 \ell }{4 \pi a}, \quad F_y=F_z=0,$$ as to be expected.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/751894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why is electromotive force in magnetohydrodynamics a vector quantity? In the mean-field dynamo theory in magnetohydrodynamics, I frequently came across a quantity;
$\langle v'\times B' \rangle$, which is termed as the mean electromotive force. I want to know that why is it termed as electromotive force, if it is a vector.
Everywhere else I have seen emf is just the potential difference and hence a scalar. Is this emf different than the emf used in mean-field dynamo theory?
| The usual EMF in circuits refers to a closed oriented path: it is the integral of net motional force per unit charge acting on current. Physical unit of this EMF is Volt. Sometimes EMF for a non-closed path is discussed, which is based on the same idea, only the integration path is not closed but has starting point and ending point.
The MHD electromotive force $\mathbf E^*$ is clearly a different concept, but still related: it is the motional electromotive force (due to motion in magnetic field) acting on current in medium, per unit charge. So the actual force on current in volume element $\Delta V$ would be
$$
\Delta \mathbf F = \rho_m \Delta V\mathbf E^*
$$
where $\rho_m$ is density of mobile charge.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/752083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Uncertainty Calculation: Applying Product Rule instead of Power Rule I use $\delta$ to represent absolute uncertainty. The power rule for the calculation of relative uncertainty in $t^2$ is
$$\frac{\delta (t^2)}{(t^2)}=2\left(\frac{\delta t}{t}\right).$$
But if I treat the power as a product and apply the product rule, I have
$$\frac{\delta (t \times t)}{(t \times t)} = \sqrt{\left(\frac{\delta t}{t}\right)^2 + \left(\frac{\delta t}{t}\right)^2} = \sqrt{2\left(\frac{\delta t}{t}\right)^2} = \sqrt{2}\left(\frac{\delta t}{t}\right).$$
Am I making a mistake? If not, how is this inconsistency reconciled?
| As was pointed out by Sandejo's answer, uncertainties only add in quadrature if the two quantities being measured are uncorrelated. The formulas for propagation of uncertainty have additional terms that must be included if the variables' uncertainties are correlated.
Specifically, suppose we have two quantities $A$ and $B$ with uncertainties $\sigma_A$ and $\sigma_B$ and covariance $\sigma_{AB}$. Then the uncertainty in $f = AB$ is [given by]
$$
\frac{\sigma_f}{f} = \sqrt{\frac{\sigma_A^2}{A^2} + \frac{\sigma_B^2}{B^2} + 2 \frac{\sigma_{AB}}{AB}}.
$$
Uncorrelated uncertainties have $\sigma_{AB} = 0$, and so you end up with the "addition in quadrature" formula you're familiar with.
On the other hand, suppose that $A$ and $B$ are perfectly correlated (as they would be if they were secretly the same variable, which you're calling $t$.) In this case, it can be shown that $\sigma_{AB} = \sigma_A \sigma_B$, and so
$$
\frac{\sigma_f}{f} = \sqrt{\frac{\sigma_A^2}{A^2} + \frac{\sigma_B^2}{B^2} + 2 \frac{\sigma_{A}\sigma_{B}}{AB}} = \frac{\sigma_A}{A} + \frac{\sigma_B}{B}.
$$
In particular, if $A = B = t$, then you have
$$
\frac{\sigma_f}{f} = 2 \frac{\sigma_t}{t},
$$
as one would expect from the power rule.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/752312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Projective representations of the Lorentz group can't occur in QFT! What's wrong with my argument? In flat-space QFT, consider a spinor operator $\phi_i$, taken to lie at the origin. Given a Lorentz transformation $g$, we have
$$\tag{1} U(g)^\dagger \phi_i U(g) = D_{ij}(g)\phi_j$$
where $D_{ij}$ is some spinor projective representation of $SO^+(3,1)$, and $U$ is an infinite-dimensional unitary, perhaps projective, representation of $SO^+(3,1)$, satisfying
$$\tag{2} U(gh)= e^{i\alpha(g,h)}U(g)U(h)$$
for some real phases $\alpha(g,h)$. Now, using (1) and (2) we have:
\begin{align}
D_{ij}(gh)\phi_j &\stackrel{(1)}{=} \tag{3}U(gh)^\dagger\phi_iU(gh) \\ \tag{4}
&\stackrel{(2)}{=}e^{-i\alpha(g,h)}U(h)^\dagger U(g)^\dagger\phi_iU(g)U(h)e^{i\alpha(g,h)}\\ \tag{5}
&\stackrel{(1)}{=} U(h)^\dagger \big((D_{ik}(g)\phi_k\big)U(h)\\ \tag{6}
&\stackrel{(1)}{=} D_{ik}(g)D_{kj}(h)\phi_j.
\end{align}
Hence $$\tag{7} D_{ij}(gh) = D_{ik}(g)D_{kj}(h).$$ That is, $D$ is a true, non-projective, representation. Clearly this is false for spinors, so where did I go wrong?
| Your observation is correct, which is why careful statements of the Wightman axiom eq. (1) (see e.g. Wiki, nLab) stipulate that $D$ is a representation of $\mathrm{SL}(2,\mathbb{C})$, the universal cover of the Lorentz group, not a representation of the Lorentz group itself.
Note that classical fields - the things on which $D$ acts - are not necessarily complex-valued, so the notion of a projective representation on them does not even make sense in general.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/752443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Horizon related question I'm a student taking an undergrad course in physics and we were discussing light and how it works in relation to vision. This question stumped us. What altitude would you need to reach on the globe before the horizon vanished from your view provided you maintained a line of vision parallel to the surface of the earth.
| If I understand the question correctly, to be fully defined you would need to specify the field of view. The horizon would disappear from a viewpoint "parallel to the surface of the earth" right at the earth unless you have a nonzero angle for your field of view. If your field of view is 180 degrees, you could technically "see" the horizon not matter how high you get, although in reality it would become small and faint.
If you draw a diagram, you should find that the horizon appears at the angle below horizontal given by:
$$
\cos(\theta) = \frac{R}{R+h}
$$
for radius of the earth $R$ and height above the ground $h$. Thus, if your field of view is angle $2\theta$
$$
h = \frac{R(1-\cos(\theta))}{\cos(\theta)}
$$
from which you see that when $h = 0$, $\theta = 0$, and as $h$ goes to infinity, $\theta$ goes to a right angle. The limitations on the field of view are thus needed to answer this question exactly.
Note that this is an ideal analysis that assumes no topography and no atmospheric refraction. As you may know if you have ever treated yourself to flat earth material (I actually enjoy solving their "proofs" from time to time), it is actually difficult to see the horizon since the curvature of light due to atmospheric refraction is often of a similar magnitude, and even greater than, the curvature of the earth.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/753614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is sound and how is it produced? I've been using the term "sound" all my life, but I really have no clue as to what sound exactly is or how it is created. What is sound? How is it produced? Can it be measured?
| Sound is a form of energy. And what does sound do? It provides the sensation of hearing. But that's just what the definition tells us. Sound is a mechanical wave. Now just by seeing that you must realise it needs a medium to propagate and must also ask yourself how does a wave propagate in a medium. Well; whenever there's a disturbance in a medium, a wave is produced. Therefore when we apply energy to anything, it disturbs the medium which in turn disturbs the air which helps sound to travel. And i do now undundrstan what you mean by measuring sound, we can measure characteristics of it but do specify what do you mean by 'measuring sound'
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/13",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
Does leaning (banking) help cause turning on a bicycle? I think it's clear enough that if you turn your bicycle's steering wheel left, while moving, and you don't lean left, the bike will fall over (to the right) as you turn. I figure this is because the bike's momentum keeps it moving in the direction you were going, and since your wheels have friction against the ground, the top of the bike moves forward relative to the bottom of the wheels. The top of the bike going north while the bottom of the wheels go northwest will understandably cause you to topple.
So to counteract this and keep you from falling over, leaning into the turn is necessary.
But is there also a direct causal relationship -- that leaning will cause the bike to start to turn? If I start leaning left, I will turn left... but maybe that's because I know that if I don't turn the steering wheel left, the bike will fall over (to the left). I experimented with unruly turns of the steering wheel when I was a kid, and got my scrapes and bruises. Now that I'm a cautious and sedate adult I'm not anxious to experiment that way. :-)
(I also want to ask why airplanes bank into a turn... they don't have the same issues as a bike, i.e. the bottom part has no special friction against the ground. But that would probably make the question too broad.)
| Yes. A lean induces a side force that makes the bike follow a circular path.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/24",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 10,
"answer_id": 8
} |
Mnemonics to remember various properties of materials I'm trying to figure out how to remember that
*
*hardness: how resistant it is to deformation
*toughness: how resistant it is to brittle failures
*stress: force on a surface area
*strength: ability to withstand stress without failure
*strain: measurement of deformation of a material
Does anyone know of a mnemonic or easy way? I only know these from googling them, and I'm finding it tricky to remember them.
| For Shiny, Malleable, Ductile, Conductors of heat and electricity and forming Ionic compounds, there is a mnemonic which goes like
Silver
Made
Desert i.e
Cupcakes and
Ice-creams for after dinner
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is it possible that all neutron stars are actually pulsars? I'm assuming that what I've been told is true:
We can only detect pulsars if their beams of electromagnetic radiation is directed towards Earth.
That pulsars are the same as neutron stars, only that they emit beams of EM radiation out of their magnetic poles.
So, isn't it possible that neutron stars emit EM radiation in the same fashion as pulsars, just not in the right direction for us to detect it?
|
Do there exist neutron stars without relativistic jets? Also, could jets be locked in alignment with the spin axis, resulting in a beam that does not pulse for any line of sight? For some reason discussion has been focused on Earthbound detectability of these jets. Instead, I'm looking for an answer using astrophysics that deals with all lines of site, not just those pointing toward us.
I think the expectation here is a radio-quiet neutron star. Though most of the neutron stars are pulsars, these are the special types that are more likely to satisfy the constraints. Either they don't emit relativistic jets, they have their magnetic axis aligned to the rotational axis, or the radio beams are always directed away from Earth. There's also another possibility that we haven't detected any emissions yet (I mean, we haven't swept the whole sky). For instance, the fact that Geminga is a pulsar was quite unknown for 20 years. Later, it was discovered to have a periodicity of 237 milliseconds.
As far as I've known, these radio-quiet neutron stars haven't been declared as a non-rotating neutron star yet. Instead, their periodicity and a few other details have been listed as unknown. Examples include RX J0822-4300 and RX J185635-3754 (it was accounted as a candidate for quark star, however Chandra and Hubble observations excluded it from the list)
There are a few papers related to these species, which I fear is quite beyond my knowledge...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/90",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
Book about classical mechanics I am looking for a book about "advanced" classical mechanics. By advanced I mean a book considering directly Lagrangian and Hamiltonian formulation, and also providing a firm basis in the geometrical consideration related to these to formalism (like tangent bundle, cotangent bundle, 1-form, 2-form, etc.).
I have this book from Saletan and Jose, but I would like to go into more details about the [symplectic] geometrical and mathematical foundations of classical mechanics.
Additional note: A chapter about relativistic Hamiltonian dynamics would be a good thing.
| This isn't explicitly about just mechanics as it tries to hit a lot of different areas in physics but it covers the material you are asking for:
http://www.amazon.com/Differential-Forms-Applications-Physical-Sciences/dp/0486661695
It will absolutely provide you with a firm basis in the geometrical considerations you mentioned in your question (tangent bundle, cotangent bundle, 1-form, 2-form, etc.).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 9,
"answer_id": 4
} |
Law for tap water temperature I was wondering if anyone put together a law to describe the rising temperature of the water coming out of a tap.
The setup is fairly simple: there's a water tank at temperature T, a metal tube of length L connected to it and a tap at the end where temperature is measured. The water flows at P l/s.
Given that the metal tube is at room temperature initially, what law describes the temperature of the water at any instant? What is the limit temperature of the water?
Thanks.
| The answer is going to depend on the heat transfer coefficient between the tube and the surrounding room (unless you specify that the tube is held at constant temperature), the heat transfer coefficient between the tube and water, the outside & inside diameter of the tube, and the length of the tube. This is a moderately involved heat transfer problem, unless additional constraints are provided to simplify it.
An excellent resource for understanding the mathematics behind this sort of heat transfer problem can be found over here. It's very similar to the solution posted above by @Cedric, but may be a little easier for some to follow.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is acceleration an absolute quantity? I would like to know if acceleration is an absolute quantity, and if so why?
| I've finally figured it out.
First, let's define precisely what it means for some quantity to be absolute or relative. In the context in question, it has to do with whether a quantity is absolute (that is, has the same value) or relative (that is, has different values) when measured by two inertial observers moving with respect to one another.
Of course, first we need to define what an inertial observer is: it's an observer for which Newton's laws are applicable without having to resort to adding fictitious forces.
Ok, so now we have two observers, Alice and Bob, both of which are inertial. They both observe the motion of some object. Let the index 1 correspond to quantities measured in A's reference frame and the index 2 correspond to quantities measured in B's reference frame. The position of the object is clearly a relative concept, since
r₂ = r₁ + u t
(where u is the velocity of Bob with respect to Alice, and is constant since they're both inertial observers). Note that the time, t, is the same for both observers, as it must be according to Newtonian Mechanics. The object position is a relative concept because r₂ ≠ r₁.
Now, take the time-derivative of both sides and we get
v₂ = v₁ + u
that is, the velocity of the object with respect to one observer is different than the velocity of the same object with respect to the other observer. Hence, velocity is a relative quantity in Newtonian Mechanics.
Next, take the time-derivative of both sides once again, and we obtain
a₂ = a₁
(since u is constant). Thus, the acceleration of the object is the same in both reference frames. Acceleration, therefore, is absolute in Newtonian Mechanics.
When we take into account the theory of relativity, then time flows at different rates for different inertial observers and the result above for the acceleration is no longer true.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
Why can you "suck in" cooked spaghetti? We all know that there is no "sucking", only pushing. So how are cooked spaghetti pushed into your mouth? The air pressure applies orthogonal on the spaghetti surface. Where does the component directed into your mouth come from?
| When you perform the sucking action, a pressure difference is clearly created and maintained by your lungs between the surrounding air and the air inside your mouth. An important point to notice here is that the mouth must not too far open (a bit lets it work still), else the pressure gradient between inside and outside of the mouth cannot be maintained by the expanding lungs over any significant time period.
Firstly, observe that there is a normal force on the surface of the spaghetti due to (some arbitrary) air pressure. As I have just explained, there is a pressure gradient between the inside and outside of the mouth, directed inwards, hence there is a driving force inwards. Combining the two effects gives a net force directed at some angle to the surface of the spaghetti. Hence, it is an angled force, with some component pointing inwards towards the mouth, that pushes it in.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 1
} |
What property of objects allow them to float? I used to think that the shape of an object determines its ability to float (boat-shaped objects are more likely to float, and spheres tend to sink). But my friend, who is fond of making me look stupid, took me to the local lake showed me a sphere that floated and a boat-shaped object made out of iron that sunk.
Is it based on the mass? I'm not sure that is possible, because I've seen really heavy things (like airplane carriers) float, and really light things (like my friend's iron boat) sink.
What property of certain objects allow them to float, if any?
| Actually, the answer is a bit more subtle than just density. The principle that is behind floating objects is Archimedes' principle:
A fluid (liquid or gas) exerts a buoyant force, opposite apparent gravity (i.e. gravity + acceleration of fluid) on an immersed object that is equal to the weight of the displaced fluid.
Thus, if you have an object fully immersed in a fluid, the total force it feels is given by (positive sign means down):
$$F = \text{gravity} + \text{buoyancy} \\= \rho_\text{object} V g - \rho_\text{fluid} V g \\= (\rho_\text{object} - \rho_\text{fluid}) V g$$
Thus, if the average density of the object is lower than that of the water, it floats. If the object is partially immersed, to calculate the buoyant force you have to consider just the immersed volume and its average density:
$$F = \rho_\text{object} V g - \rho_\text{fluid} V_\text{immersed} g$$
Note that when I was talking about density, I was talking about the average density of the object. That is its total mass divided by its volume. Thus, a ship, even if it is made out of high-density iron it is full of air. That air will lower the average density, as it will increase the volume considerably while keeping the weight almost constant.
If you want to understand this better you can give the following problem a try :)
What is the height an ice cube of side L floats in water?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 9,
"answer_id": 0
} |
Why don't spinning tops fall over? One topic which was covered in university, but which I never understood, is how a spinning top "magically" resists the force of gravity. The conservation of energy explanations make sense, but I don't believe that they provide as much insight as a mechanical explanation would.
The hyperphysics link Cedric provided looks similar to a diagram that I saw in my physics textbook. This diagram illustrates precession nicely, but doesn't explain why the top doesn't fall. Since the angular acceleration is always tangential, I would expect that the top should spiral outwards until it falls to the ground. However, the diagram seems to indicate that the top should be precessing in a circle, not a spiral. Another reason I am not satisfied with this explanation is that the calculation is apparently limited to situations where: "the spin angular velocity $\omega$ is much greater than the precession angular velocity $\omega_P$". The calculation gives no explanation of why this is not the case.
| From your linked article:
Spin a top on a flat surface, and you will see its top end slowly
revolve about the vertical direction, a process called precession. As
the spin of the top slows, you will see this precession get faster and
faster. It then begins to bob up and down as it precesses, and finally
falls over.
The drawing shows a circle instead of a spiral due to leaving out variables like friction and gravity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 7,
"answer_id": 5
} |
Why does water make a sound when it is disturbed? When I disturb a body of water, what causes the familiar "water moving" sound?
| If you throw a stone in a large body of water, this is what you should hear:
*
*a high pitched slapping sound when the stone makes contact with the water. This is due to the air between the stone and the water being pushed out, as well as the surface ripples
*a low pitched sound growing in pitch (like "Doo-eeeee" ;-) due to the stone sinking and leaving an air "hole" in the water in its trail; this is then filled gradually with water which makes the sound raise in pitch
*a bubbling sound due to air bubbles trapped under water by the stone, emerging.
*a random noise due to waves colliding on the surface of the water.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
} |
Planet orbits: what's the difference between gravity and centripetal force? My physics teacher says that centripetal force is caused by gravity. I'm not entirely sure how this works? How can force cause another in space (ie where there's nothing).
My astronomy teacher says that gravity is (note: not like) a 3D blanket and when you put mass on it, the mass causes a dip/dent in the blanket and so if you put another object with less mass it will roll down the dip onto the bigger mass. Is this true and is this what causes the centripetal force.
| Gravity is a force.
Gravity is directed towards the center of the orbit i.e. the sun.
That makes gravity the centripetal force.
Imagine a ball attached to a string and you are holding the other end of the string and moving your hand in such a way that the ball is in circular motion. Then tension in the string is centripetal force.
Now,
ball = earth
you = sun
tension in the string = gravity
Hope it helps. I have no idea about general relativity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Home experiments to derive the speed of light? Are there any experiments I can do to derive the speed of light with only common household tools?
| There is a trick I have heard about before but never tried. The basic idea is to put a mars bar in a microwave oven for a short amount of time. First you remove the turntable, so the chocolate bar stays stationary. Then you turn the microwave on just long enough for the chocolate to start to melt. It should melt at the nodes of the standing field. You simply measure the distance between the nodes, and multiply by the frequency of the microwave oven to obtain the speed of light. There is a YouTube demonstration (by a kid) here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "58",
"answer_count": 11,
"answer_id": 1
} |
Positrons versus holes as positive charge carriers From Wikipedia: [The Dirac sea is a theoretical model of the vacuum as an infinite sea of particles with negative energy. It was first postulated by the British physicist Paul Dirac in 1930 to explain the anomalous negative-energy quantum states predicted by the Dirac equation for relativistic electrons. The positron, the antimatter counterpart of the electron, was originally conceived of as a hole in the Dirac sea, well before its experimental discovery in 1932.]
and:
[Dirac's idea is completely correct in the context of solid state physics, where the valence band in a solid can be regarded as a "sea" of electrons. Holes in this sea indeed occur, and are extremely important for understanding the effects of semiconductors, though they are never referred to as "positrons". Unlike in particle physics, there is an underlying positive charge — the charge of the ionic lattice — that cancels out the electric charge of the sea.]
It always confused me to think of holes as positive charge carriers in semi-conductors as not being real: real electrons move from one lattice-position to another lattice-position, which effectively looks like a positive hole in the lattice that is moving in the other direction, but in reality a real electron moves, the hole is kind of an "illusion".
On the other hand the positrons are always introduced as real hard-core particles.
The quotes from the Wikipedia article make me unsure: how should I look upon these phenomena?
Edit: holes in a Dirac sea give rise to real pos. entities in one case and to unreal pos. entities in another - how can we distinguish, is it a matter of formalism?
| Holes in semi-conductors are considered to be quasiparticles. A quasiparticle is a group of particles that end up acting and having many of the properties of a single particle for a period of time. They are normally considered to have a lifetime tau and decay exponentially.
Are positrons and electrons that we obverse quasiparticles too? QPs with a very long decay time? That goes beyond our current level of physics. I have not heard any evidence that they are but the conjecture has been considered.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
} |
Why is it thought that normal physics doesn't exist inside the event horizon of a black hole? A black hole is so dense that a sphere around it called the event horizon has a greater escape velocity than the speed of light, making it black. So why do astronomers think that there is anything weird (or lack of anything Inc space) inside the event horizon. Why isn't simple the limit to where light can escape and in the middle of event horizon (which physically isnt a surface) is just a hyper dense ball of the matter that's been sucked in and can't escape just like light. Why is it thought that the laws of physics don't exist in the event horizon?
| If you are inside the event horizon of a spinning black hole, there are closed timelike curves--paths through spacetime which, when you progress them into the future, they end up in their past. These curves clearly violate the principle of causality, and thus, the region inside the black hole can be considered unphysical in a sense.
Also, of course, the spacetime singularity also lives inside of the black hole's horizon. Classically, it is a point or ring of infinite density at which the curvature of spacetime is infinite. Spacetime histories intersecting the singularity must end if you believe in classical mechanics. So, there is clearly some new physics there--most relativity researchers believe in something called the Cosmic Censorship conjecture, which says that nearly all* physically reasonable solutions of Einstein's equations hide their black hole singularities behind a horizon.
*There are some known exact solutions where physically reasonable matter distributions collapse to horizonless singularities, but they are believed to form a set of measure zero in the space of matter distributions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Notation of plane waves Consider a monochromatic plane wave (I am using bold to represent vectors)
$$ \mathbf{E}(\mathbf{r},t) = \mathbf{E}_0(\mathbf{r})e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)}, $$
$$ \mathbf{B}(\mathbf{r},t) = \mathbf{B}_0(\mathbf{r})e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)}. $$
There are a few ways to simplify this notation. We can use the complex field
$$ \tilde{\mathbf{E}}(\mathbf{r},t) = \tilde{\mathbf{E}}_0 e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)} $$
to represent both the electric and magnetic field, where the real part is the electric and the imaginary part is proportional to the magnetic. Often it is useful to just deal with the complex amplitude ($\tilde{\mathbf{E}}_0$) when adding or manipulating fields.
However, when you want to coherently add two waves with the same frequency but different propagation directions, you need to take the spatial variation into account, although you can still leave off the time variation. So you are dealing with this quantity:
$$ \tilde{\mathbf{E}}_0 e^{i\mathbf{k} \cdot \mathbf{r}} $$
My question is, what is this quantity called? I've been thinking time-averaged complex field, but then again, it's not really time-averaged, is it? Time-independent? Also, what is its notation? $\langle\tilde{\mathbf{E}}\rangle$?
| "Space-dependent phasor" makes some sense. Electrical engineers will often just say that's the "complex amplitude of the time-harmonic field" or "time-harmonic phasor".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the temperature of the surface and core of a neutron star formed 12 billion years ago now equal to? In what part of the spectrum is it radiating? In the infrared, in the microwave? Or is not radiating anymore at all?
In russian:
Чему сейчас равна температура поверхности и ядра нейтронной звезды, которая образовалась 12 миллиардов лет назад?
В каком диапазоне она сейчас излучает? В инфракрасном, микроволновом? Или не излучает вообще?
| You'll have to wait for a real physicist to jump in for a numerical answer. As far as I know, it doesn't burn but is simply radiating the energy it started out with. I suspect the
thermal conductivity is quite high. If you assume the thermal conductivity is so high that there is essentially no temperature difference between the photosphere and the core you should be able to use the Stefan Boltzmann law to determine the rate of heat loss. But even then you'd need to know the heat capacity of matter in that form, so any physicists willing to jump in with actual theory?
If I assume heat capacity is independent of temperature, the cooling equation (after throwing out all the constants) would look like:
$$\frac{dT}{dt}=-T^4$$
That has a solution, (again throwing out constants) like $T=t^{-1/3}$.
There could be other sources of energy, initial magnetic field, rotation, gravitational
(it would probably shrink at least somewhat as it cools), and it there is any matter nearby
the potential for stuff to fall onto it. {That's about as far as a dare venture, again hopefully an astrophysicist will jump in}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why does kinetic energy increase quadratically, not linearly, with speed? As Wikipedia says:
[...] the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $\frac{1}{2}mv^2$.
Why does this not increase linearly with speed? Why does it take so much more energy to go from $1\ \mathrm{m/s}$ to $2\ \mathrm{m/s}$ than it does to go from $0\ \mathrm{m/s}$ to $1\ \mathrm{m/s}$?
My intuition is wrong here, please help it out!
| One way to look at this question of yours is as follows:
$$ E(v) = \frac{m v^2}{2} \; . $$
So, if we multiply the velocity by a certain quantity, i.e., if we scale the velocity, we get the following,
$$ E(\lambda v) = \frac{m (\lambda v)^2}{2} = \lambda^2 \frac{m v^2}{2} = \lambda^2 E(v)\; . $$
That is, if you scale your velocity by a factor of $\lambda$, your Energy is scaled by a factor of $\lambda^2$ — this should answer your question (just plug in the numbers).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "294",
"answer_count": 18,
"answer_id": 17
} |
How do neutron stars burn? Is it decay or fusion or something else?
*
*What makes a neutron star burn, and what kind of fusion/decay is happening there?
*What is supposed to happen with a neutron star in the long run? What if it cools, then what do the degenerated matter looks like after it cools? Will
the gravitational equilibrium be ruined after some burn time? How does it explode if it can explode at all?
| I think this is a good almost-popular introduction to neutron stars, the processes expected to occur therein and their evolution.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 0
} |
Equilibrium and movement of a cylinder with asymmetric mass centre on an inclined plane A cylinder whose cross section is represented below is placed on an inclined plane. I would like to determine the maximum slope of the inclined plane so that the cylinder does not roll. The mass centre (CM) of the cylinder is at a distance r from the central axis. The cylinder consists of a cylindrical shell with mass $m_1$ and a smaller cylinder with mass $m_2$ placed away from the axis and rigidly attached to the larger cylinder. What is the influence of friction? Is it possible to establish the law of the movement? I think that the piece may roll upwards until it stops.
The figure was copied from Projecto Ciência na Bagagem -- Cilindro desobediente
EDIT: Depending on the initial conditions is it possible to find the highest point the cylinder rolls to, before stopping?
EDIT2: From Institute and Museum of the History of Science -- Cylinder on inclined plane [another cylinder]
"When placed on the inclined plane,
[another] cylinder tends to
roll upward, coming to a halt at a
well-determined position."
| The centre of mass (CM) is at distance $r=\frac{m_2 d}{m_1+m_2}$ from the geometrical centre of the cylinder.
In diagram (a), if the centre of mass lies within the red circle then it will be always on the downhill side of the point of contact P. It will always exert a clockwise torque about P, causing the cylinder to roll continuously down the incline. So the asymmetric cylinder cannot be placed in a position of stability unless
$r \ge R\sin A$.
In diagram (b) the cyinder rocks on the incline between P and Q. The cylinder is stationary at these two positions, and mechanical energy is conserved, so the PE must be the same - ie the CM must lie on the same horizontal line.
The distance PQ is $R(B+C)$ so the centroid has risen a vertical distance of
$R(B+C)\sin A=r\cos C-r\cos B$.
Given an initial orientation angle B, this transcendental equation gives the final orientation angle C. The distance PQ can then be found.
The following resources deal with the dynamics of the rocking/rolling motion :
Rolling motion of non-axisymmetric cylinders by Carnevali & May 2004
A jumping cylinder on an inclined plane by Gomez, Hernandez-Gomez & Marquina 2012
Librational motion of asymmetric rolling bodies and the role of friction force, Pavia University, undated
Rolling of asymmetric disks on an inclined plane by BYK Hu 2011
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
logarithmic wind speed profile Under some atmospheric stability condition, over flat terrain, it has been observed for a while that the ratio between wind speed at height $h_1$ above the earth and the wind speed at height $h_0$ is $\log\frac{h_1}{h^*}/\log\frac{h_0}{h^*}$ where $h^*$ is related to the terrain (called roughness length). (see for example http://en.wikipedia.org/wiki/Log_wind_profile)
What are the theories (with some details or references please) that explain this rule. Please put only your prefered theory (and hence one per post).
Thanks in advance
| Logarithmic profile for wind speed regards the bottom part of atmospheric boundary layer (say, about the bottom 100 m, on a boundary layer about 1000 m high). It can be deducted doing some non obvious but reasonable assumptions.
A) Vertical flux of horizontal momentum due to turbulence must be uniform in the lowest part of the atmosphere. Let's consider a reference frame where the average velocity $\overline{u}$ is directed along x axis. Let's decompose velocity in its average and random (turbulent) parts, according to Reynolds decomposition: x component of velocity is given by
$u = \overline{u} + u'$
Vertical component is:
$w = w'$
where $\overline{u'} = 0$, and $\overline{w'} = 0$, but in general $\overline{u'w'} \neq 0$: u' and w' are covariant.
Vertical flux of horizontal momentum is given by $\overline{u'w'}$. Thus the first assumption can be expressed as follows:
1: $\overline{u'w'} = constant$
B) Prandtl hypothesis: random part of horizontal velocity u' is proportional to vertical wind shear:
2: $u' = l' \frac{\partial \overline{u}}{\partial z}$
where l' is the "mixing length": we can suppose that an air particle maintains its original horizontal speed during its random motion for a length l', before mixing with the surrounding air.
C) Vertical length scale of turbulent eddies is comparable to their horizontal length scale, thus the random part of vertical velocity is of the same order as the horizontal one:
3: $w' \approx l' \frac{\partial \overline{u}}{\partial z}$
Using expressions 2 and 3 in 1:
4: $\overline{u'w'} = \overline{l'^2} \left(\frac{\partial \overline{u}}{\partial z}\right)^2$
D) At the bottom part of the atmosphere the absolute value of mixing length l' is proportional to high z: this is reasonable because random motion are limited at the bottom by the earth surface. The hypothesis is: $(|l'| = kz)$ where k is Kàrmàn constant. Substituting l' in expression 4 we obtain:
5: $\overline{u'w'} = (kz)^2 \left(\frac{\partial \overline{u}}{\partial z}\right)^2$
Extracting the square root of 4 and separating the variables we obtain:
6: $d\overline{u} = \frac{\sqrt{\overline{u'w'}}}{k} \frac{dz}{z}$
Integrating we obtain the logarithmic profile:
$\Delta \overline{u} = \frac{\sqrt{\overline{u'w'}}}{k} \log{\frac{z}{z_0}}$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Learning physics online? I'm thinking of following some kind of education in physics online. I have a master degree in Computer Science and have reasonable good knowledge in physics. I would like a program of 1-2 years and I'm more interested in particle physics.
Is there any good online program that offer something similar?
| Stanford has posted a bunch of theoretical physics courses by top physicist Leonard Susskind:
Classical Mechanics
Quantum Mechanics
Special Relativity
General Relativity
Cosmology
Statistical Mechanics
I'm a physicist but I still watched all these, just because Susskind is such a great teacher of physics. This is as good as it gets!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 16,
"answer_id": 4
} |
How many atoms per light year does light encounter when traversing interstellar space? Interstellar space is pretty empty but there a small number of of atoms (mostly hydrogen?) floating around. How many atoms per light year would a photon encounter while traversing interstellar space?
| The number of scattering events ("encounters") per unit path length is equal to the density of scatterers times the scattering cross section (summed over all species of scatterers).
The density of particles (mostly hydrogen atoms) in the interstellar medium (ISM) is on the order of 1/cm^3. If we assume that the photon scattering cross section of a hydrogen atom is the square of a common atomic length scale, 1 angstrom^2, then the number of scattering events per unit path length is 10^-14 per meter, or ~100 per light year, give or take a factor of 10.
Take this with a grain of salt, however! The scattering cross section for photons is strongly energy-dependent. For visible photons, say, the cross section is greatly increased if the photon is sufficiently near an atomic transition energy. The hydrogen atoms "look much larger" to these photons, so there are much more collisions.
Lower in the EM spectrum, you run into the fact that the ISM is actually a plasma (it contains ionized hydrogen and free electrons). I believe the plasma frequency is on the order of 1 kHz (strongly depends on temperature...), so radio waves significantly below 1 kHz cannot propagate through the ISM at all. They scatter from the plasma as a whole, rather than individual particles, so the ISM is actually opaque to them.
So the real answer is "it entirely depends on what energy photons you're talking about".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Books that every physicist should read Inspired by How should a physics student study mathematics? and in the same vein as Best books for mathematical background?, although in a more general fashion, I'd like to know if anyone is interested in doing a list of the books 'par excellence' for a physicist.
In spite of the frivolous nature of this post, I think it can be a valuable resource.
For example:
Course of Theoretical Physics - L.D. Landau, E.M. Lifshitz.
Mathematical Methods of Physics - Mathews, Walker. Very nice chapter on complex variables and evaluation of integrals, presenting must-know tricks to solve non-trivial problems. Also contains an introduction to groups and group representations with physical applications.
Mathematics of Classical and Quantum Physics - Byron and Fuller.
Topics in Algebra - I. N. Herstein. Extremely well written, introduce basic concepts in groups, rings, vector spaces, fields and linear transformations. Concepts are motivated and a nice set of problems accompany each chapter (some of them quite challenging).
Partial Differential Equations in Physics - Arnold Sommerfeld. Although a bit dated, very clear explanations. First chapter on Fourier Series is enlightening. The ratio interesting information/page is extremely large. Contains discussions on types of differential equations, integral equations, boundary value problems, special functions and eigenfunctions.
| "Récoltes et Semailles" by Alexander Grothendieck, however it's not for a physicist, but for a scientist in general. I've read it at roughly the same time as "Surely, you are joking Mr. Feynman" and I appreciate RS much more.
I think that book is just invaluable. It is translated in English somewhere and around 25% in Russian, which I read. Though formally per Grothendieck's will all his works shouldn't be distributed, there is little trouble to get them.
I'm not so sure there are books on physics topics itself that can be recommended for everyone. It greatly depends on the interests and the level of the reader.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 24,
"answer_id": 9
} |
Lightning strikes the Ocean I'm swimming in - what happens? I'm swimming in the ocean and there's a thunderstorm. Lightning bolts hit ships around me. Should I get out of the water?
| My girlfriend and I were swimming in the ocean about 30 feet from a beach in Costa Rica as a storm was approaching. We saw occasional flashes of lightning, and I was counting the time it took to hear the thunder, which was at least 7-8 seconds at the fastest, and often 10 or more seconds. I figured as it was at least a mile or more away we were fine, but after a delay with no lightning or thunder, we had a sudden flash, that appeared closer, with thunder a split second later. I felt a slight buzz in my body and I almost didn't mention it to her as we scrambled to shore, but then she said "I just felt zapped!" I would describe it as similar to static shock of touching a door knob after walking across a dry rug, but more diffuse and softer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 6,
"answer_id": 5
} |
Explanation: Simple Harmonic Motion I am a Math Grad student with a little bit of interest in physics. Recently I looked into the Wikipedia page for Simple Harmonic Motion.
Guess, I am too bad at physics to understand it. Considering me as a layman how would one explain:
*
*What a Simple Harmonic motion is? And why does this motion have such an equation $$x(t)= A \cos(\omega{t} + \varphi)$$
associated with it? Can anyone give examples of where S.H.M. is tangible in Nature?
| Simple harmonic motion (SMH) describes the behavior of systems characterized by a equilibrium point and a restoring "force" (in some generalized sense) proportional to the displacement from the equilibrium.
Example system
A simple mechanical system with this behavior is a mass on a spring (which we will consider in one dimension for ease). There is some point where the mass is stable: it can be left alone with no external forces on it and will not accelerate. That's the equilibrium, call it $x_0$. If you move the mass from that point, the spring exerts a restoring force $F = -k(x-x_0)$. Here $k$ is a property of the spring called the "spring constant"; a stiff spring hs a high spring constant and a weak spring has a low value for $k$. Lets consider the special case where $x_0 = 0$ (with no loss of generality, but just to keep the number symbols to a minimum).
So the acceleration of the mass at any point in time is
$$a = \frac{d^2x}{dt^2} = \frac{F}{m} = -\frac{kx}{m} $$
which is a second order linear differential equation, the solutions to which are of the form you exhibit above (where $A$ and $\phi$) take the part of the two constants of integration needed to make the solution agree with an arbitrary set of boundary conditions.
Let's prove it by explicit substitution:
$$x(t)=A \cos(\omega t + \varphi )$$
$$v(t) = \frac{d}{dt} x(t) = A \sin(\omega t + \varphi ) \omega $$
$$a(t) = \frac{d}{dt} v(t) = \frac{d^2}{dt^2} x(t) = -A \cos(\omega t + \varphi ) \omega^2 $$
so substituting back in we get
$$
-A \cos(\omega t + \varphi ) \omega^2 = \frac{F}{m} = -\frac{k x}{m} = -\frac{k A}{m} \cos(\omega t + \varphi )
$$
which implies that
$$ \omega^2 = \frac{k}{m} $$
which relates the spring constant and the mass of the object to the frequency of oscillation.
General discussion
The thing that is important about SHM is that these restoring "forces" linearly proportional to "displacement" (which we allow that there may be generalized meanings for both "force" and "displacement") are very common in the universe. So a great many phenomena may be described this way (and even more may be if we limit ourselves to small perturbations).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 0
} |
Common false beliefs in Physics Well, in Mathematics there are somethings, which appear true but they aren't true. Naive students often get fooled by these results.
Let me consider a very simple example. As a child one learns this formula $$(a+b)^{2} =a^{2}+ 2 \cdot a \cdot b + b^{2}$$ But as one mature's he applies this same formula for Matrices. That is given any two $n \times n$ square matrices, one believes that this result is true: $$(A+B)^{2} = A^{2} + 2 \cdot A \cdot B +B^{2}$$ But eventually this is false as Matrices aren't necessarily commutative.
I would like to know whether there any such things happening with physics students as well. My motivation came from the following MO thread, which many of you might take a look into:
*
*https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics
| I would say that for most people, the quadratic scaling of kinetic energy with speed is somewhat of a mystery.
People don't understand how if you go twice as fast, a car accident actually four times as energetic, hence the high number of reckless drivers and deadly accidents.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 49,
"answer_id": 23
} |
How do contact lenses work? I understand how telescope, microscope and glasses work.
But how do contact lenses work?
| The thing that confused me at first about contact lenses was that they didn't look like "normal" lenses, in which each side of them curved in the opposite direction to each other.
As I understand them, contact lenses are Meniscus lenses - wherein the sides are curved in the same direction as each other, but more curved on one of the sides. You can see some examples here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Home experiment to estimate Avogadro's number? How to get an approximation of Avogadro or Boltzmann constant through experimental means accessible by an hobbyist ?
| Well, addressing only $N_A$, and allowing that it only gets you halfway there: Carl Sagan did a bit on Cosmos for estimating the size of a oil molecule. Put a known amount (both volume and mass) of cooking oil on a calm body of water and wait for the slick to diffuse to it's maximum contiguous area. Estimate the area.{*} Divide the original volume by that area and you have a rough value for the linear size scale of the molecule and thus how much volume it occupies. (We're neglecting issues of aspect ratio and polarity here so this is only a order of magnitude value.)
Now, if you have the molecular mass (from a mass spectrometer or some such) you're got all you need.
{*} A challenging step, but I'd use a photograph with some know length reference for scale and a planimeter. If you're not familiar with these instruments get ready to geek out. Very cool.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 2
} |
What's the difference between helicity and chirality? When a particle spins in the same direction as its momentum, it has right helicity, and left helicity otherwise. Neutrinos, however, have some kind of inherent helicity called chirality. But they can have either helicity. How is chirality different from helicity?
| Chirality and helicity are exactly the same thing in the massless limit. By this, I mean that either term can be used interchangeably in the massless limit (recall that a condition for massless particles, is that they necessarily move at velocity c). And yes you are correct in that there is no frame of reference where we can hypothetically boost beyond say a photon, and find that helicity has flipped. Having said that, there also exist massive particles with definite chirality. This has to do with spin and a process called quantum interference. The explanation has its roots tied deep in the understanding of QM and field theory. It's long winded and mathematical. But it's crucial to having a full understanding of the Standard Model. Look it up and research! And if you have anymore questions we will be happy to oblidge. Good luck.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 5,
"answer_id": 2
} |
What is the difference between "kinematics" and "dynamics"? I have noticed that authors in the literature sometimes divide characteristics of some phenomenon into "kinematics" and "dynamics".
I first encountered this in Jackson's E&M book, where, in section 7.3 of the third edition, he writes, on the reflection and refraction of waves at a plane interface:
*
*Kinematic properties:
(a) Angle of reflection equals angle of incidence
(b) Snell's law
*Dynamic properties
(a) Intensities of reflected and refracted radiation
(b) Phase changes and polarization
But this is by no means the only example. A quick Google search reveals "dynamic and kinematic viscosity," "kinematic and dynamic performance," "fully dynamic and kinematic voronoi diagrams," "kinematic and reduced-dynamic precise orbit determination," and many other occurrences of this distinction.
What is the real distinction between kinematics and dynamics?
| You should think of it in terms of programming a computer to simulate the physical system. Kinematics is the data structure you need to simulate the general situation, what variables with what range of values. Dynamics is the actual algorithm that simulates the motion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "84",
"answer_count": 14,
"answer_id": 10
} |
Doppler's effect use While i was in high-school i learn't the Doppler's Effect which if i remember correctly is:
*
*The Apparent change in the frequency of sound caused due the relative motion between the sound and the observer.
This phenomenon seems obvious, but what i would like to know is, what use does Doppler Effect have in real life. Why is it useful?
| To my knowledge,following are some uses of Doppler effect:
*
*Measurement of the speed of approaching automobiles.
*Ships are equipped with an instrument called SONAR which works on the principle of Doppler effect. It detects the pressence of under water rocks and submarines...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
} |
What sustains the rotation of earth's core (faster than surface)? I recently read that the earth's core rotates faster than the surface.
Well, firstly, it's easier to digest the concept of planetary bodies, stars, galaxies in rotation and/or orbital motion.
But, what makes a planet's core rotate? And in the earth's case, faster than its surface?
Secondly, I am aware that the core's rotation is what lends the earth its magnetic field but.. what keeps it going in the first place?
| According to our current understanding of the formation of planets, they are created from dust which originates in a previous supernova (or other large) explosion. This is called a nebula.
http://en.wikipedia.org/wiki/Nebular_hypothesis
As this nebula contracts into planets, there are generally two possible cases: that the dust that is eventually going to form a planet is spinning, overall, or not. The most general and likely case is that it is spinning in some form. There is no overall reason why it shouldn't, short of some "magical" combination of canceling factors.
What happens when this spinning dust cloud contracts under gravity? It spins faster (think of an ice skater spinning), due to the conservation of angular momentum.
Therefore, you end up with planets with a relatively fast spinning core.
Now, what keeps this spin going? Fundamentally, conservation of angular momentum. Nothing is keeping it going, and it's actually spinning down. It should eventually stop - in million or billions of years. That's how much momentum is stored in it :-)
Why does the crust spin slower than the core? For two reasons: since things spin faster as they contract, you naturally end up with a variation of speed between the outer layers and the inner core; secondly, the core is actually dragging the crust around, but the Earth is not really a solid so the drag cannot be thought of as a rigid motion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
} |
Best example of energy-entropy competition? What are the best examples in practical life
of an energy-entropy competition which favors entropy over energy?
My initial thought is a clogged drain -- too unlikely for the
hair/spaghetti to align itself along the pipe -- but this is probably
far from an optimal example. Curious to see what you got. Thanks.
| Well, as a physicist working mainly in the field of statistical physics, I obviously have to mention Ising model. In this case the model is actually tractable (at least in two dimensions) and can tell us a whole lot about (not just) energy-entropy battle. It's obvious that the ground state (for ferromagnetic case) is all spins pointing one way (say up). Now, if you point some spins the other way then you are losing in terms of energy (something like number of neighbors times number of wrong spins) but actually you gain hugely in entropy (because of translational invariance of lattice models). This argument can be made very precise when working with polymer model (which is isomorphic to Ising model) and considering low-temperature cluster expansion.
I am sorry, but I am not really able to provide nice references. Wikipedia articles are pretty bad. Maybe I should spend some time bringing them up to the current knowledge (that is to say, knowledge since like 1970s) about cluster expansion. For now, if anyone is interested, just read the basic paper on the topic.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Is it possible to separate the poles of a magnet? It might seem common sense that when we split a magnet we get 2 magnets with their own N-S poles. But somehow, I find it hard to accept this fact. (Which I now know is stated by the magnetic Gauss's Law $\vec{\nabla}\cdot \vec{B} =0.$)
I have had this doubt ever since reading about the quantum-field-theory and I know I might sound crazy but is it really impossible to separate the poles of a magnet?
Is there some proof/explanation for an independently existing magnetic monopole?
| Magnetic monopoles certainly exist. This does not require a GUT, they exist in any theory where the electromagnetic U(1) is compact (i.e. where charge is quantized). This follows only from the semiclassical behavior of black hole decay, and so does not require unknown physics.
The reason is essentially the one you state--- you can polarize a black hole in a strong magnetic field, and let it split by Hawking radiation into two oppositely magnetically charged black holes of opposite polarities. Magnetically charged black holes exist in classical General Relativity, as are arbitrary electric-magnetic charge ratio holes, and you can't forbid them, at least not for macroscopically sized black holes, without ruining the theory.
When you let the monopolar black holes decay, you find relatively light monopoles. The lightest monopoles will be lighter than its magnetic charge, so that two such monopoles will repel magnetically, not attract. Presumably, the monopole you find will be a (small multiple of) Dirac's magnetic monopole quantum.
To me, this is as certain as the existence of the Higgs. We haven't observed either one, but the theoretical argument is completely convincing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 8,
"answer_id": 0
} |
Can I parameterize the state of a quantum system given reduced density matrices describing its subparts? As the simplest example, consider a set of two qubits where the reduced density matrix of each qubit is known. If the two qubits are not entangled, the overall state would be given by the tensor product of the one qubit states. More generally, I could write a set of contraints on the elements of a two-qubit density matrix to guarantee the appropriate reduced description.
Is there is a way to do this more elegantly and systematically for arbitrary bi-partite quantum systems? I'm particularly interested in systems where one of the Hilbert spaces is infinite dimensional, such as a spin 1/2 particle in a harmonic oscillator.
| Given an n-partite system and observables $\hat{T}_i$ with exspectation values $<\hat{T}_i> = t_i$, you can write your state as a maximum entropy state:
$$\rho = \frac{1}{Z}exp\left(\sum_i \theta_i \hat{T}_i\right)$$
see Theorem 2 in http://arxiv.org/abs/quant-ph/0603012
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to obtain gold through nuclear decay? Is there a series of transmutations through nuclear decay that will result in the stable gold isotope ${}^{197}\mathrm{Au}$ ? How long will the process take?
| Natural gold exists, so the answer to the first part of your question is unambiguously "Yes". 'Cause all those heavy elements get made by transmutation in supernovae.
I can't answer the time scale thing because I haven't a table of the isotopes in front of me right now.
Checking with http://ie.lbl.gov/education/isotopes.htm I find that $^{197}$Pt has a beta decay branching fraction of about 3% and a halflife of about 95 minutes, and $^{197}$Ir decays by beta with 5 minute halflife and $^{197}$Os decays by beta with a 64 hour halflife...
Anyway, you can chase this as far as you care to.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
Which experiments prove atomic theory? Which experiments prove atomic theory?
Sub-atomic theories:
*
*atoms have: nuclei; electrons; protons; and neutrons.
*That the number of electrons atoms have determines their relationship with other atoms.
*That the atom is the smallest elemental unit of matter - that we can't continue to divide atoms into anything smaller and have them retain the characteristics of the parent element.
*That everything is made of atoms.
These sub-theories might spur more thoughts of individual experiments that prove individual sub-atomic theories (my guess is more was able to be proven after more experiments followed).
| I would say that one experiment that demonstrates the atomic nature of things is the observation of Brownian motion. But it is not the experiment itself that convinces that things are made of atoms, rather its theoretical explanation given by Einstein in one of his 1905 papers (actually Einsteins work for his PhD was on the subject of atomic theory and there are several publications in the period 1903-1905). Of course there is also the observation of Rayleigh who calculated Avogadro’s number by the distance from which he could make out the figure of Mount Everest, assuming that light is scattered by atoms and that is why far away objects look fuzzy (1,2). Also scattering experiments demonstrated the atomic nature of things.
(1) Rayleigh, On the transmission of light through an atmosphere containing small
particles in suspension, in Scientific Papers by Lord Rayleigh Vol. 4, pp. 247–405,
New York: Dover, 1899/1964.
(2) P. Pesic, Eur. J. Phys. 26, 183 (2005).
(3) Patterson, G. Jean Perrin and the triumph of the atomic doctrine (2007) Endeavour, 31 (2), pp. 50-53.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 3
} |
Spectral Line Width and Uncertainty principle so I've been at this for about 3 - 4 hours now. It is an homework assignment (well part of a question which i've already completed). We did not learn this in class. All work is shown below.
An atom in an excited state of $4.9 eV$ emits a photon and ends up in the ground state. The lifetime of the excited state is $1.2 \times 10^{-13} s$.
(b) What is the spectral line width (in wavelength) of the photon?
So lets look at what I have done so far. I have done the following:
$$\Delta E \Delta t = \frac{\hbar}{2} $$ but
$$E = h f$$ so
$$\Delta f = \frac{1}{4\pi \Delta t}$$
but if I take $\Delta f$ and convert it into wavelength using $\lambda f = c $ then it gives me the wrong answer. I've tried MANY variations of the above formulas.
The correct answer is $0.142 nm $
Can anyone give me a hint?
| Hint: Your problem is in the "take $\Delta f$ an convert it to wavelength using $\lambda f = c$" part. The equation $\lambda f = c$ does not imply $\Delta \lambda \Delta f = c$.
Answer: Rather it implies,
$\lambda f = c$
$\lambda = \frac{c}{f} $
Now differentiate:
$d\lambda = -c\frac{df}{f^2}$
$df \ll f$ so treating the $\Delta f$ as a differential works fine. If this was not the case, you'd want to integrate from $f_{min}$ to $f_{max}$ (not given in the problem, just how you'd have to do it otherwise). This formula gives you the right answer. The sign opposite signs just indicate the higher frequency corresponds to the lower wavelength and is ignored in the final answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What's the difference between running up a hill and running up an inclined treadmill? Clearly there will be differences like air resistance; I'm not interested in that. It seems like you're working against gravity when you're actually running in a way that you're not if you're on a treadmill, but on the other hand it seems like one should be able to take a piece of the treadmill's belt as an inertial reference point. What's going on here?
| What if the runner gets on her bicycle on the treadmill? Does she have to bike harder on an inclined treadmill than a horizontal one, both running at the same speed?
(In this case, biking on a horizontal treadmill is almost effortless because on a bike, the real effort is the wind resistance, which would be non-existent in this case.)
If the treadmill were turned off, and she stopped pedalling, she would start to accelerate backwards down the slope due to a component of her weight $mg$ acting down the hill, namely $m g \sin(\theta)$.
The feeling on the (running) inclined treadmill at angle $\theta$ would be the same as biking on the flat (again with no wind resistance), but with an elastic strap or spring exerting a backwards force on the bike (equivalent to the component $mg \sin(\theta)$ of her weight, contingent on the required angle of ascent). Unlike the case on the flat without the elastic, she now has to press harder on the pedals to do work applying the appropriate force at the point of contact of the driving wheel at the appropriate speed.
The arguments & conclusions for the bicycle carry over to the runner. But it is easier to imagine the situation for the bike, because it is less confused by the complicated motion of running.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 10,
"answer_id": 8
} |
Why does the (relativistic) mass of an object increase when its speed approaches that of light? I'm reading Nano: The Essentials by T. Pradeep and I came upon this statement in the section explaining the basics of scanning electron microscopy.
However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. At such velocities, one needs to do relativistic correction to the mass so that it becomes[...]
We all know about the famous theory of relativity, but I couldn't quite grasp the "why" of its concepts yet. This might shed new light on what I already know about time slowing down for me if I move faster.
Why does the (relativistic) mass of an object increase when its speed approaches that of light?
| The mass (the true mass which physicists actually deal with when they calculate something concerning relativistic particles) does not change with velocity. The mass (the true mass!) is an intrinsic property of a body, and it does not depends on the observer's frame of reference. I strongly suggest to read this popular article by Lev Okun, where he calls the concept of relativistic mass a "pedagogical virus".
What actually changes at relativistic speeds is the dynamical law that relates momentum and energy depend with the velocity (which was already written). Let me put it this way: trying to ascribe the modification of the dynamical law to a changing mass is the same as trying to explain non-Euclidean geometry by redefining $\pi$!
Why this law changes is the correct question, and it is discussed in the answers here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 14,
"answer_id": 0
} |
Ice skating, how does it really work? Some textbooks I came across, and a homework assignment I had to do several years ago, suggested that the reason we can skate on ice is the peculiar $p(T)$-curve of the ice-water boundary. The reasoning is that due to the high pressure the skates put on the ice, it will melt at temperaturs below $273 K$ and thus provide a thin film of liquid on which we can skate. It was then mentioned as fun fact that you could ice-skate on a planet with lakes of frozen dioxide because that gas has the $p(T)$-curve the other way round.
My calculations at that time told me that this was, pardon my french, bollocks. The pressure wasn't nearly high enough to lower the melting point to even something like $-0.5$ degrees Celsius.
I suppose it is some other mechanism, probably related to the crystal structure of ice, but I'd really appreciate if someone more knowledgeable could tell something about it.
| Yup, this is true that the pressure is too small, but the true explanation is not justified yet. Nevertheless the common sense is that there is a lubricating film of water or at least anomalous ice.
For an overview, see: http://lptms.u-psud.fr/membres/trizac/Ens/L3FIP/Ice.pdf
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 7,
"answer_id": 1
} |
Hawking radiation and quark confinement The simple picture of Hawking radiation is that a pair-antiparticle pair is produced near the event horizon, then one falls into the black hole while the other escapes. Suppose the particles are quarks-antiquarks, which experience quark confinement thanks to QCD. If one of them is swallowed by the black hole, its partner is left alone. Eventually the quark gains enough energy and turns into a hadronic jet.
Is my line of thinking correct? If yes, is it (or QCD in general) taken into account when calculating Hawking radiation?
| I think your reasoning should be correct.
However, as I understand the Hawking radiation (i.e. not much) the effect is derived independently of the microscopic theory. You just assume that particle (which is to escape from around the horizon) needs to accelerate (a lot) in order to overcome black hole's gravity and then notice that accelerating observer will register a black body radiation. This is called Unruh effect.
Now, you can certainly try to derive the effect from the first (i.e. QFT) principles. E.g. this paper claims that it is indeed possible to derive the results from the free QFT. But for realistic QFT (like QCD) this doesn't seem technically feasible and the results you obtain (i.e. Hawking temperature) can depend on the precise theory used.
This paper talks about similar effect but instead from tunneling from under the black hole horizon it discusses tunneling out of the confinement horizon (this is called white hole because confinement means white hadrons; not to be confused with white holes from GR).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
What is terminal velocity? What is terminal velocity? I've heard the term especially when the Discovery Channel is covering something about sky diving. Also, it is commonly known that HALO (Hi-Altitude, Lo-Opening) infantry reaches terminal velocity before their chutes open.
Can the terminal velocity be different for one individual weighing 180 pounds versus an individual weighing 250 pounds?
| Terminal velocity is the (asymptotic) maximum velocity that you can reach during free-fall. If you imagine yourself falling in gravity, and ignore air resistance, you would fall with acceleration $g$, and your velocity would grow unbounded (well, until special relativity takes over). This effect is independent of your mass, since
$F = ma = mg \Rightarrow a = g$
Where terminal velocity arises is that air resistance is a velocity-dependent force acting against your free fall. If we had, for example, a drag force of $F_D=KAv^2$ ($K$ is just a constant to make all the units work out and depends on the properties of the fluid you're falling through, and $A$ is your cross-sectional area perpendicular to the direction of motion) then the terminal velocity is the velocity at which the forces cancel (i.e., no more acceleration, so the velocity becomes constant):
$F = 0 = mg - KAv_t^2 \Rightarrow v_t=\sqrt{mg/KA}$
So we see that a more massive object can in fact have a larger terminal velocity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/1989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Can I levitate an object without using an electromagnet? I know that it's possible to make an object levitate using an electromagnet to hold it up.
But is it also possible to do this with regular magnets? Is there a special kind of magnet I need in order to have one powerful enough to hold an object up?
I'm asking because I have this idea in mind where I want to make a decorative item that is levitating an inch or so above its container.
A is the object I want to levitate.
B is the container.
C shows the magnetic field that is repelling A and B from each other achieving the levitation.
The size of this would be as small as a food plate, maybe even smaller. B is barely a pound or two in weight.
Is that possible?
| yes, you can. Not easily though.
My favorite is the Meisner effect. It would require a superconductor for B, which would most likely require liquid nitrogen. Basically, the levitating magnet (A) would create an current in the superconductor (which has "zero" resistance). The current would then go for as long as the superconductor superconducts, creating a magnetic field, which would repulse A.
For more explanation, check out wikipedia. Basically, a I have no idea how much this would cost. Although, the liquid nitrogen needed would most likely make it impractical. By the way, there are no room temp superconductors currently known.
more methods are shown here: http://en.wikipedia.org/wiki/Magnetic_levitation. Some methods shown there do in fact use electricity, but some don't. Check out the spin-stabilized levitation. Very cool.
The last would be to build a bowl of magnets. If you arrange it right, you should be able to get it stable.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/2029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
} |
How can one find the energy eigenfunctions of a particle in a finite square well via the Klein-Gordon equation? It is said that Klein-Gordon equation is a relativistic version of the Schrodinger equation. In Schrodinger equation, it is straightforward to include potential energy. But for K-G eqn things seem to be more complicated.
To put it specifically, how can one find the energy eigenfunctions of a particle in a finite square well via Klein-Gordon equation?
| K-G is a relativistic equation. So if you want to couple it to external fields you have to use a vector potential $A_\mu$ which will modify the partial-derivative, so that the Klein-Gordon equation:
$$ \left( \partial^\mu \partial_\mu + m^2 \right)\phi = 0 $$
becomes:
$$ \left ( D^\mu D_\mu + m^2 \right) \phi = 0 $$
where $ D_\mu = \partial_\mu + \imath A_\mu $.
Then it remains to figure out what $A_\mu$ will correspond to a given potential - finite square well or otherwise. For this you will likely have to fix a gauge before you can identify the potential as such.
A square well potential will correspond to $ A_\mu = (V(x), 0,0,0)$, where $ V(x) = V_0 \,\forall |x| > \pm a $ and $V(x) = 0, x \in [-a,a] $. The resulting behavior is a little counter-intuitive, yielding something known as the Klein paradox.
For further details see ch. 4 for this excellent thesis on this problem by D. Schaffer.
Edit: This page also has excellent information along these lines.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/2077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
How long a straw could Superman use? To suck water through a straw, you create a partial vacuum in your lungs. Water rises through the straw until the pressure in the straw at the water level equals atmospheric pressure. This corresponds to drinking water through a straw about ten meters long at maximum.
By taping several straws together, a friend and I drank through a $3.07m$ straw. I think we may have had some leaking preventing us going higher. Also, we were about to empty the red cup into the straw completely.
My question is about what would happen if Superman were to drink through a straw by creating a complete vacuum in the straw. The water would rise to ten meters in the steady state, but if he created the vacuum suddenly, would the water's inertia carry it higher? What would the motion of water up the straw be? What is the highest height he could drink from?
Ignore thermodynamic effects like evaporation and assume the straw is stationary relative to the water and that there is no friction.
| I have an argument that the water in the straw will rise to twice the equilibrium height.
David and Martin's answers consider the system of water in the straw. I will consider the system of the water in the straw plus the water in the reservoir.
As water goes into the straw, the water level in the reservoir drops, and the atmosphere does work on the system. If a volume $V$ of water enters the straw, the work done on the system is $PV$, with $P$ the atmospheric pressure. Assume that the reservoir has a large surface area so that the level the reservoir drops is negligible.
When the water is at its peak in the straw, the kinetic energy of the system is zero, so the potential energy is $PV$. The potential energy is also $\rho g V h/2$. So the maximum height of the water is
$$h = \frac{2P}{\rho g}$$
This answer is different from Martin and David's. I think this might be because when the water starts moving, the pressure at the entrance to the straw may not be $P$ any more.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/2111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "66",
"answer_count": 12,
"answer_id": 7
} |
Why did this glass start popping? I remember a while ago my father dropped a glass lid and it smashed. It looked something like this. When that happened, for about 5 minutes afterwards, the glass parts were splitting, kind of like popcorn, and you could hear the sound. I was just wondering why this happened, and the particles didn't just sit quietly in their own original parts?
| Here is an idea. You could have cracking (fracture) behaving as a critical process. From the dropping of the glass you could have micro-fractures forming. In time and because of the amount of tension (and energy) concentrated in the glass from the fall, these micro-fractures slowly could organize to form big fractures, which would cause the popping sounds you continued to hear for some time after the fall of the glass lid. If it was some kind of crystal that broke I would be more confident that that was the case, but since it is some kind of glass, I am not so sure that it can happen something like that, since glass is amorphous, but I am not sure that it can't either.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/2156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Is it possible for information to be transmitted faster than light by using a rigid pole? Is it possible for information (like 1 and 0s) to be transmitted faster than light?
For instance, take a rigid pole of several AU in length. Now say you have a person on each end, and one of them starts pulling and pushing on his/her end.
The person on the opposite end should receive the pushes and pulls instantaneously as no particle is making the full journey.
Would this actually work?
| The answer is no. The pole would bend/wobble and the effect at the other end would still be delayed.
The reason is that the force which binds the atoms of the pole together - the Electro-Magnetic force - needs to be transmitted from one end of the pole to the other. The transmitter of the EM-force is light, and thus the signal cannot travel faster than the speed of light; instead the pole will bend, because the close end will have moved, and the far end will not yet have received intelligence of the move.
EDIT: A simpler reason.
In order to move the whole pole, you need to move every atom of the pole.
You might like to think of atoms as next door neighbours If one of them decides to move, he sends out a messenger to all his closest neighbours telling them he is moving. Then they all decide to move as well, so they each send out messengers to to their closest neighbours to let them know they are moving; and so it continues, until the message to move has travelled all the way to the end. No atom will move until he has received the message to do so, and the message won't travel any faster than all the messengers can run; and the messengers can't run faster than the speed of light.
/B2S
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/2175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "117",
"answer_count": 16,
"answer_id": 7
} |
Anti-gravity in an infinite lattice of point masses Another interesting infinite lattice problem I found while watching a physics documentary.
Imagine an infinite square lattice of point masses, subject to gravity. The masses involved are all $m$ and the length of each square of the lattice is $l$.
Due to the symmetries of the problem the system should be in (unstable) balance.
What happens if a mass is removed to the system? Intuition says that the other masses would be repelled by the hole in a sort of "anti-gravity".
*
*Is my intuition correct?
*Is it possible to derive analytically a formula for this apparent repulsion force?
*If so, is the "anti-gravity" force expressed by $F=-\frac{Gm^2}{r^2}$, where $r$ is the radial distance of a point mass from the hole?
Edit:
as of 2017/02 the Video is here (start at 13min):
https://www.youtube.com/watch?v=mYmANRB7HsI
| I think that your initial intutiion is right--before the point particle is removed, you had (an infinite set of) two $\frac{G\,m\,m}{r^{2}}$ forces balancing each other, and then you remove one of them in one element of the set. So initially, every point particle will feel a force of $\frac{G\,m\,m}{r^{2}}$ away from the hole, where $r$ is the distance to the hole. An instant after that, however, all of the particles will move, and in fact, will move in such a way that the particles closest to the hole will be closer together than the particles farther from the hole. The consequence is that the particles would start to clump in a complicated way (that I would expect to depend on the initial spacing, since that determines how much initial potential energy density there is in the system)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/2196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
} |
What is the definition of colour (the quantum state)? I heard somewhere that quarks have a property called 'colour' - what does this mean?
| Let me give a straightforward and basic answer here; perhaps someone else can elaborate.
The best way to think of colour is something analogous to charge in electromagnetism. (Indeed, colour is often called colour charge). It is the fundamental property of particles relating to the strong force, and like electric charge, comes in discrete values (termed red, green, and blue - don't take them literally though). Note however that strong integration works in a far more complex way than electromagnetism (Maxwellian, or even quantum electrodynamics).
The Wikipedia article states:
The "color" of quarks and gluons is
completely unrelated to visual
perception of color.[1] Rather, it is
a whimsical name for a property that
has almost no manifestation at
distances above the size of an atomic
nucleus. The term color was chosen
because the abstract property to which
it refers has three aspects, which are
analogized to the three primary colors
of red, green, and blue.[2] By
comparison, the electromagnetic charge
has a single aspect, which takes the
values positive or negative.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/2228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Why are materials that are better at conducting electricity also proportionately better at conducting heat? It seems like among the electrical conductors there's a relationship between the ability to conduct heat as well as electricity. Eg: Copper is better than aluminum at conducting both electricity and heat, and silver is better yet at both. Is the reason for this known? Are there materials that are good at conducting electricity, but lousy at conducting heat?
| What links the two conductivities is that they both depend on how transparent the material is to electrons traveling around the Fermi energy.
Thermal conductivity also has a contribution from lattice vibrations, but for metals the contribution from electrons dominate.
In an analogy, imagine two recevoirs of water connected by a channel. The two recevoirs and their height correspond to the leads of the conductor at different potentials and the width for the channel at the surface corresponds to the transmission of electrons at the Fermi surface.
Changing the height of the water in one recevoir in relation to the other will create a net flow of water that depends linearly on the width of the channel. This is the analogy of the flow of electrons through the conductor.
Temperature in this picture correspond to the amount of waves in the recevoirs. And how fast wave energy in one recevoir moves to the other recevoir also strongly depend on the width of the channel.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/2245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Polar vs non-polar fluid In the book "Vectors, Tensors, and the Basic Equations of Fluid Mechanics" by Rutherford Aris I read the following:
If the fluid is such that the torques
within it arise only as the moments of
direct forces we shall call it
nonpolar. A polar fluid is one that is
capable of transmitting stress couples
and being subjected to body torques,
as in polyatomic and certain
non-Newtonian fluids.
Can someone help me understand this? In particular, it would be helpful if someone could give me another definition of polar and nonpolar fluids.
| A polar fluid is just a fluid where the constituent molecules have a polarization -- it could be a fluid of molecules that have a magnetic spin moment, or something like H2O where each individual molecule has a nonzero electric dipole -- and at the macroscopic level, as you average over all of the microscopic moments, you get a net polarization for the whole fluid.
I'm no expert on fluid mechanics, but I imagine the polarization somehow couples to the stress tensor in a way that generates torques in whatever equations of motion the author is interested in.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/2364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.