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Weight distributions If a man is standing on two weighing machines (scales), with one foot on each, Will both machines show equal weight or his weight will be distributed in two machines?
| If the person in concern is standing perfectly, such that his weight distribution is equal over both legs, and if both the weights are calibrated perfectly, then yes, both the weights will show equal readings. (The readings each being half the weight of the person.)
In a realistic case (without 100% perfection), however, the weight would be unequally distributed over both the machines. The sum of these two readings would equal the weight of the person within a tiny error margin.
| {
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"url": "https://physics.stackexchange.com/questions/183324",
"timestamp": "2023-03-29T00:00:00",
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Does lithium-6 "decay" when hit by a neutron? I am talking about the nuclear reaction
$$
^6\text{Li} + n \rightarrow\ ^4\text{He} +\ ^3\text{H} + 4.78\text{MeV}
$$
A neutron hits a lithium-6 nucleus and together they form an alpha and triton particle. Is it valid to say that the lithium nucleus "decays" when hit by a neutron? Is there any other verb which better describes the change of the lithium nucleus?
I am interested in the correct terminology.
| The process by which the lithium becomes fissile due to neutron capture is called neutron activation. The subsequent decay is simply a fission reaction.
There seems to be a precedent on various sites for such a process to be called a 'neutron capture induced fission reaction', although most of the Google results for the term refer to the more usual fission of uranium and other heavy nuclei.
| {
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Is true black possible? Black is the absence of light because it absorbs light, but when we create black paint or black objects, light is always reflected, either in all directions in matte or smoothly in shiny black objects, making it never a true black. Would it be possible to use polarization to create an object that does not reflect any light, creating a truly black substance, without any shadows or reflection of light?
| Just to add to the above answers, and since to did not limit your question to the visible range - if you define black as absence of light (photons emitted or reflected), then there is no such substance, because according to black body radiation model, everything with a temperature above absolute zero (which is essentially truly everything in the universe:) radiates, meaning it's always emitting photons, and thus is not black.
| {
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How to calculate the period of the movement from a potential? I have an assignment, where I have an object moving in 1-D with a given mass and energy, and the potential V(x), and I'm supposed to calculate the period of the movement as a function of the energy
$$
V(x)=\begin{cases}\infty &x < -a \\
0 &-a < x < 0\\
\alpha x^2 & x>0
\end{cases}
$$
Should I find 3 Lagrangians for the 3 separate parts of the potential? And then how would I come to the period of the movement? Thanks in advance!!
| You can calculate the action integral as a function of energy
$$
J(E) = \oint p_x dx = 2\sqrt {2m} \int_{x_0(E)}^{x_1(E)} \sqrt {E-V(x)}dx
$$
where $x_0$ and $x_1$ are the turning points. (In your case: $x_0 = -a$ and $x_1 = \sqrt{E/\alpha}$). The Period is then given by the derivative
$$
T(E) = dJ/dE
$$
| {
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Can we quantitatively understand quark and gluon confinement in quantum chromodynamics and the existence of a mass gap? Quantum chromodynamics, or QCD, is the theory describing the strong nuclear force. Carried by gluons, it binds quarks into particles like protons and neutrons. According to the theory, the tiny subparticles are permanently confined. A quark or a gluon cannot be taken from a proton because the strong force gets stronger with distance. How much do we understand of this phenomenon? and what do we hope to find about it by studying M-theory?
| Confinement cannot be rigorously shown in QCD with current techniques, because all analytic results in QCD are perturbative and the perturbative expansion breaks down at low energies where the coupling becomes strong.
QCD has a negative $\beta$-function, i.e. the Yang-Mills coupling grows at lower energies and becomes weaker at high energies. But the $\beta$-function itself is only known perturbatively, that is as a power series in the coupling strength. This expansion makes sense when the coupling is weak not when it is strong. This perturbative calculation of the $\beta$-function therefore teaches us two lessons:
*
*The theory is asymptotically free, the coupling becomes weaker and weaker at high energies. Since the perturbative computation is justified in this regime, we can conclude that asymptotic freedom has been shown rigorously.
*At low energies the coupling becomes strong due to the $\beta$-function, but the calculation of the $\beta$-function itself relies on weak coupling. So we cannot say anything definite about this regime. However, the growing coupling is seen as a hint of confinement. One cannot, for example, compute meson masses in perturbative QCD.
Lattice computations however have been able to compute some meson masses, but they are of course numerical. Maybe somebody with more knowledge of lattice QCD can add some details.
There is the hope to study the strongly coupled regime of QCD using the holographic duality which maps QCD to a weakly coupled gravity theory (more precisely a string theory, which is closely related to M theory). There are some qualitative results, see e.g. here and here, but the holographic dual of QCD, if it exists, is not known and the quest for is has been unsuccessful.
| {
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Notation of vectors It's very common to see $\text{F} = 30 \text{ N}$ when the problem is unidimensional. Yet, force is a vector. Shouldn't I write $|\overrightarrow{F}| = 30 \text{ N}$? Because if I write $\overrightarrow{F} = 30 \text{ N}$ I'm saying that the vector is equal to an scalar. On the other hand, I rarely see $\overrightarrow{F} = (30, 0, 0)$.
| Some people use $\mathbf{F}$ instead of $\vec{F}$ or even $\overrightarrow{F}$. I agree that often $F=\| \vec{F} \|$ is a convenient shortcut. So for example
A force $\mathbf{F}=(10 \mbox{ N},0,0)$ has magnitude $\|\mathbf{F}\|=10 \mbox{ N}$.
The components of $\mathbf{F}$ are $F_x = 10\mbox{ N}$, $F_y=0$ and $F_z=0$
So the subscript is used to designate which component, and the italicized variable indicates it is a scalar.
| {
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Filming light in slo mo So I was thinking of a cool video I could make when I thought of filming a video of me turning on a light. I thought that if you film yourself turning on a light and slo mo that video enough times (theoretically I you had a camera that could record that many frames per second)will you be able to see light moving through your videos view, or will it just turn into a slide show?
Thanks in advance for any answers
(PS if this has already been done please give me a link to the video.)
| Light moves at about a foot per nanosecond, or a meter every three nanoseconds. In order to capture it propagating across a room over a few frames, you would need to gather something like a billion frames per second. No consumer camera -- indeed no camera on Earth -- is capable of this.
Now there have been people playing with "fempto-photography," but they use a hack. They switch on the light many, many, many times, in exactly the same way, and each time record just one frame at a very precise time. They then put all the frames into a single movie. That is, they leverage precision timing rather than raw speed, the former of which is achievable.
Note though that the equipment is still beyond the consumer level. You need a shutter speed fast enough so as to not blur a single frame, as well as nanosecond-resolution triggering.
| {
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Temperature effects on lead against radiation I would like to know if bringing lead to near absolute zero temperatures would have any affects on how resistive it is against gamma radiation. It takes 40 centimeters of lead to reduce gamma radiation effects by a factor of a billion (medium energy levels). Since atoms come closer to each other at colder temperatures, I would imagine this would increase the amount of atoms in a given surface area, thus increasing the chances a gamma ray will interact with an atom. Does anyone know if this could possibly reduce the thickness of lead needed to block gamma rays? I have been looking everywhere and have not been able to find this answer. Any help would be much appreciated. I am still very new to physics and have so many questions that I have no one to talk with about, now that I have gotten involved with physics on this level. Thanks, everyone's help is appreciated.
| It is the mass of material more than the thickness that determines the stopping power (which incidentally is a function of energy - so you can't simply state "40 cm reduces gamma flux one billion times" without specifying the energy).
Lead has a positive coefficient of thermal expansion - so the same amount of lead will become slightly thinner at colder temperatures. If you take into account that the lead sheet shrinks in all three dimensions, then the number of atoms per unit area goes up. This increases the probability of an interaction.
So yes - the same sheet of lead, cooled down, will be a slightly better shielding material.
At room temperature the coefficient of thermal expansion is approximately $3\cdot 10^{-5}/\mathrm{K}$ so if you cool it by 300 degrees it will shrink by about 1% in all directions. At that point it will be 3% denser - a sheet of the same thickness will have 3% better attenuation. The same sheet (which got thinner) will have about 2% greater attenuation.
By contrast, changing to denser materials (eg tungsten, gold or uranium) would give a much bigger jump in shielding effectiveness per unit thickness.
| {
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How electrons move so fast in a electric circuit? Whenever we switch on a bulb......it takes almost no time to glow up.....But we know that the atoms of a solid are tightly packed and there is a very little space between them.
So how the electrons travel through them irrespective of so much blockages in the conductor???
| Electrons can sneak pass all the atoms because of their wave function. They behave like waves not like particles. In short, because of quantum mechanics. In a periodic assembly of atoms like metallic solid they should not feel any resistance when moving through but because it is not perfectly periodic they feel aperiodic potential and this is why they scatter. They scatter on vibrations of atoms!
And, they move really fast but they have no definite direction. When you apply electric field to that metal, signal travels through the wire and all electrons practicly at once start moving in one direction, but slowly. This is known as drift speed. While they still move chaotically and with great speed, they slowly drift in one definite direction and it is this speed that that adds up to the original and still existing chaotic motion that is related to current and our harvesting energy of moving electrons.
| {
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Entanglement of Tripartite States Is there any simple algorithm to determine the entanglement of a tri-partite state?
In particular, what is the proof for entanglement of $ |GHZ\rangle $ and $ |W\rangle $ states?
$ |GHZ\rangle =\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle) $
$ |W\rangle =\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle) $
| $\newcommand{\ket}[1]{\left| #1 \right>}$A state $ \ket \psi \in H_1 \otimes H_2 \otimes H_3$ is said to be entangled if there exist no coefficients $a_i,b_i,c_i$ such that:
$$\ket \psi = \sum_{ijk} d_{ijk} \ket{e^1_i} \otimes \ket{e^2_j} \otimes \ket{e^3_j} = \sum_i a_i \ket{e^1_i} \otimes \sum_j b_j \ket{e^2_j} \sum_k c_k \ket{e^3_k} \tag{1}$$
where $\ket{e^i_j}$ is the $j$th basis vector of $H_i$.
If there exists such coefficients than the state said to be separable and the state is not entangled.
People tend to condense the notation by defining $\ket{e_1e_2e_3}\equiv \ket{e_1} \otimes \ket{e_2} \otimes \ket{e_3}$ so I shall do so in what is next to come.
Assuming that your states are doublet states, let's start calculating. I'll only do the first one, which is easier to do and leave the second one to you, which is analogous to what I'll do now.
Assume that you can write your state in the separable form (1)
\begin{align}
\ket{GHZ}&= \frac{1}{\sqrt{2}} (\ket{000}+\ket{111})\\&\overset{?}{=} \big(a_1 \ket 0 + a_2 \ket 1 \big) \otimes \big(b_1 \ket 0 + b_2 \ket 1 \big) \otimes \big(c_1 \ket 0 + c_2 \ket 1 \big) \\
&= a_1b_1c_1\ket{000}+ a_2b_2c_2 \ket{111} + \text{other linear combinations}
\end{align}
Notice that in order to get your state you cannot have any of the $a_i,b_i$ or $c_i$ to be zero, if this would be the case you cannot get either $\ket{000}$ or $\ket{111}$. However these are all the coefficients you have, therefore if none of them zero, then you cannot get any of the other linear combinations to vanish, which is a contradiction to the assumption that you can write your state in the separable form (1). Thus this state is an entangled state.
Note that the intuition that I talked about in my comment above saves you a lot of time and effort but as Ellis Cooper said, "Rigor cleans the window through which intuition shines.", so it is up to you how much rigour you want.
| {
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Polarization and Reflection The polarization is a property of waves that can oscillate with more than one orientation. Given this, when the light is reflected from a surface, does the reflection change the property/orientation of waves?
| Yes it does. Since the direction of the light beam changes with reflection also the direction of polarization. This is mostly because the observer is in a fixed coordinate system and the light beam changes its local coordinate system during reflection.
For an idealized reflector and an observer which moves along with the light beam, the direction of the E-field oscillation would stay the same.
For almost all real and hence non-ideal materials the reflectivity is different for the different polarization components (i.e. different absorption). This induces a true change of the polarization of light in addition to the changing coordinate system effect.
| {
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Why is the electric field inside a charged conductor zero in the electrostatic case? I am trying to understand the idea (or the fact) that most books introduce which is about the electric field inside a charged solid conductor.
Books tell that the field has to be zero everywhere inside solid conductor, otherwise charges will move around. Using this idea and Gauss's law, the charges inside the solid conductor is zero.
Now let us take for example four extra positive charges (each =1.2x10^-10 coulomb) inside a solid conductor of radius 1.
According to the idea of charge at the surface and due to the symmetry, the charges will distribute as follows:
I have plotted the electric potential (V=Q/(4πε0r)) and electric field (E=-∇V) using principle of superposition and the plot is:
Clearly the electric potential inside the conductor is not constant and the electric field is not zero.
How can this issue be explained?
| We assume that the the electric field is uniform for a charged solid sphere. It follows that the electric charge of the sphere is equal to
$$ Q = \rho V$$
Where $\rho$ is the charge density and $V$ is the volume. Therefore,
$$ Q = \rho V = \frac{4}{3}\rho R^3$$
We create a Gaussian surface in the form of a sphere of radius $r <R$. Thus, using Gauss's Law,
$$ E(4\pi r^2) = \frac{Q}{\epsilon_0} \implies E = \frac{1}{\epsilon_0}\frac{4\pi}{3}R^3\rho \implies E(r) = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^3}r$$
From the center of the sphere, the electric field is $0$. As you move away from the center of the sphere to $R$, the electric field increases in linear fashion that is proportional to $r$.
| {
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Inductance of two parallel wires I've been asked to calculate the inductance per unit length for two wires or radius $a$ separated by $2d$ where $2d>>a$.
Starting from $\int_{s} B.dS = LI$ Im not sure what surface to take?
For each wire the field at a distance r away is given by $\int_{l} B.dl = \mu_0 I$ and by superposition $B$ in the first integral is their sum. And so $$LI= \int_{s} B_1.dS + \int_{s} B_2.dS $$ where $B_1$ and $B_2$ are the contributions from the two wires.
|
Answer if anyone is interested. In the end the areas outside the inner edges of the wire cancelled by symmetry and so the surface i was looking for was the area enclosed.
| {
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Is it possible to write the fermionic quantum harmonic oscillator using $P$ and $X$? The Hamiltonian of the quantum harmonic oscillator is
$$\mathcal{H}=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2$$
and we can define creation and annihilation operators
$$b=\sqrt{\frac{m\omega}{2\hbar}}(X+\frac{i}{\omega}P)\qquad{}b^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(X-\frac{i}{\omega}P)$$
where the following commutation relations are fulfilled
$$[X,P]=i\hbar\qquad{}[b,b^{\dagger}]=1$$
and the Hamoltonian can be written
$$\cal{H}=\hbar\omega\left(b^{\dagger}b+\frac{1}{2}\right).$$
Now, it is also known that we can define a fermionic quantum harmonic oscillator with the Hamiltonian
$$\cal{H}=\hbar\omega\left(f^{\dagger}f-\frac{1}{2}\right)$$
where $f$ and $f^{\dagger}$ satisty the following anticommutation relation
$$\{f,f^{\dagger}\}=1.$$
What I am trying to get is a Hamiltonian for the fermionic harmonic oscillator using $P$ and $X$. I have tried defining
$$f=\sqrt{\frac{m\omega}{2\hbar}}(X+\frac{i}{\omega}P)\qquad{}f^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(-X-\frac{i}{\omega}P)$$
because after imposing the anticommutation relation $\{X,P\}=i\hbar$ for $X$ and $P$ (as I guess would suit a fermionic system) these definitions of $f$ and $f^{\dagger}$ imply $\{f,f^{\dagger}\}=1$. Nonetheless, for the Hamiltonian I get
$$\mathcal{H}=\frac{P^2}{2m}-\frac{1}{2}m\omega^2X^2$$
where I get an undesired minus sign. My question is then the following: is it possible (with an appropriate definition of $f$ and $f^{\dagger}$ in terms of $X$ and $P$) to obtain the first hamiltonian I have written from the fermionic oscillator Hamiltonian written in terms of $f$ and $f^{\dagger}$?
| Let's start from
$$H = \hbar \omega \left(f^\dagger f - \frac{1}{2}\right),$$
with $\{f, f^\dagger\}=1$, $\{f, f\} = 0$ and define fermionic position and momentum coordinates by
$$ \psi_1 = \sqrt{\frac{\hbar}{2}} \left(f + f^\dagger\right) \\
\psi_2 = i\sqrt{\frac{\hbar}{2}} \left(f - f^\dagger\right) $$
with the following anticommutation relations:
$$
\{\psi_i, \psi_j\} = \hbar \delta_{ij}.$$ So the operators anticommute with each other and square to $\hbar/2$.
We the find the Hamiltonian formulated in the new coordinates
$$H = -i \omega \psi_1 \psi_2,$$
which clearly gives rise to oscillatory motion, as can be seen by calculating the Heisenberg equations of motion:
$$\dot \psi_1 = -\omega \psi_2 \\
\dot \psi_2 = +\omega\psi_1.
$$
This doesn't have the form you expected it to have, but that just shows the weirdness of fermionic degrees of freedom.
| {
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Difference between fusion plasma and fluorescent lamp plasmas How is the plasma in a compact fluorescent lamp (CFL) different from a plasma in say ITER or the sun? Why does ITER need 100MK and a CFL can work at practically room temperature (apart from the filament)?
Or could ITER also create a plasma by charging the gas inside the reaction chamber but not have enough energy for the reaction, so they heat it directly (microwaves) and charging it would be of no use?
Or is it the degree of ionization the volume of gas has achieved? Like, a CFL has around $x$ ions and a sun plasma has only ions?
| ITER needs very high ion temperatures (100M K) so the deuterons and tritium nuclei are fast enough to overcome electrostatic repulsion and undergo thermonuclear fusion. A CFL only needs to have a conductive plasma in order to have an electron current exciting atoms in the gas.
| {
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Is Chern-number for free fermion system always limited by total band number, i.e. number of orbits with a unit cell? If so, how to see that?
Also I think it has been proven that the total Chern-number for free fermion system is 0?
If you know how to prove it, please make some comment or hopefully a sketch of proof.
I encounter this in a funny way. In fact, I just want to confirm the maximum Chern-number on Honeycomb lattice is 2 which I conjectured from a very peculiar way.
(I know people have already developed method to design bands with arbitrary Chern number. However that does not violate the statement since what they've achieved is to have multilayered system which essentially have N orbits in one unit cell to have a band with Chern-number N)
| https://arxiv.org/abs/1205.5792 The first example in the paper is $C=3$ on a triangular lattice with two orbitals per site. It is essentially three-layers of Haldane's honeycomb lattice model, but stacked together in a clever way so the translation symmetry is restored.
UPDATE: In fact two-band free fermion Hamiltonian on a square lattice can realize higher Chern number. Write $H=\sum_{\mathbf{k}}c_{\mathbf{k}}^\dagger h(\mathbf{k})c_{\mathbf{k}}$, where $h(\mathbf{k})=n(\mathbf{k})\cdot{\sigma}$. $n(\mathbf{k})/|n(\mathbf{k})|$ is a map from $T^2$ to $S^2$ and the Chern number is just the winding number (or the degree of the map). So obviously there is no bound on the winding number. And it is not hard to write down representatives for each winding number. The only problem is that to realize higher Chern number, you need longer and longer range hopping. So the limitation is really how short-range you want the hopping terms to be, not the number of orbitals.
| {
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Does Time change over temperature? I am not a physicist, I am just an engineer.
But I dared to ask whether the temperature changes the perception of time.
Let's consider a particle that "stops" at absolute zero. I was thinking as a hypothesis, that our perception of time changes and the particle actually does not stop at all.
| Your idea does not seem to work if you have two particles at different temperatures. Assume you "stop" one of them but not the other. Then does the time slows down for only one particle and not the other? or how would you explain that?
| {
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Phase added on reflection at a beam splitter? If we have light of a particular phase that is incident on a beam splitter, I assume the transmitted beam undergoes no phase change. But I thought that the reflected beam would undergo a phase change of $\pi$. I have, however, read that it undergoes a phase change of $\pi/2$.
Which is it, and why?
| It actually depends on what kind of beam spitter you have.
I'll give a general treatment and shows that the conclusions of both Emilio Pisanty and Steven Sagona are basically correct, corresponding to different specific beam splitters, which are all common in the laboratory. For simplicity we don't consider loss in this answer.
First of all, the definition of "phase shift" in this specific answer is chosen as the relative phase between reflected and transmitted light from the same port. More specifically, we let the transmission coefficient to be real number $t$, and the reflection coefficient will thus carries the information about relative phase shift $r e^{i\theta_\alpha}$, where $\alpha=1,2$ representing the light coming from beam splitter port 1 or 2 (see picture below).
Instead of assuming symmetrical phase shift, we allow any possible phase shift, as we are discussing general beam splitter which is not necessarily symmetric.
Then we write the physics procedure happening in beam splitter into the following matrix form:
\begin{eqnarray}
\begin{pmatrix}
E_3 \\ E_4
\end{pmatrix}
= \underbrace{
\begin{pmatrix}
t & r e^{i\theta_2} \\ r e^{i\theta_1} & t
\end{pmatrix}}_{ M}
\begin{pmatrix}
E_1\\ E_2
\end{pmatrix}
\end{eqnarray}
The conservation of energy requires $|E_3|^2 + |E_4|^2 = |E_1|^2 + |E_2|^2$, which is equivalent to the mathematical statement that the beam splitter matrix is unitary. That gives us
\begin{eqnarray}
M^\dagger M =
\begin{pmatrix}r^2 + t^2 & rt(e^{i\theta_2}+e^{-i\theta_1}) \\ rt(e^{i\theta_1}+e^{-i\theta_2}) & r^2 + t^2
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}
\end{eqnarray}
It can be seen that $r^2 + t^2 =1$ is automatically true as we deal with lossless beam splitter here. The rest of the above matrix equation gives us
\begin{eqnarray}
e^{i\theta_2}+e^{-i\theta_1} \equiv 2 e^{i(\theta_2-\theta_1)/2} \cos \frac{\theta_1+\theta_2}{2} =0
\end{eqnarray}
from which we conclude that
\begin{equation}
\theta_1 + \theta_2 = \pi
\end{equation}
If we have a beam splitter with symmetric phase shifts, $\theta_1 =\theta_2 = \pi /2$, then
\begin{equation}
M=
\begin{pmatrix}
t & ir \\ ir & t
\end{pmatrix}
\end{equation}
This is consistent with Steven Sagona's argument based on the same assumption.
If we have a beam splitter that is not only symmetric in phase shifts ($\theta_1 = \theta_2=\pi/2$), but also symmetric in reflection and transmission ($r=t=1/\sqrt 2$), then we have
\begin{equation}
M=\frac{1}{\sqrt 2}
\begin{pmatrix}
1 & i \\ i & 1
\end{pmatrix}
\end{equation}
Then we basically arrive at Emilio Pisanty's second equation.
If we have a beam splitter that is not symmetric in phase shifts ($\theta_1 = 0, \theta_2=\pi$), which is also very common in the lab, then we have
\begin{equation}
M=\frac{1}{\sqrt 2}
\begin{pmatrix}
1 & -1 \\ 1 & 1
\end{pmatrix}
\end{equation}
Remember I have been putting the relative phase factor to reflection instead of transmission so that we have $\{t, r e^{i \theta_\alpha}\}$. Now, we can also put the phase factor to transmission so that we have $\{t e^{-i\theta_\alpha}, r\}$, without losing any physics. What we got after is Emilio Pisanty's first equation, namely
\begin{equation}
M=\frac{1}{\sqrt 2}
\begin{pmatrix}
1 & 1 \\ 1 & -1
\end{pmatrix}
\end{equation}
| {
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Crane Balancing, Center of mass I am working on the ICPC 2014 Problem C "Crane Balancing"
The initial idea was to calculate the center of mass of the polygon, which I did via this equation:
Where the Area A:
Now, the solution is to binary search over the mass and look for th e maximum mass M where the crane is still balanced.
But I have a problem figuring out how the mass added affects the position of the polygon center of mass,
Thanks in advance.
| Assume that your crane is lying on the side set L on the axis c, where L can be just a point, can be just a line section [Ai, Bi], or can be unity of points and line sections.
You don't need to know how the mass affects the crane, you just need to now the minimal and maximal mass, for which the x coordinate of the center of mass is between infinum(L) and supremum(L).
Now, just brute force for all the vertices, and for each of them find the mass that it can handle.
And, the x coordinate of the centroid of the new system crane - lifted object, where the object is attached to the vertice (xi, yi), can be calculated by the following formula (knowing the mass of both objects):
Cx_new = Cx + (xi - Cx) * Mo/(Mc + Mo)
Where Cx is the x coordinate of the centroid of the crane, Mo is the mass of the object, Mc is the mass of the crane (can be calculated as Area * 1 kg/m^2).
And knowing the infinum(L) and supremum(L), you can easily find the mass that your crane can handle.
So, finding the Cx and Cy can be done in O(n), finding the MinMass and MaxMass for fixed vertice (xi, yi) can be done in O(1), and looping through all these vertices can be done in O(n), the final complexity is O(n^2).
| {
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What would be the rate of acceleration from gravity in a hollow sphere? Lets say the Earth is hollow and you are in the center of it (Same mass except all of it is on the outside like a beach ball) If you move slightly to one side now your distance is closer to that side therefore a stronger gravitational force however at the same time you have more mass now on the other side. At what rate would you fall? Which direction?
Also, is there a scenario where depending on the radius of the sphere you would fall the other direction or towards the empty center?
| If the mass/charge is symmetrically distributed on your sphere, there is no force acting on you, anywhere within the sphere. This is because every force originating from some part of the sphere will be canceled by another part.
Like you said, if you move towards on side, the gravitational pull of that side will become stronger, but then there will also be "more" mass that is pulling you in the other direction.
These two components cancel each other exactly.
| {
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Particle Physics Decay Question - Eta Prime Decay Parity/Angular Momentum Conservation I was hoping someone could clarify why the following decay does not occur:
$ \eta ^{'0} \rightarrow \pi ^{0} + \rho ^{0}$
The quark compositions and spin parity are as followed:
$ \eta ^{'0} : (u\bar{u}+d\bar{d}+s\bar{s}) / \sqrt{3} ;J^{P} = 0^{-} $
$ \pi ^{0} : (u\bar{u}-d\bar{d}) / \sqrt{2} ;J^{P} = 0^{-} $
$ \rho ^{0} : (u\bar{u}-d\bar{d}) / \sqrt{2} ;J^{P} = 1^{-} $
In order to conserve parity and angular momentum I thought that the two final particles states would have to be produced with angular momentum $l = 1$ between them (as parity of angular momentum 'part' is $ (-1)^{l}$ this would conserve parity and we can couple 0,1 and 1 to give 0 which conserves angular momentum). Does anyone know what is wrong this approach or alternatively a more straight forward reason why this does not occur.
| This decay (occurring via the strong interaction) violates the charge conjugation since $J^{PC}(\pi^0) = 0^{-+}, J^{PC}(\rho^0) = 1^{--}, J^{PC}(\eta'^0) = 0^{-+}$.
The charge conjugation transforms a particle in its anti-particle. In the case of the 3 particles involved in this decay, they are all their own anti-particle, and the effect of the charge conjugation operator $C$ is therefore (taking as an example the pion) $C|\pi^0> = \eta_C |\pi^0>$, meaning that the $\pi^0$ is eigenstate of the charge conjugation with eigenvalue $\eta_C = +1$. The $\rho^0$ has $\eta_C=-1$ and the $\eta'^0$, +1 (remark: $\eta_C$ is necessarily $\pm 1$ because when you apply twice the charge conjugation you should recover the initial state). The requirement of the charge conjugation conservation by the strong interaction would imposes:
$\eta_C(\eta'^0) = \eta_C(\pi^0) \times \eta_C(\rho^0)$ which is not the case $+1 \ne (+1) \times (-1)$. Thus this reaction is forbidden.
| {
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Density of states of 3D harmonic oscillator Consider the following passage, via this image:
5.3.1 Density of states
Almost all of the spin-polarized fermionic atoms that have been cooled to ultralow temperatures have been trapped by magnetic fields or focused laser beams. The confining potentials are generally 3D harmonic traps. So let's consider this case in more detail. You might be interested to note that Fermi's original paper on fermionic particles considered this case, not the 3D box case above. As we saw previously, ignoring the zero-point energy in each dimension the eigenvalues (accessible energy states) are given by $\epsilon(n_x, n_y, n_z)=n_x\hbar\omega_x + n_y\hbar\omega_y + n_z\hbar\omega_z$. In order to evaluate the various integrals, we first need to obtain the density of states per unit energy. A rough way to do this is to simply set $k_i=n_i$, so that
$$\epsilon^2 = k_x^2(\hbar\omega_x)^2 + k_y^2(\hbar\omega_y)^2 + k_z(\hbar\omega_z)^2 \equiv k^2(\hbar \overline\omega)^2,$$
where $\overline \omega = (\omega_x\omega_y\omega_z)^{1/3}$ is the mean frequency, and $dk_i/\epsilon_i=1/\hbar\overline \omega$. Because $k_i=n_i$ now rather than $k_i=\pi n_i/L$, th 3D density of states is given by
$$g(\epsilon) = \frac{k^2}{2} \frac{dk}{d\epsilon} = \frac{\epsilon^2}{2(\hbar\overline\omega)^3}.$$
for the first displayed equation,
shouldn't be $\epsilon^2 =\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 + 2\epsilon_{n_x}\epsilon_{n_y} + 2 \epsilon_{n_x}\epsilon_{n_z} + 2\epsilon_{n_y}\epsilon_{n_z}$..?
if I assume $\omega_i=\omega$ for $i=x,y,z$
by
$\epsilon_{n_x}=\hbar \omega n_x $
$\epsilon_{n_y}=\hbar \omega n_y $
$\epsilon_{n_z}=\hbar \omega n_z $
$\epsilon_{n_x,n_y,n_z}=\hbar \omega(n_x +n_y +n_z)$
let $\vec{k}=(k_x,k_y,k_z)$ where $k_i=n_i$
$$\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 = \hbar^2 \omega^2 (k_x^2 + k_y^2 +k_z^2 ) = \hbar^2 \omega^2 k^2 \not=\epsilon^2~?$$
And for second displayed equation, why it's not $$\frac{\pi k^2}{2} = \frac{1}{8}4\pi k^2~?$$
| Absorbing the irrelevant ħω constants into the normalization of the suitable quantities, for the 3D isotropic oscillator, $\epsilon=n+3/2$, while for each n the degeneracy is $(n+1)(n+2)/2$; (see SE ). Scoping the power behavior of a large quasi-continuous n, leads you to the answer.
The number of states then goes like $N\propto n^3 \propto \epsilon^3$, and hence the density of states like $dN/d\epsilon\propto \epsilon^2$.
| {
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Why is the shadow of a wind turbine a bit slow at first then very quickly, etc. and not equivalent to the wings? When I stand at a wind turbine and look at the shadow on the ground, the shadow is a bit slow at first then very quickly, etc. Very strange.
| Depending on the angle of the sun, the shadow becomes elongated so that it traces out an ellipse rather than a circle. This means that the shadows of the blades must traverse a different distance across the ground but within the same time period, thus giving rise to the periodic variation in velocity.
If the sun is at an angle $\alpha$ in the sky (see the diagram below) then there are four intervals of interest:
*
*$\alpha<\frac{\pi}{4}$: the sun is low in the sky and the shadow is longer than the actual turbine. As a result, the shadow must occasionally speed up.
*$\alpha=\frac{\pi}{4}$: the dimensions of the turbine are preserved and the shadow moves at a constant speed.
*$\alpha>\frac{\pi}{4}$: the sun is high in the sky and the shadow is shorter. It must therefore occasionally slow down.
*$\alpha=\frac{\pi}{2}$: the sun is directly overhead and no shadow is cast (at least, the blades cannot be distinguished).
This is simple to tackle mathematically. When the sun is directly behind/in front of the turbine, the coordinates of a particular blade are transformed as follows when projected onto the ground:
$$ R(\cos(\theta),\sin(\theta))\mapsto R(\cos(\theta)/\tan(\alpha),\sin(\theta)) $$
as is shown above. The elongation is even greater when the sun is at a horizontal angle with respect to the normal of the blades, but I shall ignore this effect for simplicity.
While the velocity of an actual blade ($R\dot{\theta}$) is constant, its projection has a velocity that depends on the angle $\theta$,
$$ v=R\dot{\theta}\sqrt{\frac{\sin^2(\theta)}{\tan^2(\alpha)}+\cos^2(\theta)} $$
and so it follows that the shadow speed varies.
For example, here is a plot of the relative tangential velocity of the shadow as a function of $\theta$ when the sun is relatively low in the sky ($\alpha=\pi/6$):
| {
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Is the Planck length the smallest length that exists in the universe or is it the smallest length that can be observed? I have heard both that Planck length is the smallest length that there is in the universe (whatever this means) and that it is the smallest thing that can be observed because if we wanted to observe something smaller, it would require so much energy that would create a black hole (or our physics break down). So what is it, if there is a difference at all.
| There is a tiny bit more going on than the otherwise excellent answer by zeldrege suggests. Imagine that you wish to probe an unspecified object to examine its structure. If we use light to look at the structure of an object, we need to have its wavelength smaller than the size of the details we wish to look at. Probing an object that has a (linear) size equal to the Planck length, requires that the energy of the photon be greater than the mass of a black hole of that "size". So, a classical black hole would be formed by our energy probe, thus preventing us to see details inside the object we wish to investigate. We are lead to an apparent contradiction, which suggests an incompatibility between Relativity and Q.M.
| {
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Meniscus attached to an inclined plate To be more specific, suppose a hydrophilic infinite plate is stuck into a semi-infinite region of water, above the water is a semi-infinite region of air, when the plate is stuck into the water vertically, the contact angle is $\alpha$, as shown in the figure below:
Needless to say, the menisci on both sides are symmetric, but what will happen when the plate is inclined for an angle $\beta$? Will the contact angle $\alpha$ remain unchanged? The meniscus on which side will be higher?
| Section 2 of the paper below uses geometric & variational techniques to show that the Young-Dupre relation still holds for a meniscus formed by a solid of rotation that makes an arbitrary angle with the vertical:
http://www.unisanet.unisa.edu.au/staffpages/stanmiklavcic/cm_anziam2.pdf
I can't imagine that the result would be all that different if you restricted things to planar symmetry instead of the cylindrical symmetry instead. On a fundamental level, Young's relation is just a statement about the three surface tensions (solid-liquid, liquid-gas, and solid-gas), specifically that their components tangent to the surface are balanced.
| {
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Number of Nodes in energy eigenstates I have a question from the very basics of Quantum Mechanics. Given this theorem:
For the discrete bound-state spectrum of a one-dimensional potential let the allowed energies be $E_1<E_2< E_3< ...$ with $E_1$ the ground state energy. Let the associated energy eigenstates be $ψ_1,ψ_2,ψ_3,...$. The wavefunction $ψ_1$ has no nodes,$ψ_2$ has one node, and each consecutive wavefunction has one additional node. In conclusion $ψ_n$ has $n−1$ nodes.
What is the physical interpretation for the number of nodes in the concrete energy eigenstate? I understand that the probability of finding the particle in the node point is $0$ for the given energy. However, why does the ground state never have a node? or why does every higher energy level increments number of nodes precisely by 1?
| I guess there is not that much to grasp, unless you can really understand dark spots on an electron diffraction pattern. Very roughly explanation would be to interpret wave functions of a particle in a potential well as "standing waves", or as two interfering waves reflected from the walls of the well. Increasing the energy leads to higher harmonics, which leads to additional nodes. Nodes' numbering is the same as in the case of a classical string.
| {
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Why do electric sparks appear blue/purple? Electric sparks tend to appear blue or purple or white in color. Why?
| Air is normally a bad conductor of electricity, but with enough voltage it can be converted to plasma, which is a good conductor. In a plasma, the electrons constantly bind to and leave atoms. Each time an electron binds to an atom, it emits the energy in light. As a result, the plasma glows the color of a photon with that energy. There are a few different energy levels that get involved, so the spectrum has a few different peaks. The final color depends on the gas you use. For example, neon looks red or red-orange. Air ends up looking blue, so electricity passing through air makes it glow blue.
| {
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If I shoot a hockey puck on ice, is the force of me shooting it applied throughout its travel, or is it a one time force? For example, if I take a slap shot on a hockey puck, from what I understand, the forces acting on the puck are friction, the normal force, and the puck's weight. And, since I'm not constantly either pushing the puck, or pulling the puck there are no other forces acting on the puck, because me shooting the puck was just a one time force. Am I thinking about this correctly?
| Yes, you are thinking about it correctly. No force is required to keep the puck in motion. This is an important idea in physics. It is actually a common misconception among physics students that a force is required to keep an object in motion, so it is good you do not have this misconception.
| {
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Are spherical coordinates distances or angles? I've become confused about spherical coordinates when dealing with electric fields.
The way I always understood spherical coordinates is something like the below picture. To define a vector, you give it a distance outwards (r), and two angles to get a final position. Below, the $\theta$ and $\phi$ components are measured in radians.
(Courtesy Wikipedia.org)
However, you can also have, say, an electric field in spherical coordinates. In this case, the unit vectors $\theta$ and $\phi$ don't define angles but rather values of the vector fields. So, in the case of electric fields, we might have $E_\theta = 10\text{ Vm}$. That is, at every point there will be this electric field component in the theta direction.
So, it seems there are two different ways of dealing with spherical coordinates. One, where the $\theta$ and $\phi$ components represent angles, and one where they represent values of the components in those directions.
This would then give you two different measures of lengths of the vectors. In the first case, the length of the vector is always given by the r component. In the second case, you take $|\vec{E}|=\sqrt{E_r^2+E_\theta^2+E_\phi^2}$.
What am I mixing up here?
| For an electric field usually you have three components like $(E_x, E_y, E_z)$ in Cartesian coordinate system. Now you want to rewrite the same vector in a spherical coordination, what you should do is as follows:
first you write the vector like the electric field as $\mathbf{E}=|E|\mathbf{e_r}$ where $|E|$ is given by the $|E|=\sqrt{E_x^2+E_y^2+E_z^2}$ and $\mathbf{e_r}$ is a function of $\theta$ and $\phi$. Then One should have $\theta=\text{acos}(E_z/|E|)$ and $\phi=\text{atan}(E_y/E_x)$. If you like to write in a three component form, it is always (|E|, 0, 0).
| {
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True randomness? I am a physics high-school student so my knowledge is not very deep on the subject.
We started learnning about quantum mechanics and on some processes that my teacher described as random. I began to think on the concept of randomness and question it, thinking how can a process or an outcome be determined without any cause, how an outcome be determined at all in complete randomness?
I searched the internet and figured the scientific community does not agree with me. I'd really like to understand how can true randomness exist? Why the scientific community rejects the idea that ''random'' events may just have a cause we are not aware of?
What am I missing?
Thanks for answers and sorry for bad English.
| Bell proved that, if there exists such an unknown cause, then it surely must violate special relativity (Information must travel faster than the Speed of Light).
Taken from Wikipedia
"Realist interpretations of quantum mechanics are possible, although, such interpretations must reject either locality or counter-factual definiteness."
See
http://en.wikipedia.org/wiki/EPR_paradox
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Obtaining a copy of Hawking's Ph.D thesis - Properties of Expanding Universes Due to its popularity, I am interested to know the 4 chapter titles and topics covered in S.W. Hawking Ph.D, Properties of Expanding Universes. I also ask this because that thesis is hardly available.
| Every thesis submitted for a PhD in Cambridge is archived at the Cambridge University Library. They should be able to get you a copy (for a fee).
See http://www.lib.cam.ac.uk/deptserv/manuscripts/dissertations.html
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What about a surface determines its color? Light falls on a surface. Some wavelengths get absorbed. The other are reflected. The reflected ones are the colors that we perceive to be of the surface.
What is the property that determines, what wavelengths are reflected and what are absorbed? Is it electronic configuration of the object on which the light falls?
If yes, then if we know the electronic configuration of a surface can we make a model, which will predict the color it will show?
| Paul G. Hewitt has a great non-mathy description of this in his book Conceptual Physics.
Strike a tuning fork and it vibrates at a characteristic frequency, its Natural Frequency. The tuning fork might put out other frequencies but they are dampened faster than the target frequency. Dampening sucks energy out of vibration and becomes thermal if it doesn't become sound.
You can think of many things as being made up of tiny tuning forks. So hitting a metal sounds different from hitting wood.
Thes stiffness of a spring can determine the frequency of its vibrations. Again it has a natural frequency.
Like these examples, you can think of atoms and molecules as tiny optical tuning forks that re-emit some of the light and absorb the rest. Some light passes through.
The color you see depends on various features of the atomic and molecular interactions. For example, how strongly bounded an electron is to its parent atom is roughly analogous to the stiffness of a spring.
Vibrations in general can often be approximated through the analysis of Simple Harmonic Motion.
This is why Planck was able to analyze the interaction of Black Body Radiation with matter by assuming the radiation interacting with tiny "Harmonic Oscillators" in the surface of the black body.
| {
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Is there any tension in a massless spring that connects two free falling bodies in different horizontal planes? Two bodies A and B of same mass $m$ are attached with a massless spring and are hanging from a ceiling with a massless rope. They are in same vertical plane but not in same horizontal plane.
Now the string that connected A with the ceiling is cut and the system is experiencing free fall.
1. Is there any tension in the spring?
My attempt:
Now the whole system should descend with the acceleration $g$ and the body B (and also A) experiences a gravitational pull $mg$.
Let the tension in the spring be T.
Therefore, from the free body diagram of B, $mg - T = mg$,ie. $T=0$.
*But A also moves downwards, so puts a force on B, how to take account of that? Will there be an relative acceleration between A and B? I am confused about the free body diagrams of A and B.
*Will the tension change if the mass of A and B are different?
| The answer is that it depends on how your initial spring loaded mass is moving. But, the fascinating (but not too fascinating once you phrase it like this) part is that until the compression wave from the top interacts with bottom out on the slinky the dynamics of the bottom half won't change.
If we assume it was at rest, essentially the top mass will move fast enough that the center of mass will accelerate at $9.8 m/s^2$. When it reaches the true equilibrium length of the string without gravity it will start accelerating the bottom mass. At this point if you looked from the COM frame the spring would appear to be oscillating like it normally does. This is because the oscillations occurring here is called an eigenfrequency. The other eigenfrequency (since this problem has 2 independent variables) is the motion of the COM. With the motion of the COM and the motion of both masses about the COM you have all the information needed to reconstruct the dynamics of your masses.
A great demonstration of this is the slinky in this video, which is like a spring with an equilibrium length of zero:
https://www.youtube.com/watch?annotation_id=annotation_314765&feature=iv&src_vid=eCMmmEEyOO0&v=uiyMuHuCFo4
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Why do momentum and position have the same direction in space in the Heisenberg Uncertainty Principle? It is known that $\Delta p \Delta x \geq \frac{h}{4\pi}$. I read that the two uncertainties must be along same axes. Why is that so?
| On the one hand, the uncertainty principle is in fact a postulate of quantum mechanics, so the question why it is postulated the way it is does not make a lot of sense. On the other hand, few thing can be said with respect to the question.
First, the exist a more general form of the uncertainty principle usually refered to as the Robertson–Schrödinger uncertainty relation, which states that for any pair of observables $A$ and $B$ represented by self-adjoint operators $\hat{A}$ and $\hat{B}$ respectively,
$$
\sigma_{A}\sigma_{B} \geq \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|,
$$
where $[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$ is the commutator between the operators.
Secondly, for the specific case of position $\hat{q}$ and momentum in the same direction $\hat{p}_q$ operators, the commutator is known to be
$$
[\hat q,\hat p_q] = i\hbar \, ,
$$
where $q$ is $x$, $y$, or $z$. This is also a postulate, known as the canonical commutation relation, so the question why is again not quite appropriate here. However, one must note specifically that the canonical commutation relation is the relation between the so-called conjugate variables, the variables such that one is the Fourier transform of another. Only position and momentum in the same direction are such variables, while any other combination is not. In fact, commutators for any other pair ofobservables are all zero, so that in summary we have,
$$
[\hat q,\hat p_{q'}] = i\hbar \delta_{q q'} \, , \quad [\hat q,\hat q'] = [\hat p_q,\hat p_{q'}] = 0 \, ,
$$
where $\delta_{q q'} = 1$ only if $q=q'$ and is zero otherwise. Physically it means that only position and momentum in the same direction could not be measured simultaneously, while all other pairs of observables $q$ and $p_q$ could.
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What is the physical interpretation of second quantization? One way that second quantization is motivated in an introductory text (QFT, Schwartz) is:
*
*The general solution to a Lorentz-invariant field equation is an integral over plane waves (Fourier decomposition of the field).
*Each term of the plane wave satisfies the harmonic oscillator equation.
*Therefore, each Fourier component is interpreted as a harmonic oscillator in ordinary QM
*The $n$'th energy level of each Fourier component is now interpreted as $n$ particles.
Everything in 1-3 looks like a sensible application of ordinary QM to a field. But how did 4 come about? What is the justification?
| In the statistical mechanics of the grand canonical ensemble, one needs to allow for
superpositions and mixtures of of states with different particle number. Thus one is
naturally led to considering the tensor product of the $N$-particle spaces with arbitrary $N$. It turns out (and is very relevant for nonequilibrium statistical mechanics) that one can reinterpret the resulting any-number-of-particles quantum mechanics as a nonrelativistic field theory, in which the number operator is defined to have the eigenvalue $N$ on $N$-particle space. (If one considers a single Fourier mode, this explains your 4.)
The resulting field formalism is called the second quantization (of the first quantized 1-particle space). You can read about this e.g., in the appendix of Reichl's statistical physics book.
If one replaces the 1-particle Schroedinger equation by the Klein-Gordon or Dirac equation one gets (after normal ordering) the relativistic version.
| {
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On energy levels and emission of photons This is a very basic question but I cannot seem to find the answer anywhere.
Say we have an atom in ground state. Its first energy level is 2 eV. An incoming photon of energy 2.5 eV hits an electron in the atom (with the lowest energy level) which is excited and moves up one enery level. However it does not have enough energy to reach the next energy level. When it loses energy it emits a photon of energy 2 eV. What has happened to the remaining 0.5 eV? Is it some other type of energy in the atom such as kinetic energy?
| The probability of a photon of energy $E$ — or corresponding wavelength $\lambda = h c / E$ — being absorbed by an atom and bringing an electron from level $i$ to level $j$ is given by the cross section $\phi_{ij}(\lambda)$, which is a sharply peaked function of wavelength (a so-called Lorentzian function). For instance, to bring a hydrogen atom from its ground state to its first excited state, a photon of $\lambda$ = 1215.67 Å, or $E$ = 10.2 eV is needed. If the energy is just a little smaller, or a little larger, say 10.21 eV, corresponding the a wavelength half an Ångström shorter, the probability of being absorbed is many orders of magnitude smaller. The photon is said no longer to be in resonance with the energy level.
Thermal broadening
In reality, when a photon enters an ensemble of atoms, the temperature of the gas causes the atoms to have a (Gaussian) distribution of velocities. This means that although the incoming photon has a "wrong" energy of 10.21 eV, there are plenty of atoms that happen to be moving away from the photon at exactly the right speed, such that in the reference frame of the atom, the photon is redshifted into resonance. That is, the probability of being absorbed is large.
The effect is that the absorption cross section of the energy level is broadened, becoming a convolution of the "natural" (Lorentzian) and the thermal (Gaussian) profiles, called a Voigt profile.
| {
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Why do higher modes propagate more in the cladding of an optical fiber than lower modes? I am trying to understand the theory of inter-modal dispersion in optical fibers. It seems quite obvious that if higher modes have a greater angle of incidence in the fiber than lower modes, the path length of higher modes through the fiber is larger. This is because higher modes undergo more reflections, but they also have a greater part of the light wave traveling is the cladding. Here the speed of light is a little higher than in the core and therefore the higher modes are moving faster than lower modes.
The theory of a part of the light wave traveling in the cladding has to do with the evanescent field I think, but why do higher modes have a greater part of the wave traveling in the cladding in comparison with lower modes?
| For optical fibers with cylinder geometry there are an orbital angular momentum (OAM) mode number $\ell\in\mathbb{Z}$, and a radial mode number $m\in\mathbb{N}$. The average radial position grows with $m$. A fixed OAM $\ell$ leads to a centrifugal term in the radial equation, which diminishes the number of radial modes. See e.g. the RP photonic encyclopedia.
| {
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Are Fluted Barrels More Rigid Than Standard Ones? This seems to be quite a debated topic in the shooting community and is something I'm not sure about.
A fluted barrel is a barrel that has grooves milled into it to reduce weight, help it cool faster and supposedly to make it less flexible (see picture below)
I don't see how removing material like this would make it more rigid, because surely by removing mass the barrel would have less inertia to help keep it rigid when under the stress of firing, unless it had the same effect as corrugated sheet metal. For example if you had a flat strip of material compared to a corrugated piece the corrugated piece would have a lot less flex.
So if you had two barrels made of the same material being the same length and width with one being fluted and the other being a smooth cylinder which would be more rigid under the stress of firing?
(Sorry if this is not how you should post on here, I normally only use StackOverFlow.)
| A 'more rigid' barrel is actually a relative question. If two barrels start with the exact same profile and then one is fluted, and only the resistance to bending is measured, then the difference is extremely small. One barrel will be significantly lighter, but generally not more rigid since the rigidity is largely a function of the exterior cross section.
The reason barrels are fluted isn't to make them more rigid, but reduce the weight of an already-rigid barrel that would otherwise be too heavy for its application. The enhanced cooling of additional surface area and shorter distance from the heat source (the bore) and the outer barrel surface is another reason,
I recently fluted a barrel that started at 5 lbs 3 ounces. 6 flutes (1/4" ball endmill, 1/8" deep, 13" long) took about 7 ounces off the barrel, weight savings is bigger that it looks.
| {
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Turning points of particle
A particle of mass $m$ and energy $E<0$ moves in a one-dimensional Morse potential:
$$V(x)=V_0(e^{-2ax}-2e^{-ax}),\qquad V_0,a>0,\qquad E>-V_0.$$
Determine the turning points of the movement and the period of the oscillation of the particle.
I have started learning for my exam and this is one of the exercises in my textbook.
Never dealt with this type of question, so these are my thoughts so far:
To get the turning points I was thinking of solving the equation $E=V_0(e^{-2ax}-2e^{-ax})$. I was doing the arithmetics with the absolute value of $E$. But still I couldn't seem to find the values for $x$. At the end I used Wolfram Alpha to find the values but it gave me results with complex values. Is there a simple way to solve this type of equations for $x$?
Anyway, about the period, I assume it's the time it takes for the particle to get from $x_1$ to $x_2$. But how am I supposed to approach this? How would I get a time value just out of the equation for the potential?
I hope someone can help me out here.
| nephente's answer solves for the period. My answer is just made to make you see how to go on from the point you were, solving:
$$ e^{-2ax}-2e^{-ax}=\frac{E}{V_0} $$
We make the change $y=e^{-ax}$, which yields:
$$y^2-2y-\frac{E}{V_0} = 0$$
Second grade equation. Solve for $y$, and have in mind that $E/V_0$ is a negative number. It may give you complex solutions otherwise.
| {
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Force of an ideal spring Suppose you have an ideal spring (constant of the spring $k$) attached to a uniform disc of radius $R$ as in the picture below:
The force $F$ in red is from the spring.
My question is the following:
How should I decompose the force $F$ into its $x$ and $y$ components??
My intuition would tell me to multiply the module of the force by the sine and cosine of the angle between $F$ and the $x$ axis or something similar, however I noted in some exercises and exams, the solution simply states that $F_x = -kS(u_x)$ where $S$ is the distance from the y axis, $u_x$ the unit vector of the $x$ axis and $F_x$, of course, the $x$ component of $F$. What is the correct approach? I could not find anything useful on the internet so far...
| You have a correct answer, but I'm not sure if you have it for the right reason. Hooke's law states that
$$ \vec{F}_{\text{spring}} = -k\vec{r}$$
where $\mathbf{r}$ is the displacement of the spring from its equilibrium length (which you have not provided in the problem). If you take the equilibrium length to be 0, the spring force then has a magnitude
$$|\vec{F}_{\text{spring}}| = kR$$
where $R$ is the distance between the center of the disk and the origin.
Taking $\theta$ to be the angle between the angle the spring makes with the x axis, then,
$$\vec{F_x} = - kR \cos{\theta} \hat{u_x}$$
$$\vec{F_y} = - kR \sin{\theta} \hat{u_y}$$
High school trig tells you that $\cos{\theta} = \text{adjacent}/\text{hypotenuse}$ and $\sin{\theta} = \text{opposite}/\text{hypotenuse}$. In this case $\cos{\theta} = S/R$ and $\sin{\theta} = \sqrt{R^2-S^2}/R$. Therefore,
$$\vec{F_x} = - kR\Big(\frac{S}{R}\Big)\hat{u_x} = -kS\hat{u_x}$$
$$\vec{F_y} = - kR \Big( \frac{\sqrt{R^2-S^2}}{R} \Big) \hat{u_y} = -k\sqrt{R^2-S^2}\hat{u_y}$$
| {
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How can we be sure the Maxwell speed distribution equation is always a rational number? The Maxwell speed distribution equation is given as
$$f(v) = 4\pi \biggl(\frac{m}{2\pi kT}\biggr)^{3/2}\exp\biggl(-\frac{mv^2}{2kT}\biggr)v^2.$$
The left hand side gives the fraction of molecules and is a rational number. But the right hand side is a product of several kinds of numbers, such as integers, fractions, irrational numbers, and transcendental numbers.
Since a rational number can only be equal to a rational number, the right hand side must also be a rational number. Thus, in order that an inconsistency/impossibility does not arise, how do we ascertain that we always get a rational number on the right hand side?
| A few issues with this argument. First of all, you're clearly right that the RHS can yield an irrational/transcendental number. It would require insane mathematical coincidence for this not to be so. (It's really hard to have $e^x$ be algebraic, for instance.) And there's nothing in the formula about how many particles there are; we could imagine $N =2$, which would imply that the LHS can only be $0, .5, 1$!
First of all, let's recognize that this is a probabilistic formula like many in statistical mechanics. It holds on average, and yields the probability that an atom has velocity $v$. In general, you can interpret this as "fraction of atoms with that velocity" but clearly there are mathematical difficulties with that interpretation.
Next, many formulas in statistical mechanics hold only in the thermodynamic limit, which means that they assume $N \to \infty$. This then allows for any real number to appear on the LHS, since we have a continuum of values rather than just something that is $m/N$.
Finally, remember that physics is emphatically not math, and really never can be. Any real experiment inevitably brings in complications that we neglect so as to make the problem analytically tractable, so the physical systems we experiment with are never the mathematical systems we calculate with. We can aim for a close congruence between the two, but let's not forget that they really are different things.
| {
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Why do our ears pop? Have you ever been on a train going through a tunnel or plane and your ears pop?I was wondering why this happens and I know it relates to pressure but don't know exactly the reason
| When a train rushes through a tunnel, it tries to push the air out of the way, but the narrow confines of the tunnel force the air to be compressed in front of the train, as though the train were a piston in an air compressor.
The compressed air tries to find extra volume wherever it can, and since the train does not make as tight a fit to the tunnel walls as a piston, some air escapes compression by flowing between the train and the tunnel walls, eventually filling the void behind the train.
Air rushing past the train is forced into the narrow area between train and tunnel, just as though it were entering the venturi of a carburetor. So it accelerates and can flow at greater velocity than even the train's forward motion. The Bernoulli effect (http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html) causes the fast-flowing air to lower the pressure between the railcars and the tunnel walls. This creates suction that draws air out of the train, effectively lowering the pressure inside the railcars.
There are two kinds of energy between the railcars and the tunnel walls: Kinetic energy of fast flowing air molecules, and kinetic energy of chaotically compressed air molecules. The laminar flow of the air is more powerful and more organized than the compression of air in the imperfectly sealed space, so the organized kinetic energy of laminar flow increases as the chaotic kinetic energy of compressed gas decreases. Energy per unit volume remains constant and conserved as air velocity increases and air pressure decreases.
Your ears pop because your eustacian tubes are trying to equalize your internal pressure fast enough to compensate for the air being sucked out of the railcars. The popping sound is air bubbles from your eustacian tube entering the middle ear.
| {
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How can there be entropy change in this system? How can there be an entropy change in this system?
Suppose if I have a system consisting of liquid water, $1\, \mathrm{kg}$ at $290\,\mathrm{K}$, I stir it, and do say, $10\, \mathrm{J}$ of work on it, I can calculate the temperature change of the system given that:
$$U = cT \quad\mbox{ and }\quad S = c \ln \Omega$$
for $c$ constant.
From the fundamental equation of thermodynamics:
$$dU = dQ + dW = 0 + dW = 0 + 10 = 10\,\mathrm{J}$$
Hence:
$$dT = \frac{dU}{cM} = \frac{1}{410}\,\mathrm{K}$$
But how can there be a change of entropy in the universe when $dQ = 0$. I understand that we can calculate it using the formula for $S$ given, but I don't understanding how the fundamental equation allows this?
$$dQ = 0$$
and
$$dS = T^{-1}\,dQ$$
Hence, it may be concluded that:
$$dS = 0$$
Can someone tell me where my understanding is lacking, because obviously the entropy change is not zero in this case?
| There are two points here:
*
*Those equations above from $dU=\delta Q+dW$ are for gas (ideal gas), not liquid. You cannot use them to calculate the entropy of liquid.
*Work done in the first law of thermodynamics is defined as: $dW=pdV$ (quasistatic process), where p is generalised force (pressure) and V is generalised displacement (volume). When you stir the system, you do not do a work on the system in term of the first law.
| {
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Explanation of Michelson Interferometer Fringe Shift I have been working on an experiment where 2 glass microscope slides are pinched together at one end (so that there is a "wedge" of air between them) and placed in the path of a laser in one leg of a Michelson interferometer. When I move the glass slides (fractions of a mm at a time) so that the path of the laser is closer or further from the place where the slides are pinched, a fringe shift occurs. I cannot seem to explain why this is happening! Any help with explaining this phenomenon would be greatly appreciated! If any more specifics about the setup or dimensions of the slides are needed, please let me know.Also, a full "light to dark" fringe shift occured roughly every 4mm of moving the slides.
| At a guess, the effect rises from the fact that your interferometer is not properly aligned. The presence of linear, rather than circular, fringes suggests that there is an angular misalignment. Then moving the wedge causes a lateral shift in the intersection point of the beam and the angled slide, which results in a shift in the apparent position of the beam at the target.
Try aligning the system to produce a bull's eye fringe pattern, and see if the anomaly persists.
| {
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Force between two point charges moving parallel to each other When we observe two point charges moving parallel to each other we can see two forces acting on each of the charges:
*
*the Coulomb force
*the magnetic force ($\mathbf{F}=q\mathbf{v}×\mathbf{B}$) (similar to the force between two parallel wires with the same current)
However, if we change to the charges' frame of reference, there will only be one force - the Coulomb one, and moreover, the amplitude of this force should be the same as the previous one's (the distance between the charges is constant and the force is independent of velocity). The resulting force should be independent of the frame of reference, yet it isn't and I can't find the missing piece. I know that a magnetic field is just an electric field viewed from a different frame of reference, but it doesn't help in this specific case.
I found this question to be similar: Two electron beams exert different forces on each other depending on frame of reference?, but in one charge scenario the charge density has no meaning.
| If the charges are moving parallel to each other, in the reference frame of of the one charge, the charges aren't moving. So the magnetic field is zero and the force is $\bar F=q \bar E $. For the charge to be still, the force must be zero, and thus the electric field also. But if the electric field can't be zero, then there is another force that holds the charges in their positions. In the general reference frame we have that(for the one charge): $F=0 --> \bar E = -\bar u \times \bar B $.
Hope this helps.
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Obtaining quantum Hamiltonian for charged particle from path integral formulation I was working on Shankar 8.6.4, which is about obtaining in one dimension the Hamiltonian operator of a charged particle from the path integral formulation.
First, I get the propagator over a time slice $\epsilon$, which is
$$
U(x,\epsilon;x') = \sqrt{m \over 2\pi i\hbar\epsilon} \exp\left({i\over\hbar}\left( {m\eta^2\over 2\epsilon} - {q\eta\over c}A(x+\alpha\eta,0) - \epsilon q\phi(x+\alpha\eta,0)\right)\right)
$$
where $\eta = x'-x$, $A$ is the vector potential, and $\phi$ is the scalar potential.
Then, I obtain an expression for the time evolution of the initial wavestate in a time slice:
\begin{align}
\psi(x,\epsilon) &= \int_{-\infty}^\infty U(x,\epsilon;x') \psi(x',0) \, dx'\\
&= \sqrt{m \over 2\pi i\hbar\epsilon} \int_{-\infty}^\infty \exp\left({i\over\hbar}\left( {m\eta^2\over 2\epsilon} - {q\eta\over c}A(x+\alpha\eta,0) - \epsilon q\phi(x+\alpha\eta,0)\right)\right) \psi(x+\eta,0) \, d\eta
\end{align}
I expand the following expressions into series:
\begin{align}
\exp\left(-{iq\eta\over\hbar c}A(x+\alpha\eta,0)\right) &= 1 - {iq\eta\over\hbar c}A(x+\alpha\eta,0) - {1\over 2}\left({q\eta\over\hbar c}A(x+\alpha\eta,0)\right)^2+\cdots\\
&= 1 - {iq\eta\over\hbar c}A(x,0) + \left(- {1\over 2}\left({q\over\hbar c}A(x,0)\right)^2-{i\alpha q\over\hbar c}{\partial A(x,0)\over\partial x}\right)\eta^2+\cdots
\end{align}
\begin{align}
\exp\left(-{i\epsilon q\over\hbar}\phi(x+\alpha\eta,0)\right) &= 1 - {i\epsilon q\over\hbar}\phi(x+\alpha\eta,0) + \cdots\\
&= 1 - {i\epsilon q\over\hbar}\phi(x,0) + \cdots
\end{align}
\begin{align}
\psi(x+\eta,0) &= \psi(x,0) + \eta{\partial\psi(x,0)\over\partial x} + {\eta^2\over 2}{\partial^2\psi(x,0)\over\partial x^2} + \cdots \\
\end{align}
The wavestate after the time slice is
\begin{align}
\psi(x,\epsilon) =& \sqrt{m \over 2\pi i\hbar\epsilon} \int_{-\infty}^\infty \exp\left(-{m\eta^2\over 2i\hbar\epsilon}\right) \\
& \cdot \exp\left(-{iq\eta\over\hbar c}A(x+\alpha\eta,0)\right) \exp\left(-{i\epsilon q\over\hbar}\phi(x+\alpha\eta,0)\right) \psi(x+\eta,0) \, d\eta \\
=& \sqrt{m \over 2\pi i\hbar\epsilon} \int_{-\infty}^\infty \exp\left(-{m\eta^2\over 2i\hbar\epsilon}\right) \\
& \cdot \left( \left(1-{i\epsilon q\phi_x\over\hbar}\right)\psi_x + \eta^2\left(- {1\over 2}\left({q A_x\over\hbar c}\right)^2\psi_x-{i\alpha q\over\hbar c}{\partial A_x\over\partial x}\psi_x-{i q\over\hbar c}A_x{\partial\psi_x\over\partial x}+{1\over 2}{\partial^2\psi_x\over\partial x^2}\right) \right) \, d\eta\\
=& \left(1-{i\epsilon q\phi_x\over\hbar}\right)\psi_x + \left({i\epsilon\hbar\over m}\right)\left(- {1\over 2}\left({q A_x\over\hbar c}\right)^2\psi_x-{i\alpha q\over\hbar c}{\partial A_x\over\partial x}\psi_x-{i q\over\hbar c}A_x{\partial\psi_x\over\partial x}+{1\over 2}{\partial^2\psi_x\over\partial x^2}\right)\\
=& \psi_x - {i\epsilon\over\hbar}\left( q\phi_x + {1\over 2m}\left({q A_x\over c}\right)^2+{i\hbar\alpha q\over m c}{\partial A_x\over\partial x}+{i\hbar q\over m c}A_x{\partial\over\partial x}-{\hbar^2\over 2m}{\partial^2\over\partial x^2} \right)\psi_x
\end{align}
where for fields and states, $\omega_x=\omega(x,0)$.
According to the above,
$$
\newcommand{\ket}[1]{\left| #1\right\rangle}
i\hbar\dot{\ket{\psi}} = \left(q\phi + {1\over 2m}\left({q A\over c}\right)^2 - {\alpha q \over m c} P A - {q \over m c} A P + {P^2\over 2m} \right)\ket{\psi}
$$
but after the midpoint prescription $\alpha={1\over 2}$, the Hamiltonian should be
$$
q\phi + {1\over 2m}\left({q A\over c}\right)^2 - {q \over 2 m c} P A - {q \over 2 m c} A P + {P^2\over 2m}
$$
What happened with the coefficient of $AP$?
| As Tom pointed out, I made a mistake with assuming that the operator PA is -ih(∂A/∂x) when it should be -ih(∂A/∂x+A∂/∂x). This means that there is a (1-α) coefficient for the AP operator.
| {
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Is this definition of orthohelium and parahelium incorrect?
"One electron is presumed to be in the ground state, the 1s state. An electron in an upper state can have spin antiparallel to the ground state electron ($S=0$, singlet state, parahelium) or parallel to the ground state electron ($S=1$, triplet state, orthohelium)."
From HyperPhysics
When they say "parallel to the ground state electron" then is it assuming that they are both spin up, or both spin down? If so, isn't it then ignoring the $S=1$ state with spin up and spin down:
$$|1\rangle |0\rangle = \frac{|+\rangle|-\rangle + |-\rangle|+\rangle}{\sqrt 2}$$
Therefore, if one electron is presumed to be in the ground state, 1s, state, if the spins can be opposite, a second electron can also occupy the ground state in $S=1$ orthohelium. Is this correct?
| With such a two-electron spin wave function, both electrons cannot be in the 1s state due to the electron parity. (The wave function should be antisymmetric relative to exchange of two particles.) When both electrons are in the 1s orbital, the only possible spin function reads
$$\frac{1}{\sqrt{2}} \left( |\uparrow \downarrow \rangle-| \downarrow \uparrow\rangle \right)$$
| {
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What would put a harddisk drive (HDD) under 350G's of force? I always see the label and it says 350G's withstandable. What would put this over 350G's? Is it even possible to hit 350Gs of force to a hard drive?
| You are not the first person to ask this question.
https://superuser.com/questions/925826/what-would-put-a-hdd-under-350gs-of-force claims that 350 g of force is slightly more than a soccer player kicking a football. What this means is that you basically can kick your case, and it shouldn't brick your hard drive. It might cause other issues though, so don't go around kicking your computer case.
And I just noticed that you also posted the linked question. So technically, you are the first person to ask this question on the SE network.
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Relationship between phase velocity and group velocity using the de Broglie postulates If I have to show that the group velocity of a free particle is twice the phase velocity, is the following argument correct (avoiding to use the wave function and the momentum operator):
For a particle with energy $E$ and momentum $p$ we have the circular frequency
$$\omega = \frac{E}{\hbar}$$
and the wave length
$$\lambda = \frac{2 \pi \hbar}{p}\, .$$
As usual for waves we get the phase velocity by the formula $$v_{p} = \frac{\omega \lambda}{2\pi} \, .$$
This gives us $$v_{p} = \frac{\omega \lambda}{2\pi} = \frac{E \cdot 2\pi \hbar }{\hbar p \cdot 2\pi} = \frac{E }{p} = \frac{ p^2 }{2 m p} = \frac{ p}{2 m} = \frac{v}{2}.$$ But the speed of the particle, $v$, is nothing but the group velocity $v_g$ of the corresponding wave function. Therefore $v_g = 2 v_p$.
| This is a perfectly correct derivation that uses the correspondence principle nicely: we can identify the group velocity with the classical velocity because a classical particle corresponds to a quantum particle whose wavefunction is a sharply peaked wavepacket, whose velocity is the group velocity.
If you want to do it more formally, you can also start from the usual definitions of the group velocity and phase velocity,
$$v_g = \frac{d \omega}{dk}, \quad v_p = \frac{\omega}{k}.$$
The simplest form of the de Broglie relations are
$$E = \hbar \omega, \quad p = \hbar k.$$
Your forms are perfectly right too, but a little more complicated because there are $2\pi$'s all over the place. Next, we know that for a free particle, the energy is
$$E = \frac12 mv^2 = \frac{p^2}{2m}.$$
Combining this with the de Broglie relations, we have
$$\omega = \frac{\hbar k^2}{2m}.$$
Using the definitions of the group and phase velocity
$$v_g = \frac{\hbar k}{m}, \quad v_p = \frac{\hbar k}{2m}.$$
Then $v_g = 2 v_p$ as desired. Incidentally, another way to that the classical velocity is $v = v_g$ is to note
$$p = \hbar k = m v$$
and comparing with our previous expression gives $v = v_g$.
| {
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Explosion in a sphere and the Gravitational field outside Take a hollow sphere and conduct a process on the inside, which transfers mass into kinetic energy (e.g. we let a big nuclear bomb detonate or something like that). For simplicity, assume that this will happen in a spherically symmetric way. How will the gravitational field look at the outside? In particular, how do the "new" components of the Stress-Energy tensor affect the field?
My (very, very vague) physical intuition says, that in the metric tensor, calculated from the Einstein tensor, the additional contributions from the Stress-Energy tensor will somehow compensate for the loss in mass. However, far out, in the asymptotically flat case, the gravitational field will get weaker due to Newton's Theorem. But how does this relate to Birkhoff's theorem? The weak field limit is obviously not static. The problem is somehow, that the $T^{\mu\nu}$ is in this example not mass-dominated, therefore most of the stuff one generaly does will not work. So, what happens here?
| According to Birkhoff's theorem, any spherically symmetric gravitational field should be produced by a mass at its center. Conversion of mass to kinetic energy inside a spherical shell should not affect the gravitational field outside the shell. Even if the interior of the shell becomes an intense hot group of photons of energy M*c^2, they will gravitate the same as a mass E/c^2.
Since your interior process is conducted "in a spherically symmetric way", the Schwarzschild metric describes the gravitational field as long as the electric charge is zero and the angular momentum is zero. Even if the hollow sphere expanded, the gravitational field would not change - there would be no gravitational waves.
| {
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Are wavelength and the distance same thing? Can you clarify for me the following question: are wavelength and distance same?
I know wavelength is measured in terms of distance but when we have a look at the two equations:
$$
c=f\,\lambda\\
v=d/t
$$
it actually explains the same thing where $v=c$=velocity and $1/t$ is frequency. So $\lambda$ should be equal to $d$. So if $\lambda = d$, then why do we have two equations existing instead of one. Can we use any equation to calculate velocity?
| A wavelength is a particular distance, corresponding to the length travelled during a period, which is a special time. Since $v=d/t$ holds good for the distance $d$ travelled by a constant velocity object over any given time interval $t$, a fortiori this relationship holds for the special, particular time known as the period. So, yes, $v=d/t$ is how you derive $c=f\,\lambda$, but of course not every distance travelled by a disturbance has the particular significance of the wavelength.
| {
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Is net work and total work same? According to my text book
Total Work = Delta Kinetic Energy = KEf - KEi
But then work is defined to be
dot product of Force (vector) and Displacement (vector).
Also to my knowledge work is positional.
So, if we assume an object running in circle, and it completes one cycle,
Is it correct to say net work = 0? or is net work = total work?
I have one more question,
if gravitational force is only thing acting on the system, in which
the object is moving downward vertically,
do we say Work is Kinetic Energy, and Gravitational Force is Potential Energy?
or the opposite of what I think it is?
| I am trying to go to a bit basic level. The formula work=Force*Displacement works only if the force is constant and not changing its direction or magnitude. When an object moves in circle,the force continuously changes its direction. So to calculate it we have to use integral of F with dl,assuming that force remains constant for a very short displacement dl. And net work and total work are same,just two different words of english.Also if there is a conservative force in space,the work done by that force does not depends upon in what path is object moving. It just depends upon final displacement in direction of force.
| {
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Why is probability of finding the electron at a certain point when one of the slits is closed $|\Psi|^2 $ & not $|\Psi|^2 dx$? Let in a given physical condition, the wave-function to a particle be assigned as $|\Psi (x_i,0,0,t)|^2 dx$.
Now, at the double-slit experiment , the probability of finding the particle at any $x$ when hole 2 is closed is given as $P_1 = |\Psi|^2$.
My problem is here. At the first, I learnt that $P_i = |\Psi_i|^2 dx$. But in the double-slit experiment, I found only $P_1 = |\Psi_1|^2$ devoid of $dx$. Why is it so?
| Given the wavefunction $\psi(x)$, the probability to find any particle within an interval $[x_0,x_0+\Delta x]$ is
$$ P([x_0,x+\Delta x_0]) = \int_{x_0}^{x_0+\Delta x} \lvert\psi(x)\rvert^2\mathrm{d}x$$
i.e. $\lvert\psi(x)\rvert^2$ is not a probability, but a probability density that has to be integrated over a set of non-zero measure to yield a notion of probability.
Formally, it is not clear what the object "$\lvert\psi(x)\rvert^2\mathrm{d}x$" is when not appearing behind an integral sign, but it may intuitively be thought of as the "probability of finding the particle in an interval of infinitesimal width $\mathrm{d}x$ around $x$".
| {
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Typo in Griffiths' electrodynamics We are referring to the second equality in equation (9.24) in section 9.1.3 Boundary Conditions: Reflection and Transmission.
$$ \frac{\lambda_1}{\lambda_2} = \frac{k_2}{k_1} = \frac{v_1}{v_2}. \tag{9.24}$$
Shouldn't this be $ \frac{k_2}{k_1} = \frac{v_1}{v_2} \frac{\omega_2}{\omega_1}$?
Namely, using the substitution $\omega = 2\pi \nu = kv$.
Context. We are considering the boundary conditions at the knot at which two strings are tied to each other. $k_1,\lambda_1$ and $v_1$ are the parameters for the first string. We look at the incident, reflected and transmitted waves.
Supplement. Introduction to Electrodynamics, Fourth Edition, Griffiths
| This is not a typo. In this situation both waves must have the same frequency, $\omega_1=\omega_2$, so the extra factor is simply unity. The equality of both frequencies is due to the fact that, at the interface itself, the waves must remain in step for all time, which can only happen if they oscillate at the same frequency. (In contrast, the wavelengths can differ, because neither medium can "see" what happens in the interior of the other one. The wavelength - i.e. the wavevector $k$ - happens internally to the medium; the interface only sees the frequency and the transverse component of $\mathbf k$.)
| {
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Difference between a reversible change and a reversible process? Question
In thermodynamics what is the difference between a reversible change and a reversible process?
Additional information
I am new to the topic of thermodynamics and getting confused about the difference, if any, between a reversible change and a reversible process. It seems to me that the difference is to do with the equilibrium of the system with the surroundings.
| This is simply about words. A process can cause a change.
For example: A (reversible) adiabatic process can cause a (reversible) temperature change.
| {
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How is the Lagrangian defined in GR?
*
*Reading about the Schwarzschild metric in general relativity I see that sometimes $$L=g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}$$ and sometimes $$L=\sqrt{g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}.$$
Which is the right way?
*Also how is the energy $E$ defined as $$E=-\frac{\partial{L}}{\partial{\dot{t}}}=\left(1-\frac{2M}{r}\right)\dot{t}~?$$
Because $E$ here doesn't have units of energy. Am I missing something here?
| The correct way is to define the reparametrization-invariant action
$$ S[X] = \int d\tau \sqrt{g_{\mu \nu} (X(\tau)) \cdot \frac{dX^{\mu}}{d\tau} \frac{dX^{\nu}}{d\tau} }. $$
Note that the choice of $\tau$ is arbitrary. The system has a large group of gauge symmetries - those are reparametrizations of the worldline (different choices of $\tau$).
One way to deal with this is to gauge-fix the system. For example, we can choose to set $\tau$ to be a proper time along the geodesic (induced by the metric):
$$ d\tau = \sqrt{g_{\mu \nu} (X(\tau)) dX^{\mu} dX^{\nu}}. $$
But this immediately implies that the square root in the action is constrained to be equal to $1$. This is why it is convinient to drop the square root ($\sqrt{1} = 1$, right?) and write
$$ L = g_{\mu \nu} (X(\tau)) \cdot \frac{dX^{\mu}}{d\tau} \frac{dX^{\nu}}{d\tau}. $$
But this only works if $\tau$ is the proper time.
Also, you should multiply the Lagrangian by the overall factor of $m$. This will restore the units of energy.
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How and when are the relativistic corrections applied to GPS satellites? It is known that there is a need to correct the onboard clocks to reduce the time difference from 38μs to 50ns. Where is relativity playing its role here? Why cant the clocks be simply synchronised with the ground clocks through telecommunication? If these are not possible then how are the clocks corrected?
|
Where is relativity playing its role here?
This part is well explained in Wikipedia: Effects of relativity on GPS
Why cant the clocks be simply synchronised with the ground clocks
through telecommunication?
They can, and in fact that's exactly part of what the GPS Control Segment does: measuring the satellite clock errors; except that the GPS satellite clocks are not forced to tick in synchrony; instead, the amount of their error is broadcast to users so that it can be subtracted as a correction in software calculations.
If the onboard relativistic hardware correction didn't exist, the satellite clock error would just be larger; the correction accounts for the bulk of the expected deviation between nominal carrier frequency and what is actually received on the ground, as per GPS ICD:
"The carrier frequencies for the L1 and L2 signals shall be coherently
derived from a common frequency source within the SV [satellite
vehicle]. The nominal frequency of this source -- as it appears to an
observer on the ground -- is 10.23 MHz. The SV carrier frequency and
clock rates -- as they would appear to an observer located in the SV
-- are offset to compensate for relativistic effects. The clock rates are offset by ∆f/f = -4.4647E-10, equivalent to a change in the P-code
chipping rate of 10.23 MHz offset by a ∆ f = -4.5674E-3 Hz. This is
equal to 10.2299999954326 MHz. The nominal carrier frequencies (f0)
shall be 1575.42 MHz, and 1227.6 MHz for L1 and L2, respectively."
| {
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Quantum entanglement and the big bang Prior to the Big Bang all matter was compressed into a point of high density. Why isn't all matter already entangled?
| A particle can only be maximally entangled with exactly one other particle.
If it helps, you can think of being maximally entangled as having a perfect relationship between two particles rather than either particle having a perfect property in the slightest.
If you had a perfect spin up (in a particular direction) then obviously you could have some (that direction) correlation with anything and everything to the degree that the other particle has a probability of giving spin up in that direction. But that is 100% not what entanglement is. Entanglement is having the correlation instead of the two things having there own properties.
When you are maximally entangled, then every measurement gives every result equally likely so measurements reveal nothing about what you were, they just force a particular actualization of the relationship between the two particles into existence, and once that actualization is actualized, they are not entangled any more.
So an entanglement is an ability to actualize a relationship along a yet to be determined basis (you can measure in many different directions). The measurement causes the relationship to appear along the directions measured, so that's want entanglement is.
And you can be partially entangled with many particles, but when you are measured your entanglement with everything else is realized and then gone (and you become maximally entangled with the measurement device, because measurements actually create new maximal entanglement).
| {
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Can we find actual rest mass of things on Earth Earth moves around the Sun and the Sun moves around the galaxy and the galaxy moves with unknown speed and direction. We have speed so the mass of us all altered.
Can we know the real rest mass? If so, can we deduce our speed in the universe?
| We do have a rough idea of the relative speed and direction of our galaxy, with respect to the other galaxies around us, the so called local group.
In general relativity, which is our best theory of the universe to date, there is no such thing as absolute speed, as it depends on which frame of reference you use to measure things in.
Our Earth and the objects on it are composed of atoms, as I am sure you know, and these atoms in turn are composed of electrons, protons and neutrons.
The "real" rest mass of these objects is determined by experiment, as we do not currently have a theory that predicts their mass.
To keep this answer from going on too long, (and to avoid trying to explain things that other people can do better than I can), I am skipping over areas such as renormalization (with respect to elementary particles) and dark matter, (with respect to our galaxy).
| {
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Find the force needed to accelerate body to a certain velocity for a certain time with respect to drag force So, the problem is straightworward when we suggest that air resistance force is constant:
$$ \vec F = \frac {\vec V_1 - \vec V_0} {t} m / b $$
$$ \vec V_0, \vec V_1 - \text {initial and final velocities respectively},\\ t - \text {time, during which velocity will become its final value}, \\b - \text {some constant for drag force} $$
The problem arises when using quadratic air resistance: $$m a = -c v^2$$
I don't have any clue for solving this and asking for help.
What I must do is to accelerate plane to the certain velocity in a certain period of time. So, I must find needed acceleration or force, itsn't matter, because can be easily recalculated.
Thanks in advance!
P.S. It's sad, but I'm not familiar with calculus, I'm more programmer, not mathematician.
P.P.S. I know what differentiating and integration is, what derivative and displacement is, but really not familiar with differential equations, but guess it is here.
| If your object has mass $M$ and you want to accelerate it with acceleration $a$ to a specific end-velocity $v$ you have to keep in mind that the energy
$$e = \frac{M\cdot v^2}{2}$$
and also
$$e = F\cdot x$$
where $x$ is the distance over which the force $F$ which equals $M\cdot a$ is applied. Knowing that you can solve for the distance over which you have to accelerate:
$$\frac{M\cdot v^2}{2} = M\cdot a\cdot x \to $$
$$ \to x = \frac{v^2}{2\cdot a}$$
Now you have the distance and can easily solve for the time you need to reach your velocity $v$:
$$t = \sqrt{\frac{2\cdot x}{a}}$$
Example: If you have a drag force of $Fd = -4 \frac{kg \cdot m}{s^2}$, your object has a mass of $M = 3 kg$ and you want to reach $v = 10 m/s$ in $t = 5 s:
$$a = \frac{v}{t} \to a = \frac{10 m/s}{5 s} = 2 \frac{m}{s^2}$$
You multiply that with yor mass and get a force of
$$F = 6 \frac{kg \cdot m}{s^2}$$
This is the force you would need without drag force. Now you simply substract the drag:
$$F_{total} = F-Fd = (6+4)N = 10 N$$
| {
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Weather Meridians Are there meridians that effect cloud movement? While observing radar images of moisture bearing cloud movements (at my location in Florida), I've noticed a splitting of the cloud formations a majority of the time. It has become apparent and predictable.
| No, meridians won't affect cloud movement because there is no meridian which is special due to any geophysical feature. Latitudes may have an effect due to the 23.5$^o$ tilt of the Earth. The 23.5$^o$N and 23.5$^o$S latitudes mark the maximum latitudes for the Sun to be directly overhead, and the Equator marks the great circle along which there is (nearly) always 12 hours of daylight and 12 hours of dark.
All that said, I would guess that what you see in the clouds is more of a local effect of land and sea causing some type of regular up-draft of warm air which causes the water droplets that make up the cloud to evaporate at the locality. This would have the appearance of the cloud formation ``splitting.'' Or the up-draft could be so strong that it actually pushes the moist, cold air of the cloud apart.
Clouds are dynamic systems, always reforming as moist bundles of air hit cold or warm bundles. The moisture will condense into tiny droplets in the cold air and the droplets will evaporate into water vapor in the warm air.
| {
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Is it possible for man to break Earth into 2 parts? Many countries have extreme devastating nuclear weapons. Also they have weapons in very large numbers.
I want to ask that
Is it possible for man to break earth into 2 parts with the help powerful weapons like nuclear weapons or any other technology?
| No, because the vast majority of the planet has a molten interior and where it is not in the liquid phase it is held in solid phase by the internal pressure. You could maybe disperse it into space with a big enough bomb, but not actually break it into two parts.
| {
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Why is the introduction of a quantization volume necessary for quantization of the EM field I have been working through the quantization of the electromagnetic field, and every source I find introduces a quantization volume with periodic boundary conditions in the process, in which we fit the general solution of $A(\boldsymbol{x},t)$. Why is this necessary? I understand that this allows us to consider a countably infinite sum over wave vectors, rather than an uncountable one, as the wave vectors are made to satisfy the periodic boundary conditions.
I have the vague impression that it has something to do with the orthogonality of the wave functions (solutions to the wave equation of the field before quantization), so that integration as follows yields a delta function
$$\sum\limits_{kk'} \int dx \, e^{i(k-k')x} = \delta(k-k')$$
but then I think this should work equally in the continuous case
$$\int\int {dk \,dk'} \int dx \, e^{i(k-k')x} = \delta(k-k')$$
What am I not understanding? Thanks in advance for any help!
| One reason for the box is the Fourier expansion of field in stable macroscopic condition (thermal radiation, cavity oscillations) works well only for finite volume. For infinite volume, the Fourier integral of such stationary field is problematic, because the field function is not L2 integrable.
| {
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$\phi^4$-theory: nested two-loop contribution _8_ Wherever I see calculations of two-loop contributions to the $\phi^4$ propagator (such as Peskin, page 328, on the bottom), only the sunset diagram (aka the Saturn diagram) is considered, but not, say, the two-loop diagram involving a loop on top of a loop (looks like this: _8_). Does it not contribute? As far as I can tell, it does and the loop integral for it is
$$\int\frac{d^4k}{(2\pi)^4}\frac{d^4q}{(2\pi)^4}\frac{1}{(k^2−m^2)^2}\frac{1}{q^2−m^2}$$
with a high degree of divergence ($Λ^2$). Am I correct or am I missing something?
| As far as I know, I think you mean this diagram,
where,
And the divergent part of $I_1$ is $I_1^{\text{div}}=-\frac{m^2}{8\pi^2\epsilon}$, where $\epsilon$ is from $d=4-\epsilon$. At least we can see that $I_1^{\text{div}}\frac{\partial I_1^{\text{div}}}{\partial m^2}$ will have a divergent term like $1/\epsilon^2$. Thus, we can see this $\_8\_$ diagram will contribute to the two-loop for the correction of mass.
| {
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Advantages of a deadblow hammer? A deadblow hammer is a type of mallet with a hollow head filled with shot or sand. When you hit with it, say on an anvil, the mallet does not rebound, but just falls flat and heavy.
I don't understand the advantage of this from a physics point of view. Force is force, right? Normally when I compute the force of a hammer I multiply the velocity and the weight (mass). How is the deadblow hammer any different?
| The hammer can be thought as a mean to deliver enough energy to the nail to deform the underlying material and let it penetrate deeper.
Ideally, all the energy the hammer gets from your arm as kinetic energy is transferred to the material and results in its (hopefully) permanent deformation, but as always, that's not the case in practice.
What should be a totally anelastic collision (e.g. your hammer should stay on the nail after the impact) is instead slightly elastic: part of the kinetic energy is not used for the deformation, thus some of the hammer momentum "comes back" as a recoil.
While in a normal hammer this remaining momentum is transferred back to the entire tool, if you fill the mallet with sand or with something which can move independently from the rest of the hammer, part of this momentum is transferred to this detached mass which thus recoils instead of your hammer, leaving less momentum for the bouncing of the rest of the tool.
The principle is the same used in large cannons: the barrel is detached from the rest of the weapon to let it recoil without displacing the whole cannon.
Another way you can figure it: take a plastic bottle full of sand (but water should work as well) and drop it, you'll see that the bouncing is heavily dampened by the recoiling of its contents, that you should see going around and markedly "jump" notwithstanding the movement of the bottle which contains them.
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Cylindrical capacitor in an electric circuit I've come across a tricky question and would appreciate some hints or explanations as to why the given solution is the way it is. The question reads as follows:
A coaxial cable consists of a wire with radius $a$ (the core of the cable), which is wrapped with insulating material with dielectric constant $\varepsilon$, until radius $b$ (called the insulator). Around the cable there is a layer of conducting material (radius $c$ from the center of the cable and is called the wrapper).
The wire's length is $d$ such that $d \gg a,b,c$. At one side of the cable, a voltage source $V_0$ with inner resistance $R_0$ is connected to both the wire and the wrapper, and at the other side, a resistor $R$ is connected instead of a voltage source.
It asks to find the magnetic and electric fields $B(r)$, $E(r)$, where $r$ is the distance from the center of the cable (from $z$-axis in the picture), when $t\rightarrow+\infty $.
In the solution, they said that when $t\rightarrow +\infty$, no current will pass through the cylindrical capacitor so: $I=\frac{V_0}{R_0+R}$ therefore $V\left(\text{final}\right)=V_0 \frac{R}{R_0+R}$.
I do not get this, how can one imagine how this circuit works? Is there an equivalent and more simple circuit? According to what they said, after infinite time, no current passes through the capacitor, but the wires are connected to the wrapper so how can there be current at all in the circuit? All I know is when an uncharged capacitor is charged, it will act as an open switch in the circuit after a long time.
Possible equivalent Circuit?:
| I think the solution is incorrect.
If the current becomes zero after an infinite time, the potential of the rod and the wrapper would become equal ( because they are connected by a wire of $R$ resistance, and if $I=0$, the $V$ across the wire $=0$. )
Also, your diagram seems incorrect. You assume that there are different wires connecting the capacitor and the resistance $R$ but, in the question, the wire IS a part of the capacitor.
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Does isotropy imply homogeneity? This question comes from exercise 27.1 in Gravitation by Misner, Thorne and Wheeler. They required the following:
Use elementary thought experiments to show that isotropy of the universe implies homogeneity.
I know homogeneity as the universe is the same everywhere at a given time, and isotropy is related to direction.
I wonder how the isotropy of the universe implies homogeneity.
| This is closely related to the fact that in a Euclidean space, coordinate translations can be generated by performing two successive rotations around different points, as isotropy is essentially rotation invariance and homogeneity translation invariance. Suppose we have a rotation $R(\vec{r}_0)$ respect to $\vec{r}_0$ defined through the action on any point $\vec{r}$ as
$R(\vec{r}_0)\vec{r}=\vec{r}_0+R(\vec{r}-\vec{r}_0)$
Then clearly $R(\vec{r}_0)\vec{r}_0=\vec{r}_0$. For simplicity, we simply use $R$ to denote a rotation respect to the origin. Then for any point $\vec{r}$, two successive rotations around origin and $\vec{r}_0$ respectively would give
$R^{-1}(\vec{r}_0)R\vec{r}=\vec{r}_0+R^{-1}(R\vec{r}-\vec{r}_0)=\vec{r}+(I-R^{-1})\vec{r}_0$
Then for any translation $\vec{a}$, we can choose the coordinate system such that $\vec{a}=(a,0,0)$, then set
$\displaystyle R^{-1}=\left(\begin{array}0 &-1&\\1&&\\&&1\\\end{array}\right)$
and $\vec{r}=(a/2,a/2,0)$, we get
$\vec{r}+\vec{a}=\vec{r}+(I-R^{-1})\vec{r}_0=\vec{r}_0+R^{-1}(R\vec{r}-\vec{r}_0)=R^{-1}(\vec{r}_0)R\vec{r}$
Then if the space is invariant under rotations with respect to any point, it will be invariant under translation. In curved spacetimes, instead of global rotations, we need to consider Killing vectors. And similarly, existence of Killing vectors for isotropy at every point implies the existence of Killing vectors for homogenity. For details, see Chapter 13 of Weinberg's extraordinary book, Gravitation and Cosmology.
| {
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Electric Motors: why do they draw more current when stalled, and less when moving? I'm familiar with how an electric motor works, and also familiar with what stall current and free current refers to in an electric motor. I'm also somehow familiar with the notions of energy being transformed from electromagnetic to kinetic and to heat.
I've also reviewed questions like these:
*
*https://robotics.stackexchange.com/questions/613/what-is-stall-current-and-free-current-of-motors
*Why do electric motors draw current even when they are not moving?
What I'm yet to fully understand is why exactly, from a physics point of view, does the current increase when the motor is stalled, and more intriguingly why does the current drop when the motor is moving.
I'm guessing that the coil resistance increases/decreases when the motor is moving/stalled, but I can't grasp the idea of why does that happen and what the explanation at an electric/electromagnetic level is.
I guess another way to put it's why isn't current always flowing at it's maximum in an electric motor and actually drops when the motor is moving?
thanks!
| When a motor moves it also acts as a generator and the current trough the windings is given by the difference of the external voltage and the induced voltage. When the motor stands still, though, the generated voltage is zero and the windings will draw the max. current they can based on their DC resistance.
In other words, the faster the motor runs, the higher a voltage it generates and the smaller the difference becomes. There are, of course, electrical and mechanical losses, so there has to be a finite amount of power that is being supplied by the power source, which means that the motor never quite reaches the rpms at which the generated voltage would be exactly the same as the supply voltage.
| {
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Coulomb collision I was reading an article by N. Bohr and came upon the following problem (the following wording is actually taken from a book by Thompson - Conduction of Electricity Through Gases):
Let $M_1, M_2$ be the masses of the corpuscles $A$ and $B$ respectively; we shall suppose the velocities of the colliding corpuscles so great that in comparison the corpuscles in the atom may be regarded as at rest. Let $V$ be the velocity of $B$ before the collision, $b$ the perpendicular let fall from $A$ on $V$. Then if $2\theta$ be the angle through which the direction of relative motion is deflected by the collision, we can easily show that, taking the force between the corpuscles to be $e^2/r^2$,
$$
\sin^2\theta=\frac{1}{1+\frac{b^2V^4}{e^4}\left(\frac{M_1M_2}{M_1+M_2}\right)^2}.
$$
Original text.
So I wanted to show the formula but as both Bohr and Thompson say that it is easy to show, I assume it is rather complicated. Anyway, can someone help me with graphical representation of the situation? Namely, what $b$ is?
I had expected that the situation is like:
But after reading the text "perpendicular let fall" I doubt it.
| I believe b is the impact parameter. So b should be the perpendicular distance between the asymtote you drew and the fixed ion. The Wikipedia link has a picture that is pretty clear, although the picture in the link is illustrating Coulomb REPULSION, while your problem involves Coulomb attraction..
| {
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Are there limits for the speed of sound? A maximum or a minimum only? The speed of sound in materials of various states of matter differs a lot.
But does it have fundamental limits?
*
*Is there a maximal possible speed of sound?
*Is there a minimal possible speed of sound?
*Is the speed of sound in a material a multidimensional function of many uncorrelated dimesions of material properties, and current state states, such that the function ends up as a vast unknown n-dimensional surface where we not even know in which direction to look for a maximum, and are already happy to find a local maximum at all?
*Obviously, the speed of light in vacuum is a upper limit for the speed of sound in general. But that does not imply that, for a given Material, the material specific speed of light is an upper limit for the speed of sound in the material.
Of course, in the set of speeds of sound in all materials where it can be measured, there is a maximum and a minimum.
But there are some materials where we can not currently measure the speed of sound, say short lived isotopes available in small numbers of atoms, neutron stars, and other things your university can not order for the laboratory.
From a theoretical perspective, maybe one yould reason about speed of sound independent of existing or potentially existing materials?
| Fresh out of the oven: Physicists have discovered the ultimate speed limit of sound. Quote:
Sound is a wave that propagates by making neighbouring particles
interact with one another, so its speed depends on the density of a
material and how the atoms within it are bound together. Atoms can
only move so quickly, and the speed of sound is limited by that
movement.
Trachenko and his colleagues used that fact along with the
proton-electron mass ratio and the fine structure constant to
calculate the maximum speed at which sound could theoretically travel
in any liquid or solid: about 36 kilometres per second.
| {
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Is magnetic reconnection reconcilable with magnetic field lines neither starting nor ending? According to Maxwell's equations, magnetic fields are divergence-free: $\nabla \cdot \mathbf{B} = 0$. If I understand this correctly, this means that magnetic field lines do not start or end. How can we reconcile this with magnetic reconnection?
| Magnetic reconnection comes from a cartoon picture of what magnetic field line motion may portray. This is not based on any physical law. Field lines are not real entity - just a means to display the lines of force when magnetic field is present in space. Field line motion is non-unique also, which is a fundamental flaw for people relying on field line motion to understand physics. That is why the magnetic reconnection is so misleading. It is unphysical to cut and join magnetic field lines. The exact equation for describing the time rate of change of B is not what is given in Eq. (2) by Kyle Kanos. The generalized Ohm's law is far more complicated than representing all the other missing terms in the simplified Ohm's law by a scalar resistivity. One should learn more about how magnetic reconnection is entrenched in space plasma physics - see, e.g., the following publications: note that Hannes Alfvén is a Nobel Laureate and is the one who invented MHD, Alfvén, H., On frozen-in field lines and field-line reconnection, J. Geophys. Res., 81, 4019-4021, 1976; Alfvén, H., Electrical currents in cosmic plasmas, Rev. Geophys., 15, 271-284, 1977; Akasofu, S.-I., Auroral substorms as an electrical discharge phenomenon, Progress in Earth and Planbetary Science, 2:20, doi:10.1186/s40645-015-0050-9, 2015; Lui, A. T. Y., Comparison of current disruption and magnetic reconnection, Geosci. Lett., 2:14, doi 10.1186/s40562-015-0031-2, 2015.
| {
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Force acting on a dipole placed in a non-uniform electric field For an electric dipole is placed in a non-uniform electric field, why does the net force act in the direction of increasing electric field?
| This is best understood by approximating the dipole as a pair of finite charges $\pm q$ separated by a finite distance $d$. In a uniform electric field, the electrostatic forces on each of the charges will cancel out exactly, but in a non-uniform one the forces on the two will be slightly different, leading to a slight imbalance and therefore a non-zero net force. As you take the distance to zero, the difference in electric field goes to zero, but the charge also grows to exactly cancel it out.
To be more quantitative, suppose the negative charge is at $\mathbf r$ and the positive charge at $\mathbf r+d\mathbf n$. The total force is then
$$
\mathbf F=q\left[\mathbf E(\mathbf r+d\mathbf n)-\mathbf E(\mathbf r)\right].
$$
To get the correct form for the limit, change from the charge $q$ to the electric dipole $p=qd$, to get
$$
\mathbf F=p\frac{\mathbf E(\mathbf r+d\mathbf n)-\mathbf E(\mathbf r)}{d}.
$$
The true force on a point dipole is the limit of this as $d\to0$,
$$
\mathbf F=p\lim_{d\to0}\frac{\mathbf E(\mathbf r+d\mathbf n)-\mathbf E(\mathbf r)}{d},
$$
and this is exactly the directional derivative along $\mathbf n$, typically denoted $\mathbf n\cdot \nabla$, so
$$
\mathbf F=p\mathbf n\cdot \nabla\mathbf E=\mathbf p\cdot \nabla\mathbf E.
$$
| {
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Is air infiltration a type of convective heat transfer (convection) I have a building / physics question...
A major source of heat loss in homes and buildings is infiltration through cracks (warm air from inside seeping out). Wondering if this falls in the category of convection as a mode of heat transfer?
| My guess, if the source of heat is in the centre of the room, is a mixture of radiation directly to the walls, plus convection of the air to the walls.
When the heat gets to the walls, it's conduction through the walls, unless there are holes, cracks or vents in the walls, when it's convection of air through the walls.
Warm air moves outwards, from the room to the outside, so maybe I would not use the word infiltration in this question.
| {
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Simple Harmonic Motion in Special Relativity I was trying to see what results I would get if I were to incorporate relativistic corrections into the case of a harmonic oscillator in one dimension. I thought that if the maximum velocity of the oscillating body were to approach relativistic velocities, the measurement of the time period of motion and related measurements should change. I assumed a lab frame $(t,x)$.
The equation of motion should be
\begin{equation}\frac{dp}{dt}+kx=0
\end{equation}
where $t$ is the time measured in the lab frame. $\frac{dp}{dt}$ is calculated to be $\gamma^3m_0\ddot{x}$(since only $a_{\mid\mid}$ exists for this motion). Naively, I took this as my solution in the following way:
\begin{equation}\gamma^3m_0\ddot{x}=-kx
\end{equation}
\begin{equation}\implies\ddot{x}=-\frac{k}{\gamma^3m_0}x
\end{equation}
\begin{equation}\implies\ddot{x}=-\omega^2x
\end{equation}
So I took the angular frequency $\omega$ simply like that. However, I forgot that $\gamma$ is a function of $\dot{x}$ and this obviously complicates things... for instance the period becomes a function of velocity. I'm not sure if my approach is correct and I don't have the tools to solve the differential equation I would get if i were to open up $\gamma$, so my questions are the following:
*
*Am I doing this right?
*Am I missing something conceptually?
*Is there a better way to approach the problem?
*And if, by some miracle, my last expression for $\ddot{x}$ is correct, could I have
some help solving the differential equation?
EDIT: If you are posting the relativistic expression, please post the corresponding evaluation of the Classical limit of your expression.
|
I was trying to see what results I would get if I were to incorporate relativistic corrections into the case of a harmonic oscillator in one dimension. [...]
$ \gamma^3~m_0~\ddot{x} = -k x $
Since your question is specificly about harmonic motion, we might (instead) insist on this solution
$$x[~t~] := x_{\text{max}}~\text{Sin}[~\omega~(t - t_0)~],$$
and ask about the expression, as a function of $x$, of a corresponding suitable conservative force (or eventually, the "shape" of a corresponding potential).
Consequently, we'd insert the "harmonic solution" into the left-hand side of your equation:
$$ \gamma^3~m_0~\ddot{x} \mapsto -m_0~\omega^2~x_{\text{max}}~\text{Sin}[~\omega~(t - t_0)~]~/~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega~\text{Cos}[~\omega~(t - t_0)~]}{c}\right)^2 }$$
and ask how to express it as a function of the variable $x$ (instead of variable $t$). That's of course straightforward by inserting the harmonic solution:
$$
\begin{align*}
\gamma^3~m_0~\ddot{x} ~~~ & \mapsto -m_0~\omega^2~x~/~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega~\text{Cos}[~\text{ArcSin}[~x/x_{\text{max}}~]~]}{c}\right)^2 } \\
~ & = -m_0~\omega^2~x~/~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega~\sqrt{1 - (x/x_{\text{max}})^2}}{c}\right)^2 } \\
~ & = -m_0~\omega^2~x~/~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega}{c}\right)^2 + \left(\frac{x~\omega}{c}\right)^2 }.
\end{align*}
$$
Of course we should require $0 \lt x_{\text{max}}~\omega \lt c$, whereby the square root is always defined, and real.
In the ("non-relativistic") limit that $x_{\text{max}}~\omega \ll c$ (and since $0 \le |x| \le x_{\text{max}}$) this force expression approaches the right-hand side of your equation: $-m_0~\omega^2~x \equiv -k~x$.
It is also easily integrated to obtain the corresponding "relativistic harmonic potential form"
$$V[~x~] := m_0~c^2 ~\sqrt{ 1 - \left(\frac{x_{\text{max}}~\omega}{c}\right)^2 + \left(\frac{x~\omega}{c}\right)^2 } - V_{\text{ref}};$$
where the term with explicit $x$ dependence in the limit $x_{\text{max}}~\omega \ll c$ approaches $m_0~c^2~\frac{1}{2}~\left(\frac{x~\omega}{c}\right)^2 = \frac{1}{2}~m_0~\omega^2~x^2 \equiv \frac{k}{2}~x^2$.
| {
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Do electrostatic fields really obey "action at a distance"? In an electromagnetic theory class, my professor introduced the concept of "action at a distance in physics".
He said that:
If two charges are at some very large distance, and if any one of the charge moves, then the force associated with the charges changes instantaneously.
But according to Einstein, no information can travel faster than the speed of light.
So photons (the information carriers in electromagnetic force) cannot instantaneously deliver information.
So that we associate a field with the two charges and if any charge moves, there is a deformation in that field and this deformation travels with the speed of light and conveys the information.
If the field deformation information cannot travel more than the speed of light, how does the force instantaneously change at very large distances?
| The force is not propagated instantly. It takes time for the information to get from one point to another.
You can treat that as an instant if you are working with small enough distances and velocities, but it's not. If you'll ever study field theory you'll meet retarded potentials that are just this: the field propagates at the speed of light and it's no longer seen as instant.
| {
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Do neutrinos refract? The most benign of interactions is refraction. While neutrinos rarely interact with matter in a sense like the photoelectric effect, does that mean that they don't refract either?
| Neutrinos are weakly interacting quantum mechanical point particles, with very small mass.
Refraction is a classical mechanics phenomenon, happens to waves traveling in a medium and it is a collective synergy of many photons impinging on the field of the atoms and molecules of the medium. Individual photons are not refracted but are scattered. In synergy with the zillions of photons in the light wave refraction appears as an emergent phenomenon.
Neutrinos have very small probability of scattering with the atoms and molecules but there do exist neutrinos coming from the sun in great numbers and thus the question is not irrelevant. There exist studies that derive a refraction index of the neutrino flux within various models for neutrinos for the sun, as for example here.
It is pointed out that, if neutrinos are to maintain coherence over the required distance for the Mikheyev-Smirnov-Wolfenstein solutions to the solar-neutrino problem, effects arising from neutrino multiple scattering must be considered. We give a simple derivation for the neutrino index of refraction that takes into account this effect. The same method is also shown to be useful for situations with varying matter densities and neutrino mixing. We also examine the question whether the coherence of propagating neutrinos in matter will be affected by switching on an external magnetic field, assuming neutrinos have a large magnetic moment.
| {
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Eigenstates of sum of creation and annihilation operators Does the operator $a+a^\dagger$ have eigenstates? If yes, what are they?
| No, it has not discrete spectrum (on $L^2(\mathbb{R}^d)$). In fact $a+a^*$ is proportional to the position operator (or the momentum one, depends on your definition of $a$ and $a^*$; by the usual one the position operator $x$ is proportional to the real part $a+a^*$ and the momentum $p$ to the imaginary part $\frac{1}{i}(a-a^*)$). Both position and momentum operators have purely continuous spectrum, so there are no eigenstates that are square integrable (but there are the usual "generalized eigenvectors"; i.e. delta functions for the position operator).
| {
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Free fall into circular motion If I'm on a roller coaster free falling from height $h$ and then suddenly start going into horizontal motion with a radius $r$ of turn what is the $g$-force I experience?
I worked out the equation like this but am not sure if it is correct:
*
*(1) instant velocity of free-fall $v=\sqrt{2 g h}$
*(2) uniform circular motion acceleration $a = \frac{v^2}{r}$
*(3) $g$-force $ gf = \frac{a}{g} = \frac{v^2}{g r}$
My doubts are:
*
*I don't know if I can use uniform circular motion equation since $v$ is not constant
*Where is the g-force directed towards? The center of the turn?
| You're good.
Yeah, you can pretty much assume that it's a constant velocity, as long as $h \gg r.$ As I recall, the expressions involved are extremely simple as long as you don't try to figure out exactly what's happening in time: actually solving the Euler-Lagrange equations gives you some sort of $\int d\theta / \sqrt{a + b \sin\theta} = t$ equation for $\theta(t)$, or something awful like that.
So, let's have some coordinates and geometry: You start out at $[x, y] = [0, r + h]$, then at $[x, y] = [0, r]$ you enter a circle centered at $[r, r]$: and your progress along that circle I'll denote with $\theta$ as $\vec r = [x, y] = [r, r] - r\cdot[\cos\theta, \sin\theta]$. You then emerge after $\theta = \pi/2$ at position $[r, 0]$, moving forward. We'll express time derivatives as dots, and I'll define $\omega = \dot\theta$.
Your net acceleration during this arc is $\vec a(\theta) = - g\cdot[0, 1] + c(\theta)\cdot [\cos\theta, \sin\theta]$ for some $c(\theta)$, since that's the direction the constraint force pushes. However we know that this must also have a very special shape: $$\ddot{\vec r} = \frac{d}{dt} \left(r~\omega~[\sin\theta, -\cos\theta]\right)= r~ \dot\omega~ [\sin\theta, -\cos\theta] + r~ \omega^2~ [\cos\theta, \sin\theta].$$The $x$-component gives the simpler version of this constraint, $c(\theta) = r~ \dot\omega~ \tan\theta + r~ \omega^2.$
Since $\dot{\vec r} = \vec v = r~\omega~[\sin\theta, \cos\theta]$, we can quickly state that from energy conservation$$v^2 = r^2\omega^2 = 2g(h + r\sin\theta),$$as the constraint force does no work. With a time-derivative this also gives$$r^2~2~\omega
~\dot \omega = 2 ~g ~r~\omega~ \cos\theta $$so we have $r \omega^2 = 2g(\sin\theta + h/r)$ and $r~\dot\omega=g\cos\theta,$ so if I've done all of the mathematics right,$$c(\theta) = g~\left(3~\sin\theta + 2~\frac hr\right).$$In the limit $h\gg r$ you get an approximately-constant $c(\theta) = 2 h g / r$, and so your g-force on the turn is maximized at $\theta\approx0$ where it is $a \approx \sqrt{g^2 + c^2} \approx c.$
This solution corresponds exactly to $v^2 / r$ with $v = \sqrt{2 g h}$, as you're inclined to do.
If you don't have $h \gg r$, then you'll have to instead either space-average the force (as I said, time-averaging is probably a nightmare) or maximize the force with$$\frac d{d\theta} \big\lVert[c(\theta)~\cos(\theta),~~c(\theta)~\sin(\theta) - g]\big\rVert = 0.$$In general you will find that your acceleration points "almost" at the center of rotation but a little bit lower than it, because the gravitational force is also in there.
| {
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What happens if the load on the electrical generator exceeds its generation power? And why? What happens if the load on the electrical generator exceeds its power generation? and why?
To be more precise, suppose we have a standard induction generator operating at frequency $\nu=50\:\mathrm{Hz}$ and voltage $V_0$, and rated to produce a maximum power $P_0$, and that we connect this to a load $R<V_0^2/P_0$, which will try to draw more power than the generator's capacity. Obviously the details will depend on the type of generator, but, generally speaking: what will be the generator's response, and what physical processes are involved?
| As the current drawn by the load increases the torque opposing the motion of the prime mover on its coils increase. This opposing torque is a result of the force acting on the coil since it's a current carrying conductor moving in a magnetic field. Hence its rpm reduces and so does its voltage output.
| {
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Why are bandstructures plotted only along certain symmetry points? Why is it that bandstructures are usually represented along certain symmetry points ? What determines these symmetry points ?
| 3-D bandstructures are still very useful. (2 dimensions for x and y and 3rd for Energy).
For example, 3-D bandstructure plots allow us to see the Dirac cones that are found in the bandstructure of Graphene
Note that the x and y direction cover every single point for a given z (which in this case is kz = 0). So in this sense this plot is more 'traditional' and doesn't follow an arbitrary path through the BZ, but covers every single point in a given plane.
| {
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With a machete, why is a diagonal cut more effective than a right angle one? When cutting back some thick growth in the garden a question that always nagged me. Why is cutting diagonally seemingly more effective than cutting at right angles? Part of the answer is obviously to do with the ease of cutting vertically as opposed to horizontally (with vertical stems), but this also seems to be true at most other growth angles as well.
| Simple answers:
When you cut diagonally, less material is being moved aside during a given bit of time. The time of the cut is longer, the force the same, therefore you are applying the same force to less material in a given moment. Therefore it's easier to cut the longer arc, even though the force is unchanged.
Also, gravity is assisting your diagonal blow, increasing the force of your blow, while a horizontal blow is unaffected by gravity.
Also, your musculature is better suited to a 45 deg. angle than to a horizontal.
| {
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Why are angles dimensionless and quantities such as length not? So my friend asked me why angles are dimensionless, to which I replied that it's because they can be expressed as the ratio of two quantities -- lengths.
Ok so far, so good.
Then came the question: "In that sense even length is a ratio. Of length of given thing by length of 1 metre. So are lengths dimensionless?".
This confused me a bit, I didn't really have a good answer to give to that. His argument certainly seems to be valid, although I'm pretty sure I'm missing something crucial here.
| Because length is relative, but angle is absolute.
(There is such a thing as a maximum angle against which you can compare, but not a maximum length.)
| {
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Does body weight affect the speed when going downhill on a mountain bike? We know heavier objects fall faster when dropped at certain height. I was wondering if I am going downhill on my mountain bike without any peddling, will I travel faster or slower because I am fat?
| Heavier objects do not fall faster per se. But for heavy objects the influence of the air resistance will be smaller, if they have a similar surface area compared to the light objects.
The answer depends on the properties of your tyres and the road. But on an even road the air resistance will typically dominate once you reach a certain speed (the friction of the wheels $F_W$ will be more or less independent of speed, but not of weight as a heavier person deforms the tyres more, generation more friction, but as it is not the dominant part we will ignore it for now).
The air resistance of a person will vary approximately like $m^{2/3}$ or $m^{1/3}$ in dependence of the mass. The air resistance in turbulent flow is given by $F_R = \frac 1 2 \rho c_D A v^2$, where $\rho$ is the density of the fluid, $c_D$ is the dimensionless drag coefficient depending on the form, $A$ is the area of the object perpendicular to the flow and $v$ the velocity relative to the fluid. Your mass scales like $L^3$, so your area scales like $L^2 = m^{2/3}$ assuming isotropic growth, the drag coefficent $c_D$ will be roughly independent of your weight but highly dependent on your position and clothing, which also influence your surface area).
Your acceleration will be given by:
$$ a = g \sin(\theta) + F_\text{W} - \frac 1 2 c_D \rho v^2 \frac{A}{m} = \text{const} - O(m^{-1/3}). $$
This means you are at an advantage if you are heavier (or rather: larger and therefore heavier), as the influence of the drag scales like $m^{-1/3}$.
If we assume that your weight is not distributed equally in all directions you gain even more. But still, as the range of typical human weights which a bike can support is from about $50\,\mathrm{kg} \ldots 150\,\mathrm{kg}$ a light person in a aerodynamic position with tight clothes will probably still be faster than a heavy person sitting upright (as they will reduce their area to a fraction and lower their $C_D$).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Do black holes violate the uncertainty principle? If black holes have mass but no size, does that imply zero uncertainty in position? If so, what does that imply for uncertainty in momentum?
| General relativity is a classical theory. The Heisenberg uncertainty principle does not apply to it.
The research frontier in physics now exists in quantizing gravity and unifying it with the other three forces (strong , weak, electromagnetic). Once that is done the solution for the black hole will become a probability distribution and the Heisenberg principle will apply. The macroscopic classical solution of a point singularity will become a quantum mechanical uncertainty locus which will not change the macroscopic description. h_bar is a very small number and is already easily satisfied by the classical mechanics solutions all our constructions and engineering depend on.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/193954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minus sign in the time ordering operator The time ordering operator is usually defined as
$$\mathcal{T} \left\{A(\tau) B(\tau')\right\} := \begin{cases} A(\tau) B(\tau') & \text{if } \tau > \tau', \\ \pm B(\tau')A(\tau) & \text{if } \tau < \tau'. \end{cases}$$
The minus sign applies when $A$ and $B$ are fermion operators.
My question is now: why appears there a minus sign? I think the answer lies somewhere in the argument, that one can write the operators in second quantization and use the canonical (anti) commutation relations for the fermionic creation and annihilation operators:
$$\{\hat{c}_\nu,\hat{c}_{\mu}^\dagger\} = \delta_{\mu \nu}\\
\{\hat{c}_\nu,\hat{c}_{\mu}\} =0 \\
\{\hat{c}^\dagger_\nu,\hat{c}_{\mu}^\dagger\} = 0.$$
Can someone explain this more explicitly? I really don't see how this only changes the sign and does not introduce any constants coming from the $\delta_{\mu \nu}$ term which may appear.
| Perhaps the easiest way to see that there should be a Grassmann sign factor $(-1)^{|A| |B|}$ in the definition of time ordering
$$\tag{1} {\cal T} \left\{ A(t_A) B(t_B)\right\}
~:=~ \theta(t_A-t_B) A(t_A) B(t_B)
+ (-1)^{|A| |B|} \theta(t_B-t_A) B(t_B) A(t_A), \qquad $$
is to go to the classical limit $\hbar\to 0$. Here $|A|$ denotes the Grassmann parity, which is $0~{\rm mod}~2$ if $A$ is a boson, and $1~{\rm mod}~2$ if $A$ is a fermion. Moreover, $\theta$ is the Heaviside step function. In the classical limit, all fields should supercommute, which mean that the supercommutator
$$\tag{2} [A,B]~:=~ AB - (-1)^{|A| |B|}BA~=~0 \qquad\leftarrow \text{classically}\qquad$$
should vanish. This follows from the correspondence principle between QM and classical mechanics:
$$\begin{array}{ccc} \text{Operator} &\longleftrightarrow& \text{Symbol/Super-function}\cr
A& \longleftrightarrow& a \cr B& \longleftrightarrow& b \cr
[A,B]& \longleftrightarrow&i\hbar \{a,b\}_{PB}+ {\cal O}(\hbar^2)\cr \text{Supercommutator}&\longleftrightarrow& \text{Super-Poisson bracket.}\end{array} \tag{3}$$
In particular, time-ordering should not matter in the classical limit $\hbar\to 0$:
$$\tag{4} {\cal T} \left\{ A(t_A) B(t_B)\right\}~=~A(t_A) B(t_B)~=~(-1)^{|A| |B|}B(t_B)A(t_A).\qquad\leftarrow \text{classically}\qquad $$
But this will only be the case if we include the Grassmann sign factor $(-1)^{|A| |B|}$ in the definition (1).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Did the Sun form around a solid core? When Jupiter formed I assume like the other planets it started as tiny clumps of matter that eventually came together, became gravitationally bound and then eventually captured a lot of gas. I've also heard it was capable of collecting a lot of solid ice due to its distance from the Sun. Anyway, if Jupiter were larger we might be living in a binary star system. So, my question then becomes, did the Sun have a similar beginning to Jupiter and in what way was it different? Did the Sun form around a solid core?
| Star formation isn't completely answered, but it is well believed that a solid core is not necessary. However if the sun did form around a planetary-sized solid core we would not know the difference. Due to the very high temperature of the sun, the result is not meaningfully different from colliding with planetary bodies early on (which is plausible given the number of planetary body collisions that are invoked to explain the solar system).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Where does particle borrow energy from to tunnel? Where does particle borrow energy from to tunnel? It is implied that particle can borrow energy and leaped over to the other side wherever that is, the shorter the gap the more energy it borrows my question is where does a particle borrow its energy from and what criteria allows it to do the borrowing, last but not least how does the particle return the borrowed energy? Is it linked to quantum foam?
| On a quantum level, particles don't really have "momentum". They're waves. The way the Schrodinger equation works, they move faster if they have a shorter wavelength. So we defined momentum based on the wavelength. Kinetic energy also is part of the whole conservation of energy thing, so we have a very good reason to define it how we did, but again, it's based on the wavelength. But kinetic energy doesn't have to be positive. If it is, you get a sine wave, and if it's not, you get an exponent. The end result is that particles don't propagate very far if they have negative kinetic energy. But they can still get somewhere. So we call it quantum tunneling.
They don't literally borrow energy. They just have negative kinetic energy for a bit. But sometimes it's easier to think of particles as little billiard balls bouncing around like Newton would predict, and somehow getting extra energy to get past those hills. So sometimes people say the particles borrow energy.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why must heat supplied in the definition of entropy be reversible? Can't it be irreversible after all it is a state function? The definition of entropy contains the term $Q_\text{rev}$ which means the heat supplied or taken out reversibly. I thought yes it can be after all only the initial & final states are important as entropy is a state function irrespective of the process heat is transferred. However I was baffled when I first read Clausius' theorem where it is written that $dS \geq \dfrac{Q}{T}$. If $Q$ is transferred irreversibly, then $dS$ is greater than $\dfrac{Q_\text{irrev}}{T}$; if the heat transfer is reversible, then only $dS$ equals $\dfrac{Q_\text{rev}}{T}$. So, does that mean entropy depends on the process heat energy is transferred?? Then, how can it be a state function? Where am I mistaking? Please explain.
| Suppose you start with a system in some state $P_1, V_1, T_1$ and you add some quantity of heat $\Delta Q$ to it so the system changes to a different state $P_2, V_2, T_2$. The final state will depend on how you added the heat $\Delta Q$. Adding the heat $\Delta Q$ in a reversible process will result in different values for $P_2, V_2, T_2$ compared with adding the same amount of heat $\Delta Q$ in an irreversible process.
Entropy is indeed a state function, so if you know $P_2, V_2, T_2$ you can calculate the entropy change. Since reversible and irreversible processes will result in different values for $P_2, V_2, T_2$ they will also result in different values for the entropy change.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is double-slit experiment dependent on rate at which electrons are fired at slit? I am a mathematician and I am studying string theory. For this purpose I studied quantum theory. After reading Feynman's book in which he described the double-slit experiment (Young's experiment) I was wondering if I send one electron per day or per month (even more), could I see the interference pattern?
| Yes, the electron is discribed not by a path, like a macroscopic object, but by a wavefunction. And if an undisturbed electron (we better say an undisturbed wavefuction associated with an electron) goes through the slit it, just like a normal wave, interferes with itself, producing an interference pattern that will become visible if you only wait long enough.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Where does the $(\ell + x)^2\dot\theta^2$ term come from in the Lagrangian of a spring pendulum? I am reading some notes about Lagrangian mechanics. I don't understand equation 6.9, which gives the Lagrangian for a spring pendulum (a massive particle on one end a spring).
$$T = \frac{1}{2}m\Bigl(\dot{x}^2 + (\ell + x)^2\dot{\theta}^2\Bigr)\tag{6.9}$$
I don't understand where the component $(\ell + x)^2\dot{\theta}^2$ is coming from. If we say the $x$-component is radial and $y$ is tangential, so we have according to this $\vec{v}^2 = v_{x}^2 + v_{y}^2$, then $y = (\ell + x)\sin\theta$ by small angle approximation we have $y = (\ell + x)\theta$, but then if we choose this coordinate system then $V(x,y)$ equation doesn't make sense specifically the potential from gravity! If someone could shed some light into this that would be nice.
| Velocities in the kinetic part of Lagrangian
The variable $\;x\;$, that represents the displacement of the string from its position at rest, has been replaced by the variable $\;s\;$ in order not to be confused with the coordinate $\;x\;$ of a Cartesian system.
The velocity of the particle $\:\mathbf{v}\:$ is analysed as follows
\begin{equation}
\mathbf{v}=\mathbf{v}_{s}+\mathbf{v}_{\theta}
\tag{01}
\end{equation}
where $\:\mathbf{v}_{s}\:$ the component along the string line and $\:\mathbf{v}_{\theta}\:$ that normal to it.
Now,
\begin{equation}
v_{s}=\dfrac{d\left(\ell+s\right)}{dt}=\underbrace{\dot\ell}_{=0}+\dot{s}=\dot{s}
\tag{02}
\end{equation}
\begin{equation}
v_{\theta}=\left(\ell+s\right)\omega =\left(\ell+s\right) \dfrac{d\theta}{dt}=\left(\ell+s\right)\dot{\theta}
\tag{03}
\end{equation}
\begin{equation}
v^{2}=v_{s}^2 + v_{\theta}^2=\dot{s}^2 + (\ell + s)^2\dot{\theta}^2
\tag{04}
\end{equation}
| {
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Is Veritasium's "Shadow Illusion" caused by Image formation or Diffraction? I was watching this video on the YouTube channel Veritasium. In this episode, the host shows people paper containing holes of different shapes in the middle. So there is a paper which has a hole in the shape of a triangle, another one containing a square, a circle and so on.
Every time you place a cardboard against the sun and observe the shadow, a circle is obtained,at the centre of the shadow, regardless of the shape of the hole. The host explains that this is the image of the sun that we are seeing on the shadow.
I find this difficult to comprehend.
My question is:
Aren't 'lenses' required to converge the rays to make an image? How can a hole in the centre of a cardboard form 'images'. When the rays from the sun reach the cardboard they are going to be parallel. How can a hole in the center of a paper form an image from parallel rays? Is the host correct? Is this caused due to diffraction, where in the light bends around the hole, blurring the edges and hence forming a circle regardless of the shape of the hole?
PS: I am sorry, I couldn't summarise the question into a nice title.
| It's a pinhole camera image of the sun - as DJohnM's comment said.
My question is: Aren't 'lenses' required to converge the rays to make
an image? How can a hole in the centre of a cardboard form 'images'.
No - all that is required is an aperture (hole) to restrict the range of rays that reach the screen to form an image. All a lens does is allow a wider bundle of rays to be used to make the image = more light. In terms of making the image it is the hole formed by the center of the lens that is important.
| {
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Why under red light the additive synthesis of primary colors gives white where one primary color is absent? Under White Light
Under Red Light
For instance, red & green under red light gives white.
| To add to WhatRoughBeast's answer, this is because you are not really viewing the image under pure red light. The "red light" you used to create the image has small amounts of green and blue wavelengths, causing the white areas to appear more "white" than the red areas, even though they are both just slightly different shades of red. To quote WhatRoughBeast, I suspect that if you were to use really high-quality monochromatic red light, such as a collimated laser beam, and you made sure that there was no other source of light available, your "white" would go away.
To simulate this, I took the original image and removed the green and blue channels. (Just a note, I made some small modifications to the original image, such as white-balancing and small saturation changes to make this affect more apparent.) The only difference between the first image and the second image is that the Blue and Green Channels were removed - no additional retouching. As you can see, the red and white areas become identical shades of red.
(I noticed these images have slightly different appearances on different monitors. On one of my monitors, you can still see a slight difference between the "red" and the "white" portions, but on my other monitor, they are identical. In an ideal world, they would be identical)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does anti-matter increase or decrease in entropy over time? Antimatter is matter going backwards through time. From the perspective of a matter-based observer does antimatter:
*
*Increase in entropy (and therefore decrease in entropy in its own time)
OR
*
*Decrease in entropy (and therefore increase in entropy in its own time)
Option 2 would seem to explain why we don't see much anti-matter (it all went into energy.)
Which is correct? Does antimatter increase or decrease in entropy over time?
| Antimatter increase in entropy over time. We can verify this with a thought experiment. Take ten positrons. Put five in one side of a chamber with a barrier and then the other 5 on the other side of the barrier in the same chamber. The chamber and barrier are also made of antimatter. The positrons repel each other and so each have a certain amount of kinetic energy due to changes in their potential energy. Now, remove the barrier and see if they tend to randomly assort themselves throughout the chamber. Because they are more likely to be found in disordered arrangements over time we know that the entropy of this system is increasing with time.
What we wouldn't observe, is all of the positrons moving back to where they started and not mixing at all. Not that it's impossible, just extremely unlikely.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The Makeup of the Pentaquark Why is it that when they have the artist's rendition of the Pentaquark it shows two downs, two ups, and one anti-strange quark? Is this or is this just for show? Follow up to this question: if this configuration is just for show, what is the Pentaquark truly made of?
| The recent high-energy experiments have found the true pentaquark model of the proton: four quarks and an anti-quark. In fact these quarks may be highly-energetic electrons. Proton is pentaquark.
https://www.researchgate.net/publication/340741231_The_Geometry_of_the_Proton_and_the_Tetryen_Shape
https://vixra.org/abs/1708.0146
| {
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Is the event horizon of black holes visibly sharp, or blurry? As you come in closer to a black hole, how do you see the event horizon? Is it always like a clear-cut surface? Or it only looks clear-cut from a distance, but as you come closer to the black hole, you start seeing it's a blurry layer, and everything around you gets gradually darker as you fall through?
I know that the Schwarzschild radius would define an exact (clear-cut) sphere around the singularity (the surface of no return), but what actually happens to light itself? Anyway, my question might be lacking in imagination, since light gets bent badly already outside the event horizon.
| Calculating what you would see as you fell into a black hole is straightforward but tedious. Fortunately there are lots of sites that have done this for you. Actually, if you've been to the cinema recently the film Interstellar does a pretty good job of it.
Less spectacularly, have a look at this site that has videos of what the journey would look like. There is a sharp cutoff between the light and dark areas, though the bending of the light means the cutoff isn't simply the edge of the black hole.
| {
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"url": "https://physics.stackexchange.com/questions/195427",
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Engine fuel consumption vs power Something that has bothered me for a long time is why a 600 hp engine uses more fuel per kilometer than a 80 hp engine. Let pretend I have two equal cars (same shape same weight etc) except for the engine, one is 600 hp and other is 100 hp. If I manage the throttle to accelerate both with the same acceleration, my knowledge says that will require the same energy to get to a certain speed, so why will the 600 hp will use more fuel/energy?
Also if I am right is it possible that in the future cars will be so efficient that a big engine will only consume more if pushed to the limit?
| It’s all about engine's efficiency. According to wikipedia:
gasoline engine's efficiency = 1/(BSFC × 0.0122225)
(Also Actual efficiency can be lower or higher than the engine’s average due to varying operating conditions.)
To calculate BSFC (Brake specific fuel consumption) use the formula: BSFC=r/P (where: r is the fuel consumption rate in grams per second (g/s) and P is the power produced in watts.)
And after merging 2 formula we have:
r (fuel consumption)=BSFC×P=P/( gasoline engine's efficiency× 0.0122225)
So for those two car ( car A with 600hp power and car B with 80hp power) if we consider:
r(A)=r(B) --> engine's efficiency(A)=7.5×engine's efficiency(B)
So for each engine with more power you should have more engine's efficiency in order to fuel consumption be at fixed level.
Now real example:
Car A: Audi 2.5 litre TDI (1990) power=88kW efficiency=42.5 --> fuel consumption= 198 (g/kW·h)
Car B: Volkswagen 3.3 V8 TDI (2000) power=165kW efficiency=41.1 --> fuel consumption=205
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Force division of moving pulleys? I am a second grade at a middle school and I was reading a physics workbook to prepare for a test. And I was solving pulley problems and one problem made my brain stop.
The problem asked me what would the minimum force of F would be when the weight of the pulleys were 30N. I checked the answers and the way to solve it. The workbook told me that the force on each string holding the moving pulley equals to 1/n. (n = the numbers of string holding the pulley) Why is it true?
| So within a cord/string there is a property called tension which is a measure of the force exerted along the string. If the string stretches homogeneously (the same at all parts of the string) then it turns out that this tension is the same at all parts of the string: you pull with force $m$ Newtons, then everywhere you see the string you need to think of it as a force of $m$ Newtons.
Pulleys, because they "roll" perfectly, allow the tension to come to the same value between the two sides of the string. (In other words, if there is a tension imbalance, then it will pull the rope in one direction, so that the rope will just roll along the wheel of the pulley from low to high tension. This will stretch out the low-tension side and relax the high-tension side, bringing the tension difference closer to 0, until the tension difference is 0 and they're both the same tension.)
Because of this, it's as simple as looking at the big wheel that's suspended in midair and doing a force-balance on that. It is being pulled downward by this weight 450 N and its own gravity 30 N; it is being pulled upward by $3 T$ where $T$ is the force of tension within the string; and $T = F$ according to the diagram, so $3 T = 480\text{ N}$ or $F = T = 160\text{ N}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is Gauss' law valid for time-dependent electric fields? The Maxwell's equation $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$ is derived from the Gauss law in electrostatics (which is in turn derived from Coulomb's law). Therefore, $\textbf{E}$ must be an electrostatic field i.e., time-independent. Then how is this equation valid for the electric field $\textbf{E}(\textbf{r},t)$ which is time-dependent (for example, the electric field of an electromagnetic wave)? Can we prove that $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ ?
EDIT: I have changed $\boldsymbol{\nabla}\cdot \textbf{E}=0$ to $\boldsymbol{\nabla}\cdot \textbf{E}=\frac{\rho}{\epsilon_0}$ in the question.
| You need to watch what you mean by the ambiguous term "derive", which can mean either "was derived historically" (i.e. was motivated by or is a derivative of, in the non-mathematical sense) or "is derived logically/mathematically".
Historically, I think you are correct that $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ was "derived" by Maxwell from the electrostatic version $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$, which in turn was "derived" from Coulomb's law.
Logically, it's the other way around. $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ is a fundamental law of the universe (at least it is in classical electromagnetism; in reality it is "derived" from quantum electrodynamics). The electrostatic version of it can be "derived" mathematically from this law as a special case. The same is true of Coulomb's law.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Why do semiconductors remain neutral outside the depletion region? Why there is a sharp cut off of the charged region outside the depletion region, like on this image?
For example why don't electrons on the conduction band in the n-type side rush towards the positively charged area making the whole piece positively charged somewhat, not just at the area near the depletion region?
The source of the confusion is that I know if you charge up a regular conductors the internal currents will uniformly distribute the charge along the whole piece, while insulators are only locally charged up, since they cannot carry current.
Semiconductors here seem to act like insulators, but diodes do carry current when used. How?
| From your response to @JonCuster:
The question is. when the pieces are in equilibrium why don't the remaining charge carriers distribute themselves uniformly in each half.
The remaining charge carriers within the depletion region are not electrons but semiconductor ions (both positive and negative) locked into their positions within the crystal so they can't move. If the lattice ions could move, then they would redistribute themselves to eliminate or minimize the electric field across the junction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/195928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Where does the equation $p=\frac{1}{c}\sqrt{T^2 +2mTc^2}$ come from? Where does the relativistic formula
$$p~=~\frac{1}{c}\sqrt{T^2 +2mTc^2}$$
come from? What is the derivation from Einstein's formula? $T$ is the kinetic energy $m$ is the mass $p$ is the momentum.
| Hint: $T = E - E_0 = m\gamma c^2 - mc^2 = mc^2(\gamma -1)$ and $p = |\vec p| = m\gamma |\vec v| = m\gamma v$
| {
"language": "en",
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