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Does a planar object balance on a unique point? Consider a horizontal planar convex 2D object (say lying on x-y plane) with uniform density. Under constant gravitational force (say in -z direction), does it always balance on a unique point lying on the object (i.e. sum of the torques vanishes with respect to a unique point)?
I guess the answer is yes and if so then I want to conclude that the point must be the object's center of mass (that is the object will balance in any orientation w.r.t. that point), assuming the uniqueness of center of mass.
Possibly the question is trivial but I have been confused over this for some time. Any comments or answer will be appreciated.
EDIT: Added the assumption of convexity, as otherwise the point of balance may not lie on the object.
Let me add that, one may assume existance of the point. I am more interested in showing that there can not be two or more points of balance.
| If there is a point in the planar object from which it can be balanced, the net torque is zero from this point. Choosing the point as origin:
$0 = \tau = \int_v \mathbf r \times d\mathbf F$
Considering the object in the $xy$ plane, the weight in the direction $-z$, density and $g$ constants, and its thickness = $t$; the position vector $\mathbf r = (x , y , 0)$, and $|d\mathbf F| = \rho g t dxdy$
$\rho g t \int_S (x , y , 0) \times (0 , 0 , -1)dxdy = 0$
Solving the cross product:
$\int_S (-y , x , 0)dxdy = 0$
For the vector resulting from the integral limited by the boundary be zero, all its components must be zero:
$\int_S -y dxdy = 0$ and $\int_S x dxdy = 0$
If it is true for the point , by the same argument (that all components must be zero), it is also true:
$\rho t \int_S (x , y , 0) dxdy = \int_v \mathbf r \rho dv = 0$
The above is the definition of center of mass. So, the balance point is the COM.
| {
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$P=\epsilon_o \chi E$ or $\epsilon_o \chi E_o$ Suppose the polarisation inside a dielectric is given by $P$, then is it related to the electric field as $\vec{P}=\epsilon_o \chi \vec{E}$ where $E$ is the field inside the dielectric or is is $E$ the original field that would have been present in that region in absence of the dielectric?
| It is a matter of definition. The usual definition of $\chi$ implies the electric field $\bf E$ actually present inside the dielectric.
However, one has to notice that in the special case of a dielectric completely filling a large parallel-plates capacitor at a fixed difference of potential between the plates, the resulting field $\bf E$ is the same present in the vacuum when the same difference of potential exists. In this way, it is possible to control experimentally the field $\bf E$, which in general depends on the geometry of the sample. Moreover, since $\chi$ in the usual linear regime is not depending on the field, its value can be extracted in this favorable set-up, remaining the same for more complicated geometries of the external field and of the sample. Therefore, if one is referring to this special situation it may happen that the field $\bf E$ could be called the external field $\bf E_0$, although the definition contains the field $\bf E$.
| {
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How can we discern so many different simultaneous sounds, when we can only hear one frequency at a time? As I understand it, the eardrum works like any other kind of speaker in that it has a diaphragm which vibrates to encode incoming motion into something the inner ear translate to sound. It's just a drum that moves back and forth, so it can only move at one rate or frequency at any given time.
But humans have very discerning ears and can simultaneously tell what instruments are playing at the same time in a song, what the notes in the chord of one of those instruments is, even the background noise from the radiator. All of this we can pick apart at the same time despite that all of these things are making different frequencies.
I know that all of these vibrations in the air get added up in a Fourier Series and that is what the ear receives, one wave that is a combination of all of these different waves. But that still means the ear is only moving at one frequency at any given time and, in my mind, that suggests that we should only be able to hear one sound at any given time, and most of the time it would sound like some garbled square wave of 30 different frequencies.
How can we hear all these different frequencies when we can only sense one frequency?
|
But that still means the ear is only moving at one frequency at any given time
No, it doesn't mean that at all.
It means the eardrum is moving with a waveform that is a superposition of all the frequencies in the sound-wave it is receiving.
Then, within the inner ear, hair cells detect the different frequencies separately. It is entirely possible for several hair cells to be stimulated simultaneously so that you hear several frequencies at the same time.
| {
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Determining the curvature by symmetries of the metric Given the Kahn-Penrose metric:
$$
ds^2=2dudv-(1-u)^2dx^2-(1+u)^2dy^2
$$
I calculated the Riemann Tensor and found that all elements equal 0.
Is there some symmetry principle by which I could have easily deduced zero curvature or other properties directly from the metric without actually doing the calculation?
EDIT: I would be grateful for an explanation about the connection between the curvature and symmetries of any metric, not just the Kahn-Penrose example.
| One way to see that the curvature tensor is zero is to start with the fact that there is no dependence of metric components on null coordinate $v$. So performing Kaluza–Klein reduction along the Killing vector $\partial_v$ we would obtain a Newton–Cartan spacetime with two spatial coordinates ($x$ and $y$), while $u$ now becomes Galilean time coordinate.
The time evolution of spatial metric corresponds to the following law of motion for comoving observers:
$$
X(u)=(1-u)\, X_0,\qquad Y(u)=(1+u)\,Y_0.
$$
But these are equations for linear motion along a straight line. So there is no Newtonian gravity here and $(u,x,y)$-spacetime is just a reparametrization of an empty Galilean spacetime with zero NC curvature and correspondingly the original Kahn–Penrose metric (which would be the Bargmann lift of NC solution) would also have zero Riemann tensor.
| {
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Do photons "lose energy" when they are absorbed? Recently in my biology class I learned about an experiment in which isolated and illuminated chlorophyll pigments fluoresce in the red part of the spectrum, but also, the solution of the pigments gets hotter. Are the photons that are reflected as the electrons fall back to their ground states lower in energy than the photons absorbed?
| Strictly speaking: when a photon is absorbed, it ceases to exist, so it doesn't make sense to ask whether it loses its energy.
In the experiment that you describe the green light photons are absorbed, and red photons (having lower energy) are emitted as fluorescence. The likely reason for this is that the energy of an absorbed photon is distributed between the electron and other degrees of freedom of the molecule (e.g., its internal vibrations or the overall kinetic energy - these are expressed as heating of the material). So, when the electron relaxes to its ground state, it reemits as a photon only the part of the energy of the original green photons.
| {
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Can sound be used for propulsion? I'm no physicist so this might seem absurd.
I Remember watching a cartoon as a kid where the character uses a powerful speaker to propel his cart and I was wondering if this was actually possible.
Being a highschooler I am aware to propel forward you shoot something backward.
So maybe in the case of a speaker it could "shoot out" sound waves?
| This is a difficult question, even for people who are physicists. I say this after reading:
https://www.physics.princeton.edu/~mcdonald/examples/hidden_sound.pdf
which discusses the concept of hidden momentum, a concept I've never heard of. Moreover, based on that article, it appears experts argue whether it does or does not exists in various circumstances.
The general idea is that sound waves move energy (density), $u$, at the speed of sound ($v$) in some direction, and that leads to an energy flux:
$$ \vec S = u\vec v $$
Following $E=mc^2$, we can turn that energy flux into a momentum:
$$ \vec p = \frac{\vec S}{c^2} = \frac{uv}{c^2} $$
which is tiny. It is in fact a relativistic effect, and is so tiny, that the author writes:
"It appears that the fluid-dynamics community generally considers that the laws of relativity are not relevant to this branch of physics, and that the momentum density can and should be ignored"
So the answer is "yes" if you're pedantic, and "no" if you are practical.
| {
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Polarity in a magnetized Möbius strip When a flat iron or Alnico washer is magnetized one of the faces develops a north polarity and the other, south. The geometric shape here is simple.
However, when a standard Möbius strip (or one of given thickness, radii of curvature and torsion of edges) is magnetized, which regions develop a north polarity and which regions south and according to which geometrical or other mass distribution criterion/law?
I am curious to know because such a Möbius strip (of rectangular section) has only one surface and only one edge.
Is magnetic polarity and strength distribution after magnetization influenced by changed geometry ( by homeomorphism ) ?
It may be easy to make a flattened thin Möbius strip looking like a recycling symbol to apply a magnetizing current. Thanks in advance for references.
| The answer is easily obtained by remembering what creates the permanent magnetic field. All subatomic particles are magnetic dipoles and in some materials - natural or man-made - subatomic particles are aligned by their magnetic dipoles in such a way that a stable macroscopic magnetic field is present.
The stability of permanent magnets is a relative thing. An external magnetic field can destroy or "remap" the magnetic field of a permanent magnet. And the same thing happens to the areas of a magnetised flat material when it is joined to a Möbius strip. Initially, the electrons at the junction surface are reoriented by their magnetic dipoles, but since a Möbius strip has no beginning and no end, the reordering should occur over the entire strip.
Theoretically, therefore, the magnetic dipoles should all be oriented along the stripe. In practice, however, the outer electrons of the atoms are part of the magnetic interactions with the other electrons and the nucleus (and also intermolecular interactions) and it would be an interesting area of research - if not done so far - to see what happens in reality in different materials and at different temperatures.
| {
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What happens to the rest of the 95 percent in quarks? Quarks are bound by gluons. Gluons have a mass of 0, while mass of quarks is only 5%.
Where is the missing 95%?
| In the level of quarks and gluons one is in the realm of quantum mechanics and special relativity. Special relativity assigns to each particle a four vector , whose "length" is the invariant mass of the particle .
The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.
Thus all particles are of fixed mass, but mass is not an additive quantity, in the same way that in three vectors the length of the new vector from the addition of two vectors is variable. Thus the rest of the mass of the proton comes from the added four vectors of the multitude of quarks antiquarks and gluons. See this.
This is a rough argument, as quantum mechanics has to be involved,
, and the the particles in the picture are off mass shell, and one needs a theory to get the bound proton, as lattice QCD, but the four vector algebra is true and gives a feeling for how although zero mass and low mass particles can compose the proton.
| {
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Volumetric Dilatation Rate, Material derivatives, and Divergence in class we derived the following relationship:
$$\frac{1}{V}\frac{dV}{dt}= \nabla \cdot \vec{v}$$
This was derived though the analysis of linear deformation for a fluid-volume, where:
$$dV = dV_x +dV_y + dV_z$$
I understood the derived relation as:
$$\frac{1}{V}V'(t) = \nabla \cdot \vec{v}$$
However, my professor recently told me that the $d/dt$ operator before V, stood for the material derivative and not the common derivative. I am very confused as to how is that the case, given that we did an infinitesimal analysis of linear deformation, in a way I could call analogous to any other infinitesimal analysis that results in the common derivative.
I also tried deriving the equation by taking the material derivative of $V$, and dividing by $V$:
$$ \frac{1}{V}\frac{DV}{Dt} = \frac{1}{V}\frac{\partial V}{\partial t} + \frac{1}{V}(\vec{v} \cdot gradV)$$
but I was unable to.
| The continuity equation reads $$\frac{\partial \rho}{\partial t}+v\centerdot \nabla \rho+\rho \nabla \centerdot v=0$$where $\rho$ is the fluid density. Dividing this by $\rho $ gives $$\frac{1}{\rho}\left(\frac{\partial \rho}{\partial t}+v\centerdot \nabla \rho\right)+\nabla \centerdot v=0$$But, since the density is the inverse of the specific volume V, we have $$\frac{1}{V}\left(\frac{\partial V}{\partial t}+v\centerdot \nabla V\right)=\frac{DV}{Dt}=\nabla \centerdot v$$
| {
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If a jet engine is bolted to the equator, does the Earth speed up? If a jet engine is bolted to the equator near ground level and run with the exhaust pointing west, does the earth speed up, albeit imperceptibly? Or does the Earth's atmosphere absorb the energy of the exhaust, and transfer it back to the ground, canceling any effect?
| Yes, but it is so minuscule that it wouldn't be noticed. It's the same as if your friend was driving a car and you stick your head out of the window and blew in the opposite direction of the car's movement
| {
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If the sea surface were absolutely calm should the Sun reflection be the area of a circle instead a ribbon? Although waves produced on the sea can cause different points of the sea surface to reflect sunlight towards the same observer, how is that kind of ribbon image produced? Why isn't the reflection stretched also perpendicularly to the line of sight by wavyness of the sea?
| Yes, the image of the sun or moon in a perfectly flat lake or ocean would be round. The "ribbon" is due to ripples in the surface of the water. To understand why it's a ribbon instead of a broad expanse of reflected light, all you need to do is shine a flashlight, at the top section of a reflective ball, move the laser in a small circle parallel to itself, and see where the reflected light goes. It's spread into a beam that lies in a vertical plane. The "parallel to itself" requirement is to make sure you're simulating a collimated beam. You can accomplish the same thing by forming a collimated beam using a telescope.
| {
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Third-order Feynman diagrams of 2-point function in $\phi^4$-theory $\newcommand{\Braket}[1]{\left<\Omega|#1|\Omega\right>}$
Hello,
I am currently studying QFT and have a problem concerning the 2-point correlation function in $\phi^4$-theory. When I draw all the Feynman diagrams contributing to $\left<\Omega|\phi(x)\phi(y)|\Omega\right>$ (so without the vacuum bubbles) up to $O(\lambda^3)$, I get the following:
However, I am not sure about the ones in the curly brackets. Are those the same diagrams? If so, how does the symmetry factor account for them?
I tried to find an answer to this, but nobody draws the third order diagrams.
| If you denote the external points by $x,y$ and the internal points by $1,2,3$ than the diagrams of the last line in your picture are $ x1, 11,12,23,23,23,3y$ and $x1,12,12,12,23,33,3y$ with integration over the internal points and what is between brackets is contracted. Yes, they are the same.
| {
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What is the propagation direction of plane wave? As far I know, plane wave equation is given by:
$$\vec{E}(\vec{r},t)=\vec{E_0} \cdot e^{i(\vec{k}\cdot \vec{r}-\omega t)} \hspace{2cm} \tag{1}$$
In some textbook propagation direction of $(1)$ is given as $\vec{k}$;
while in some other books, i found propagation direction of $(1)$ as $-\vec{k}$
so, can anyone tell me what is the propagation direction of $(1)$ and how can we determine it from $(1)$?
| $\vec E(\vec r,t)$ is at a peak when $e^{i(\vec k \cdot \vec r-\omega t)} = 1$, and a minimum when it is $-1$. Maxima happen when $i(\vec k \cdot \vec r-\omega t) = 0$ or $2 \pi i$. It is at a minimum when the exponent is $\pi i$.
One peak is found where $\vec k \cdot \vec r =\omega t$. As t gets larger, the peak will move to where $\vec k \cdot \vec r$ is larger. To make the dot product larger, choose an $\vec r$ farther in the direction that $\vec k$ points.
That is to say, the wave will propagate in the direction that $\vec k$ points.
| {
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Hypothetically, why can't we wrap copper wires around car axles and turn them into electromagnets to help charge the batteries? We already have a magnetic core, why can't we use it to recharge the batteries? The only problems I see with it are potentially wiping magnetic data, but doesn't the electromagnet have to be revolving around the damageable device?
| This is basically what happens in the alternator. The car's engine, which turns the wheels, also turns the alternator's rotor. The magnetic rotor is surrounded by coils of wire, and induces a current that charges the battery. It is important to note, though, that this does take energy from the engine: nothing is for free.
| {
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In QFT why does the degree of the interaction terms in Lagrangian start from 3? I'm new to QFT so it's not obvious to me why there is no quadratic interaction terms in Lagrangians.
For example, the Lagrangian for a real scalar field is
$$L=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-\frac{1}{2}m^2\phi^2-\sum_{n\geq 3}\frac{\lambda_n}{n!}\phi^n.$$
What's the reason that we can't add terms like $g\phi^2$ to the free field Lagrangian?
| Contrary to what the other answer's claim, we do add these $g\phi^2$ terms as interaction term.
Check any QFT text book, specifically the section on mass renormalization and field renormalization counter terms.
| {
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Are all atoms spherically symmetric? If so, why are atoms with half-filled/filled sub-shells often quoted as 'especially' spherically symmetric? In my atomic physics notes they say
In general, filled sub-shells are spherically symmetric and set up, to a good approximation, a central field.
However sources such as here say that, even for the case of a single electron in Hydrogen excited to the 2p sub-shell, the electron is really in a spherically symmetric superposition $\frac{1}{\sqrt3}[2p_x+2p_y+2p_z]$ (which I thought made sense since there should be no preferred direction in space).
My question there now is, why is the central field approximation only an approximation if all atoms are really perfectly spherically symmetric, and why are filled/half-filled sub-shells 'especially spherically symmetric?
| Your quote references "filled sub-shells". You then write in your question "all atoms are really perfectly spherically symmetric".
Not all atoms have only filled sub-shells. Noble gasses are strong examples of atoms with only filled sub-shells and the small ones behave approximately spherically symmetrically.
Most atoms do not have all of their sub-shells filled. It is rather common for the "last" one or two sub-shells to be partially filled. Look at the electron configuration of the transition metals for many, many examples. As one such example, Chromium's configuration ends with $\mathrm{[Ar] 3d^5 4s^1}$, so neither of the last two sub-shells is filled and we should not expect (and do not find) that Chromium is spherically symmetric.
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Question regarding Lorentz Transformation and Space Contraction- Contradiction I stumbled upon this question regarding Special relativity- and have seemed to reach a contradiction.
I am trying to find the distance that the ball travels
I am obviously not looking for the numerical answer, but I'm trying to understand what should be my intuition when looking at these types of problems.
The train is moving at a velocity of $c/2$ relative to earth, while the ball is moving at a velocity of $c/3$ relative to the train.
The train has a proper length of $L_0$.
Now, when trying to find the distance that the ball travels regarding earth, I see two approaches:
*
*If we set $t=t'=0$ as the time where the ball is at the back if the train ($x=x'=0$), we find that the time he travels in the train's frame of reference is $3L_0/c$ and the distance is $L_0$. Using Lorentz transformation we find the distance traveled is $5L_0/\sqrt{3}$ in the earth frame of reference.
*In the train's frame, the ball is moving at a speed of $c/3$ a distance of $L_0$. Using the proper length we are able to calculate the train's length in the earth's frame as $\sqrt{3}L_0/2$. Seeing as the ball starts at the back of the train (in both frames) and reaches the front of the train, this is the balls distance traveled.
What am I missing here? Which approach should I use and where does this contradiction come from?
Thanks a lot in advance!
| You would end up in the same contradiction (wich is not) in Newtonian mechanics. Position is relative in both newtonian mechanics and Special Relativity. If i am moving relative to you you would see me covering a distance, but i would see myself standing still covering no distance.
| {
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Luttinger Liquid Parameter Physical Meaning - Attractive or Repulsive For 1+1D systems at long wavelength, it is known that the (Tomonaga)-Luttinger Liquid can be rewritten in terms of bosonic parameters, where the Hamiltonian densities can be written
\begin{align*}
H =& \left[\psi_L i\partial_x \psi_L - \psi_R i \partial_x \psi_R + 2\pi g_2 \rho_R \rho_L + \pi g_4 (\rho_R^2 + \rho_L^2) \right]\\
\rightarrow H_\text{boson} =& \frac{v}{2}\left[\frac{1}{g}\Pi^2 + g(\partial_x \phi)^2\right]
\end{align*}
where $\Pi = \partial_x \theta$ is the momentum conjugate to $\phi$.
I have been confused about the relationship of $g$ to the parameters $g_2, g_4$, where either
\begin{align*}
g = \sqrt{\frac{1 + g_4 \pm g_2}{1+ g_4 \mp g_2}}
\end{align*}
In particular, for repulsive interactions (i.e. $g_2,g_4 > 0$), the choice of sign determines whether or not $g>1$ or $g<1$. For instance, in the seminal work by Kane and Fisher, they write that $g>1$ for attractive interactions, while in the lecture notes by Senechal and lecture notes by Fradkin, they write the opposite.
Which one is correct, or am I just missing something silly here? Perhaps a more physical question is if there can be an unambiguous physical meaning assigned to g?
| Unfortunately the literature on Luttinger liquid theory and bosonization suffers from a lack of consistent notation. This is just one of many inconsistencies you will find between different authors. Generally any single author will (hopefully) be self consistent, but comparing between different authors can require quite some care and effort. The situation is so bad that I know of at least 2 authors (Giamarchi's textbook Quantum Physics in One Dimension and this by von Delft and Schoeller) which have sections explaining how to convert their notation into the notation found in a number of other notable works. (Not the same notable works naturally).
So to answer your question both are correct and you just need to pick a convention and stick to in in your own work, whilst being very careful when comparing to others.
In terms of the physical interpretation, $g$ can be directly related to the compressibility of the system. Obviously I cannot say the exact nature of this relationship without going through exactly how $g$ and the bosonic fields have been defined, but it should be either directly proportional or inversely proportional (notice that your 2 sign conventions for $g$ amount to swapping $g$ with $\frac{1}{g}$)
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How warm are radioactive metals? I read that radium is warm to the touch -- is that because of actual heat or is that because, for example, the radiation it emits creates the sensation of warmth? How high of a temperature can a radioactive element or isotope actually have?
| Yes radioactive materials are warmer than they would otherwise be. There are two reasons for this.
One is that the decayed atom will have picked up some kinetic energy in recoil and that is heat.
Another is that some of the radiation does not make it outside but hits another atom and transfers its energy to it as more heat. That includes any low-energy photons emitted by the decayed nucleus as it stabilises.
But how hot it gets depends on a variety of factors, such as the half-life, the total energy released in each decay, the concentration or richness of radioactive atoms in the material, the size vs surface area of the lump (the bigger it is, the proportionately less area it has to lose excess heat), and any cooling or insulating effects in its immediate environment.
At one end of the scale, such as a small piece of uranium ore, the heat is not enough to be measurable above background noise. At the other extreme you get a meltdown.
Natural radium is almost entirely radium 226. Its refined form appears to represent a borderline case. The following calculation is off-the-cuff and ignores any impurities, radioactive in themselves or otherwise, somebody please correct any mistakes! Radium 226 has a half-life of 1,600 years. One mole (gramme molecule) weighs 226 g and contains 6 X 1023 atoms. There are 5 x 1010 seconds in 1,600 years, so in the mole's first second of existence over 1013 atoms will decay. Its decay energy is 4.9 MeV, yielding say 4.9 x 10^23 eV or around 10^4 joules, enough to raise the temperature of 226 g of water by 10 degrees or so. The molecular weight of water is 20, so the temperature of the radium will increase by 10 x 20/226 or around 1 deg. That assumes no heat outflow from the radium; its actual temperature rise after an hour will be significantly lower, say 0.5 deg. Note that this is a rise above ambient, it is not an absolute value. Given a human hand several degrees warmer, and you may find that material properties like thermal conductivity may distort your subjective sense of warmth to the touch.
| {
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Aharanov-Bohm Effect Gradient of Line Integral In Griffiths' Quantum Mechanics 2nd edition section 10.2.3 the phase
$$g(\mathbf{r}) = \frac{q}{\hbar}\int_{O}^{\mathbf{r}}\mathbf{A}(\mathbf{r}')\cdot d\mathbf{r}'$$
is defined. It is noted that this integral is only defined if $\nabla\times\mathbf{A} = 0$ throughout the region and the region is simply connected (noted in the 3rd edition). The gradient of this function is then found to be
$$\nabla g = \frac{q}{\hbar}\mathbf{A}.$$
If my understanding is correct this is due to the Fundamental Theorem of Line Integrals along with the above comments implying $\mathbf{A}$ is conservative. Hence,
$$\mathbf{A} = \nabla a$$
so that
$$\begin{split}\nabla g &= \nabla\left(\frac{q}{\hbar}\int_{O}^{\mathbf{r}}\mathbf{A}(\mathbf{r}')\cdot d\mathbf{r}'\right)\\ &= \frac{q}{\hbar}\nabla\left(a(\mathbf{r})-a(O)\right)\\ &= \nabla a(\mathbf{r})\\ &= \mathbf{A}.\end{split}$$
Assuming this is correct we come to my question. Griffiths introduces the Aharanov-Bohm Effect for a particle in a box, with $\mathbf{r}$ denoting its position and $\mathbf{R}$ that of the centre of the box. He then defines
$$g = \frac{q}{\hbar}\int_{\mathbf{R}}^{\mathbf{r}}\mathbf{A}(\mathbf{r}')\cdot d\mathbf{r}'$$
for this system. He then calculates
$$\nabla_{\mathbf{R}}\left(e^{ig}\right) = -i\frac{q}{\hbar}\mathbf{A}(\mathbf{R})e^{ig}.$$
I don't understand how this was found since the particle's position changes as $\mathbf{R}$ is moved so should also contribute to the derivative. Also, I feel that the conditions for $\mathbf{A}$ to be a conservative field aren't satisfied in this case. The region enclosed by the solenoid having non-zero curl.
Would someone be able to clarify these points for me?
| I think $\mathbf{r}$ is just an arbitrary position in space where the particle could be for the integral in the definition of $g$. So it doesn't depend on $\mathbf{R}$ in $g$. It is the potential $V$ used in other parts of the setup which confines the particle's position $\mathbf{r}$.
We can treat the space outside the solenoid as two simply connected parts: $$H_1 := \{r>a,\,0\leq\phi\leq \pi,\,-\infty\leq z\leq \infty\}\quad\text{ and }\quad H_2 := \{r>a,\,-\pi\leq\phi\leq 0,\,-\infty\leq z\leq \infty\}$$
specified in cylindrical coordinates, where in both $H_1$ and $H_2$ we have that $\mathbf{A}$ is curl free. So by the simply connectedness $\mathbf{A}=\nabla\varphi_1$ on $H_1$ for a potential field $\varphi_1$ on $H_1$ and $\mathbf{A}=\nabla\varphi_2$ on $H_2$ for a potential field $\varphi_2$ on $H_2$.
It follows by the same theorem that $\varphi_1(\mathbf{R}) = g(\mathbf{R})$ for $\mathbf{R}$ in $H_1$ and similarly for $\varphi_2$. To be completely precise there should really be a $g_1$ defined for $\mathbf{R}$ in $H_1$ and a $g_2$ defined for $\mathbf{R}$ in $H_1$. This is since $g$ is not well defined due to the non-simply connectedness of the space outside the solenoid.
Hence by the fundamental theorem of calculus for line integrals
$$\nabla_{\mathbf{R}}g(\mathbf{R}) = -i\frac{q}{\hbar}\mathbf{A}(\mathbf{R})$$
on both $H_1$ and $H_2$. Thus also everywhere outside the solenoid, which gives the result.
| {
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Meaning of adding a term to the Hamiltonian in a quantum harmonic oscillator Let $H$ be the Hamiltonian in a harmonic oscillator,
$$
H = \sum_{n=0}^{\infty} \hbar \omega \left (n+\frac{1}{2} \right ) |n\rangle \langle n|.
$$
Suppose we introduce the interaction
$$V = \sqrt{2} \hbar \omega
(|0\rangle \langle 1|+|1\rangle \langle 0|
$$
and the new Hamiltonian is $H+V$. How can we understand the physics under this interaction? Is there any classical interpretation?
| You can think of the initial Harmonic oscillator with the Hamiltonian $H=\sum_{n}\hbar\omega(n+1/2)|n\rangle\langle n|$ as a free system, that is under time evolution a state prepared in any one of the $|n\rangle$ remains in that state. However, the effect of adding an interaction term of the form that you proposed now allows for transitions among the states $|0\rangle$ and $|1\rangle$.
For example, consider the time evolution of an initial state $|0\rangle$ under the full new Hamiltonian $H'=H+V$.
$|\psi(t)\rangle = e^{-iH't/\hbar}|0\rangle \approx |0\rangle + (-\frac{it}{\hbar})H'|0\rangle$
where I have taken $t$ to be small. We see that $H'|0\rangle=\frac{\hbar\omega}{2}|0\rangle + \sqrt{2}\hbar\omega|1\rangle$ and so
$|\psi(t)\rangle \approx (1 - \frac{it\omega}{2})|0\rangle - it\sqrt{2}\omega|1\rangle$
So, we see that now the system has a finite probability of being in the state $|1\rangle$. This would not be possible when there was only $H$. For now, we just added the interaction term by hand, but such terms usually come into the picture when you couple your original system to another system. For example, if we couple the original Harmonic oscillator to say a laser, one could consider the following interaction.
$V = \sqrt{2}\hbar\omega ( a \otimes |1\rangle\langle0| + a^{\dagger} \otimes |0\rangle\langle1|)$
Where $a,a^{\dagger}$ are the photo creation and annihilation operators. So, when the system absorbs one photon, the system goes from $|0\rangle$ to $|1\rangle$ and the reverse when it emits a photon.
| {
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Why is energy lost here? Let's say a $1 \ \text{kg}$ block is moving.
With a speed of $1 \ \text{m/s}$ so its kinetic energy is $\frac{1}{2} \ \text{J}$. Now let's gently place a block of mass $3 \ \text{kg}$. Now as linear momentum is conserved due to lack of external forces on the system the blocks move together with velocity $1/4 \ \text{m/s}$ but the energy is now $\frac{1}{8} \ \text{J}$ which is lesser than it used to be.
Where has the energy gone?
| Energy is lost due to work done by friction . Try to analyze each block individually.
| {
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Non-Analytic Equations and Chaos Could anyone please tell me an example of an equation with no analytic solution(s) that is not a chaotic one? And what is the physical meaning of having analytic solution? For instance, the three body problem does not have in general analytic solution and it leads to chaos. But I don't know if this is a general statement. I have absolutely no idea. Could anyone explain me, please?
|
Could anyone please tell me an example of an equation with no analytic solution(s) that is not a chaotic one ??
A simple fifth order polynomial ($k_5 x^5 + k_4 x^4 + k_3 x^3 + k_2 x^2 + k_1 x + k_0 = 0$) has no analytic solution, but is not chaotic.
And what is the physicall meaning of having analytic solution ??
There is no physical meaning. Nature doesn’t care if we have nice functions to describe something.
| {
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Noether's Theorem and Liouville's Theorem Liouville's theorem states that for Hamiltonian systems the phase space volume $V(t)$ is a conserved quantity, i.e., $\frac{d}{dt}V(t)=0$. This is related to the fact that trajectories in phase space do not cross and a point in phase space has a unique time evolution.
Noether's theorem tells us that conserved quantities correspond to continuous symmetries/cyclic coordinates, and vice versa.
My question is: what is the continuous symmetry/cyclic coordinate corresponding to the conservation of phase space volume?
| There are several versions of Liouville's theorem. One version states that a Hamiltonian vector field (HVF) $X_H=\{H,\cdot\}_{PB}$ on a symplectic manifold $(M,\omega)$ is divergence-free
$$ {\rm div} X_{H}~=~0.$$
One may view the above HVF as the underlying symmetry.
| {
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Is projectile motion an approximation? Doesn't the acceleration vector points towards the center of the Earth and not just downwards along an axis vector. I know that the acceleration vector's essentially acting downwards for small vertical and horizontal displacements but if the parametrization of projectile motion doesn't trace out a parabola, what is the shape of projectile motion?
| It is an approximation, as everything in physics is an approximation based on mathematical/statistical modeling. And by definition, a model is an imperfect representation of reality. Models employ simplifying assumptions in order to make problems tractable so we can explain what is happening and hopefully make predictions. How good your explanation/predictions are depends on 1)how good the model is and 2)how close your situation is to the assumptions made by the model.
In the case of projectile motion, the standard equations work quite well if you throw a small dense object inside a windless room. It's a much different story if you throw a light big object during a windy day. But no matter what, the model will never exactly explain what is going on because reality never exactly matches the assumptions of a model.
Paraphrasing a quote often attributed to the statistician George Box "all models are wrong, some are useful". For projectile motion, you lose accuracy because the earth is not a perfect sphere with uniform gravitation, there's no such thing as a perfect vacuum , etc etc etc. Because of these things there is no exact shape that describes such motion, much like there is no shape that exactly describes orbital motion. Parabolas and ellipses are very close, but as you suspected they will always be approximations.
| {
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Will liquid nitrogen evaporate if left in an unopened container? SOS! I left work today and got a horrible feeling that I forgot to put the lid back on a large container of liquid nitrogen which contains many racks of frozen cells in it. If this did happen, how long would it take liquid nitrogen to evaporate? Does it start to evaporate as soon as it is exposed to oxygen? Will all the liquid nitrogen be gone from the container when I go back tomorrow?
| This is because the temperature at which liquid nitrogen changes phase to gas is below room temperature. I suppose you could apply Newton's law of cooling differentially to the surfaces of the canister. The open end is exposed to the atmosphere, and you can imagine volume elements that cool before the liquid nitrogen at the bottom of the open canister.
So basically you might see it evaporate slowly from the top.
| {
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Is it even theoretically possible for a perfect clock to exist? I have heard that even atomic clocks lose a second every billion years or so. That raises the question, is it even theoretically possible for a perfect clock to exist, one that never gains or loses time?
| Well, there is no concept of absolute time or a perfect ' tick tock ' in the universe. Phenomena happen at their own rate.
You can't quantify their 'perfection'. You can quantify the errors you made while measuring their physical aspects.
| {
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How long does it take an electron to emit (or absorb) a photon? A photon is emitted (or absorbed) by a transitioning electron. How fast is this process?
| The emission process always takes some time. How much depends on the kind of transition.
If the transition is spontaneous dipole transition, like when excited electronic state of an atom decays, the rate of spontaneous transition as found using quantum electrodynamics is
$$
\Gamma = \frac{\omega_{12}^3|\mu_{12}|^2}{3\pi\varepsilon_{0}\hbar c^3}
$$
where $\omega_{12}$ is emission frequency ($(E_1-E_2)/\hbar$, $\mu_{12}$ is transition dipole matrix element magnitude $|\langle 1 |\boldsymbol{\mu}|2\rangle|$. So mean time needed for one transition is
$$
1/\Gamma = \frac{3\pi\varepsilon_{0}\hbar c^3}{\omega_{12}^3|\mu_{12}|^2}.
$$
For hydrogen transition $2P \to 1S$, this turns out tobe around 1.6 nanoseconds [1].
[1] http://farside.ph.utexas.edu/teaching/qmech/Quantum/node122.html
| {
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Problem deriving entropic uncertainty relation In this paper the authors state that the inequality near the bottom of page 2 reduces to inequality (1) when $N=1$. However, I am struggling to get that result, as I have an extra minus sign in front of the integrals. Can anyone try this for themselves and see if they get the correct result?
$$\frac{n}{4}N(1+\ln{\pi})-\frac{1}{2}N^{-1} \int d^n \mathbf{r}|\Psi(\mathbf{r})|^2 \ln{|\Psi(\mathbf{r})}|-\frac{1}{2}N^{-1} \int d^n \mathbf{k}|\tilde{\Psi}(\mathbf{k})|^2 \ln{|\tilde{\Psi}(\mathbf{k})}|+N \ln{N \geq 0}$$
should reduce to
$$-\langle \ln{\rho}\rangle - \langle \ln{\tilde{\rho}}\rangle \geq n(1 + \ln{\pi}) $$
where
$\rho (\mathbf{r})= |\Psi(\mathbf{r})|^2$, $\tilde{\rho}(\mathbf{k})= |\tilde{\Psi}(\mathbf{k})|^2$ and $\langle \rangle$ denotes mean value, so that $\langle \ln{\rho} \rangle = \int d^n \mathbf{r} \,\rho (\mathbf{r})\ln{\rho(\mathbf{r})}$
| Due diligence first. The first term of the first equation you wrote is missing a sign, as you should have easily checked yourself, from the coefficient of $\ln \pi$.
That is, write down the parent $W(q)$ right above the equation you start with, around your minimum q =2, i.e. $q\equiv 2(1+\epsilon)$, so that $p= 2(1-\epsilon) +O(\epsilon^2)$, dismissing any higher orders of $\epsilon$.
The relevant term for the derivative w.r.t. $2\epsilon$ then is $\pi^{-n2\epsilon/4} N$, so you may catch that paper's typo.
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Explosion of an asteroid in orbit I've learned from an answer here on this site that if a body were to split apart in orbit the center of mass will continue to be on the same orbit. (Couldn't find the post)
But let's say an asteroid is to blow up in two pieces such that the smaller piece reverses its direction and velocity and thus stays in the same orbit, but going backwards.
But now the bigger piece must be faster than the velocity of the center of mass, this higher velocity changes the orbit of the bigger mass.
Now the center of mass is not in line the smaller piece continues the same orbit while the bigger piece changes the orbit. This doesn't seem right. What went wrong?
For example this question https://www.toppr.com/ask/question/an-asteroid-orbiting-around-a-planet-in-circular-orbit-suddenly-explodesinto-two-fragments-in-mass/
An internal explosion causes the center of mass to move farther and farther away from the Earth.
I'm very certain I've made astronomical blunders but I would like to know my mistake.
|
if a body were to split apart in orbit the center of mass will continue to be on the same orbit.
This is not always true. It would be true if we were talking about a projectile in a uniform gravitational field, but for an orbiting satellite the gravitational field is non-uniform. That means that for large displacements from the center of mass the split portions may experience very different gravitational fields than they would have experienced on the original orbit. The center of gravity is not the same as the center of mass for a non uniform gravitational field.
| {
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Why the formula of kinetic energy assumes that the object has started from an initial velocity of zero? According to my physics textbook, the formula of kinetic energy is:
$$
W = \frac{1}{2}mv^2
$$
Where $m$ is is mass of the object and $v$ is the velocity of the object. The equation is calculated from this (according to that book as well): ($a$ is acceleration, $s$ is displacement, $t$ is time and $F$ is force whose value is $ma$)
$$
\begin{align}
W &= Fs \\
&= mas \\
&= ma \cdot \frac{1}{2}at^2
\end{align}
$$
Here comes the problem. According to that book: ($u$ is initial velocity)
$$
s = ut+\frac{1}{2}at^2
$$
But in the equation of kinetic energy, $s$ is replaced by $\frac{1}{2}at^2$, which is only possible when the initial velocity ($u$) is zero ($s = ut+\frac{1}{2}at^2 = 0t+\frac{1}{2}at^2 = \frac{1}{2}at^2$). Why the initial velocity is assumed to be zero here? What if the object has a non-zero initial velocity?
| The $s$ in equation $W=mas$ can be replaced from 3rd equation of motion.
$$W=\frac{ma(v^2-u^2)}{2a}$$
$$W=\frac{1}{2}m(v^2-u^2) \rightarrow K.E.=\frac{1}{2}m(v^2-u^2)$$
Or else, if you still want to substitute $s$ from 2nd equation of motion:
$$W=ma(ut+\frac{1}{2}at^2)$$
$$W=ma(ut)+\frac{1}{2}m(a^2t^2)$$
From 1st equation of motion; $t=\frac{(v-u)}{a},\,\, at=v-u$
$$W=mau\cdot \frac{(v-u)}{a}+\frac{1}{2}m(v^2+u^2)-mvu$$
Rearranging and simplifying the terms we get:
$$W=K.E.=\frac{1}{2}m(v^2-u^2)$$
| {
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Is a wave function a ket? I just started with Dirac notation, and I am a bit clueless to say the least. I can see Schrödinger's equation is given in terms of kets. Would I be correct to assume if I were given a wavefunction, say $\Psi(x)=A\exp(-ikx)$, would I be able to just use the notation $\lvert \Psi\rangle =A\exp(-ikx)$?
| Yeah, as a beginner you can basically do this with no problems, but a physicist would tend not to do that because it's mixing notations. $|Y\rangle$ is a state, i.e. a vector. $Y(x)$ is a function that eats a value of $x$ and spits out the value of the wave function $Y(x)$. So whereas $Y(x)$ "depends" on an input, $|Y\rangle$ does not, and the expression $|Y\rangle = Y(x)$ isn't "balanced" on both sides. A slightly more correct expression would be to write
$$
\langle x | Y\rangle = \int dx' \delta(x-x')Y(x') =Y(x)
$$
But really this is a minor quibble.
| {
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Oscillation coil: where is the electric field? Let assume a simple RF coil fed with an alternating current at RF frequencies, say 100MHz.
I believe that no one doubts that the coil will radiate RF energy in the form of radio waves.
A radio wave is classically composed of an electric vector and a magnetic vector orthogonal one to the other.
Now, let assume a low frequency big coil, say, for magnetic induction, and assume it is fed by a low frequency current (e.g. 1 KHz). An oscillating magnetic field will be perceived near the coil, but I've never heard that the coil also radiates a sensible electric field. Yet, there is no difference with the former RF coil, but the frequency. So, my question is: why is there no sensible electric field radiated by the low frequency coil?
| Since the magnetic field is varying with time, by Maxwell-Faraday's induction law there is a rotation of the electric field. One can then distinguish the propagating and the non-propagating part but that is outside the scope of the question.
| {
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Definition of Ensemble Im studying statistical mechanics and came across the ensembles.
*
*Now system of large number of particles can be defined by an ensemble which contains elements (infinite of them) where each element is the mental copy of system at a particular time and time average of any quantity of system can be assumed as same as ensemble average (avg over these mental copies)
*In another book I saw that an ensemble can also be defined as a collection of a very large number of assemblies which are essentially independent of one another but which have been made macroscopically as identical as possible.
Now my doubt is that the element in the 1 case is same as assembly in case 2 or different?
How they differ from one another?
| They are just the same. This ensemble definition is a bit archaic and was useful back in the days in order to visualize the Ergodic principle ($\lim_{T\rightarrow \infty}\frac{1}{T}\int^{t=T}_{t=0} dt \ ...=\frac{1}{Z}\int_{\Gamma} \ d\alpha \ ...$ ).
| {
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Doppler Effect and Radar Sensors Radar sensors make use of the Doppler effect to measure the radial velocity of an object.
The radar's Tx antennas emit an electromagnetic wave which travels to the moving objects, is reflected and the radar's Rx antennas detect the incident wave.
Due to the movement of the measured object, the received wave has a different frequency.
According to wikipedia (https://en.wikipedia.org/wiki/Doppler_effect#Radar), the shift in frequency $\Delta f$ is given by:
$$\Delta f = \frac{2 \Delta v_r}{c} f_0,$$
where $\Delta v_r$ is the relative radial velocity between radar sensor and object, $c$ is the speed of light and $f_0$ is the radar's frequency.
A derivation of this equation is for example given in this script about radar sensors (page 21 and following):
https://www.ei.ruhr-uni-bochum.de/media/ei/lehrmaterialien/39/a715b063167d904ec4a9a5cea2a1a54d4defc115/RuhrUni_Scriptum.pdf
What bugs me know is the following: The derivation is entirely classifcal. No time dilation or relativistic effects enter the picture. My understanding is that for electromagnetic waves (which do not need a medium for propagation), the equation for the relativistic Doppler effect has to be used:
$$f_r = \sqrt{ \frac{1-\beta}{1+\beta} } f_s$$
with $\beta = v/c$.
But this equation (with its square root) does not match at all with the mentioned formula for the Doppler radar. For me it appears like that a classical method is used for the derivation of the radar's frequency shift, just like it would have been done for a sonar sensor (which uses sound waves and not EM waves).
Why is this correct?
| In the case of radar velocity measurement by doppler shifts, note that the velocity of the object being measured is of order ~10 to 1000 m/s. This is so slow compared to the speed of the radar pulse itself that relativistic corrections do not have to be made in order to get a velocity measurement accurate enough to justify writing someone a speeding ticket or to shoot down a fighter jet.
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Circuit (Pseudo-Homeworks) I'm studying Kirchoff's Laws and how to find $\Delta V$ between two points in a circuit (CC). I already know how to use it but I look on internet for more exercise to learn more and find this circuit
I'm a little confuse about it, because there is a wire in between that don't understand exactly. In this case I wanna find $\Delta V_{AB}=V_A-V_B$. I start to do this:
I write
$$\Delta V_{AB}=V_A-V_B = \Delta V_{AC} +\Delta V_{CD} +\Delta V_{DB} $$
Then if I apply Kirchhoff's voltage law in right part
$$ -30 + 6 I_3 + 4 I_3 =0$$
and I have the relations
$$ I_1 =2 A \quad \quad \Delta V_{AC}=5 I_1 \quad \quad \Delta V_{CD} =5-10 I_2 \quad \quad \Delta V_{DB}= 4 I_3 $$
But I'm not sure how to find $I_2$
| First, notice that the supplied current of the current source, must be equal to the current that flows through the 'back' of it, and therefore must be equal to the current that flows 'upward' from node $C$:
Then, applying KCL on node $C$:
Yields:
$$I_1 = I_1 +I_2$$
$$I_2=0A$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Is $\phi^4$ theory an attractive or repulsive force?
*
*Does the well known $\phi^4$ theory, with Lagrangian $$\mathcal{L}=\frac{1}{2}
\partial_\mu \phi\partial^\mu \phi-\frac{m^2}{2}\phi^2 -\frac{\lambda}{4!}\phi^4,$$
yield an attractive or repulsive force for the $\phi$ particles? (Here we use the $(+,-,-,-)$ sign convention.)
*And does there exist also a potential in the Born Approximation (as we have for the Yukawa interaction)?
| If $\lambda>0$, the force is repulsive. If it $\lambda<0$, the force is attractive, but the system is unstable to vacuum decay.
To be more specific $\lambda \phi^4$ interaction is the relativistic version of the Schrodinger delta-function potential.
| {
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Will a changing $E$ field induce a current in a loop similar to a changing $B$ field? An induced current in a wire loop that is caused by a changing B field is a common EM question. However, I couldn't find examples online where the B field was substituted for a changing E Field.
The following question was given on a test and the goal was to find the current flow caused by a varying B Field first, then a varying E Field. My answer is illustrated below.
While it was simple to deduce the direction of the current with a changing B field (clockwise), when the E field was subbed in below, my answer was completely different. Instead, I ended up with an induced B field that was counterclockwise on the outside of the loop and clockwise on the inside of the loop.
| $\nabla\times E = -\frac{\partial B}{\partial t}$ says there is "circulation" of E around the loop. E will push charges along the loop.
This follows from Stoke's Theorem: $\int_{loop} E \cdot \space dl= \int_{surface} \nabla \times E \space\text{ds}$
$\nabla \times B = \frac{\partial E}{\partial t}$ says there is "circulation" of B around the loop. That isn't going to push charges along the loop.
Both conclusions, if the charges are pushed along the loop or not, follow from $F=q(E+v\times B)=qE +q (v\times B ).$ The first term is the force due to the electric field. The second term is the force due to the magnetic field.
The charges must be moving, with velocity, $v_{\perp \space to \space B}$, perpendicular to B to feel the magnetic force $F=q(v \times B)$. The charges in the loop can't do that because they are constrained by the loop. And, if they did feel a force, that force would be perpendicular to the loop and $B$.
Feynman Lectures. Vector Integral Calculus
| {
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How do you make more precise instruments while only using less precise instruments? I'm not sure where this question should go, but I think this site is as good as any.
When humankind started out, all we had was sticks and stones. Today we have electron microscopes, gigapixel cameras and atomic clocks. These instruments are many orders of magnitude more precise than what we started out with and they required other precision instruments in their making. But how did we get here? The way I understand it, errors only accumulate. The more you measure things and add or multiply those measurements, the greater your errors will become. And if you have a novel precision tool and it's the first one of its kind - then there's nothing to calibrate it against.
So how it is possible that the precision of humanity's tools keeps increasing?
| Greeks used a compass and straight edge. A compass is as precise as an A-frame anchored with a needle, and a straight edge is some wood that is sanded flat by a carpenter. One wooden straight edge can be used to create a straighter straight edge because the sanding grit is small and it evens out the bumps.
Distances were an issue until the micrometer which gave birth to nano precise engineering. There's very good documentaries about early micrometers. Later micrometers and straight edges become laser precise, as reliable as photons, i.e. relatively accurate, not perfectly.
| {
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Determining the partial derivative of a metric tensor Im new to the Tensor Calculus and General Theory of Relativity, and I have one question. I want to determine the Christoffel symbols in FRW metric.
This is the general equation of Christoffel symbols:
$$\Gamma^{\mu}_{\hphantom{\mu}\alpha\beta}=\frac{1}{2}g^{\mu\nu}\bigg[\frac{\partial g_{\alpha\nu}}{\partial x^\beta}+\frac{\partial g_{\beta\nu}}{\partial x^\alpha}-\frac{\partial g_{\alpha\beta}}{\partial x^\nu}\bigg]$$
So the $g_{\mu\nu}$ in the expanding FRW universe equals to:
$$g_{\mu\nu}=\begin{bmatrix}
-1 & 0 & 0 & 0\\
0 & a^2(t) & 0 & 0 \\
0 & 0 & a^2(t) & 0 \\
0 & 0 & 0 & a^2(t)
\end{bmatrix}$$
How can I calculate the partial derivatives of the metric with respect to coordinates, maybe time (In this situation It's $g_{\alpha\nu,\beta}, g_{\beta\nu,\alpha}, g_{\alpha\beta,\nu}$)? I did not find a normal explanation in the books. I am a beginner, for me it is quite difficult, I will be very glad to hear your feedback.
| FRW metric is written as
$$-c^2 d\tau ^2 = -c^2 dt^2 +a(t)^2 d\textstyle{\sum}^2$$
where for simplicity take $k=0$ in Cartesian coordinates which refers to zero curvature then $d\textstyle{\sum}^2 = dx^2 +dy^2 + dz^2$. Where the metric tensor $g_{\mu \nu}$ is
$$g_{\mu\nu}=\begin{bmatrix}
-1 & 0 & 0 & 0\\
0 & a^2(t) & 0 & 0 \\
0 & 0 & a^2(t) & 0 \\
0 & 0 & 0 & a^2(t)
\end{bmatrix}$$
Keep in mind that scale factor $a(t)$ doesn't have any directional dependence since
The relative expansion of the universe is parametrized by a dimensionless scale factor $a$..
All the off-diagonal components of the metric is zero, their derivatives as well. So you only need to consider the diagonal terms. For time derivatives you need to calculate $g_{00,0}$, $g_{11,0}$, $g_{22,0}$, $g_{33,0}$. For coordinate derivatives calculate the following:
$$g_{\mu \nu , j}$$
$g_{00,1}$, $g_{00,2}$, $g_{00,3}$, $g_{11,1}$, $g_{11,2}$, $g_{11,3}$, $g_{22,1}$, $g_{22,2}$, $g_{22,3}$, $g_{33,1}$, $g_{33,2}$, $g_{33,3}$ where Greek letters go from $0,1,2,3$ and Latin letter from $1,2,3$.
After comma it denotes w.r.t what you are taking the derivative for instance $g_{22 , 0}$ means you're taking the derivative of second row second column (count from zero not one) w.r.t time and $g_{33 , 1}$ means you're taking the derivative of third row third column (count from zero not one) w.r.t $x-$ component. Rest is just taking the partial derivative of the scale factor squared since $g_{00}= -1$.
| {
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What does it mean that a free particle has no definite energy in quantum mechanics? In the quantum mechanics case of the infinite square well, the general solution to the Schrodinger equation is a linear combination of solutions with definite energy states. When you measure the particle, it will take one of these energy values.
I am now looking at the case of the free particle and I see that (because the stationary states are not normalizable) a free particle cannot exist in a stationary state. In his textbook, Griffiths states "there is no such thing as a free particle with a definite energy".
I'm struggling to see what this actually means; if you measure the energy of a free particle you will still get a value right? And won't this energy necessarily correspond to one of the infinite solutions for the free particle?
| Short answer: physically no plane waves really exist. Mathematically, we use them all the time. If we are desperate for normalization, we normalize them by particle flux, rather than the total probability.
Let me first deviate into discussing photons, snce this situation has been extensively coveregd on this site: in order to have a truly plane wave we have to be able to observe it during an infinitely long time and at infinitely long lengths. This is impossible: firstly, because our observations are necessarily made over finite time and distance, which impose the finite width on the photon spectrum. Moreover, a photon has been emitted at some time that is at a finite distance from now (certainly not earlier than the creation of the unievrse), which again imposes finite width of the spectrum.
The situation is exactly the same for particles, although it might appear less avident. Our ability to observe plane waves is equally limited, and they necessarily interact with their surroundings. Yet, in many problems we can neglect these effects as small, and use the plane waves (typically as a Fourier expansion).
| {
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Matrix Representation of Lorentz Group Generators Let $\Lambda^{\alpha}{}_{\beta}$ denote a generic Lorentz transformation.
Then, an infinitesimal transformation can be written like
$$\Lambda^{\mu}{}_{\nu} = \delta^{\mu}{}_{\nu} + \omega^{\mu}{}_{\nu} $$
where
$$\omega^{ij} = \epsilon^{ijk}\theta_k$$
$$\omega^{i0} = - \omega^{0i} = \delta^i$$
where $i,j,k$ run from 1 to 3 and $\delta^i$ is a parametre related with boosts. Then, an infinitesimal transformation has a matrix representation
\begin{pmatrix}
1 & -\delta_1 & -\delta_2 & -\delta_3\\
-\delta_1 & 1 & \theta_3 & -\theta_2\\
-\delta_2 & -\theta_3 & 1 & \theta_1\\
-\delta_3 & \theta_2 & -\theta_1 & 1
\end{pmatrix}
However, we can also write
$$\Lambda^{\mu}{}_{\nu} = \delta^{\mu}{}_{\nu} + i\frac{\omega^{\alpha \beta}}{2}\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} $$
where $J_{\alpha \beta}$ are the generators of the group. I want to prove that $J_{01}$ is of the form
\begin{pmatrix}
0 & -i & 0 & 0 \\
-i & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
My problem is in understanding the notation in
$$\Lambda^{\mu}{}_{\nu} = \delta^{\mu}{}_{\nu} + i\frac{\omega^{\alpha \beta}}{2}\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} $$
For example, I tried to compute $\left(J_{01} \right)^{0}{}_{1}$ by doing
$$\Lambda^{0}{}_{1} = \delta^{0}{}_{1} + i\frac{\omega^{01}}{2}\left(J_{01} \right)^{0}{}_{1} $$
$$\Leftrightarrow - \delta_1 = 0 -i \frac{\delta_1}{2}\left(J_{01} \right)^{0}{}_{1}$$
which yields $\left(J_{01} \right)^{0}{}_{1} = -2i$, which is not correct. What am I doing wrong?
| Thanks to @Charlie and @Cosmas Zachos I was able to find the correct answer.
It simply suffices to develop the sum
$$\frac{\omega^{\alpha \beta}}{2}\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} = -\delta_1 \left(J_{01} \right)^{\mu}{}_{\nu} - \delta_2 \left(J_{02} \right)^{\mu}{}_{\nu} - \delta_3 \left(J_{03} \right)^{\mu}{}_{\nu} + \theta_3\left(J_{12} \right)^{\mu}{}_{\nu} - \theta_2 \left(J_{13} \right)^{\mu}{}_{\nu} + \theta_1\left(J_{23} \right)^{\mu}{}_{\nu} $$
where I used
$$\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} = - \left(J_{\beta \alpha } \right)^{\mu}{}_{\nu} $$
and the other properties mentioned above.
| {
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Free body force diagram with 3 pulleys I am certain this has already been asked and answered, but I have not been able to find it. It has been about 25 years since I last sat in a physics class, and some parts are a bit rusty! I am planning to hang some heavy gear in my garage and am trying to determine the weight at each mounting point. (Yes, I know I need to add in a large safety factor for overhead rigging). I have tried to simplify the plan in the attached drawing (which I have learned is called a free body diagram!...you can teach an old dog new tricks!)
At point "A" the rope will be firmly fastened. It then travels down to a series of 2 pulleys on my load (200 pounds for this example)...then back up to point "B" which is another pulley...and finally down to a crank on the wall @ point "C"
My basic understanding is that the 200# are split evenly between point A and B...each with 100# pulling down(ish) on them. But then the wall crank ("C") is pulling presumably with 100#...which might then double the force at "B"?...or am I overthinking this? I know things get trickier given that none of the lines are perfectly vertical, but I am just trying to get "close enough" in this case.
Seems like:
A = 100 pounds
B = 200 pounds??
c = 100 pounds
THANKS!
| Let T be the tension in the rope. The force acting on weight is 2T upwards. So T is half the weight of the box. Well I can't say T is 100lb as pound is unit of mass not force but anyways I hope you understand what I am trying to convey.
The hinge force at A is T, at B is 2T and at C is T; hence hinge forces are calculated accordingly.
| {
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What does "commuting with the Hamiltonian" mean? In quantum mechanics an observable or an attribute to a particle (like spin) is conserved if and only if it commutes with the Hamiltonian. What does this mean? What observables do not commute with the Hamiltonian?
| Two operators $A$ and $B$ commute if (and only if) their commutator $[A,B]$ vanishes
\begin{equation}
[A,B] \equiv AB - BA = 0 \implies A,B\ {\rm commute}
\end{equation}
Consider a Hamiltonian operator for a single particle in 1 dimension
\begin{equation}
H = \frac{p^2}{2m} + V(x)
\end{equation}
where $x$ the position operator, $p$ is the momentum operator, and $m$ is the mass of the particle (which is just a number).
$x$ and $p$ have a commutator
\begin{equation}
[x,p]=i\hbar
\end{equation}
Using this, it is easy to see that $[H,x]\neq 0$, and $[H,p]\neq 0$ unless $V(x)$ is a constant. Therefore, in general, neither the position nor the momentum are conserved. The momentum is conserved only if the potential is constant.
| {
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Why is mist gray but water clear? I was walking outside one cold afternoon with my mask on and my glasses began fogging up. The mist was initially gray.
I kept walking without cleaning my glasses and eventually enough mist collected that that it transformed into clear water droplets.
This got me thinking: why is mist gray but water clear? Or perhaps more specifically, why are smaller water droplets gray and larger droplets clear? I couldn't find any explanation online. What is the physics behind such shenanigans?
| Regarding water droplets collecting on the surfaces of your glasses:
Those water droplets backscatter the incoming light in random directions, including ones away from your eyes. This means that any glass lens surface populated with water droplets will appear less bright than it would without the droplets, and the random scattering will obliterate anything you might otherwise be able to discern through the lens. In addition, small droplets crowded closely together cannot be resolved by your eye as individual droplets. Result: uniform smooth gray appearance.
As the water droplets begin to merge, you start being able to see objects through the lens and through the water droplets and the result is less dim (gray).
| {
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Why does water cast a shadow even though it is considered 'transparent'? If you pour water from a container, the flowing water stream seems to cast a shadow. I am not sure you can call it a shadow, but it definitely is not letting all light through it. How is this possible and what uses can it have?
| This is just to add an illustration to noah's and Ralf Kleberhoff's answers which correctly point out that refraction is the main reason.
Note that although most of the light rays do make it through the water drop, most of them do not continue on the path with the rest of the light bundle, but end up somewhere else. As a result, right behind the drop, the light intensity is much lower than it would be without the drop, whereas around this shadow it'll actually be slightly higher.
| {
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What do the different "levels" in a Kac-Moody algebra tell us physically? I'm reading a book on conformal field theory and I am on a section which derives the current algebra:
$$[j_m^i,j_n^j]=\sum_l f^{ijl}j_{m+n}^l+k \;m \delta^{ij}\delta_{m,-n} $$
We can prove this result assuming we are working with chiral fields of conformal dimension 1 using some rather abstract arguments, but I would like to relate some on these things back to something more tangible. In particular this factor of $k$ which refers to the level of a conformal field.
This $k$ comes about by diagonalising a structure constant $d_{ij}=k\delta_{ij}$. The constant can be realized by re-scaling fields. What does this represent physically? I'd guess it has to do with the scale we elect to describe our theory with, but I am not sure. Any intuition would be nice!
| The scale of the algebra is always chosen so that the length of the longest root in the underlying finite Lie algebra is 2. This choice simplifies a number of formulae, and in particular ensures that the level $k$ is an integer.
| {
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Can the Auger effect cause a second electron to be just excited instead of ionised and emitted from the atom? From what I understand, the Auger effect is usually defined as when an electron deexcites but instead of releasing its change in binding energy as a photon, it transfers it as kinetic energy to another electron which, if greater than its binding energy, will cause this second electron to be emitted from the atom.
My question is why is this process defined with the second electron being emitted from the atom instead of just excited to a higher energy state sometimes.
My guess is maybe it has something to do with entropy and the fact that there are so many more possible states for the second electron final state if it is emitted that maybe only in this case will this process actually occur (instead of just emitting a photon as usual).
|
My question is why is this process defined with the second electron being emitted from the atom instead of just excited to a higher energy state sometimes.
The atom is a unit tied up quantum mechanically . To observe transformations of an atom, there must be an interaction that can be measured. An emitted photon can be measured. An emitted electron can also be measured. If the whole process happens within an atom, there is no measurable/observable effect. An electron just going to a higher energy level can emit a photon when de-exciting , but there is no measurable way to determine that it comes from a transfer from a different electron, the way there is for an ejected electron to be identified with a different energy level:
Upon ejection, the kinetic energy of the Auger electron corresponds to the difference between the energy of the initial electronic transition into the vacancy and the ionization energy for the electron shell from which the Auger electron was ejected.
| {
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Interdependence of $P,V$ and $T$ Can we prove that the thermodynamic state of a system is completely determined by any two out of the three factors $P,V$ and $T$ ? (Without using statistical mechanics. Only using Thermodynamics)
NB: I have not learnt the axiomatic formulation of Thermodynamics.
| There exist systems with other degrees of freedom other than pressure temperature and volume, for example in magnetic systems you can also talk about the degree of magnetization and the applied field. So the completely general answer is no, because it is not always true.
Once you have limited yourself to systems that only have these degrees of freedom, we can point to the equation of state
$$
p = -\left(\frac{\partial F}{\partial V}\right)_{T}
$$
where $F(T,V)$ is the Helmholtz free energy. This links the three quatities, so at most 2 of them can be independent.
In terms of why at least 2 of them must be independent, this is again really a matter of definition. If we have a system in contact with a heat bath at a fixed temperature or in a thermally insulated container then we do have an extra equation linking the variables (the fixed temperature or the adiabatic equation respectively) and so only have 1 degree of freedom. We tend to think of these relations, however, as constraints on a more general $pV$ system, rather than fundamentally different systems.
| {
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Quantum Field theory, solving delta / green function I have read an equation as follow
$$[-(k^2-m^2)g^{\mu\nu}+k^{\mu}k^{\nu}]D_{\nu\lambda}(k)=\delta^{\mu}_{\lambda}$$
The solution is given as:
$$D_{\nu\lambda}(k)=\large{\frac{-g_{\nu\lambda}+k_{\nu}k_{\lambda}/m^2}{k^2-m^2}}$$
I can only verify the answer but do not know how to yield the result step by step... are there any standard procedures to yield the solution?
| We can make use of the projection operators
\begin{align}
& P^{T}_{\mu \nu} = g_{\mu \nu} - \frac{k_{\mu}k_{\nu}}{k^{2}}, \\
& P^{L}_{\mu \nu} = \frac{k_{\mu}k_{\nu}}{k^{2}},
\end{align}
where $P^{T}_{\mu \nu}$ and $P^{L}_{\mu \nu}$ are the transverse and longitudinal projectors. $P^{T}_{\mu \nu}$ and $P^{L}_{\mu \nu}$ are the projection operators onto vectors orthogonal and parallel to $k^{\mu}$, respectively. You can verify by yourself that they satisfy the properties of projectors.
We then write
\begin{align}
-(k^{2}-m) g^{\mu \nu} + k^{\mu}k^{\nu} = A \cdot P^{T,\mu \nu} + B \cdot P^{L,\mu \nu}
\end{align}
where we can find
\begin{align}
A = -(k^{2} - m^{2}),\ B = m^{2}.
\end{align}
Now, to find the inverse $D_{\mu \nu}(k)$ we simply invert the $A$ and $B$ and then lower the $\mu$ and $\nu$ in $A \cdot P^{T,\mu \nu} + B \cdot P^{L,\mu \nu}$, where we yield
\begin{align}
D_{\mu \nu}(k) & = \frac{1}{A} P^{T}_{\mu \nu} + \frac{1}{B} P^{L}_{\mu \nu} = \frac{1}{-(k^{2}-m^{2})}P^{T}_{\mu \nu} + \frac{1}{m^{2}} P^{L}_{\mu \nu} \\
& = \frac{1}{-(k^{2}-m^{2})}\left( g_{\mu \nu} - \frac{k_{\mu}k_{\nu}}{k^{2}} \right) + \frac{1}{m^{2}} \frac{k_{\mu}k_{\nu}}{k^{2}} \\
& = \frac{1}{k^{2}-m^{2}}\left( -g_{\mu \nu} + \frac{k_{\mu}k_{\nu}}{m^{2}} \right).
\end{align}
When solving this sort of equations it is always very helpful to use $P^{T}_{\mu \nu}$ and $P^{L}_{\mu \nu}$, since the inverting process is very simple. For more reference you can see "Quantum Field Theory: Lectures of Sidney Coleman". In chapter 28.6 he actually present the same calculation that I did here. He also gave the identities verifying that $P^{T}_{\mu \nu}$ and $P^{L}_{\mu \nu}$ are indeed projectors.
| {
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Why do bullets shoot through water but not through sand? There are a few questions only on this site about this but none of them answer my question.
Can cannonballs go through water?
Why does a bullet bounce off water?
I find it hard to understand why bullets shoot through water at longer distances but stop in sand almost right away:
*
*Both water and sand are made up of smaller droplets/grains and both are relatively heavy elements (sand is only 1.5 times heavier per volume). Water molecules are bound by Van der Waals force into droplets, sand molecules are bound by covalent bonding into crystals
*At slow speed, I can put my hand through water and sand both. The droplets and grains can roll over and accommodate an object easily.
*At high speed, an airplane crashing onto water will fall into pieces because water acts in this case like a solid, because the molecules and droplets don't have enough time to rearrange to accommodate the object. Same with sand.
*Now in the case of a bullet, this argument seems not to work. In air, bullets reach speeds over 1800 mph. Bullets penetrate water, and can keep high speeds up to 10feet. On the other hand, bullets can't penetrate sand at all, they stop completely almost with no real penetration.
Bullets can keep high speeds up to 10 feet in water.
https://mythresults.com/episode34
Bullets in sand are completely stopped after 6 inches.
https://www.theboxotruth.com/the-box-o-truth-7-the-sands-o-truth/
Question:
*
*Why do bullets shoot through water but not through sand?
| The shape
The mechanical displacement of water in water requires less energy than the displacement of sand in sand. This is because the movement of a water molecule through the conglomeration of water molecules requires less energy for the displacement and rotation of each molecule than the displacement of grains of sand.
https://de.wikipedia.org/wiki/Sand
The density and weight
Even if it would be able to mill sand into its atoms, the weight of the sand atoms is twice the weight of water. The displacement and the related to it energy losses will be higher.
| {
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Gauss' law from Hamiltonian density of electromagnetic field I am going through David Tong's QFT course, for which lecture notes and exercises are available online at http://www.damtp.cam.ac.uk/user/tong/qft.html.
In Question 1.8 we have the Lagrangian (density)
$$L = -\frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \frac{1}{2} m^2 C_\mu C^\mu,$$
$$F_{\mu \nu} = \partial_\mu C_\nu - \partial_\nu C_\mu,$$
which is like the standard electromagnetic field in the case $m=0$.
I eventually derive the conjugate momenta $\Pi_\mu$ to $C_\mu$ and convert the Lagrangian to a Hamiltonian
$$H = -\frac{1}{2} \Pi_i \Pi^i + \frac{1}{4}F^{ij}F_{ij} - \frac{1}{2} m^2 C^{\mu}C_{\mu} - \Pi_i \partial^i C_0,$$
answering the question.
However in a pdf of tutor's solutions I came across online (which I maybe shouldn't link), the tutor comments and interprets further:
they rearrange the last term,
$$H = -\frac{1}{2} \Pi_i \Pi^i + \frac{1}{4}F^{ij}F_{ij} - \frac{1}{2} m^2 C^{\mu}C_{\mu} - C^0(\partial_i \Pi^i) - \partial_i (\Pi^i C^0),$$
and comment
[the term] involves an irrelevant three-divergence term. Since the remainder of the Hamiltonian contains no derivatives in $C^0$, $C^0$ may be regarded as a multiplier that, in the $m=0$ theory, imposes the constraint $\nabla \cdot \Pi = m^2 C^0 = 0$, which is precisely Gauss' law.
Since we are back to examining the $m=0$ case, this is a statement about the standard electromagnetic field.
I don't understand either statement here.
How is $\partial_i (\Pi^i C^0)$ "irrelevant"? Can we just ignore this divergence, which as far as I can see has a nonzero value?
$- C^0(\partial_i \Pi^i)$ could be a (Lagrange) multiplier, how is it rearranged to include the $m^2$ term and (together) constrain to $\nabla \cdot \Pi=0$?
| Main points:
*
*A total spacetime divergence in the Lagrangian (or Hamiltonian) does not change the EOMs, cf. e.g. this Phys.SE post.
*If we know that the fields vanishes on the boundary, e.g. by imposing pertinent boundary conditions, we can use the divergence theorem to argue that a divergence term cannot contribute to, say the EM energy.
*The EOM for $C_0$ reads $\nabla \cdot \Pi = m^2 C^0$.
*In the massless limit $m=0$, this EOM becomes Gauss's law in vacuum.
| {
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Why is the electric flux through a surface between a dipole zero
In surface $S_3$, as per the Gauss Law, the net electric flux is zero as there is no charge enclosed in the surface. Also the other reason the book mentions for this that since the amount of flux entering is equal to the amount leaving, the net flux is zero. Also the book say or every closed surface in an external field is zero. But flux is the measure of the field lines (electric field intensity) passing through a surface. Even if the Gaussian surface near a charge that itself doesn't contain the charge has the electric field lines passing through it. So shouldn't it have some flux rather than it being zero?
|
Even if the Gaussian surface near a charge that itself doesn't contain
the charge has the electric field lines passing through it. So
shouldn't it have some flux rather than it being zero?
Yes there is electric flux. But there's a difference between electric flux and net electric flux. Gauss' law says that the net electric flux across a closed surface equals the (net) charge enclosed divided by the electrical permittivity of the space.
While there is electric flux across surface $S_3$, the net flux across surface $S_3$ is zero. Similarly, there is electric flux across the surface $S_4$, but the net flux across surface $S_4$ is zero, assuming the magnitude of the positive and negative charge is the same for a net charge of zero.
The magnitude of the electric flux, $\Phi_{E}$, through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field, or
$$\Phi_{E} = \overrightarrow E\overrightarrow Acos\theta$$
Where $\overrightarrow E$ is the electric field vector (an electric field line), $\overrightarrow A$ is a vector normal to and of magnitude equal to the area facing outward from the area, and $\theta$ is the angle between. For further details, see the following: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html
So looking at surface $S_3$, since the direction of the the electric field lines coming in at the top is opposite to the direction of the outward facing area vector, the electric flux is negative. At the bottom, each line exits in the same direction as the outward facing area vector, so the electric flux is positive. Since every line that enters the enclosed space exits it as well, the sum of the positive and negative flux, the net flux, is zero.
Hope this helps
| {
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Meaning of 1+1 dimensions I came across the notion of 1+1 dimensions in a condensed matter context and, in particular, while studying bosonization, which relates to 1D quantum systems. Indeed, the Wikipedia article about it makes reference to 1+1 dimensions in the very first sentence, though the hyperlink redirects to the more general articles about dimensions. What is meant with $d+1$ dimensions? Why is in this case 1 or 2 less appropriate than 1+1?
| To simplify a problem and perhaps gain some insight in how to solve it in its full form, physicists sometimes create a "toy universe" which is missing one or two spatial dimensions, and recast their problem in that toy universe. Sometimes the problem can be solved in that universe and sometimes that solution is of use.
Our universe is (3+1), for 3 dimensions of space and one of time. Toy universes could be (2+1), in which one dimension of space is omitted, or (1+1) in which two dimensions of space are omitted.
In addition, as d_b points out in his comment below, there are real-world systems in condensed matter physics which have effectively less than 3 spatial dimensions, and that these systems are not "toy universes" as described above.
| {
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Why is the radial velocity considered zero? I had recently come across a question which is stated as below:
A disc placed on a large horizontal floor is connected from a vertical cylinder of radius $r$ fixed on the floor with the help of a light inextensible cord of length $l$ as shown in the figure. Coefficient of friction between the disc and the floor is $\mu$. The disc is given a velocity $v$ parallel to the floor and perpendicular to the cord. How long will the disc slide on the floor before it hits the cylinder?
I thought hard for a few days but I couldn't solve it as the mathematics was terrible. Finally, while trying many other things, I tried considering the radial component of velocity to be zero and it worked! I got the answer.
But I am not able to understand the logic behind considering the radial velocity zero. Would someone please help me to understand it?
Edit: The figure is given as below:
| Because the cord is inextensible, its tension does no work on the disc (in other words, the disc always moves perpendicular to the cord). Therefore we can model the situation as a linear deceleration under friction. It would be the same if the cylinder's radius were zero and the motion were circular.
| {
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Standing waves - why do wavelengths fit perfectly? When reading about standing waves it is always said that only certain wavelengths are "allowed". I understand that these wavelengths are a requirement for there to be a standing wave due to the boundary conditions, but what does "allowed" mean in this context?
When creating a wave on a fixed string, does this mean that the wavelength will always be an appropriate fraction of the length of the string such that a standing wave exists - i.e it is not possible to create waves with a wavelength that would not create a standing wave on a fixed string?
Or are the wavelengths completely dependent on the source that created the wave, and standing waves are simply the special case/coincidence when the wavelength is appropriate?
| The reason they are "standing" is precisely because the wavelength fits perfectly. If the wavelength didn't fit, it wouldn't be standing but moving along.
So your thinking is good enough, just the wrong way round.
Why a given piece of string (or, for that matter, a bridge) has a tendency to create standing waves is another issue, but I am certain others would give you a better explanation than I.
| {
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How does Inertial force arise? Consider the following scenario$-$
A ball sits on the floor of a bus, which was originally at rest w.r.t the ground. Suddenly it accelerates forward, and we observe the ball moving backwards. Well, originally the ball does not move, it is only the bus that moves forward. Why doesn't the ball simply accelerate with the bus? A common answer might be $-$
The body tries to maintain it's inertia of rest.
In this case, my question would be, how is this force generated that tries to oppose the motion of the ball with the bus?
P.S.- Please consider frictional force too, since I feel like I know the frictionless case.
| Because the rolling resistance of the ball is less than the force of acceleration transferred to the ball from the floor of the bus
| {
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Kepler's laws for circular orbits Kepler's first law states that planets revolve around the sun in an ellipse with the sun at one focus of this ellipse. (a special case would be a circular orbit with the sun at the center).
The second law states that the areal velocity is a constant. Thus we can write ($dA = c dt$). If we integrate over one complete cycle we find that the area of the orbit, which is proportional to the square of the radius, is proportional to the time period.
The third law states that the square of the time period is proportional to the cube of the semi-major axis of the elliptical orbit.
My question is, should we swap out "semi-major axis" and replace it with "radius", or is there something missing? if we can, that leads to a contradiction with the second law.
however, how can the result obtained from the second law be wrong for circular orbits? What am I missing?
|
My question is, should we swap out "semi-major axis" and replace it with "radius"...
We can always do that. As you noted, circle is a special case of ellipse.
The second law states that the areal velocity is a constant. Thus we can write (dA=cdt).
Correct. Here c is areal speed.
If we integrate over one complete cycle we find that the area of the orbit, which is proportional to the square of the radius, is proportional to the time period.
Integrating we will get $\pi r^2 = cT$
From this we cannot conclude that $T$ is proportional to $r^2$. The reason is areal velocity $c$ is also dependent on $r$. I hope the source of confusion is clear.
| {
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Application of the principle of conservation of angular momentum and the principle of conservation of energy I came across this question and was left confused.
A satellite is launched in a direction parallel to the surface of the earth with a velocity of $36900 \; \mathrm{\frac{km}{hr}}$ from an altitude of $500 \; \mathrm{km}$. Determine the maximum allowable error in the direction of launching if the satellite is to go into orbit and come no closer than $200 \; \mathrm{km}$ to the surface of the earth.
I got confused because when the direction of velocity is parallel to the surface of the earth then the direction of velocity is also perpendicular to the radius vector. Then from the principle of conservation of angular momentum
$$r_1 m v_1 \sin \phi_1 = r_2 m v_2 \sin \phi_2$$
where
$r =$ distance of the satellite from the center of the earth,
$m =$ mass of the satellite,
$v =$ velocity of the satellite, and
$\phi =$ the angle the velocity makes with the radius vector.
The altitude of the satellite is either maximum or minimum. This means that $500\; \mathrm{km}$ is either a minimum or maximum altitude of the satellite. After using the principle of conservation of energy I figured that $500\; \mathrm{km}$ was the minimum altitude but despite that I am given $200 \;\mathrm{km}$ which also seems to be the minimum altitude (I mean how can we have two minimum altitudes). Can someone please tell me what's going on?
| Starting with the given velocity and altitude, you can use conservation of energy to find the velocity at the point of nearest approach. (This will be parallel to the surface.) Then you can use conservation of angular momentum to find the (error) angle (measured from the parallel to the surface) at the launch point.
| {
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Disintegration of the deuteron Considering the scattering of gamma rays on a deuteron, which leads to its break up acording to:
$$ \gamma+ d \longrightarrow p +n $$
we can use the conservation of energy and momentum in order to determine the minimum photon energy in order to make this reaction possible, which happens to be very close to the binding energy of the deuteron nuclide. We assume that the speed of the proton and
neutron after scattering are highly non-relativistic.
So, the conservation of energy:
$$ p_{\gamma}c + m_dc^2=m_pc^2+m_nc^2+ \frac{1}{2}m_pv_p^2+\frac{1}{2}m_nv_n^2
\tag 1$$
and of momentum:
$$ \vec p_{\gamma} = m_p \vec v_{p}+m_n \vec v_{n}
\tag 2$$
Squaring equation $(2)$ we obtain:
$$ p_{\gamma}^{ \space 2} = m_p^{ \space 2}v_{p}^{ \space 2}+m_n^{ \space 2}v_{n}^{ \space 2} + 2m_pm_n \vec v_{n}\cdot \vec v_p \tag 3$$
But if we are looking for the minimum photon energy, then we must find the case where
the proton and neutron speeds are also minimum. We assume $m_p \approx m_n$ and $v_n \approx v_p$.
With these approximations and the deuteron mass formula: $m_d = m_n + m_p -\frac{B_d}{c^2}$ we should be able to get to this equation:
$$ p_{\gamma}^2 = 2 m_d \left( \frac{1}{2}m_pv_p^2 + \frac{1}{2}m_nv_n^2 \right)
\tag 4$$
but I'm having trouble getting there. Can someone show me how to do it?
| The mass of a system is constant in reactions and is a Lorentz invariant quantity, where the mass is defined to be $$E^2-p^2c^2.$$
$E$ is the total energy of the system, and $p$ is the magnitude of the net momentum of the system. For your initial system, in the lab, you have
$$E=p_{\gamma }c+m_d c^2\text{ and } p=p_{\gamma}.$$
In the center-of-momentum frame, after minimal energy reaction, the neutron and proton are both at rest so $$E_{cm}=(m_n+m_p)c^2 \text{ and } p_{cm}=0.$$
Set these two mass calculations equal to each other and you can solve for $p_{\gamma}c$.
Now have a minimum photon energy in terms of the masses only without having to worry about the lab velocities with the binding energy concept built in because you will have terms in which it is important that $m_d \ne m_n+m_p$. This will also show you that if you approximate $m_d=m_n+m_p$, the minimum energy is zero. That's because the Q of this reaction with the approximation is zero, when in reality it is not.
If you want the lab velocities, at minimum photon energy, the neutron and proton will have identical velocities (they have zero velocity in the CoM), and their total momentum will be $(m_n+m_p)v=p_{\gamma}$. You can use that to solve for $v$. If you use the mass equality appoximation above, $v=0$.
| {
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Interesting inertia problem Consider the following.
A car is accelerating with acceleration $a$. A string is attached to the roof of the car and to the bottom of the string, an object of mass $m$ is attached. Given $\theta$, the angle between the vertical and the string (which is not $90^\circ$ due to inertia of the object).
How to derive an expression for the acceleration $a$ of the car given $\theta$ and $m$?
And when does $\theta$ remain constant?
I found a similar question, but the answers to that post were too low quality (as is also evident by the fact that the user didn't accept any of those as solutions); so don't flag this post as a duplicate of that.
| So if we consider the x-component of the tension $Tsin(\theta)$ and given the car moves with acceleration $a$ then
$$ma - Tsin(\theta) =0$$
and so
$$a=\frac{Tsin(\theta )}{m}$$
for a mass $m$. Remember that the mass experiences an inertial force and so $a$ is the acceleration of the car and mass for an observer inside the car. We can write this in terms of theta as
$$\theta=sin^{-1}(\frac{ma}{T})$$ This tells us that $\theta$ will continue to increase if $a$ increases, and if the rope does not break, it appears that $\theta$ can take on any value from $0\le \theta \le 90$ degrees.
It could snap at some point $\theta$ but to calculate this we would need a maximum value for $T$. If $\theta =90$ degrees then you’d have the condition $$T-ma=0$$
| {
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Is there a material which opposes electric fields? I was wondering if a material existed which opposed applied electric fields, analogously to a diamagnetic material, which opposes magnetic fields. So the flow of charge would be against the field, rather than toward the field. In other words, instead of electrons flowing or atoms polarizing towards a positive electric field, they move away from it and towards the negative, like a diamagnetic material aligning itself against the magnetic field. Maybe a big stretch, but I was wondering if that existed in any form.
Edit: Maybe I was unclear in my description of opposing electric fields. In a dielectric, the atoms and molecules polarize so that the negatively charged parts face the positively charged electrodes, and vice-versa. What I would be looking for would be the opposite of a dielectric. Would a negative permittivity material be an example of that?
| If charge flows opposite the direction of the applied field, you have a material with negative resistivity. No such material exists.
However, it is possible to build a circuit that, within a limited range of applied voltage, delivers an opposing current (i.e. a negative resistance circuit). Such a circuit must necessarily include active devices and have a power source of some kind (i.e. it delivers power back to the input source rather than absorbing power from it).
It is also possible to build a circuit or device that has a negative differential resistance. That is, although the overall current flows in the usual direction associated with the applied voltage, a small increase in applied voltage produces a decrease in the current, or vice versa. Such devices don't necessarily require a second power source. Examples include tunnel diodes and Gunn diodes. The input of a DC-DC switching regulator circuit with a fixed load also exhibits negative differential resistance.
| {
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Does the magnetic field produced by a current carrying wire, exert a magnetic force on the wire itself? I have to calculate the pressure on a current carrying wire. Since there is a pressure on the wire, there must be a force on it, which is a magnetic force. Does the magnetic field produced by the wire, exert a magnetic force on the wire itself? If this is true, why?
|
I have to calculate the pressure on a current carrying wire. Since there is a pressure on the wire
This is not relevant to the main question, but what do you mean by “pressure”? Pressure normally means the ratio of force to area. What area would you be using here?
Does the magnetic field produced by the wire, exert a magnetic force on the wire itself?
No.
Perhaps the concept will be easier to understand in the context of gravity. For example, consider the Earth. Does the gravitational field exerted by the Earth exert a force on the Earth itself? Of course not; it is not possible for an object to exert a force on itself. This applies to all forces, including magnetic forces.
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Convert newton formula to acceleration formulas I am trying to understand a problem from the book Solving Problems in Scientific Computing Using Maple and MATLAB. The problem is about trajectories of tennis balls.
The author gives two formulas $D_L(v)$ for drag force and $M_L(v)$ for magnus force:
$$D_L(v)=C_D\frac{1}{2}\frac{\pi d^2}{4}\rho v^2$$
$$M_L(v)=C_M\frac{1}{2}\frac{\pi d^2}{4}\rho v^2$$
with $C_D$ and $C_M$ drag coefficient and magnus coefficient respectively.
Later he gives the trajectory vector by:
$$m\frac{d^2\vec{r}(t)}{dt^2}=-m\vec{g}-D_L\frac{\vec{v}}{v}+M_L\frac{\vec{\omega}}{\omega}\times\frac{\vec{v}}{v}.$$
So far so good. But then he derives from that the following two formulas:
$$\ddot{x}=-C_D\alpha v\dot{x}+\eta\,C_M\alpha v\dot{z}$$
$$\ddot{z}=-g-C_D\alpha v\dot{z}-\eta\,C_M\alpha v\dot{x}$$
where $v=\sqrt{\dot{x}^2+\dot{z}^2}$, $\alpha=(\rho\pi d^2)/(8m)$ and $\eta=\pm 1$ is the direction of rotation.
If I understand correctly, $\dot{x}, \dot{z}$ are the velocities and $\ddot{x}, \ddot{z}$ are the accelerations in $x$ and $z$ direction.
I do not understand how to get to these equations by plugging $D_L$ and $M_L$ in above newton formula. For example, how do I handle the cross product on the right to summarize the sum? How do I get two $\alpha$'s on the right in both equations?
And one minor thing: why $L$ in the index of $D_L$ and $M_L$? Is it for lift?
Any help is appreciated! Let me know, if I forgot an important variable or other information is missing.
| From $v=\sqrt{\dot{x}^2+\dot{z}^2}$ we get that $\dot{y}=0\Rightarrow y=constant$. With that on mind, the cross product is
$$\vec{\omega}\times\vec{v}=\omega_y \dot{z}\hat{i}+(\omega_z \dot{x}-\omega_x \dot{z})\hat{j}-\omega_y\dot{x}\hat{k},$$
where $\hat{i} , \hat{j} , \hat{k}$ are unitary vectors in $x , y , z$ directions respectively.
Now equations of motion for each component are
$$m\ddot{x}=-C_D\frac{1}{2}\frac{\pi d^2}{4}\rho v^2\frac{\dot{x}}{v}+C_M\frac{1}{2}\frac{\pi d^2}{4}\rho v^2 \frac{\omega_y \dot{z}}{\omega v}$$
$$m\ddot{z}=-mg-C_D\frac{1}{2}\frac{\pi d^2}{4}\rho v^2\frac{\dot{z}}{v}-C_M\frac{1}{2}\frac{\pi d^2}{4}\rho v^2 \frac{\omega_y \dot{x}}{\omega v}.$$
Doing $\alpha=\frac{\rho\pi d^2}{8m}$
and $\eta=\frac{\omega_y}{\omega}$
we get
$$\ddot{x}=-\alpha C_D v\dot{x}+\eta\alpha C_M v\dot{z}$$
$$\ddot{z}=-g-\alpha C_D v\dot{z}-\eta\alpha C_M v\dot{x}.$$
| {
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Approximating the angle between the trajectory I started to learn physics this semester and I found the following task:
A contestant is participating in a half-maraton tournament(straight line length $L =21095$ meter) running in a zig-zag manner (constantly surpassing other contestants), holding a stable angle $\alpha$ between the trajectory. After finishing the run, the contestant has noticed that the distance travelled was 500 meters longer than $L$. Approximate the angle $\alpha$ without using a calculator.
I have no idea how to approach this problem so any help would be useful.
| The angle is small but it is not zero. The problem relies on the fact that the angle is small when it asks to find an approximate value without using a calculator.
You find $cos(\alpha)=\frac{L}{L+\Delta L} \approx 0.977 $.
But, for small angles, you can approximate $cos(\alpha) \approx 1-\frac{\alpha^2}{2} $
So, by comparing the two relationships, you can get the angle in radians as
$\alpha \approx 0.21 rad$.
To find it in degrees you just multiply by $180/\pi$ (or about 57 or even more roughly, 60) and indeed you get an angle of aproximately $12^0$.
If you are not familiar with the Taylor expansion for cosine function, you can just use $sin^2 +cos^2 =1 $ to find $sin(\alpha)$ and then use the fact that for small angles $sin(\alpha) \approx \alpha$. This is also a result of a Taylor expansion hovewer it is used in introductory Physics in the study of the pendulum so it is more familiar for the students, I believe.
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What keeps dry flat pieces of sand crumbles together (even if you microwave them)? Originally these flat pieces get created as the top layer on the beach where water pulls back (low tide), leaving the sand to dry on the Sun. The top layer dries and creates a separate layer, making these flat solid pieces.
Sometimes these flat crumbles are extremely thin, with a thickness of only around 4 grains. I cannot understand how these stay in one piece after they dry completely:
*
*first I thought it was simply moisture, but if I let them dry completely on the Sun, they become dry throughout. When I brake them apart, even the crack looks and feels completely dry throughout.
*then I thought maybe it is salt, somehow the salty ocean water sticks the grains together, and after they dry, the salt stays in between the grains as some kind of glue
*the grains are themselves edgy crystals that can somehow lock together because there might be vacuum
*Same thing happens in microwave, the sand piece stays solid, like a rock, and won't fall apart. This makes me think it shouldn't be water that keeps it in one piece.
Will a microwave heat sand?
Question:
*
*What keeps dry flat pieces of sand crumbles together (even if you microwave them)?
| The water which was mixed with the sand contained not only salt but lots of microorganisms called plankton. Taken together, they make a simple sort of glue that tends to absorb moisture and retain it even when heated, and keep the grains stuck together (weakly).
This is testable, by taking a sand sample and scrubbing it around with dish detergent and rinsing it several times so the sand particles are as clean as possible, and then drying the sample in the sun to see if a crust forms.
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Propagator in string theory and boundary conditions I would like to understand how to compute propagators for open and closed strings.
My references are Tong's lecture notes, https://arxiv.org/abs/0908.0333 and Blumenhagen, Lust, Theisen book "Basic Concepts of String Theory".
Tong uses a path integral approach: he starts from
\begin{align}
0=\int\ \mathcal D X\ \frac{\delta}{\delta X(\sigma)}(e^{-S}X(\sigma'))\implies \langle \partial^2X(\sigma)X(\sigma')\rangle=-2\pi\alpha'\delta(\sigma-\sigma'),
\end{align}
which he solves by noticing that $$\partial^2 {\rm ln }(\sigma-\sigma')^2=4\pi\delta(\sigma-\sigma')\implies\langle X(\sigma)X(\sigma')\rangle=-\frac{\alpha'}{2}{\rm ln}(\sigma-\sigma')^2. $$
In this solution, he doesn't seem to be concerned with any boundary condition: therefore, this seems to be valid for both open and closed strings.
On the contrary, in the mentioned book, there are different expressions for this propagator, depending on the case (closed string, open string with NN, DD, DN, ND conditions). See pages 37, 38, 97, 98.
How are these computations performed?
| There is no an algorithmic way to solve the Poisson equation in two dimensions (except in the case of a disk). The problem you want to address must be done in a case by case manner.
The whole chapter six in Polchinski string theory texbook (page 170,Vol. 1) is dedicated to find scalar Green functions for wolrdsheets of different topology (boundary conditions for Poisson equation). The step-by-step procedure can be found in section 6.2.
Extra comments:
*
*A hint to derive the particular expression you wrote: Solve equation 6.2.8 in Polchinski (the solution is equation 6.2.9) using the answer to the following PSE question How to derive Eq. (2.1.24) in Polchinski's string theory book or solve the exercise 2.1.
*See Polchinski's String theory Green's function on $RP_2$; eq. (6.2.38) p. 176 for the particular derivation of the scalar propagator in the case of the two dimensional projective plane.
*Your question was asked in PSE before, see Deriving the reduced Green's functions in Polchinski's volume 1.
| {
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Why voltage is same across parallel circuit if it is work done per unit charge? Suppose we have the following circuit:
If voltage is work done per unit charge, why voltage is same across each resistor if the charge has to do more work in resistor R2 than in resistor R1?
| I will explain this in terms of the water analogy. Consider a river bed that follows the same setup as the circuit. The water level is the voltage. Higher voltage = higher water level = more energy per unit of water/charge. The resistors are dams that restrict the water flow. The battery is like a pump that works to keep the water level at a certain height.
You can imagine that any part of the river bed that is connected directly will very quickly assume the same height as the rest of the water. If any part is lower water will rush in to equalize the water height. You can translate this back to the circuit any wire that touches the supply voltage will quickly take on the same voltage. Any part of an (ideal) wire will have the same voltage because if it doesn't, charge will redistribute itself until it is. This changes when you encounter a resistor: resistors restrict the flow of charge and allow one end to have a higher voltage then the other. Similar to how the dam allows the water on one side to be higher.
So using this reasoning it makes more sense that the voltage difference over $R_1$ is the same as the difference over $R_2$. Let's go back to the water analogy one more time: we have two dams connected to the same lake. One dam has a large opening near the top (low resistance) and the other one a small opening (large resistance). When a water particle moves through the opening it loses potential energy because it falls down. It doesn't matter through which dam it moves: it loses the same amount of potential energy i.e. it falls the same height. But more water will move through the larger opening so low resistance = more current.
| {
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Do resistance have to heat? Not a physicist.
Is there a way to build resistances that do not heat when opposing current?
More generally, is it necessary to waste energy to resistance?
If I cut a wire, there will be an almost infinite resistance (unless the tension is large enough to break the air), but no energy cost.
If I had a switch with a duty cycle of 50 pc, I would oppose the flow of current similarly to a resistance without heating, maybe. Is that an actual method for loss free resistance?
| This is possible for alternating current (AC). For AC voltage sources the definition of resistance is extended to what is called impedance. If you supply a voltage of the form
$$V(t)=V_0\sin(2\pi f t)$$
then for simple elements (resistor, inductor or capacitor) the current will be of the form
$$I(t)=I_0\sin(2\pi ft+\phi).$$
Here $f$ is the frequency and $\phi$ is some phase offset. To make the math easier the current and voltage are often treated as complex numbers but you don't need to understand complex numbers to understand impedance. The impedance is then defined as
$$|Z|=\frac{V_0}{I_0}.$$
For a resistor we get $|Z|=R$ and a phase offset of zero i.e. $\phi=0$. Capacitors and inductors also have impedance (in addition to a phase offset) but they do this without wasting energy. In reality they will always waste some energy but ideal capacitors and inductors have zero losses. The impedance of capacitors is $|Z_C|=\frac{1}{2\pi f C}$ and that of inductors is $|Z_L|=2\pi f L$. Because impedance and resistance have similar formulas these components basically act like resistors but without wasting energy. For example if you want to step down the voltage from your outlet you would rather want a capacitor/inductor since a resistor would waste a lot of energy.
| {
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Could the observable universe be bigger than the universe? First of all, I'm a layman to cosmology. So please excuse the possibly oversimplified picture I have in mind.
I was wondering how we could know that the observable universe is only a fraction of the overall universe. If we imagine the universe like the surface of a balloon we could be seeing only a small part of the balloon
or we could be seeing around the whole balloon
so that one of the apparently distant galaxies is actually our own.
In the example with the balloon one could measure the curvature of spacetime to estimate the size of the overall universe, but one could also think about something like a cube with periodic boundary conditions.
Is it possible to tell the size of the overall universe?
Artistic image of the observable universe by Pablo Carlos Budassi.
| We could discover that the actual universe is larger than the observable universe when there is a non-zero constant curvature everwhere. Then, this would be like standing upon the earth and seeing as far as the horizon. In the picture drawn above what we can observe would be limited by a kind of cosmological horizon.
| {
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Do humans use the doppler effect to localize sources of sound? Consider a source of sound such as a person speaking or a party of people which makes a continual drone sound of the the same frequency. If a human shakes their head side-to-side with sufficient angular speed, they are in effect obtaining different frequencies of the same sound source and should be able to apply the Doppler effect to approximately localize (from prior experience) the sound source.
Do humans use the Doppler effect to localize sources of sound and have there been any studies proving this?
Edit: A link to the Weber-Fechner law and a link to the wiki article discussing the just-noticable-difference (JND) for music applications were added to the OP for reference, based on the accepted answer.
| Humans DO use dopler effect to estimate a sound source position, they just dont use it exactly the way you imagine.
The simplest example is the distance to a passing by object (a car, an airplane, a mosqito or even a talking human). A near flyby makes a rapidly lowering tone. An object passing away from you will change its tone slower.
Moving your head left and right gets you a direction by the phase difference between ears. Well, you can get the direction withour moving your head, but it will be exact up to the symmetry of your head.
| {
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Larger aperture for objective lens in a compound microscope A telescope has a large aperture to collect more light and hence improve the visibility (brightness) of the image. It also helps in improving the resolution as $\theta_{\text{min}}=\frac{1.22\lambda}{a}$ where $a$ is the radius of the aperture (assuming it to be circular). This led me to wonder why microscopes have small aperture. Yes, under laboratory conditions, the lighting is good; hence a large aperture is not needed to improve the brightness of the image, but from the perspective of resolution of microscope $d_{\text{min}}=\frac{1.22\lambda}{2\tan\beta}$, why don't we design a larger aperture? Is it just for the sake of compactness or is there any theory behind?
| Even under laboratory conditions, the light intensity collected by a microscope can be low, so a larger aperture is in general desirable. However, as you correctly point out, what matters is not the physical aperture (i.e. the radius of the objective lens) but the angle $\beta$. The quantity
$$ NA = n\sin \beta$$
is called numerical aperture and $n$ is the refractive index of the medium in front of the objective lens and
$$ \tan\beta = \frac{R}{f_{OL}} $$
where $R$ and $f_{OL}$ are, respectively, the radius and focal length of the objective lens.
The numerical aperture is related to the resolution of the imaging system via Rayleigh's law
$$ d_{min} = 0.61 \frac{\lambda}{NA}$$
Therefore, the resolution and the amount of collected light can be improved by using larger numerical apertures. This can be obtained by enlarging the lens or by reducing the focal length. Microscope objectives are typically designed to work with a tube lens, therefore the magnification will be the ratio of the focal lengths
$$M = \frac{f_{TL}}{f_{OL}}$$
Since the detector is typically a pixel-based camera it is pointless to increase the optical resolution more than the size of the pixel. Therefore, $Md_{min}> d_{pixel}$ or the optical resolution will be lost by the lack of digital resolution. From this last equation, it should be clear that it is more convenient to reduce the focal length instead of enlarging the size of the objective lens. Moreover, compactness is not a bad thing!
| {
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Energy of a $n$-particle system in special relativity Consider an inertial frame $S'$. With respect to this frame of reference consider a system of $n$ particles. The $k$-th particle has rest mass $m_{0,k}$ and it moves with speed $u_k$. Can we say that the mass of the system is $$\mathbf{m=\sum_{k=1}^{n}m_{0,k}\gamma_{u_{k}}}$$ with total energy $E=mc^2$ ,where $\gamma_{u_{k}}$ is given by $\frac{1}{\sqrt{1-\frac{{u_k}^2}{c^2}}}$ ?
And also, how does the relativistic energy expression contain the information about potential energy?
| $\sum\gamma_k m_k c^2$ is called the "total relativistic energy" $E_{rel,tot}$.
Although some have called $\frac{1}{c^2}E_{rel,tot}=\sum\gamma_k m_k$ the "relativistic mass",
because of misconceptions by novices, its use is discouraged.
(Further comments at Invariant rest mass vs Proper velocity )
The "invariant mass of the system" (or the "rest mass of the system")
$m_{sys}$ is essentially the magnitude of the total 4-momentum:
\begin{align}m_{sys}c^2
&=\sqrt{\tilde P_{tot}c\cdot \tilde P_{tot}c}\\
&=\sqrt{\left(\sum \tilde P_k\right)c\cdot\left(\sum \tilde P_k\right)c}\\
&=\sqrt{\left(\sum E_{rel,k}\right)^2-\left(\sum \vec p_{rel,k}c\right)^2}\\
&=\sqrt{\left(\sum \gamma_k m_k c^2\right)^2-\left(\sum \vec p_{rel,k}c\right)^2}\\
\end{align}
| {
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Is there a way to prove that different angular momentum components anticommute without using a specific matrix representation? I know spin-1/2 Pauli matrices satisfy the anticommutation relationship $\{\sigma_i, \sigma_j\}=2\delta_{ij} \mathbb{I}$. I wonder how this can be proved without writing down the matrix representation of these matrices and performing matrix multiplication. As the matrix representation of angular momentum operator (and hence Pauli matrix) can be written down just using the commutation relationship $[\hat{J_x},\hat{J_y}]=i\hbar\hat{J_z}$ and its cyclic substitutions, I think there should be a way to prove this anticommutation relationship just using the commutation relationship and without using any specific matrix representation of Pauli matrices.
I tried to follow a similar fashion as in determining matrix representation of $\hat{J_x}$ and $\hat{J_y}$, but the use of the lowering and raising operators $\hat{J_-}$ and $\hat{J_+}$ (which I believe may be useful in the proof) only occurs when evaluating the matrix entry $\langle s,m'|J_\pm|s,m\rangle$, which is something I want to avoid. As a result, I failed to finish the proof.
| The pauli matrices are simply a matrix basis which (up to a factor of i) represents a basis for Cartesian bivectors. In this sense they are isomorphic to the quaternions, meaning they form a representation of the Clifford algebra of $\mathbb{R}^{3}$ .
As another answerer has pointed out, the dimension of one's representation of this algebra determines whether or not the the generators will anticommute. If one would like to study non-relativistic spin-1/2 fermions, one must use a representation which reflects the change of sign of the wave functions of spin-1/2 under a rotation of $2\pi$, and it turns out one can use a 2×2 complex hermitian representation to get this job done. In this sense one does not derive the anticommutation relations (which tell you about the representation being used) from the commutation relations (which are determined by the space being described) because one chooses the anticommuting representation to describe what one observes.
| {
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Newtonian physics and equivalence principle: a doubt on acceleration and gravity First of all, the famous Einstein's elevator experiment is quite clear in my head, both of versions.
But now, consider the following:
Suppose then you wake up inside a car that is traveling in perfect straight path in a autoban (but you don't know that). The car have a constant velocity $v$ and is a self-driving car with totally dark-glass windows. You don't have any information about the outside world. After a time $t$ travelling in the straight path, the car enters in a curve. You then fells an acceleration (exactly with $9,8 m/s^2$) accelerating you.
Now, in my opinion, the person inside the car cannot say that the centrifugal acceleration is different from artificial constant gravitational field. The equivalence principle states something similar, since a person inside a elevator in a gravitational field is equivalent to a person inside a elevator which is accelerated with $9,8 m/s^2$. Furthermore, we can construct a ring-like structure to produce, via circular motion, a artificial gravitational field.
So, can I say that any accelerated frame, due to equivalence principle, is equivalent to a gravitational field?
| If the acceleration of the frame is due to rotation, there is, in addition to the centrifugal force which locally looks like gravity, also Coriolis force, which acts perpendicular to the motion of the test particle, and makes it possible to distinguish the frame from one which doesn’t rotate.
| {
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Christoffel Symbols of a parallel shift Regarding the transformation $p=3x+5y;q=x+y$ (where $x,y$ are cartesian coordinates) I want to look at the christoffel symbols for the covariant coordinate system for a parallel shift (using ruler and triangle).
Therefore I calculated the covariant basis vectors and then looked at coordinates of a vector $\overrightarrow{A}=A^x\overrightarrow{e}_x+A^y\overrightarrow{e}_y$ in the p-q coordinates system (calculating $A^p$ and $A^q$), because I then need to shift it parallel to another point. The coordinates of $A$ then become $A^p+\delta A^p,A^q+\delta A^q$
I now need to write $\delta A^p$ and $\delta A^q$ as a total differential of the variables $p,q$ and find a way to put the parallel shift in mathematic language. This is where I am stuck.
So far I have $\vec{e}_p=\frac{1}{2}(-\vec{e}_x+\vec{e}_y)$ and $\vec{e}_p=\frac{1}{2}(5\vec{e}_x-3\vec{e}_y)$.
I then also want to compare the solution to $\delta A^i=\Gamma^i_{ml}\cdot A^m\delta x^l $.
Can somebody help me understand what I need to do here?
| Location $(x,y,z)$ isn't a vector in most spaces. $(dx/dt,dy/dt,dz/dt)$ is a contravariant vector.
$dp = (dp/dx) dx + (dp/dy) dy$
I believe the basis vectors transform "opposite" to that.
$e_p = (dx/dp) e_x + (dy/dp) e_y =(dx/dp) \hat{x} + (dy/dp) \hat{y}$
As they said, these are all constants so the derivatives are going to be zero. The basis vectors are constant.
You should try a more interesting transformation. Your christoffel symbols will be zero and the parallel transformed vector will be identical to the original vector.
| {
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How do speakers vibrate for a complex music? I understand how a speaker could produce simple sound and constant frequency. How does it produce more complex sounds like music? How can you calculate what frequency to oscillate at when there are multiple instruments and voices in a song? There must be a limit to the complexity of the oscillations and the sound that can be produced.
| Unlike an acoustic instrument like a guitar string or a triangle that emits mostly a single frequency (and a set of its harmonics) defined by its physical characteristics (shape, tension etc.), a speaker is driven by electric signal, and its motion is controlled by this signal, rather than by speaker's shape.
At each point in time the position of a speaker's membrane is basically by how much current or voltage is applied to the speaker input. This makes it possible to reproduce an arbitrary waveform containing Fourier series components up to some cutoff frequency. The different frequencies add up following the superposition principle.
There must be a limit to the complexity of the oscillations and the sound that can be produced.
Right, as the frequency increases, the speaker's response amplitude decreases, and at some frequency it becomes too small to be practically useful. This limits the complexity of sounds that can be reproduced.
| {
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If $I \propto V$, then why is $R = V/I$ and not $I/V$? I know that the current flowing through a conductor is directly proportional to the potential difference across its ends (by Ohm's Law).
Hence,
*
*I ∝ V
*V ∝ I
*R = V/I, where R is a constant (Resistance)
But why can't it be derived this way?
*
*I ∝ V
*I = RV
*R = I/V
Won't these two derivations contradict each other?
Thank you
| Chris' answer is perfectly right, but I think it's missing a key point: try it out!
Suppose you're the one working with Ohm for his formula (but you have today's equipment for simplicity's sake). You do your reasoning, and you arrive at the two conclusions you pointed out. What to do? You try them out. Build a circuit, turn up the voltage. If you want the circuit to work and not blow up, do you need a more resistant or conducting material? Now do the same with current, and give the proper names to the two quantities you discovered.
| {
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Is the coefficient of restitution of a bouncing ball constant with respect to height? Is the rebound rate (or ratio) of a bouncing ball actually constant?
Edit: To clarify, I am considering a given ball and whether it has a different coefficient of restitution depending on the height dropped from (or velocity at impact).
It’s quite typical for high school mathematics textbooks to ask questions about a bouncing ball for the topic of geometric sequences. At this level of education, the maximum height of a bouncing ball is modelled as a geometric progression or exponential function. For instance, $h_{n}=h_{n-1}\times r=h_0\times r^n$, where $h_n$ is the maximum height after the $n$th bounce after it is dropped from a height of $h_0$, and $0< r < 1$ is the rebound rate (coefficient of restitution).
But is $r$ constant with respect to height? Why or why not?
If $r$ is not constant, how do other factors besides gravity and idealised elastic behaviour influence the $(h_0, r)$ relationship? For a typical ball dropped from a typical height, what association should we expect to see? How about in a vacuum to isolate deformation effects?
| No, it's not constant. It decreases with the drop height.
https://www.researchgate.net/profile/Khairul-Ismail-6/publication/260991737_Coefficient_of_restitution_of_sports_balls_A_normal_drop_test/links/57a021d808aec29aed214c06/Coefficient-of-restitution-of-sports-balls-A-normal-drop-test.pdf?origin=publication_detail
Why? Because the coefficient of restitution depends not only on the material of the object but also on its shape.
So when you increase the speed, some object might deform a lot at the impact so can't restitute the kinetic energy the same way than at lower speed.
| {
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Is radial time a solution of Einstein's equations? Imagine a (flat) 4D space where we measure time outwards in a radial direction from the origin.
So that 3D space at a given time would consist of a spherical shell. (As such this would be a closed Universe.)
In a far distant time the spherical shells at any given position would be essentially flat and the shells so big as to make the Universe appear infinite.
Light rays would have to only cross the spherical shells at 45 degree angles. Hence we could impose a partially ordered set on the events. And the lightcones at every point in this 4D space would be well defined.
In a sense this 4D space-time would have no boundary, but in another sense we have defined the origin as a special point at which time "begins". Light rays would kind of spiral out from the origin.
Is the space-time as I've described it a solution of GR? Is there anything special about it - to me it seems like the second most obvious way to imposee a partial order to the set of events into a space that is not simple Minkowski space.
Is this a solution of GR? In which case what would the metric be?
| I believe that this is equivalent to a Universe which is closed spatially but not in the time dimension since the point at $r=0$ will transform to $t=-\infty$, the transform will be $r=e^t$. The singularity at $r=0$ merely represents the infinite past. We have merely 'squished' the time dimension up to make it appear it has a boundary.
| {
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If electrons can be created and destroyed, then why can't charges be created or destroyed? I read on Wikipedia that electrons can be created through beta decay of radioactive isotopes and in high-energy collisions, for instance when cosmic rays enter the atmosphere. Also, that they can be destroyed using pair annihilation.
We also know that charge is a physical property which can be associated with electrons.
My question is why can't charges be created or destroyed if electrons can?
| Charges can be created and destroyed. Total charge cannot.
Whenever you create an electron, charge $-1$, you must also create a positron, charge $+1$. That gives total charge $0$.
Whenever you create a proton, charge $+1$, you have to create an anti-proton, charge $-1$. That gives total charge $0$.
As far as we're aware, the total charge in the Universe is zero. Every proton's positive charge is balanced by an electron's negative charge.
That said, there's a caveat to the above: the rules as I've described them have it that at any time you make matter, you also make anti-matter, which is more than just opposite charge: if they were strictly followed, we'd not be here, because all the matter and antimatter would have reannihilated into photons. The reason for it to be otherwise is not known. However, presumably it would not involve any processes breaching charge conservation. For example, it might be possible to convert anti-protons into neutrons and electrons. This would not violate charge conservation. Total charge $-1$ first, then $0$ and $-1$ as components - total, $-1$. That's one way it could happen. Other ways could work, too.
(Why do we presume that charge conservation cannot be violated? Simple: it's the easiest way to explain why the total charge of the whole Universe is zero.)
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Normalization factor for symmetric state As per Wikipedia,
The above discussion generalizes readily to the case of $N$ particles. Suppose there are $N$ particles with quantum numbers $n_1, n_2, ..., n_N$. If the particles are bosons, they occupy a totally symmetric state, which is symmetric under the exchange of any two-particle labels:
$$|n_1,n_2,...,n_N;S\rangle =\sqrt{\frac{\prod_nm_n!}{N!}}\sum_p|n_{p(1)}\rangle|n_{p(2)}\rangle\cdots |n_{p(N)}\rangle $$
The quantity $m_n$ stands for the number of times each of the single-particle states $n$ appears in the N-particle state.
But Kardar's book, Statistical Physics of Particles, says
The bosonic subspace is constructed as
$$|\vec k_1,\cdots \vec k_N\rangle _+=\frac{1}{\sqrt{N_+}}\sum_P P |\vec k_1,\cdots \vec k_N\rangle_\otimes $$
Proper normalization requires $N_+=N! \prod_{\vec k}n_{\vec k}!$.
A particular one-particle state may be repeated $n_{\vec k}$ times in the list.
But both of them are quite different. Why it is so?
| I am also looking for an answer to this, but I believe it is the latter. I think about it like this:
There are definitely $\frac{N!}{\prod_{n}(m_n!)}$ distinct permutations that change $|n_1 \rangle \cdots |n_N \rangle$ which are orthonormal and for each of those distinct permutations, there are $\prod_{n}(m_n!)$ permutations that dont change $|n_1 \rangle \cdots |n_N \rangle$
For notational convenience define $k = \frac{N!}{\prod_{n}(m_n!)}$ and $l = \prod_{n}(m_n!)$
So we can group all permutations of the $|n_1 \rangle \cdots |n_N \rangle$ into $k$ groups, each group containing the same state $l$ times and picking a representative state from each group, say states $\psi_1, \cdots, \psi_k$ gives us $k$ orthonormal states
So we can write
\begin{align*}
\bigg \| \sum_{\sigma \in S_N} |n_{\sigma(1)} \rangle \cdots |n_{\sigma(N)} \rangle \bigg\|^2
&= \sum_{i=1}^k \| l \psi_i \|^2 \\
&= l^2 \sum_{i=1}^k \| \psi_i\|^2 \\
&= l^2 k \\
&= N! \prod_{n}(m_n!)
\end{align*}
Let me know what you think.
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Is no acceleration a cause or consequence of no net force? If a body is moving with constant velocity, or is at rest, then the net force on it must be $0$. If the net force on a body is $0$, then it must be moving with constant velocity or must be at rest.
Is $0$ net force a consequence of being at rest or moving with constant velocity or is moving at constant velocity or being at rest a consequence of $0$ net force?
|
Is 0 net force a consequence of being at rest or moving with constant velocity
Yes. It's not a physical consequence, though. Constant velocity (which is also the case for a body at rest) implies zero force (which is connected to acceleration by a factor m), but it doesn't cause it. Force on a body is caused by some agent external to the body. Which is the other way round. Constant velocity implies zero force. Zero force doesn't cause zero velocity. There is no force so how can it cause anything. A finite force can cause a velocity to change.
This is a semantic issue on the meaning of cause, I guess though.
| {
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How much force applied to canal wall from that cargo ship given 220,000 tons and 12.8 knots? In case you've been hiding under a rock, or are reading this in the future: "that cargo ship" is a huge story right now (3/26/2021). A brief summary: well basically a few days ago one of the world's largest cargo ships somehow managed to dig its bulbous bow into the east wall of the Suez canal. The back end of the ship is resting on the west end and no other ships can pass. I read 220,000 tons and 12.8 knots https://www.baltimoresun.com/news/nation-world/ct-aud-nw-cargo-ship-stuck-egypt-suez-canal-20210324-oytkblgh5ngihlwitcy7hdsnwi-story.html and thought it might make a fun little physics question. Another one I thought of is how much volume of water that much weight displaces...
| For the volume of water displaced... Archimedes principle says that "a floating body displaces it's own weight of the liquid in which it floats".
So the cargo ship displaces 220,000 tons or 200,000,000 kg of water. For a density of water of 1000kg per cubic metre that's 200,000 cubic metres of water displaced.
Apparently some other ships are detouring around the Cape of Good Hope, some detour!
| {
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Why phonons are Goldstone modes? I read this in the lecture notes by David Tong:
"Gapless excitations often dominate the low-temperature behaviour of a system, where they are the only excitations that are not Boltzmann suppressed. In many systems, these gapless modes arise from the breaking of some symmetry. A particularly important example, that we will not discuss in these lectures, are phonons in a solid. These can be thought of as Goldstone bosons for broken translational symmetry."
Can someone clarify a little this part?
| This question is answered in detail in the paper Phonons as Goldstone Bosons. The question What is the difference between a photon and a phonon? is also closely related. Here, I'll just give some basic intuition.
Consider two solids that are identical except that one of them is shifted slightly in space compared to the other. Both of them are equally stable: the laws of physics are invariant under continuous spatial translations, so a slightly-shifted solid has the same energy as the original. In other words, the existence of the solid "spontaneously breaks" the underlying continuous translational symmetry, which just means it has selected one arbitrary state from the continuum of most-stable states that differ from each other only by overall translations.
A phonon is a wave in the locations of the atoms/molecules in the solid. In the infinite-wavelength limit, such a "wave" reduces to a rigid displacement. That doesn't cost any energy, so the energy of a phonon smoothly approaches zero in the infinite-wavelength limit. That's what we mean when we say phonons are massless.
When we call something a Goldstone mode, we mean that its masslessness can be attributed to the fact that a particular symmetry is spontaneously broken. The situation I just described fits that description: a solid spontaneously breaks the continuous space-translation symmetry, and phonons are massless because a phonon becomes a rigid translation in the infinite-wavelength limit. So phonons are Goldstone bosons of spontaneously broken translation symmetry.
| {
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Is the work done on a system always equal to the negative of that work done by the surroundings? I'm trying to conceptualize some aspects of thermodynamics for myself. Many textbooks often define the work done as the external force acting on the system, resulting in the formula that looks like:
$$W = -\int P_{ext}\, \mathrm{d}V$$
However, once the textbooks define the work in this manner, they often jump straight into PV indicator diagrams that represent the system's internal pressure! For example, an isobaric process is a horizontal line which does work, but many times this is actually the system pressure.
The questions have then are this:
*
*Is the work done by the system equal and opposite to the work done by the surroundings? If so, how can this be proven generally? Is this just effectively a restatement of Newton's 3rd law?
*For practical problems, do we decide to choose which ever is more convenient for the calculation? For example, if I had a piston in an external pressure reservoir, then it is obviously most easiest to calculate the work with the known external pressure instead of the more complex evolving internal pressure?
*How can I show that isothermal expansion/compression under a constant external surrounding pressure is equal to the work done by the internal pressure if they are equal and opposite? If I consider the above formula for work, then the work done by surroundings on system is $W = -P_{ext}\Delta V$, but how can I start with $P_{int}$ to arrive at this same result?
| Your interpretation in terms of Newton's 3rd law is absolutely correct. In an irreversible process, the gas does not pass through a sequence of thermal equilibrium states like it does for a reversible process. The ideal gas law is only applicable to thermodynamic equilibrium states, and gives incorrect values for other non-equilibrium states. So the ideal gas law cannot be used to determine the gas force on the piston in an irreversible process. So, unless you know the external pressure (and/or the gas lifts or lowers a specific external weight), you are stuck.
| {
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Question on Faraday's Induction Experiment We all know that moving a magnet through a loop of wire induces a current, like in this youtube experiment here. Similarly, we know that moving one solenoid (with a battery hooked up) through a larger solenoid (with no battery) will induce a current in the larger solenoid, just like in Faraday's original experiment setup below. Call the small solenoid S1 and the large one S2.
But for the charges in the S2's wire to move, they must experience a force from V cross B (where V is velocity of S1, and B is magnetic field of S1). Since V points along S1, there must be some B component that points radially. See image below; is it safe to say that S2's induction is due to the field lines outside S1, not inside?
If yes, then what would happen if you used truly infinite solenoids? Then you would have no field lines outside S1, so would there be no induced current at all?
| Faradays law is that the induced voltage in a coil is
$\epsilon = -N\frac{\Delta \phi}{\Delta t}$
Where $N$ is the number of turns in the coil, $\Delta \phi$ is the change of magnetic flux through it and $\Delta t$ is the change in time.
For infinite coils the $\Delta \phi$ part would be zero, so you are right that there would be no induced voltage or current at all. It's the change in the magnetic flux that matters...
| {
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What is astrophysical fluid? I am reading a book about astrophysical fluid dynamics, some basic fluid equations are mentioned, but is there any difference between astrophysics case and general case? Is there any definition about astrophysical fluid?
| The fluid dynamics equations are the same (continuity, conservation of momentum, and energy). What makes peculiar the astrophysical fluids are the huge variety of dynamical conditions (from almost stationary conditions to the need of relativistic fluid dynamics), the possible presence of important magnetic fields (magnetohydrodynamics is a specialized branch of fluid dynamics), and their effects on charged fluids (plasmas), and finally, the vast difference of microscopic regimes, requiring different treatments at the macroscopic level.
Finally, the constitutive equations, providing specific information about the material the fluid is made of, are also widely varying, going from the equation of state of rarified gases to the dense neutron fluid in a neutron star.
| {
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How to know if the error is in a law or in uncertainty of the measurement? I read these words in a (great) answer to this question:
There are errors that come from measuring the quantities and errors that come from the inaccuracy of the laws themselves
But how do we know that the errors are in the measuring or in the law about which we make measurements?
| Every measured value has errors. This is the principle stand of the Guide to the Expression of Uncertainty in Measurements. The magnitudes of the errors can be determined (quantified). They consist of offset (calibration) errors, measurement (device) errors, and random errors. A term with less negative context is uncertainty rather than error.
Suppose that we make a comparison of a measured value to a value predicted under a certain law. This comparison requires that we do include two things. First, we must include the total uncertainty in the measured value. This sets the confidence that we have that our measured value is an accurate representation of the reality (e.g. that the device is well-calibrated). It also sets the confidence that we have that we are using the best precisions in our devices and doing our experiments to the most reproducible manner possible.
The second thing we must do is to assure that the conditions of the experiment are not a prior outside the bounds required to apply the law. This validates in advance why we permit ourselves to use the law for comparison.
We compare whether a measured value is different from the prediction within the confidence level of the measurement and within the validity of the assumptions in the law.
What happens when we discover that a measured value is different from expectation? We can take one of two steps. 1) We can realize that our measurement has an as yet unrealized error. Maybe the devices were not calibrated or not calibrated properly. Maybe we have not done sufficient replicate experiments to encompass a strong confidence range on the population statistics. 2) We can realize that our experiment was not done within the full bounds of the assumptions required to apply law that we chose to use. Maybe we neglected to follow a critical assumption that must be in place to apply the law. Maybe we overstepped a limit on a limiting contingency to apply the law.
In summary, laws do not have errors (uncertainties); measured values have errors and laws have assumptions against which we must verify our measurement process. Laws are not inaccurate in and of themselves; measured values can be deemed accurate or inaccurate in reference to the law against which they are compared.
| {
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Why should $\lim_{V\to\infty} \frac{1}{V} \ln Q(z, V, T)$ have a finite limit? In the book Intro. Statistical Physics by K.Huang, on page 174, it is given that
In the thermodynamic limit $V \rightarrow \infty,$ we expect that:
$$
\frac{1}{V} \ln Q(z, V, T) \underset{V \rightarrow \infty}{\longrightarrow} \text { Finite limit. }
$$
where Q is the grand canonical partition function.
This is expected but is there any mathematical or physical reason and/or evidence/explanation for why this is/should be the case?
| In grand canonical ensemble, the partition function $Q$ and the grand potential has relation
$$
\Phi = -K T \ln Q.
$$
The grand potential (also known as Landau free energy) in thermodynamics can be derived using Legendre transformation from internal energy $U(N,V, S)$, where $S$ is the entropy:
The Helmholtz free energy
$$
F = F - TS; \text{ and } F = F(N, V, T).
$$
Then, the grand potential:
$$
\Phi = F - N\mu; \text{ and } \Phi = \Phi(\mu, V, T).
$$
The grand potential is an extended quantity, as well the volume $V$, but $T$ and $\mu$ are intensive quantity. The form of grand potential:
$$
\Phi = P V.
$$
Thus,
$$
\lim_{\text{thermal-limit}} \frac{\ln Q(\mu,V, T)}{V} = -\frac{P}{KT}.
$$
$P$ and $T$ are intensity variables will not change as the system becomes larger.
| {
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Is Rutherford scattering formula inconsistent with reality? On our way to deriving the famous Rutherford scattering formula, we get a formula for the fraction ($f$) of incident alpha particles scattered by $\theta$ or more and this formula has the form
$$f=\pi n t\left(\frac{Ze^2}{4\pi \epsilon_0 K_E}\right)^2\cot^2(\theta/2)$$
where $n$ is the number of atoms per unit volume, $t$ is the thickness of the foil etc. My issue with this is that the right handed limit of $\cot^2(\theta/2)$ as $\theta \to 0+$ is infinity. But this leads to a contradiction because we expect that the fraction of particles scattered by an angle of $0$ degrees or more should be exactly one. This formula makes the claim that the fraction is in fact infinite. So what is going on here? Is it that the formula breaks down for all angles smaller than that particular angle $\theta_0$ for which $f(\theta_0)=1$?
Any help on this issue would be greatly appreciated!
| Let's take a step back.
We begin with the impact parameter $b(\theta)$
and the total cross section $\sigma(\theta)$
for an alpha particle being scattered by an angle of $\theta$ or more
by a single atomic nucleus.
$$b(\theta) = \frac{Ze^2}{4\pi \epsilon_0 K_E}\cot(\theta/2)$$
$$\sigma(\theta) = \pi b^2(\theta) = \pi\left(\frac{Ze^2}{4\pi \epsilon_0 K_E}\right)^2\cot^2(\theta/2) \tag{1}$$
The important thing is: This formula was derived from analyzing a two-body-problem
(an alpha particle and a single atomic nucleus).
I.e. the alpha particle suffered only one single scattering event.
This approximation is fine when you consider only sufficiently
large scattering angles $\theta$, or equivalently, sufficiently
small cross section areas $\sigma(\theta)$.
Then there is only one atomic nucleus inside a cylindrical tube
of cross section area $\sigma(\theta)$.
And you can apply statistical reasoning to find the fraction
$f(\theta)$ of alpha particles being scattered by the many
atomic nuclei of a gold foil.
$$\begin{align}
f(\theta) &= n t \sigma(\theta) \\
&= \pi n t \left(\frac{Ze^2}{4\pi \epsilon_0 K_E}\right)^2\cot^2(\theta/2)
\end{align} \tag{2}$$
The situation becomes more complicated when you consider smaller
scattering angles $\theta$, or equivalently, bigger cross section
areas $\sigma(\theta)$.
Then there will be several atomic nuclei within the cylindrical
tube of area $\sigma$.
And hence the alpha particle will be scattered by more than one atomic nucleus.
That means, the requirement (single scattering event) used to derive
formulas (1) and (2) is no longer valid
and the formulas can no longer be applied.
So formula (2) is only applicable if $f(\theta)\ll 1$.
| {
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Series combination of springs When a spring mass system is connected vertically with two massless springs in series whose spring constants are $k_1$ and $k_2$ to a block of mass $m$ we know that equal forces act on both the springs. Let that force during oscillations be $F$.
When we calculate effective spring constant $k_s$, why don't we say the net force acting on the system is $2F$?
Finding net force acting on the above system:
When the block is attached,the system attains equilibrium position through displacements $x'_1$ and $x'_2$.
At equilibrium:
$2F'=mg$(Where $F'$ is magnitude of spring force initially by each spring)
So, $k_1x'_1+k_2x'_2=mg$ (equation 1)
When the system is pulled down it makes oscillations,now:
Total elongation be $x$
Elongation in spring 1 be $x_1$ and elongation in spring 2 be $x_2$.
Total spring force $= -k_1x'_1-k_2x'_2-k_1x_1-k_2x_2$
Total forces acting on the system $= -k_1x'_1-k_2x'_2-k_1x_1-k_2x_2+mg = -mg-k_1x_1-k_2x_2+mg$
(from equation 1)
So, total force $= -k_1x_1-k_2x_2 = F_1+F_2=2F$(as we know that both forces are equal)
So net force acting on the system is $2F$
The way I calculated effective spring constant is:
$x=x_1+x_2$
$2F/k_s = F/k_1 + F/k_2$
$2/k
_s = 1/k_1 +1/k_2$
But that is not a correct equation.
What's wrong in taking net force acting on system as $2F$.
| The springs are massless, so the tension at any point in either spring is the same value, $F$. For the two individual springs, with extensions $x_1$ and $x_2$, we have#
$F = k_1x_1=k_2x_2$
and when considered as a single spring they have an effective spring constant $k$ such that
$F = k(x_1+x_2)
\\\displaystyle \Rightarrow F = k\left(\frac F {k_1} + \frac F {k_2} \right)
\\\displaystyle \Rightarrow \frac 1 k = \frac 1 {k_1} + \frac 1 {k_2} $
Note that if we join two identical springs in series (so $k_1=k_2$ and $x_1=x_2$) then the spring constant of the joint spring is only half that of its two components. This is because it produces the same force $F$ with twice the extension of the individual springs, since $x=2x_1$.
| {
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Angular momentum commutation relations The operator $L^2$ commutes with each of the operators $L_x$, $L_y$ and $L_z$, yet $L_x$, $L_y$ and $L_z$ do not commute with each other.
From linear algebra, we know that if two hermitian operators commute, they admit complete sets of common/simultaneous eigenfunctions. The way I understand this statement is that the eigenfunctions of both operators are the same. So, if that were the case, that would mean that $L_x$ has the same eigenfunctions as $L^2$. The same goes for $L_y$ and $L_z$. That would mean that $L_x$, $L_y$ and $L_z$ all have the same eigenfunctions, which doesn't seem to be true since they do not commute with each other. How is this resolved?
| Since there's probably a mathematical development of this property in your textbook/notes, I'm guessing you want an intuitive approach on it. If this doesn't help you I'll develop the mathematics:
Imagine $L^2$ has having three properties, let's call them colors. $L^2$ is green, red, and blue. $L_x , L_y , L_z$ are respectively just green, red and blue. If two operators have the same color, they'll commute, and if they don't have any common color, they won't commute. You can easily see the commutation relationships you describe above are true for this "colorful" approach.
In fact, think of this funny experiment. The identity matrix surely commutes with any matrix you can think of, let's say, A and B. If it worked as you think, this would mean that any arbitrary pair of matrices A and B must commute!
| {
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What is light, a wave or a particle or A wave-particle? What is light?
And how do we know that light is an electromagnetic wave?
I asked my teacher and he said that when you place a compass in light's path, the needle of the compass rotates. Which I think is not a valid answer and thats not what actually happens when we place a compass in path of a light.
| Electromagnetic waves are solutions of the wave equation that we derive from Maxwell's equations. When we use Maxwell's equation to compute predictions of what we expect to see when making experimental observations with light, we see that the predictions agree with the observation. Therefore, we conclude that light is an electromagnetic wave.
The frequency of light is extremely high. Therefore, a compass needle cannot respond fast enough to react to the change in the magnetic field in light.
Light is absorbed and emitted in quantized bits of energy that we call photons. The probability distribution for the detection of such a photon is given by the modulus square of the wave function, which is also a solution of Maxwell's equations. So the wave function and an electromagnetic wave differs only in that the wave function is normalized.
| {
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Translation of Fock's original paper? Has the original paper on the Fock space [1] ever been translated to English? I'm not looking for things like Cook's paper [2], what I want is a faithful traslation from German (if it exists).
[1] V. Fock, Konfigurationsraum und zweite Quantelung, Z. Phys. 75, 622-647 (1932). https://doi.org/10.1007/BF01344458
[2] J. M. Cook, The mathematics of second quantization, Trans. Amer. Math. Soc. 74 222-245, (1953).
| Yes, this famous article by V.A. Fock has been translated from German to English, and quite recently. It is published in English in:
Faddeev, L.D. et al. - „V.A. Fock. Selected works: quantum mechanics
and quantum field theory”, CRC (Chapman & Hall), 2004, page 191.
Library of Congress Cataloging-in-Publication Data
Fock, V. A. (Vladimir Aleksandrovich), 1898-1974
[Selections. English. 2004]
V.A. Fock--selected works : quantum mechanics and quantum field theory / by L.D.
Faddeev, L.A. Khalfin, I.V. Komarov.
p. cm.
Includes bibliographical references and index.
ISBN 0-415-30002-9 (alk. paper)
1. Quantum theory. 2. Quantum field theory. I. Title: Quantum mechanics and quantum
field theory. II. Faddeev, L. D. III. Khalfin, L. A. IV. Komarov, I. V. V. Title.
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Proof of $\mathrm{tr}(\gamma^{5}\gamma^\mu\gamma^\nu)=0$ Using $\gamma^{5}\gamma^\mu=-\gamma^\mu\gamma^{5}$ and $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ I obtain
\begin{equation}\tag{1}
T_{\mu\nu}:=\mathrm{tr}(\gamma^{5}\gamma^\mu\gamma^\nu)=-\mathrm{tr}(\gamma^\mu\gamma^{5}\gamma^\nu)=-\mathrm{tr}(\gamma^{5}\gamma^\nu\gamma^\mu)=-T_{\nu\mu}.
\end{equation}
However, I need to prove
\begin{equation}
T_{\mu\nu}=-T_{\mu\nu}.
\end{equation}
Wikipedia gives a hint:
Simply add two factors of $\gamma^{\alpha}$, with $\alpha$ different from $\mu$ and $\nu$. Anticommute three times instead of once, picking up three minus signs, and cycle using the cyclic property of the trace.
Maybe I don't understand the hint, in any case, by inserting $\gamma^{\alpha}\gamma^{\alpha}=-1$ I simply obtain equation $(1)$.
| Following the given hint, we note that $\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\alpha} = \mathbb{I}$, for $\alpha=0,1,2,3$.
Thus, we can write
$$\mathrm{Tr}(\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5) =\mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5\,\gamma^{\alpha}\,\gamma^{\alpha}) \quad. $$
Further note that for $\alpha\neq \mu$ and $\alpha\neq \nu$ it holds that $$\{\gamma^{\mu},\gamma^{\alpha}\} = \{\gamma^{\nu},\gamma^{\alpha}\} = \{\gamma^{5},\gamma^{\alpha}\} = 0 \quad .$$
By making use of these anti-commutation relations, we can rearrange the first equation to
\begin{align} \mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5\,\gamma^{\alpha}\,\gamma^{\alpha}) &= - \mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\nu}\,\gamma^{\alpha}\,\gamma^{5}\,\gamma^{\alpha}) \\
&= \mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\alpha}\,\gamma^{\nu}\,\gamma^{5}\,\gamma^{\alpha}) \\
&= -\mathrm{Tr} (\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\alpha}\,\gamma^{\mu}\,\gamma^{\nu}\,\gamma^{5}\,\gamma^{\alpha}) \quad .
\end{align}
The cyclic properties of the trace then eventually yield
$$\mathrm{Tr}(\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5) = -\mathrm{Tr}(\gamma^{\mu}\,\gamma^{\nu}\,\gamma^5) \quad . $$
| {
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What is the exponential (or geometric) rule (or law) for uranium enrichment? Uranium ore starts at about .72% U-235...
At ~20% U-235, it is considered to be about '90% of the way' to weapons-grade uranium, which is about ~90% U-235...
Because uranium enrichment in centrifuges follows a geometric (or exponential) law...
I have read about this repeatedly when hearing about Iran's enrichment program....
Does anybody know what the 'rule' or 'equation' is for uranium enrichment...
(I am not trying to build a bomb, I swear....)
Edit: P.S.:
In the Work equation $W_extract = -T R ln(x)$ , what are T, R, and x?
I can find that equation nowhere else....
| You are basically trying to sort the atoms, reducing entropy. The minimum work per mole of extracting an element that has mole fraction $x$ is
$$W_{extract}=-TR\ln(x).$$
So, the cost difference as you go from $x$ to $2x$ is (assuming $x<1/2$) $$W_{extract}(2x)-W_{extract}(x)=-TR\ln(2).$$ Which is constant, as the exponential law says.
Now, for a mixture of two isotopes we need to account for the work of separating each. $$W_{allextract} = -TR \sum_i x_i \ln(x_i)$$ So for two isotopes the total energy cost is $$W=-TR [x\ln(x)+(1-x)\ln(1-x)]$$ which has a maximum for $x=1/2$ (maximal entropy). At this point the reader may well get suspicious: why is it getting easier as $x\rightarrow 1$? The answer is that the above formula assumes a process that keeps entropy constant and is close to reversible. This is a bad approximation as we approach the limit, unless we also want the process to take a very, very long time. For imperfect separation and irreversible processes the cost is higher.
| {
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Why is the identity not considered when expanding a $2 \times 2$ matrix in the Pauli basis? I am aware of the expansion of a two dimensional matrix $M$ in Pauli basis given by
$$ M = \sum_{\mu=0,1,2,3} c_\mu \sigma_\mu$$
with $\sigma_0 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, the Identity matrix and $\sigma_{1,2,3}$ the three Pauli matrices.
However, in this published article on page-14, below equation 28, one finds the following:
Parameterising the qubit operators as $ Q = a \cdot \hat{\sigma} $ , with $\hat{\sigma}$ the vector of Pauli matrices and $a$, a unit vector (I have dropped the subscripts $i$ on $Q$ and $a$ here which merely labels the time in this case and is irrelevant here)
My question: Why is the Identity matrix not taken into account as per this article?
| The set of $2\times 2$ complex matrices can be understood as a real vector space of dimension $2\times 2^2$, or equivalently, as a complex vector space of dimension $2^2$.
You can easily verify that a basis for the vector space is given by the Pauli matrices and the identity matrix. More precisely, we can write
$$\mathfrak{gl}(2,\mathbb C) = \mathrm{span}_{\mathbb R}(\{I,iI,\sigma_x,i\sigma_x,\sigma_y,i\sigma_y,\sigma_z,i\sigma_z\})
= \mathrm{span}_{\mathbb C}(\{I,\sigma_x,\sigma_y,\sigma_z\}),$$
where I'm denoting with $\mathfrak{gl}(2,\mathbb C)$ the space of $2\times2$ complex matrices.
You clearly need the identity in the basis, as $I$ cannot be written as a linear combination of Pauli matrices.
| {
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"url": "https://physics.stackexchange.com/questions/627719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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According to general relativity planets and Sun bend the spacetime (explaining gravity), but does this hold true for smaller objects? According to general relativity planets and the sun bend spacetime, and that is the explanation of gravity. However, does this hold true for smaller objects, like toys, pens, etc.? Do they also bend spacetime?
| It seems to be the case that all matter and all forms energy will bend spacetime. Whether or not there is some kind of matter we're not aware of that doesn't is another question. But, everything you'd encounter in your day to day life will bend spacetime, and produce gravitational waves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/628115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Muon $g-2$ experiment: is there any theory to explain the results? The nature of the experiment has been discussed here, but my main question is this: is there any theory that has predicted the results of this experiment or are we completely clueless about what's happening? In other words, have we come up with a new hypothetical interaction that could explain the results?
| A Nature paper was published on the same day, which seems to have attracted a lot less press. This presents a recalculation of the muon $g-2$ value, using standard model physics and their value is consistent with the new experimental value (Borsanyi et al. 2021). So there's one theoretical explanation of the result!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/628266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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