Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
The direction of frictional force in smooth rolling motion I want to ask about the direction of frictional force in smooth rolling motion which means the rolling object doesn't slide on the surface.
Here is the first case, the wheel rolls to the right so the rotation is clockwise. At the point P, velocity is zero. To make the wheel rolls faster, it bottom must rotate to the left faster and the frictional force direct to the right to oppose its tendency to slide as to keep v=0
However, in this case it's also smooth rolling motion but the direction is different. It is explained in the book that if the object slide, it will slide down the ramp so the frictional force must be up to the ramp
So, in the first case the frictional force have the same direction with the acceleration of the center of mass but it's not in the latter one. Can someone explain the difference between those 2.
If the object want to slide, it first must rolls faster with very big acceleration. I can understand the first case, the force act ton the point P direct to the left to make the wheel rolling faster so the frictional force must direct to the right. In the second case, the object rolls to the left so the force acting on P must direct to the right so why the frictional force direct to the right. Isn't its direction must be to the left.
Also, in both cases above, considering accelerating object but not considering object which move very fast. If the object is rolling very fast at a direction, to which direction will the frictional force be?
| Consider the sum of all external torques; calculated about the Center of Mass of the rolling body; then the angular acceleration tells you the direction of relative slipping w.r.t. contact surface. Now, the frictional force will act opposite to the relative slipping tendency.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/146039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
What is the maximum time dilation factor when orbiting a rotating black hole? Suppose one spaceship is stably orbiting a rotating black hole and another is far away from the black hole. What is the maximum time dilation factor between the two ships? Can it be made arbitrarily large, and if so does that require the black hole to be maximally rotating?
The innermost stable circular orbit (ISCO) for a maximally rotating Kerr black hole is for a prograde orbit at $r=m$. This is supposed to be the event horizon, which would make the time dilation factor infinite. However, the text around equation (22) in these lecture notes says that this "is an artifact of the coordinate system." Does this mean that the time dilation factor is not unbounded?
| You can get the time dilation factor by computing the redshift of a radial photon emitted by someone on a circular orbit, compared to the frequency measured by someone at rest at infinity. The derivation of this formula is a bit involved, but the answer is not too complicated:
$$
\frac{\omega_{emit}}{\omega_\infty}=\frac{r^{3/2}+a}{\left(r^2(r-3)+2ar^{3/2}\right)^{1/2}}
$$
This is for a prograde orbit, and I'm using units where $G=c=M=1$. For an extremal black hole, $a=1$ and the ISCO is at $r=1$, so you can see this factor diverges and you can get arbitrarily high redshifts coming from orbits near the horizon.
It's also interesting to consider the nearly extremal black hole, where $a=1-\epsilon$. In that case the ISCO is located at (again, from a somewhat involved calculation):
$$r_{ISCO} \approx 1+(4\epsilon)^{1/3}$$
Using these formulas, we can compute the time dilation factor coming from the ISCO to lowest order:
$$\frac{\omega_{emit}}{\omega_\infty}\approx\left(\frac{2}\epsilon\right)^{1/3}$$
So it is diverging as $\epsilon\rightarrow 0$, but it is doing so rather slowly. For example, say for some reason you wanted 1 hour in the orbit to correspond to 7 years at infinity, which is a factor of about $60000$. This requires an $\epsilon \approx 10^{-14}$, so it requires the black hole to be very close to extremal. Also if you consider Kip Thorne's bound that says astrophysical black holes can only ever reach $\epsilon\approx .002$, the maximum time dilation factor you can achieve is around $10$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/146105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
"answer_id": 0
} |
Complex Conjugate of Wave Function I've been reading through Griffiths QM book, and the only thing bugging me is they never fully described what $\Psi^* $ should be for any given function. I know it's the complex conjugate at the same time I think I just need concrete examples to solidify it in my head.
What is the corresponding $\Psi^*$ for
\begin{align}
\Psi_n(x,t) =& \sqrt{2\over a} \sin{n\pi x\over a} e^{-iE_nt} \qquad \text{(Infinite square well)} \\
\Psi_0(x,t) = &{m\omega\over{\pi \hbar}}^{1/4} e^{-{m\omega\over{2\hbar}}x^2-iE_0t} \qquad \text{(Simple Harmonic Oscilator)}\\
\Psi_k(x,t) =& Ae^{i(kx-{hk^2\over{2m}}t)} \qquad \text{(Free Particle)}
\end{align}
I think the part that is bugging me is that for the two prior cases the conjugate only alters the time term, but in the last equation, we are also altering the position term. How exactly should I rationalize this and come up with a good generalized concept of what $\Psi^*$ is?
| For every $x$ and $t$, $\Psi(x,t)$ is a complex number. $\Psi^*$ is the conjugate of that number, no more, no less. The reason it seems like sometimes it's only the $t$ part that gets conjugated is simply that often it is the only part of the wavefunction that is complex. Let's use your examples:
$\Psi = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a})e^{-iE_n t}$. We want to calculate $\Psi^*$. Well, since the conjugate of the product of two numbers is the product of their conjugates (that is, $(zw)^* = z^* w^*$), let's do it step by step.
First we need to conjugate $\sqrt{\frac{2}{a}}$, but since it's a real number, it is equal to its conjugate. So we leave it alone and move on. Now we need to conjugate $\sin(\frac{n\pi x}{a})$, but again, this is a real number, because $\sin x$ is real whenever $x$ is real. The last part is $e^{-iE_n t}$. This is actually complex, so we need to conjugate it, and its conjugate is $e^{iE_n t}$. So putting it all together, we have $\Psi^* = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a})e^{iE_n t}$.
Notice how at no point did I say something like "$\sin(\frac{n\pi x}{a})$ depends on $x$ so it shouldn't be conjugated". This is because I don't care what $x$ and $t$ are; all I care about is whether something is real or complex; it just so happens that in your first two examples, only the part that depends on $t$ is complex. But in the free particle wavefunction, everything is complex, so you need to conjugate everything.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/146211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Does a Michelson-Morley experiment uphold mass-energy equivalence and special relativity? If there is an experiment that best supports $E=mc^2$, is it the Michelson-Morley experiment?
| Nuclear plants (and theoretically fusion plants) work with $e=mc^2$.
For example:
The mass of 2 Protons and 2 Neutrons is bigger than the mass of Helium, which consists of 2 Protons and 2 Neutrons. The difference is emitted as energy when Helium is made by the fusion of 2 Protons and 2 Neutrons (the actual reaction in the sun are a bit more complicated, but it's the same principle), as seen in the sun and H-bombs.
The michaelson-morley experiment disproved the theory of the ether, giving experimental arguments for the theory of relativity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/146408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why do the storms of Jupiter have long life unlike that of Earth? Recently I saw How the Universe Works. In one of the episodes, concerning Jupiter, they told that the storms on Jupiter can survive many, many, times longer than those on Earth.
What is the reason behind it? They said that it is due to its big mass. But how can mass determine the survival of a storm?
I also want to know why Jupiter is so big and other planets are not.
| “Lacking any… surface” (see Autolatry’s answer) in Jupiter is, first, dubious (a metallic hydrogen mantle is conjectured) and second, not very important per se. For example, Uranus and Neptune almost certainly have a relatively dense mantle with a sharp upper boundary, that doesn’t preclude these planets to have very long-living vortices in the atmosphere. How deep such a “surface” is situated may influence longevity of storms.
Important difference between Earth and giants planets is that Earth’s atmosphere is shallow and sparse, whereas giants’ atmospheres are thick and dense. Why does greater density prolong existence of vortices? Mainly because dense fluids tend to have lower kinematic viscosity (dynamic viscosity changes weakly on isothermal compression/decompression, whereas kinematic viscosity is dynamic viscosity divided by density). Dependence on depth is because a shallow fluid layer above a stationary solid exhibits greater shear rate (for similar motion speeds) and hence faster dissipation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/146488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 0
} |
How is the scattering length in 2d defined? Scattering length is 3d is well-defined. In the literature, one can also see scattering length in 2d. How is it defined? Can we even generalize it to 1d?
| We want to find what with length the center of a incoming wave packet appreciate variations. The scattering approach prefer to talk about time and momentum in detriment of space, but we can talk about space after.
The scatering state always can be written as
$$
\psi^{(+)}_g(t)=\int d\alpha\, e^{-iE_\alpha t} g(\alpha) \phi_\alpha + \int d\alpha\,\int d\beta\, \frac{e^{-iE_\alpha t} g(\alpha)T_{\beta \alpha}}{E_\alpha -E_\beta + i\epsilon} \phi_\beta
$$
Where $g(\alpha)$ represent a suitable superposition of free-states $\phi_\alpha$. Note that when we send $ t \rightarrow - \infty $ the state $\psi^{(\pm)}_g(t)$ becomes $\int d\alpha\, e^{-iE_\alpha t} g^{(\pm)}(\alpha) \phi_\alpha$ because the pole in $\frac{1}{E_\alpha -E_\beta \pm i\epsilon}$. The term $T_{\beta \alpha}$ is responsible for include interactions. If we have $H=H_0+V$ then $T_{\beta \alpha}=\langle \beta|V|\alpha^{+}\rangle$, where
$$
|\alpha^{+}\rangle=|\alpha\rangle+\frac{V}{E_\alpha -H_0 \pm i\epsilon}|\alpha^{+}\rangle .
$$
We can calculate the integral on $\alpha$ in the second term by simple analysis of poles contained in $T_{\beta \alpha}$ as function of complex $\alpha$. This is because we can deforming the integral for a large semi circle in the upper or lower complex plane and the integral in this contour is damped by $\frac{e^{-iE_\alpha t}}{E_\alpha - E_\beta+i\epsilon}$. Then only the poles and cuts of $T_{\beta \alpha}$ in complex plane contribute to the integral. For positive times, in the lower, for negative times, in the upper.
The most close pole to the real axis define the typical energy and time of scattering.
For calculate the typical length is simple take the momentum and mass of the in-state and confront to the typical time. Typical time is the inverse of the distance of the pole in units of $\hbar$. This relation of energy and time is a result of the damping mechanism of $\frac{e^{-iE_\alpha t}}{E_\alpha - E_\beta+i\epsilon}$.
In terms of Green's Functions we can calculate this pole. We simply take the unperturbed Green's Function $ G_0(\alpha)= \frac{1}{E_\alpha-H_0 \pm \epsilon}$ and calculate the time of an in-particle stay unperturbed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/146551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Does zero free current entail zero $\vec H$? There are two kinds of magnetic fields (different authors give them different names), $\vec B $ and $\vec H$ which are related by the equation $$ \vec B = \mu_o (\vec H + \vec M)$$ where $\vec M$ is the magnetization.
Ampere's law for free currents states $$\oint_C \vec H \cdot d\vec l = I_{free} $$
This is my question: does zero free current entail zero $\vec H$?
My argument for that is this: since the contour integral of $\vec H$ is zero for all arbitrary curves C in a region of zero free current, $\vec H$ is necessarily zero. However this may not be a mathematically correct argument...
| Yes, I agree with your argument.
Don’t forget $\vec H$ and $\mathrm d\vec l$ are vectors!
$\vec H\cdot\mathrm d\vec l$ is a "dot product",
so the $\oint$ sum the same direction of $\vec H\cdot\mathrm d\vec l$,
Which means if $\vec H$ is a constant
you can take it out the $\int$
and it becomes $\vec H \oint\mathrm d l$ (here $\mathrm dl$ is scalar since they are in the same direction).
So if the free current is zero,
$\vec H$ must be zero!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/146677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If angular velocity & angular acceleration are vectors, why not angular displacement?
Are angular quantities vector?
... It is not easy to get used to representing angular quantities as vectors. We instinctively expect that something should be moving along the direction of a vector. That is not the case here. Instead, something(the rigid body) is rotating around the direction of the vector. In the world of pure rotation,a vector defines an axis of rotation, not a direction in which something moves.
This is what my book (authored by Halliday, Walker, Resnick) says. But they mentioned with a caution that
Angular displacements cannot be treated as vectors. ....to be represented as a vector, a quantity must also obey the rules of vector addition..... Angular displacements fail this test.
Why do they fail in this test? What is the reason?
| Angular displacement is define by change in angle i.e. ${\Delta \theta}=({\theta}_1 -{\theta}_2)$, where $\theta$ is taken as positive in anti-clock wise.
Hence angular displacement ${\Delta \theta}$ has both magnitude and direction. But let when a particle rotate from a point A in $\theta$ angle in anti-clock wise to point B and then back through $\theta$, it comes its initial point A but when a particle starts rotating from B through an angle $\theta$ and after reaching to A it backs to B. Since in two case the end point is different, we conclude that angular displacement is not commutative. That's why it is not a vector quantity.
Now we know that when a particle rotating with a angular velocity $\vec{\omega}$ it linear velocity $\vec{v}$ is always directed in tangential direction of the path.
And angular velocity is defined by: $\vec{\omega}={\frac{{\vec{r}} X {\vec{v}}}{{|{\vec{r}}|}^2}}$. Hence angular velocity is perpendicular to the plane containing ${\vec{r}}$ and ${\vec{\omega}}$ and also obey vector commutative law. That's why angular velocity is a vector quantity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/146897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
What's the escape velocity of Naked Singularities? Penrose's Cosmic Censorship Hypothesis doesn't hold for naked singularities which means that at least light can escape the singularity. But, if we calculate escape velocity with the given mass only, light shouldn't escape.
How to calculate real escape velocity of naked singularities? Also, tell me if there's a physical sense to that other than a mathematical solution of General Relativity.
| A naked singularity is created the same way as an ordinary singularity, the only difference is that if the angular momentum $J$ is greater than the mass squared, $M^{2}$, then the event horizons (which would be at
\begin{equation}
r = M(1 \pm \sqrt{(1 - (J/M^{2}})^{2}
\end{equation}
cannot exist. Also, There's no such thing as escape velocity for a black hole since the definition of a black hole singularity is that any line going to the singularity ends there.
Lines not going to the singularity (however close they may get) don't have to end.
The following paper seems to be very interesting,
though I can't tell for sure as I'm not a complete expert on Naked Singularity theory, or indeed Loop Quantum Cosmology.
http://arxiv.org/abs/gr-qc/0506129
Quantum evaporation of a naked singularity
Rituparno Goswami, Pankaj S. Joshi, Parampreet Singh
Abstract
"We investigate here gravitational collapse of a scalar field model which classically leads to a naked singularity. We show that non-perturbative semi-classical modifications near the singularity, based on loop quantum gravity, give rise to a strong outward flux of energy. This leads to the dissolution of the collapsing cloud before a naked singularity can form. Quantum gravitational effects can thus censor naked singularities by avoiding their formation. Further, quantum gravity induced mass flux has a distinct feature which can lead to a novel observable signature in astrophysical bursts."
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/147060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Velocity of car down an Inclined Plane I was learning to make a car game and came across this situation where my car is on an inclined plane. It's initial velocity is 0. Now, the problem is that my memory serves me this formula to calculate the final velocity
$$v_f^2 - v_i^2 = 2as$$
Since, $v_i = 0$, the equation would be
$$v_f^2 = 2as$$
For an inclined plane $a = g \sin \theta$,
$$ \Rightarrow v_f^2 = 2 g s \sin \theta $$
However, my memory doesn't serve me what to do when you do not know how far the car would have to travel on this inclined plane.
My problem in short:
$\bullet$ I need to add Velocity to the car which is on the inclined plane
$\bullet$ What is the formula I should use? does the formula mentioned above of any good in this case?
$\bullet$ If yes, how would I come over this displacement drawback (drawback in the sense, I would not know how far the car would travel)
$\bullet$ If no, what would you use to add the velocity to the car?
| If i understood your question correctly, you seem to not know the displacement of the car but you still need to increase the velocity of the car.
Velocity of the car has to be expressed in terms of a quantity. I suggest you use velocity as a function of time,
$$v_f = v_i + at$$
$a = g \sin \theta $ as you told correctly. So
$$v_f = v_i + g \sin \theta t$$
With my limited programming knowledge I'm guessing that it is easier to count time rather than displacement.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/147195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How does interpreting negative energy electrons as positrons solve the negative energy problem? How does interpreting negative energy electrons as positive energy positrons solve the negative energy problem? How does change of “interpretation” without fixing the mathematics have such a profound impact on the underlying physics? (After all, don't we still interpret positrons and electrons as excitations of the same underlying spinor field?).
| It basically boils down to the term $e^{-\frac i\hbar E t}$, where the minus can either be included in the energy, making it negative, or into the time. But a negative charge moving backwards in time is exactly the same as a positive charge moving forwards in time, and that is much more sensible than negative energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/147350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Where on Earth does the mass of 1 kg actually produce a 1 kg reading on a digital scale? Gravity on Earth varies by about 0.1% between poles and equator. If someone was buying/selling something mass critical e.g. gold, where is the standard place on Earth where a 1 kg mass produces a 1 kg reading as measured by a device like the following:
Based on some responses, I should give a very specific example. We take the International Prototype Kilogram mass from its repository in Paris. We then take it to one of the poles and measure it using this type of device (also shown above). We then take it to the equator and do the measurement again. The numbers are different.
Is there a place on Earth specified where the scales will read exactly 1kg?
| These digital scales basically measure the force $F$ required to counteract the weight force when used properly. From a measurement of the force, the scale then converts this to a mass measurement using some conversion akin to $m=F/g$. On different places on Earth, you'll get different "mass" measurements since these devices use a single value for $g$.
The answer to your question as to where on Earth will we get the standard value is: It's up to the manufacturer. Take a look at slide 19 of this link. This company chooses the latitude and altitude of own laboratory to calibrate their mass scale.
Now, as a separate issue, you might also be wondering where on Earth local $g$ takes on the value of $9.80665~\text{m/s}^2$, which is the defined value according to Wikipedia. This value occurs at a latitude of about 45 degrees.
If you're interested in the value at sea level for other latitudes, which you might want for calibration purposes, Wikipedia tells me that
$$g=9.780327\left(1+0.0053024\sin^2\phi-0.0000058\sin^2(2\phi)\right)\ \text{ms}^{-2}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/147439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 7,
"answer_id": 1
} |
About Right-Hand-Rule and Cross Projuct In "Physics for Scientists and Engineers with Modern Physics - 8th Edition", p. 842, it gives the total magnetic force on the segment of current-carrying wire of length $L$ in a uniform magnetic filed:
$$
\vec{F_{B}}= I\vec{L} \times \vec{B}
$$
According to Cross Product, it should be this:
But in Fleming's left-hand rule for motors, it says:
Fleming's left-hand rule is used for electric motors, while Fleming's
right-hand rule is used for electric generators.
And gives a picture to describe the right-hand-rule:
You see, it's index finger points to the "field", but according to cross product and the above equation, the index finger should point to the direction of the current.
Why they are not same? Am i missed something?
| The first example determines the force created by a current carrying conductor in a magnetic field. The second example determines the current created in a conductor moving through a magnetic field. They are both correct.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/147628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What does the exponential decay constant depend on? We know the law of radioactivity:
$$N=N_0e^{-\lambda t}$$
where $\lambda$ is the exponential decay constant. My question is: This constant depends of what?
| Here is a table of isotopes versus lifetimes the color code of the lifetimes on the right hand column:
Isotope half-lives. Note that the darker more stable isotope region departs from the line of protons (Z) = neutrons (N), as the element number Z becomes larger
Modeling a nucleus is a many body problem and also a many forces problem. There exists the nuclear force ( strong), the weak and the electromagnetic, leading to sequential decays. As most many body problems the models have to follow the data rather than be predictive.
The nuclear force will give short lifetimes, the electromagnetic ( electron capture for example) a bit longer and the weak the longest of all, as basic inputs. BUT the particular shells of the nucleus filled, the binding energies per nucleon and the ratio of protons to neutrons will have a strong role too, modifying the intrinsic lifetimes of the underlying interactions.
The nuclear shell model allows for the possibility to use fermi's golden rule as given in the answer by user22620, but the specifics of the nuclide under study have to be taken into account, no general solution.
Here is a power point presentation for the essentials of nuclear physics for those interested further.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/147708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is there a scientific term for star formation? It might be my stupidity to think that many laymen terms that most people use to describe some physics phenomena usually have a scientifically accepted term or name?
The process of star formation, does it have a specific name generally used in the scientific word? Or is Star Formation used widely by scientists and the rest of us?
| An ADS search for "star formation" turns up about 142,000 articles with "star formation" in the title or abstract. The first article is a 43 page review paper of Star Formation in Galaxies in the Hubble Sequence, written by Robert Kennicutt, Jr, one of the leaders of the field. He never defines anything else to mean star formation and one of the "key words" used for tagging the article is "star formation."
I think it's safe to say, "star formation" is the technical term that everyone uses. Note, though, that as an extension to Jim's comment, star formation is generally used in the context of a region of space that is currently forming stars, not for a single star that is being formed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/147806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Will an object rotate when we apply a force to it? What would happen if the axis of rotation passes through the centre of mass of an object? Will the object rotate when we will apply a force to the object?
Edit: The object is free, is not fixed to an axis of rotation and force is perpendicular to the object.
| The question (even after the edit) is not very clear. So I will make some general statements about forces, objects and rotation.
In order to cause a change in the angular momentum of an object (which is one interpretation of "rotate", although it can mean "stop rotating" too), you need to apply a torque.
A torque results from a force applied along a line of response that does not pass through the center of mass.
If you have a pair of forces that cancel each other in magnitude and direction, but that are not along the same line, you will get pure rotation. Any other combination (whose vector sum is non zero) will give rise to acceleration of the center of mass of the object - and further, if the net vector does not pass through the center, will also result in rotation.
I hope that clears up your confusion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/147870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What does it mean to say "a paramagnetic material is attracted to an external magnetic field?" I'm just having a hard time wrapping my head around what actually goes on when a paramagnetic material is exposed to an external magnetic field. I understand that the individual dipoles line up so that they point in the direction of the field, but why does that need to happen for there to be magnetic attraction? And what exactly is being attracted in the first place? If I imagine the dipoles are caused by little rings of current, are the rings themselves pulled in the direction of the field? Also, what happens if the dipole starts out pointed out in exactly the opposite direction of the external field?
| Magnetic forces are not easy to apprehend. Personally, I dislike magnets, so in a first step I will use coils. Consider two coils $S_1$ and $S_2$ along the $\vec{z}$ axis at a distance $d$ one from each other. They are fed by a current $\vec{i}=I\vec{e_\theta}$.
As you know, the magnetic field induced by each coil is like :
The magnetic field from $S_1$ on $S_2$ is $\vec{B}=B_z\vec{z}+B_r\vec{e_r}$. The force is $\vec{i}\wedge\vec{B}=-F\vec{z}$ : the second coil tends to be attracted by the first one. Reciprocally, if the material is diamagnetic, the magnetics domains will oppose the field (much like two magnets with same pole facing), so in my analogy the current of the second coil would be negative : the force would be in the opposite direction (repulsion).
Usually, the force between two magnets or a magnet and a magnetic material is calculate with regard to the energy. The first step is quantifying the energy between the magnets, and the force is deduced from the variation of energy in the volume between the magnets. The energy is lower when magnets are aligned and closer, thus the direction of force.
There is a whole article on wikipedia.
With coils, you can see the diminution of magnetic energy as the pooling of the fields of both coils.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/147930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
1-particle non-interacting Green function At $T=0$ in the non-interacting case the $1$-particle Green function for an electron in the excited state $\lambda$ (empty band) is of the form
\begin{eqnarray}
G^{(0)}(\lambda,t-t') = -i \theta(t-t') e^{-i\epsilon_\lambda(t-t')}
\end{eqnarray}
where $H_0 C_\lambda^\dagger|0\rangle = \epsilon_\lambda C_\lambda^\dagger|0\rangle$. This expression can be Fourier transformed into
\begin{eqnarray}
G^{(0)}(\lambda,E) &=& -i \int_0^\infty e^{i(E-\epsilon_\lambda + i\delta )t } dt \\
&=& \frac{1}{E-\epsilon_\lambda + i\delta} \ .
\end{eqnarray}
The question I have is related to the meaning of $E$ for a non-interacting particle. Shouldn't a non-interacting particle just have the energy $E=\epsilon_\lambda$?
In the interacting case things appear more clear. The excited state $\lambda$ has a limited lifetime and hence there is an uncertainty for the energy $E$, hence the actual energy of the particle in a state is given by a distribution (spectral function). But for the non-interacting case the excited state has infinite lifetime, hence $E=\epsilon_\lambda$.
| $E$ is not the energy per se of the particle, it is a Fourier parameter. What gives you the possible energy accessible to the particle is the spectral function
$$\rho(E)=-{\rm Im}(G(E))\propto\delta(E-\epsilon_\lambda),$$
which is peaked at the accessible energy of the free particle $\epsilon_\lambda$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What do we see while watching light? Waves or particles? I'm trying to understand quantum physics. I'm pretty familiar with it but I can't decide what counts as observing to cause particle behave (at least when it's about lights). So the question is what do we see with our eye-balls?
*
*We point a laser (or any kind of light source) to the wall. We see its way from point A to B. Do I "see" a particle or a wave?
*Let's see an average object. It pretty much looks like their "pieces" a observing that influences their behave. Does this means while we're watching a light it acts like particle in quantum level?
| This is misconception that light is some kind of 'mix' of waves and particles. Instead, It actually IS both waves and particles at the same time, you can't separate them from each other.
So probably, the answer could be: you see particles as well as waves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 5,
"answer_id": 2
} |
Capacitance of two non parallel plates What is the formula for capacitance of two non parallel plates at an angle with each other?If the plates were parallel then the value can be calculated as
(PermittivityX area of one plate)/distance between them.But what happens in case the plates are tilted at an angle?The question came to mind while trying to understand electrostatic separator.What would be the derivation of the formula for capacitance of two non parallel plates placed at an angle?
I did get a method from https://web.archive.org/web/20160417130540/http://www.davidpublishing.com/davidpublishing/upfile/12/15/2011/2011121573197833.pdf
Equation 6 from the above link above helps but it is independent of the length of the plates which doesn't seem likely.
| Assuming that the charge distribution is constant, using the knowledge that capacitance is added in parallel, you could treat your angled plate as being comprised of infinitely many parallel plates, approximating the angle of the plate you would like. You would then be able to integrate across this infinity of plates to find your answer.
As I said, this assumes charge distribution to be even across the plate, and so does not account for fringes. We can say that the charge will be even with a fair amount of certainty due to the knowledge that electric charges in conductors will position themselves to produce equilibrium.
For a more in-depth discussion: https://forum.allaboutcircuits.com/threads/capacitance-of-a-non-parallel-plates.121287/
Sorry for such a short answer!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Thermodynamics - ideal gas Question:
1 mol of a monoatomic gas at 298 Kelvin acquires a volume of 3 litres. It is expanded adiabatically and reversibly to a pressure of of 1 atm. It is then compressed isothermally and reversibly until its volume becomes 3 litres. Calculate the change in entropy.
What I tried:
As both the processes are reversible, and since entropy is a state function rather than a path function, we can ignore the middle step and only consider the initial and final state.
Considering $S$ to be the change in entropy:
Thus:
$$S = nC_vln({T_2/T_1})$$
where $C_v$ represents the heat capacity at constant volume, which for a monoatomic ideal gas is $1.5R$. However, I am unable to understand how to get the final temperature. I tried applying several concepts but still couldn't obtain the final temperature of the gas. Please help.
| The gas is expanded adiabatically and then isothermally. Thus the temperature it has at the end of adiabatic expansion stays the same even after the isothermal process.
Ideal Gas equation after adiabatic expansion: $p_aV_a=nRT_a$, where index "a" shows after.
You do not have $V_a, T_a$ in this equation.
However, another equation you can write down is the adiabatic process equation: $TV^{γ-1}=const$, where $γ=5/3$, because the gas is monoatomic and ideal.
Now you have a system of two equations: $$T_bV_b^{γ-1}=T_aV_a^{γ-1}$$ and $$p_aV_a=nRT_a$$, where index a means "after", index b means "before" adiabatic expansion. You have $V_b, T_b, n, R, p_a$, so this is solvable.
Solve for $T_a$ and you should get the correct value.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to make laser beams visible midair without smoke? My question is: Is it possible to make laser beams viable midair without smoke? I thought it would be a great idea to have a (second) smartphone or pc screen without having a solid screen. The reason why it has to be without smoke is that smoke would be too dependent on the environment.
If it is possible to make laser beams appear in some way midair (without smoke, just air), would this be possible? I am not sure if its possible with two lasers "collide" or some other way.
Wouter
| I recently saw a video of a demonstration by a Japanese researcher who came up with a method that used a pair of high-powered (presumably) infrared laser beams that, where they intersected, heated the air enough to turn it into plasma, creating a pulse of white light. It works, but it's slow, low-resolution, & requires staggering amounts of power. If you come up with a better method that actually works, don't tell anybody about it until you've patented it. :)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Time for a particle undergoing brownian motion to reach a point in a volume I was wondering how one could calculate the average time a particle needs to reach a random point in a small sphere (filled by water) with a radius of maybe $10 \mu m$. I thought of using the Stokes-Einstein-Equation, but then I just get a diffusion coefficient with the unit [m²/s]. Does anyone have an idea on how to solve this?
| First remark, the average hitting time is finite because the volume is finite. None of what I write would make sense in an infinite system.
Let us consider that the target is a ball of radius $a$ at the center of the sphere and let us call $T(\vec r)$ the average hitting time for a Brownian particle starting at position $\vec r$ from the origin. $T(\vec r)$ depends only on $r=\|\vec r\|$. Consider now a Brownian motion $\vec B$ during a short time $\mathrm dt$ and compute the variation of the hitting time
$$\mathrm dt=T(\vec r)-T(\vec r+\vec B_{\mathrm dt}).$$
Use the Taylor expansion in spherical coordinates of the right-hand side
$$\mathrm dt=-\vec\nabla T(\vec r)\cdot\vec B_{\mathrm dt}-\frac12\vec B_{\mathrm dt}\cdot H_T(\vec r)\cdot \vec B_{\mathrm dt}$$
where $H_T$ is the Hessian matrix of $T$ which contains only one non-zero element equal to $\frac1{r^2}\partial_r(r^2\partial_r T(r))$ on the diagonal. Taking the average over all realisations of the Brownian motion, one gets
$$\mathrm dt=-\frac12\frac1{r^2}\partial_r\left(r^2\partial_r T(r)\right) \; 2D\mathrm dt,$$ or in a simpler form, with $\Delta_r$ denoting the spherical Laplacian,
$$D\Delta_r T(r)=-1.\tag{1}$$
This form is quite general for the average hitting time. It is actually the Fokker-Planck equation in which we have replaced the time derivative by $-1$.
Let us solve (1). We have $$T(r)=-\frac{r^2}{6D}+\frac{A}r+B.$$
$A$ and $B$ are constants. We must have $T(a)=0$. The second condition depends on the boundary at $r=R$. Let us consider that the boudary is reflecting, so the particles bounce on the sphere and continue to diffuse inside. This is a von Neumann boundary condition, that translates mathematically into $\left.\partial_rT(r)\right\rvert_{r=R}=0$. This defines $A=-R^3/3D$. With $T(a)=0$ we find
$B=a^2/6D+R^3/3Da$.
Finally
$$T(r)=\frac{a^2-r^2}{6D}+\frac{R^3}{3D}\left(\frac1a-\frac1r\right).$$
If the starting point is uniformally distributed inside the sphere of radius $R$, the average is
$$ \int_a^R\frac{3r^2}{R^3-a^3}T(r)\mathrm dr=\frac{(R-a)^2}{15Da}\frac{5R^3+6 R^2a + 3 Ra^2 + a^3}{R^2+Ra+a^2}.$$
Therefore, for $a\ll R$ we get $$\left\langle T\right\rangle\approx\frac{R^3}{3Da}.$$
The average hitting time scales like $R^3$, so is actually proportional to the volume and inversely proportional to the radius of the target.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Separability of a Hilbert space and its implications for the formalism of QM In the text I'm using for QM, one of the properties listed for Hilbert space that is a mystery to me is the property that it is separable. Quoted from text (N. Zettili: Quantum Mechanics: Concepts and Applications, p. 81):
There exists a Cauchy Sequence $\psi_{n} \ \epsilon \ H (n = 1, 2, ...)$ such that for every $\psi$ of $H$ and $\varepsilon > 0$, there exists at least one $\psi_{n}$ of the sequence for which
$$ || \psi - \psi _{n} || < \varepsilon.$$
I'm having a very hard time deciphering what this exactly means. From my initial research, this is basically demonstrating that Hilbert space admits countable orthonormal bases.
*
*How does this fact follow from the above?
*And what exactly is the importance of having a countable orthonormal basis to the formalism of QM?
*What would be the implications if Hilbert space did not admit a countable orthonormal basis, for example?
| I usually see it in the reverse way, but it is a matter of taste. Hilbert spaces, in general, can have bases of arbitrarily high cardinality. The specific one used on QM is, by construction, isomorphic to the space L2, the space of square-integrable functions. From there you can show that this particular Hilbert space is separable, because it is a theorem that a Hilbert space is separable if and only if it has a countable orthonormal basis, and L2 has one.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Why do the high frequency waves have the most number of modes? While reading the Wikipedia page of Ultraviolet Catastrophe, I came across how Rayleigh and Jeans applied the equipartition theorem. They told that each mode must have same energy. Now as the number of modes are greatest in small wavelengths or large frequency, energy radiated will be infinite.
What is mode actually? And why do the large frequencies have the most modes? Please help me explaining these.
I need a math-free explanation.
| You need to look at the idea of Separation of Variables for Partial Differential Equations.
You consider a toy universe comprising oscillators in a box: let's think of a cuboid microwave cavity with electromagnetic fields losslessly confined within perfectly reflecting walls.
The Cartesian components of the electric field all fulfill Helmholtz's equation:
$$\left(\nabla^2 + \frac{\omega^2}{c^2}\right)\psi = 0$$
and they must vanish at all the boundaries. If you work though the standard separation of variables technique, you get a set of allowable modes with standing plane wave variation for the vibration and any general solution is superposition of these allowedd modes. So, the $x$, $y$, and $z$ variations are all of the form $\sin(k_j\,x_j)$.
The electric fields must vanish at the boundaries, this means that the $k_j$ can only be discrete values: $\frac{pi}{L_x},\,\frac{3\,pi}{L_x},\,\frac{5\,pi}{L_x},\,\cdots$ for the $k_x$ values, where we suppose the reflecting walls are at $x=0$ and $x=L_x$. Likewise, the $k_y$ and $k_z$ are restricted to $\frac{(2 n_j + 1)pi}{L_j}$ for $n_j = 0,\,1\,2,\,\cdots$. So we have modes defined by mode numbers: triplets of integers $(n_x,\,n_y,\,n_z)$, and the total variation of each mode is of the form:
$$\psi_{(n_x,\,n_y,\,n_z)}(x,\,y,\,z,\,t) = \sin\left(\frac{(2\,n_x+1)\,x}{L_x}\right)\,\sin\left(\frac{(2\,n_y+1)\,x}{L_y}\right)\,\sin\left(\frac{(2\,n_z+1)\,z}{L_z}\right)\\\,\cos(\omega(n_x,\,n_y,\,n_z)\,t + \delta)$$
But now, substitute this back into the Helmholtz equation and we get:
$$\omega(n_x,\,n_y,\,n_z)^2 = c^2 \left(\frac{(2\,n_x+1)^2}{L_x^2}+\frac{(2\,n_y+1)^2}{L_y^2} + \frac{(2\,n_z+1)^2}{L_z^2}\right)$$
and the above equation can only be fulfilled for discrete $\omega(n_x,\,n_y,\,n_z)$, given that the $n_j$ are integers. Now think of the integer triplets as points in 3D space: the number of modes of frequency $\omega$ or less is the number of these discrete points in the positive $x,\,y\,z$ eighth of the sphere with radius less than $\omega/c$ - in other words, roughly the volume of this eigth, as $\omega$ gets large. Thus the rate of increase of allowed modes with $\omega$ is roughly proportional to the surface area of the spherical sector. In other words, proportional to $\omega^2$. There are roughly $100$ times more modes between frequency $10\,\omega_0$ and $10\,\omega_0 + \Delta$ as there are between frequency $\omega_0$ and $\omega_0 + \Delta$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
What is the largest number of bosons placed in a BEC? What is the record for the largest number of bosons placed in a Bose-Einstein condensate?
What are the prospects for how high this might get in the future?
EDIT: These guys reported 20 million atoms in 2008. "Large atom number Bose-Einstein Condensate machines" (pdf).
EDIT 2: van der Straten's group reported 100-200 million.
https://doi.org/10.1063/1.2424439
https://doi.org/10.1103/PhysRevA.80.043606
https://doi.org/10.1103/PhysRevA.80.043605
But these guys don't seem to claim the largest BEC, so I wonder if there's larger.
EDIT 3: ~ 1 billion hydrogen atoms back in 2000 by Greytak et al.
https://doi.org/10.1016/S0921-4526(99)01415-5
Of course, hydrogen has only 1 nucleon while sodium (used by van der Straten et al.) has 23, so these are quite comparable in terms of mass.
| I would suspect the What is the largest number of bosons placed in a BEC would be the largest vat of liquid below boiling point 4He. I think 4He must be in the ground state to be a non boiling liquid and this is effectively a BEC.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 2,
"answer_id": 1
} |
Fourier and inverse fourier transform in QFT According to my lecture notes, the inverse Fourier transform of an operator $\phi(p)$ is given by
$$\phi(x)=\int \frac {d^4p}{(2\pi)^4}\phi(p)e^{-ip\cdot x}.$$
As @WenChern pointed out below, Peskin has a somewhat different formula on p.20:
$$\phi(\mathbf{x},t) = \int \frac {d^3 \mathbf{p}}{(2\pi)^3} \phi(\mathbf{p},t) e^{i\mathbf{p} \cdot \mathbf{x}}.$$
I am trying to see how these two formulas are equivalent and what would be the corresponding expansions of $\phi(p)$ oand $\phi(\mathbf p, t)$ as a Fourier transform of $\phi(x)$. Also, I'd like to know why one does not take these integrals over the mass shell in the same way one does in the following other definition of $\phi(x)$:
$$\phi(x)=\int \frac {d^3\mathbf p} {(2\pi)^32E_{\mathbf{p}}}[a(p)e^{-ip\cdot x}+a^\dagger(p)e^{ip\cdot x}]\biggr\vert_{p_0 = E_{\mathbf p}}.$$
| Hint:
1. $\phi(x,t)$ at different times are not independent.
2. $\int{d^4p\delta(p^2-m^2)}=\int{d^4p\frac{\delta(p^0-E_p)}{2p^0}}$. The left side of this equation is Lorentz invariant.
This time your question is much clearer.
If $\phi(x)$ is an arbitrary function of $x$, there's nothing confusing. If $\phi(x)$ is constrained by the Klein-Gordon equation, we have
$0=(\square+m^2)\phi(x)=\int{\frac{dp^4}{(2\pi)^4}(m^2-p^2)\phi(p)e^{-ip\cdot x}}$.
Since $e^{-ip\cdot x}$s are linearly independent, $\phi(p)$ must vanish everywhere except on the mass shell $p^2=m^2$. Then the most general form of $\phi(p)$ should be
$\phi(p)=\frac{2\pi}{\sqrt{2E_{\mathbf p}}}[\delta(p^0-E_{\mathbf p})a_{\mathbf p}+\delta(p^0+E_{\mathbf p})b_{\mathbf{-p}}^{\dagger}]$ .
Thus
$\phi(x)=\int{\frac{dp^4}{(2\pi)^4}\phi(p)e^{-ip\cdot x}}=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]e^{i\mathbf{p\cdot x}}}=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-ip\cdot x}+b_{\mathbf{p}}^{\dagger}e^{ip\cdot x}]}$.
Obviously this is just the last equation in your question.
Then the inverse Fourier transforms are
$\phi(p)=\int{d^4x\phi(x)e^{ip\cdot x}}$,
and
$\phi(\mathbf p,t)\equiv \frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]=\int{d^3\mathbf x\phi(x)e^{-i\mathbf{p\cdot x}}}$.
Due to the limitation fo the length of characters, I add the comments below.
The first identity in the last line is the definition of $\phi(\mathbf p,t)$. The second identity in it is the inverse 3-dimensional Fourier transform of $\phi(x)=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]e^{i\mathbf{p\cdot x}}}$. Direct comparison of $\phi(\mathbf p,t)$ and the general form of $\phi(p)$ shows that $\phi(p)$ contains aditional delta functions, while $\phi(\mathbf p,t)$ is free of delta functions. Beides, since $\phi(p)$ is the 4-dimensional Fourier transform of $\phi(x)$, it is not a function of $t$. I don't think that $\phi(p)$ can be understood as "a particle whith 4-momentum $p$". It onlly make sense mathematically. The square root is just a matter of convention which can be absorbed by $a_{\mathbf p}$ and $b_{\mathbf p}$ (see, Peskin p21).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/148929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does "normal torque" exist? Is there any force called normal torque? If a ruler is spinning, and it hits the floor, obviously it will stop. The floor must be exerting a "normal torque" on the ruler. If this exists, please tell me what it is really called.
| Torque is not a force. You can say there is a torque caused by normal forces, but there is no special name for that.
A normal force comes from acting with a force on an object resting next to a surface. The surface prevents the object from moving through it by producing a reaction force that is necessarily normal (perpendicular) to the surface (parallel forces are called friction or other such things). Thus the name "Normal" force. For torques, it would not make sense to call something a normal torque. What is it normal to? A torque is always parallel to the axis of rotation, so every torque is just as normal as any other torque.
The closest thing to this would be if you applied a torque to a fixed object. For instance, if you had a see-saw and you pushed up on the end that was in the air, the ground would prevent the other end from rotating through it. When you applied a torque, the ground applied a counter-torque.
This term, counter-torque, might be what you are looking for. Or, at least, the closest thing to it
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Where do Newton's Laws not work? I'm working on high school level project about Newton's Laws and I picked topic that describes situations, where they dont work. Can you name any practical cases where they do not work? Why do they not work in special and general relativity or quantum mechanics? Why do they not work on very light things (atoms) or super heavy (black holes) or super fast particles in accelerator?
Thanks for any advice!
| *
*Cartoon physics. For example, after running off a cliff, Wile E. Coyote does not begin to fall until he notices.
*The thing about many physical "laws" is that they are very good generalities, often with exceptions. For example.
*
*You can prove that the angles of a triangle always add up to 180-deg, except for the unwritten exception, that it doesn't work for spherical geometry.
*The Second law of thermodynamics is well known, but in 2002, the BBC reported that "experiment involving lasers and microscopic beads that disobeys the so-called Second Law of Thermodynamics"
*The law of conservation of energy was put into question when Henri Becquerel discovered uranium salts that emitted radiation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What would Earth have been without the Moon? Would Earth rotation have been more slowed down because of the tidal effect from the Sun, as seems to be the case with Mercury and Venus? Due to the giant impact hypothesis the angular momentum from the impact was increased and split.
If Earth not would have been a two part angular momentum system, but a single planet with the same rotation at that time as it have today, what would the tidal effect from sun have done to earth?
Are there any theories meant to explain the distribution of angular momentum in the Solar system? Roughly!
The reason for my interest is of course the question how important the Earth-Moon system's creation was to the development of which we are a part. It's a pretty well-shaped planet we live on.
| As shown in a previous answer,
Assuming they have the same density (the Sun's average density is not much smaller than that of the moon) , if they had the same apparent size in the sky, then the mass M of the object will grow as r3 (because M=4/3ρπR3 and R=θr), so the force actually grows linearly with r.
this implies that for same apparent size and density the tidal force is independent of distance , that is, the tidal force of an object of the same aparent size and same density doesnt change with distance. The apparent sizes on the sky of the Sun and the moon are about the same (imagine a total eclipse), although the density of the moon is about 2.3 larger that than of the Sun. Thus, asuming we were still rotating, we would still have tides, although a bit smaller in magnitude.
quoted from here:
Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon (Thurman, H.V., 1994)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Gibbs free energy intuition What is Gibbs free energy? As my book explains:
Gibbs energy is the energy of a system available for work.
So, what does it want to tell? Why is it free? Energy means ability to do work. What is so special about this energy? Can anyone simply explain?
I just want a math-free intuition.
| Short answer: Gibbs free energy $G = U + PV - TS$ combines internal energy $U$, pressure $P$, volume $V$, temperature $T$, and entropy $S$ into a single quantity that measures spontaneity. With that I mean that processes lowering the Gibbs free energy of your system will spontaneously occur, and equilibrium is reached when the Gibbs free energy reaches the lowest possible value.
Processes that increase the Gibbs free energy can be shown to decrease the entropy of the system plus its surroundings, and therefore will be prevented by the second law of thermodynamics. That's why it measures useful energy - your system may contain more energy, but entropy considerations will prevent you from spending it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 8,
"answer_id": 7
} |
Using tracking detector in a double slit experiment, what would we see? Let's say we put tracking detector (eg. a cloud chamber or a more advanced device) behind the double slits.
What would we see?
I think the interference pattern is three dimensional. So there are hyperbolic regions where waves cancel. So in these regions we wouldn't see particles. But speaking of a tracking detector, which direction the streaks would point to then?
Searching in the topic on the other hand I found a different view that says that the detector would reveal via the trajectory which slit the particle come from, so no interference pattern at all. But this way how would particles know in advance what kind of device I want to measure them?
| That's an excellent suggestion for an experiment since Claus Jönsson has performed his experiment with electrons at a double slit. Since the electrons must be in a vacuum, it is questionable whether the diffraction pattern shows in a cloud chamber.
However, if it succeeds it may be, that in a single-photon experiment one can see the gap through which electron passes. If so, the interpretation of the summary diffraction pattern by interfering with itself single electrons must be reconsidered. Then it will be legitimate to consider the influence of the ever-present electromagnetic field of the double slit on the electron. The EM field is always present because of the outer or free electrons in every material. And this EM field is quantized according to todays physics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can the relative position of two masses connected by a spring become negative?
Consider the following diagram. Two masses of 1 kilogram each are attached by a spring of 1 N/m. The $x$-axis is chosen such that $x_1(0)=0$ and $x_2(0)=L$ where $L$ is the length of the spring in relaxation (no restoring force acting on the masses).
At $t=0$ the speed of the first mass on the left is $v$ directed to the right. The objective is to find the relative position of the masses, $d(t)=x_2(t)-x_1(t)$ for $t \ge0$.
My attempt
Initial condition
$$
\begin{cases}
x_1(0)=0\\
\dot{x}_1(0)=v \\
x_2(0)=L\\
\dot{x}_2(0)=0 \\
\end{cases}
$$
Applying Newton's second law to the first mass:
\begin{align*}
\ddot{x}_1(t) &=-(x_1-x_2)\\
s^2 X_1 -s x_1(0) -\dot{x}_1(0) &= -(X_1-X_2) \\
s^2 X_1 -v &= -(X_1-X_2) \\
(s^2+1) X_1 -X_2 &= v
\end{align*}
Applying Newton's second law to the second mass:
\begin{align*}
\ddot{x}_2(t) &=-(x_2-x_1)\\
s^2 X_2 -s x_2(0) -\dot{x}_2(0) &= -(X_2-X_1) \\
s^2 X_2 -sL -0 &= -(X_2-X_1) \\
-X_1 + (s^2+1) X_2 &= sL
\end{align*}
Solving the simultaneous equations, I have
\begin{align*}
X_1 &= \frac{(s^2+1)v+sL}{s^2(s^2+2)} \\
X_2 &= \frac{(s^2+1)sL+v}{s^2(s^2+2)}
\end{align*}
Partial fraction expansion,
\begin{align*}
X_1 &= \frac{L/2}{s} +\frac{v/2}{s^2} +\frac{-sL/2+v/2}{s^2+2} \\
X_2 &= \frac{L/2}{s} +\frac{v/2}{s^2} +\frac{sL/2-v/2}{s^2+2}
\end{align*}
\begin{align*}
D &= X_2-X_1 \\
&=\frac{sL-v}{s^2+2}
\end{align*}
Inverse Laplace transform for $D$,
\begin{align*}
d(t) &= L \cos (t\sqrt{2}) -\frac{v}{\sqrt{2}}\sin(t\sqrt{2}) \\
&=\sqrt{L^2+\frac{v^2}{2}}\cos\bigg(t\sqrt{2}+\tan^{-1}\bigg(\frac{v}{L\sqrt{2}}\bigg)\bigg)
\end{align*}
Question
Why can $d(t)$ become negative? What does it mean?
| I don't think this is the correct answer to the problem. The correct answer is pretty straightforward and simple (it's a sine function times a constant). Check out Kleppner and Kolenkow (Intro to Mechanics, page 128) for the correct solution. Also, I've done this problem using the Lagrangian, but I've never seen Laplace Transforms being used for this problem before.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Cooling a satellite Satellites are isolated systems, the only way for it to transfer body heat to outer space is thermal radiation. There are solar panels, so there is continuous energy flow to inner system. No airflow to transfer the accumulated heat outer space easily. What kind of cooling system are being used in satellites?
| As an example, the International Space Station (ISS) has external thermal radiators. They looks similar to solar panels, but instead of pointing the flat side towards the sun, they point towards empty space. An ammonia loop carries heat from various parts of the space station to the radiators.
This is a picture of a radiator:
(source)
*
*External Active Thermal Control System
on Wikipedia
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 4,
"answer_id": 3
} |
Kirchhoff's current law with a nonlinear resistor It is said that by Kirchhoff's current law
$$
\frac{e - v_c}{R_1} = c\frac{dv}{dt} + f(v_c) + i\tag{1}
$$
and from Kirchhoff's voltage law
$$
v_c(t) = iR_2 + L\frac{di}{dt}\tag{2}
$$
from the following diagram:
It is easy to see equation (2) but I dont see how equation (1) was obtained.
| Kirchhoff's Voltage Law states that the voltage around a closed mesh or loop is zero. In this case, taking an 'imaginary loop current', $i$ around the 'central' mesh in your circuit, you get:
$$v_c(t)-v_{R2}(t)-v_L(t)=0$$
where $v_{R2}(t)$ and $v_L(t)$ are the voltage drop across the variable resistor $R_2(t)$ and inductor $L$, respectively.
I'll leave the rest to you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How was it proven that a quantum entanglement measurement of particle A, affects properties of particle B If I understood the wikipedia article correctly, quantum entanglement claims that information travels instantly between entangled particles. An act of measurement on one of the entangled particles changes (I think just determines) properties of other particle.
*
*How was it proven that the properties are not just predetermined?
For example, if I have 2 particles and know its charge is zero, then if I measure one particle and its charge is positive, I know that the other particle is negative cause of law (equation), their charge was always such, even before my measurement, it was predetermined. Or math equation $x+2=4$, I know the outcome, I know operation, I know one operand, I can solve it and find $x$. I can determine $x$ but I don't change it.
*Can multiple particles be entangled? What happens then?
Does the measurement of one particle effect other particles all together? If we make a math analogy with $x+y-2=0$, we measure $x$ and it is $1$, we know $y$ is $1$, but what if the equation is complex and there are multiple possible answers?
*Can we say that in an atom, the nucleus and electrons is quantum entanglement?
Suppose electrons around atom nucleus not particles but just a function, and they are not determined until we measure the nucleus. For example we found +6 protons in the nucleus, now using math equation $x+6=0$, we know that there must be 6 electrons. Then we measure one electron position, and again we can determine other electron positions using some math equations. How it is different from quantum entanglement? Why we say that in this scenario everything is predetermine and we only discover it, and in another scenario some "spooky action" takes place?
*How is it different from "wavefunction collapse" (measurement problem) and the "which way" double split experiment?
To me it looks like very close things.
| This video well explains Bell's Stern–Gerlach experiment that is cited as proving that quantum entanglement is not explained by predetermined "hidden variables".
https://www.youtube.com/watch?v=ZuvK-od647c
Basically, the experiment involves detecting the spin of two entangled particles via the use of 3 randomly determined spin-direction detectors. Each detector detects spin in a direction 120 degrees from each other detector. The theory is that if the spins were predetermined, the detected spin direction would be opposite 2/3 of the time, while in reality, the experiment sees spin direction as opposite 1/2 of the time. The video makes a clearer explanation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How do microwaves heat moisture-free items? Today I learnt that microwaves heat food by blasting electromagnetic waves through the water molecules found in the food.
Does that mean food with 0% moisture (if such a thing exists - dried spices?) will never receive heat from a microwave oven? And how in that case is a microwave able to melt plastics etc., which contain no obvious water?
| I don't think it's dielectric heating .. it's more akin to inductive heating .. the establishment of eddy currents (largely surficial) which heat through resistance heating. Best material would be moderately electrically conductive
substance .. hence ceramics containing metal atoms heat - where ceramics containing exclusively metal-oxide atoms - don't!
It's an AC environment so it would be more accurate to refer to the object's reactive impedance (and not it's ohmic resistance).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"answer_id": 2
} |
What does $m^*>m_e$ imply? (the effective mass of electron is larger than its rest mass) From what I understand, the concept of effective mass is just something people come up with to make electrons and holes obey the equation of motion
$$
\vec{F}=m^* \vec{a}
$$
without dealing with the charge carrier and the crystal at the same time. But how could $m^*$ be compared with $m_e$? They do not seem to be related at first sight. What does $m^*>m_e$ imply? Could anyone shed some light on this? Thanks!
| Second derivative of kinetic energy with respect to momentum equals inverse mass of a particle. In a metal, you have a band structure defined through the dispersion relation of the form E(k) where k is wave vector of electron. Second derivative of this expression can be also taken to be some sort of inertia of a particle, as you can see by analogy with a classical particle whose energy is described by the simple formula for kinetic energy. So, you can think of electron as moving in a cristal potential or as moving with effective mass as a free particle...Why is this mass larger then real mass? Well, I dont see why it has to be that way, derivative can diverge for some value of k, but also can become smaller why not? Simplest form of this inertia tensor is one for parabolic band which becomes constant..
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Newton's third law and punching a glass or a feather According to Newton's third law, action force equals reaction force in terms of magnitude.
When I punch a glass, the glass punches me back. If I exert a greater force on the glass, it will break.
Suppose a glass could sustain 100N force and that my muscles can exert up to 200N force: if I went all out, I couldn't punch the glass with a 200N force because the glass would break, which means it's not able to apply a 200N force on me. I apply F = 200N and the reaction is only f = 100N.
Now suppose I punch a feather in a vacuum, can you explain this?: does it matter if someone is holding the feather for you to punch or it's free?
| The question as it stands is not very clear. What you describe as a "200N force" is what you would measure if your fist hit a force probe. What is in fact physical is the momentum of your fist. If you imparted some momentum to your fist and set it in motion on a collision course with the feather, then Newton's laws would predict that your hand would keep moving almost unimpeded forever until it encountered a force. The force that your hand then encounters is a pulling force because it is attached to your arm, which doesn't get any longer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
How do you define a "universe" (in the context of multiverse)? How do you define a universe (in the context of multiverse)?
The traditional definition of universe is something like
"The Universe is all of spacetime and everything that exists therein".
But in multiverse theory one talks about "universes" and this raises the question of how one can in principle distinguish between two different universes.
A similar question is asked here and the top answer says "In addition, as can be shown by some holographic arguments, it is not really meaningful to talk about things that could exist outside our own cosmic horizon (or universe)." But here is an article in the Quanta Magazine which says
"If the universe that we inhabit had long ago collided with another universe, the crash would have left an imprint on the cosmic microwave background (CMB)"
So it seems that at least in some multiverse theories, different universes can send and receive physical information (light or gravitational force) between each other. So the question is
If light or gravitational force can travel between two universes then how would you know both are not part of the same universe (other than by a priori assumption)?
| It is a pseudoproblem of definition. Some people define universe as everything that could ever possible exist, to them the word multiverse is an oxymoron. But those who like to use the idea of a multiverse, use it encompassing different things depending on context. For instance, Max tegmark defines 4 different levels of universes/multiverses:
Level I: Beyond our cosmological horizon
Level II: Universes with different physical constants
Level III: Many-worlds interpretation of quantum mechanics
Level IV: Ultimate ensemble of every mathematically consistent structure
but you can define your own.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How do I make a Gaussian process generate a stream of events? I would like to generate a stream of "random" gaussian events (on for event - off for no-event) that obeys a set standard deviation and mean. what algorithm could I use?
thank you
| First method:
The principle
You can always transform one distribution to another through a transformation of the independent variable, so we get if $p:\mathbb{R}^N\to(\mathbb{R}^+\bigcup\{0\})^N$ is a probability density function of $N$ variables and we transform the independent variables $\mathbb{R}^N\to\mathbb{R}^N$ by some differentiable transformation $y_j(x_1,\,x_2,\,\cdots)$ then we get a new probability distribution $q(y_1,\,y_2,\,\cdots)$ defined by $q = p\, \partial\vec{x}/\partial\vec{y}$, where $\partial\vec{x}/\partial\vec{y} = (\partial\vec{y}/\partial\vec{x})^{-1}$ is the relevant Jacobian of the transformation. So you choose your transformation to transform the domains as needed, and you must choose your Jacobian to shape the transformation. Usually the input density $p$ is the uniform probability distribution generated by a random number generator.
The Practice
The Box-Muller transformation as described by:
Everett Carter, "Generating Gaussian Random Numbers"
is an excellent implementation of this principle.
Press, W.H., B.P. Flannery, S.A. Teukolsky, W.T. Vetterling, 1986; Numerical Recipes, The Art of Scientific Computing, Cambridge University Press, Cambridge
also describes this in full. In short, we take two independent random numbers $x_1$ and $x_2$, which are uniformly distributed in the interval $[0,\,1]$. Then we form:
$$y_1 = \sqrt{-2\,\log{x_1}}\,\cos(2\,\pi\,x_2)$$
$$y_2 = \sqrt{-2\,\log{x_1}}\,\sin(2\,\pi\,x_2)$$
and then $y_1$ and $y_2$ are Gaussian variables with unit variance and mean nought. So you would simply use, say $\sigma\,y_1+\mu$ to get a Gaussian distribution with mean $\mu$ and variance $\sigma^2$.
Second method: Use the central limit theorem. Add one hundred independent random numbers, each uniformly distributed in $[0,\,1]$. The sum will be very nearly Gaussian with a mean of $50$ and a variance of $25/3$ (equal to 100/12, since the variance of the uniform variable is $1/12$). Then scale and shift as in the first method to get your mean and variance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is Optimal Control Theory and Controllability Theory? I was exploring the methods to analyze the evolution of a system from one quantum state to another using a suitable Hamiltonian. Some searching led me to the keywords Controllability Theory and Optimal Control Theory. Although both these topics are concepts in Mathematics, please try to answer my questions with a quantum mechanical perspective.
What is it which we try to optimize using Optimal Control Theory?
Given an initial quantum system (say a superposition of many states), and the final state we want to reach using some unitary transformation, is it possible to calculate the minimum (optimal) time required in the time-evolution? If yes, a working example would be of great help to me.
| The quantum equivalent to the optimal control of densities is a topic of interest in the areas of large population control, optimal mass transport and mean field games (try https://arxiv.org/abs/1810.06064) in the control theory. The idea in the related literature is to transport a density given the individual micro-dynamics. The question of the minimum time required will depend on the available control authority and can technically be driven to zero, given an unbounded control authority.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why is Graphene Transparent? Graphene is always in the news now a days and its key features are that it is; very strong, conductive and transparent. It is so transparent that each layer of graphene will only absorb 2% of Light passing through it.
But what is it about the structure of Graphene which makes it (almost) transparent?
| Graphene is only transparent because it is very thin (one atom thick). If it absorbs 2% per layer then just a few hundred layers would absorb almost all light and that would still be a very thin sheet of graphite.
The question should be why does graphene absorb so much light compared to diamond which really is transparent? A simplified answer is that graphene is a very good conductor because it has only three covalent bonds per atom compared to the full four in diamond. This makes it possible for electrons to move freely over a sheet of graphene to conduct electricity. Like metals this means it will absorb or reflect light because the free electrons can absorb the small amount of energy in the photon. In diamond the photons would need to have enough energy to release an electron from the covalent bonds. For visible light this is not possible so the photons pass right through the diamond and are only stopped or deflected by impurities.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
What is the procedure (matrix) for change of basis to go from Cartesian to polar coordinates and vice versa? I'm following along with these notes, and at a certain point it talks about change of basis to go from polar to Cartesian coordinates and vice versa. It gives the following relations:
$$\begin{pmatrix}
A_r \\
A_\theta
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
A_x \\
A_y
\end{pmatrix}$$
and
$$\begin{pmatrix}
A_x \\
A_y
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
A_r \\
A_\theta
\end{pmatrix}$$
I was struggling to figure out how these were arrived at, and then I noticed what is possibly a mistake. In (1), shouldn't it read
$$A_r=A_x+A_y$$
Is this a mistake, or am I making a wrong assumption somewhere?
I'm kinda stuck here, and would appreciate some inputs on this. Thanks.
| What you need to do to derive the transformation matrix is first wrtie out all the relation between both coordinations.
For example, the relation between cartesian coordinates and polar coordinates are:
$$x=r\cos \theta \hspace{1cm} y=r\sin \theta $$
$$r=\sqrt{x^2+y^2} \hspace{1cm} \theta = \tan^{-1} \frac{y}{x}$$
From chain rule of partial differiation, we have
$$dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta$$
$$dy=\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta$$
Therefore, in matrix notation we have
$$
\begin{bmatrix}
dx\\dy
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta}\\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}
\end{bmatrix}
\begin{bmatrix}
dr\\d\theta
\end{bmatrix}
$$
By definition, an arbitary vector A must transform the same way as the componenets of the displacement do. Because in here we defined the basic vector $e_{\theta}$ to have a magnetude of r
So we have:
$$
\begin{bmatrix}e_x\\e_y\end{bmatrix}
=
\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}
\begin{bmatrix}e_r\\e_{\theta} \end{bmatrix}
$$
Do an inverse on the transfomation matrix and you shall get the inverse transformation relation.
$$
\begin{bmatrix}e_r\\e_{\theta}\end{bmatrix}
=
\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{bmatrix}
\begin{bmatrix}e_x\\e_y \end{bmatrix}
$$
An alternative way of thinking it, is to rotate the coordination system such that $e_x$ coincide with $e_r$. Therefore, The transformation matrix is coincidentally the same as rotation matrix rotating the cartesian system by $\theta$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
First law of thermodynamics In the first law of thermodynamics, we learned that $W$ and $Q$ are path-dependent quantities, but how are $Q$ and $W$ defined?
I mean $W = \int_{\gamma} p(s) ds$ would be one possibility, where $\gamma$ is a path from volume 1 to volume 2, but how can I calculate $p$ as a function of $V$ in general? So given an arbitrary system, how can I calculate the pressure as a function of $V$?
Similarly for $Q$. If $Q = \int_{\gamma} T(s) ds $ where $T$ is a path from entropy $S_1$ to entropy $S_2$, I have no idea how I get for an arbitrary system $T$ as a function of $S$?
| As you wrote $W = \int_{\gamma} p(s) ds$ you are implying that the pressure, I assume you denoted pressure by $p$, is function of only the volume $V$ and the dummy integration variable is in fact the volume. That is not the case, the thermal equation of state even for the simplest system involves the (absolute or empirical) temperature, as well. That is $p=p(T,V)$. The same consideration goes for your other formula of heat exchange in the case of reversible process $Q_{rev} = \int_{\gamma'} T(s) ds$ where now $s$ is entropy along $\gamma'$. The integrand (caloric equation of state) should be written as $T=T(S,V)$, say, if you take volume as the "configuration coordinate". Note, too, that I have written $\gamma'$ for the path in the $S,V$ to calculate $Q_{rev}$ but $\gamma$ for the $T,V$ plane to calculate $W$. The two paths are different since they are paths on different coordinate planes.
As to the question where does one get the thermal or caloric equations of state $p=p(V,T)$ and $T=T(S,V)$ phenomenological (classical) thermodynamics says that in general you should measure them. For some specific homogeneous systems, such as for gases or crystalline solids, one can actually calculate them by using statistical mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/151055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Does constant velocity of center of mass imply linear momentum is conserved? I know that if momentum is conserved for a system, you can derive that the velocity of the center of mass of that system is constant. I was wondering if the second condition also implies the first: if I can demonstrate that the velocity of the center of mass of a system is constant, does that imply that linear momentum is conserved in the system?
| yes, you assumption is correct, for an isolated system, conervation of linear momentum is equivalent to the velocity of the center of mass being constant. The term with variable mass from another answer is incorrect. You can only have variable mass in a non-isolated sysrtem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/151436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Complex comjugate of Schrodinger equation: paradox in matrix form? We can take the complex conjugate of schrodinger equation, and obtain
$$
-\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi = i \hbar \frac{\partial \psi}{\partial t}
$$
$$
-\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* = -i \hbar \frac{\partial \psi^*}{\partial t}
$$
this seems natural to me, however, does it indicate also the following matrix form is valid?(replace $E$ with $i\hbar d/dt$, the second being a little uncomfortable)
$$
H\phi=E\phi
$$
$$
H\phi^*=-E\phi^*
$$
Suppose we have a Hamiltonian $H$ in matrix form, and solve for the eigenvalue problem, then how are we supposed to know which is "$E$", which is "$-E$".
| The origin of the eigenvalue equation $H\phi=E\phi$ is the separation ansatz $$\psi(x,t)=\exp{\left(-i\frac{E}{\hbar}t\right)}\phi(x)$$
If you conjugate this, this will obviously change the sign of the exponent and therefore you will the same eigenvalue.
What you are trying to state would be something like "if $\lambda$ is an eigenvalue of $H$, so is $-\lambda$, which is obviously not true. Suppose you have an eigenfunction $\phi$, then $\phi^*$ is an eigenfunction to the same eigenvalue (not the negative) due to self-adjointness of the Hamiltonian.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/151547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What will happen to a permanent magnet if we keep the same magnetic poles of two magnets close together for a long time? What will happen to permanent magnet's magnetic field or magnetic ability if we keep same magnetic poles of two permanent magnet for long time?
Will any magnetic loss happen over the long period of exposure or does the magnetic strength remain the same?
Sorry if my logic is wrong. Please explain this.
| I'm trying this now. I have 2 magnets
(round with hole in middle). One is suspended on top of the other with a 1/2"plastic pipe.This site won't accept my pic as is. I believe energy is being used to keep the one magnete suspended. As a result, these magnets will lose their strength and eventually join together.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/152729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
} |
Why does room temperature water and metal feel almost as cool as each other? From what I've read about heat, temperature and conductivity, I understand that the reason water at room temperature feels colder than most other things at the same temperature (like wood, air, cotton) is because of its higher thermal conductivity. That is, it transfers heat quickly from my body to itself, as well as within itself.
(Assuming the thermal conductivity is the only reason why different materials feel colder or warmer), what I don't understand is why metals feel about as cold as water, while their thermal conductivities are 100-to-200 times higher than that of water (Water's is ~0.58 W/mK, the values for metals range from 50 to 400).
I suppose there is more to why materials at identical temperatures suck heat faster; what is it?
| The specific heat of air is more than the specific heat of oxygen in water. Therefore, the water is colder than air. neglect the specific heat of hydrogen in water as its density is negligible comparing to the oxygen. Specific heat of oxygen is 0.9 J/g/cCelcius, specific heat of air about 1 J/g/c
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/152812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Charged dielectric/conductor in capacitor It is a standard problem to consider a dielectric or a conductor between the parallel plates of a capacitor. But what happens to capacity, voltage, charge, inserting between the plates of an ideal capacitor a charged dielectric or a charged conductor (without contact with the plates)?
| There are two case it depends on the battery is connected with constant voltage then There are two main cases:-
1)When Battery is connected $V$=constant
a)Capacity:-
$$C_{0}=\frac{Q}{V_{0}}$$.For a medium permitivity constant will be written as In general,
$$\epsilon_{m}=\epsilon_{0}K$$($\epsilon_{m}$ the permitivity of medium)
So, Potential can be written as $$V_{0}=\frac{q}{4\pi\epsilon_{m}r}$$
I just defined the $\epsilon_{m}$ so
$$V=\frac{q}{4\pi\epsilon_{0}Kr}$$
$$V=\frac{V_{0}}{K}$$
So $$C=\frac{Q}{\frac{V_{0}}{K}}$$
So $$C=C_{0}K$$.
Hence here
$$C =KC_{0}$$
(Here $C_{0}$ is defined for the air )
b)Charge:-
$Q$ is directly propotional to $C$ Hence
$$Q=Q_{0}K$$
2) When Battery is not connected Q=constant
a) Potential
Again $$C=KC_{0}$$
So V is inversely proportional to the the capacitance so,
$$V=\frac{V_{0}}{K}$$
This is the case when the dielectric is added.when it will be remove the cases will change.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/152908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a confirmation of dark matter signal? Dark matter, as we know does not emit light, so confirmation of its presence is indirect. Are there any recent indirect confirmations of dark matter. A place one would look for in detecting dark matter would be to detect gravitational effects of this matter on other. In the astronomical scale there could be X-ray emissions from the dark matter carriers'(the sterile neutrino decay, eg) particles destruction- which could be detected, thereby indicating the existence of dark matter. Are there any updates regarding the presence of dark matter?
| To date there is nothing published (and serious) that makes a confirmed detection of dark matter particles. Thus, the only evidence in favor of its existence remains from indirect methods: calculate the mass that should be there based on visible sources (stars, galactic powder, etc), and use this mass to calculate the speed of stars about the galaxy as the radius of the galaxy goes out. The calculations do not agree and the consensus is that there is more mass than it could add up from every conventional source.
Conclusion 1: there is dark matter
Alternative conclusion (Conclusion 2): Newtonian gravity is not valid at galactic scales. This is a rather fringe topic among researchers but you can learn more about it here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/153025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Formula for the magnetic field due to a current loop I need expressions for the $\mathbf B$ field generated by a circular current loop at a point off-axis from the ring's axis of symmetry.
The ones I came across on the internet aren't very convincing. I verified them with Mathematica, and none seems to be correct ─ I'm checking whether $\nabla \times \mathbf B = I \hat{\mathbf e}_\theta$ and $\nabla \cdot\mathbf B =0$, but the examples here don't satisfy those (so e.g. the latter will have $\nabla \times \mathbf B=0$).
So, more generally: given a ring of current, what is the magnetic field it generates at an arbitrary point? Can this be calculated exactly?
| Remember, $\vec{\nabla} \times \vec{B} = \mu_0 \vec{J}$, and $\vec{J}$ will be zero anywhere except on the loop itself, where it will be singular. Do you mean, perhaps, that the line integral around the loop equals the current, a la Ampere's law?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/153141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Physical examples where changing the order of limits yields wrong result In mathematics it is generally not allowed to change order of limits. For example it is not always true for a sequence of functions $f_n$, that $\int_a^b \left(\sum_{n=0}^\infty f_n(x) \right) dx = \sum_{n=0}^\infty \left(\int_a^b f_n(x) dx\right)$. (Note that series $\sum_{n=0}^\infty\ldots$ and the integral $\int_a^b \ldots dx$ are mathematically defined via limits of sequences).
In my experience it happens a lot in physics lectures, that limits are changed in their order without any additional comment (such as mentioning Fubini's theorem or uniform convergence). It also seems to me that there are not many examples relevant for physics where changing the order of limits yield wrong results.
I'm looking for good physical examples showing to students that one has to be careful when he changes the order of limits. So for which physical example the order of the limits is important and you get a wrong result, when you change it?
| For example, in statistical mechanics, you get different results for systems with spontaneous symmetry breaking, say, for a ferromagnetic, depending on whether you first take the limit $N\rightarrow\infty$ or $H\rightarrow 0$ when calculating average magnetization (http://www.encyclopediaofmath.org/index.php/Quasi-averages,_method_of ).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/153309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Half-integer eigenvalues of orbital angular momentum Why do we exclude half-integer values of the orbital angular momentum?
It's clear for me that an angular momentum operator can only have integer values or half-integer values. However, it's not clear why the orbital angular momentum only has integer eigenvalues.
Of course, when we do the experiments we confirm that a scalar wavefunction and integer spherical harmonics are enough to describe everything. Some books, however, try to explain the exclusion of half integer values theoretically. Griffiths evokes the "single valuedness" argument, but he writes that the argument is not so good in a footnote. Shankar says that the $L_z$ operator only is Hermitian when the magnetic quantum number is an integer, but his argument isn't so compelling to me. Gasiorowicz argues that the ladder operators don't work properly with half-integer values. There are some low impact papers (most of them are old) that discuss these subjects, although they are a little bit confusing.
So, basically, my question is: Does anyone have a decisive argument on why do we exclude the half-integer values from the orbital operator spectrum?
| I had already answered this question, but my answer had a fatal flaw. This should be correct.
First, we have to make a physical demand. We demand that a rotation by $2\pi$ of spatial configuration (distinct from an internal configuration, i.e. spin) leaves physics invariant. From a study of the first homotopy group $Z_2$, we know that in general there are two possibilities for a rotation by $2\pi$ (this is usually studied in the context of topological quantization): $R(2\pi)=1$ and $R(2\pi)=-1$. For $R$ to be a physical spatial rotation, we demand $R(2\pi)=1.$ However, the rotation operator in the $x-y$ plane is $R=\exp(-i\theta L_z)$ where $L_z|\psi\rangle=m|\psi\rangle$. Thus we require $\exp(-2\pi im)=0$. This is only solved by $m\in\mathbb{Z}$, which, by the rules of ladder operators, implies $l\in\mathbb{Z}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/153369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 10,
"answer_id": 4
} |
Why is the gravitational force $10^{38}$ times smaller than the strong nuclear force? Also, why is the weak interaction force $10^7$ times smaller than the strong nuclear force?
| The Anthropic answer to this question is that if gravity were a lot stronger, then the evolution of the universe would have proceeded in a different way, it would have collapsed just after the Big Bang. One can speculate that all possiblities really exists, but we can obviously only find ourselves in those universes with laws of physics that are compatible with our existence. Moreover, the laws of physics will appear to be fine tuned to maximize the probability of our existence. In reality there is no such fine tuning, it's all an observer selection effect.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/153443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Electric field inside a diode When a voltage is applied to a diode (forward or reversed bias) the depletion zone is changed due to charges change in this region. My question is in both case (forward or reversed bias), how the electric field that is responsible of moving the charges in the P and N region is established ? Is it the same mechanism of electric field establishment inside a conductor i.e surface charges density making the electric field?
| You are talking about a p-n junction which forms a diode. Now a diode junction is very small compared to the rest of semiconductor. The diode junction can in a first approximation be taken as an insulator since it does not have any mobile charges compared to the rest of the region. When a voltage is applied at the ends the region outside the junction can be viewed as a conductor and hence in a fist approximation has no potential drop or electric field. Therefore all the applied voltage shifts to the p-n junction and an electric field is added or subtracted from original electric field arising from the built-in voltage. In case of forward voltage the electric field due to the applied voltages oppose that due to the built in voltage and opposite is for reverse voltage. Remember the built-in voltage is always from n side to p side.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/153493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Temperature of electroweak phase transition How does one estimate the temperature at which electroweak phase transition (EWPT) occurred? Somewhere I have read it is around 100GeV but the reason was not explained.
| I don't think that an unambiguous justification can be given because the dynamic of the electroweak symmetry breaking (EWSB) is still unknown. We don't have a well established theory describing how the Higgs scalar potential evolves with the temperature. When people talk about the scale of the EWSB, they usually refer to two possible things:
*
*before EWSB, the weak bosons are massless. After EWSB, they get a mass (91 GeV for $Z^0$ and 80 GeV for $W^\pm$). The scale is therefore of the order of the mass of the weak bosons, roughly 100 GeV.
*before EWSB, the Higgs vacuum expectation value (v.e.v.) is 0, the field is symmetric. After EWSB, the v.e.v. is about 246 GeV. So again, the v.e.v. value is representative of the scale of the EWSB, still of the order of 100 GeV.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/153743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
Why do floating water drops form spheres? Consider a drop of water floating in an inertial frame in STP air (e.g., the ISS). Intuitively, the equilibrium shape of the drop is a sphere.
How would one prove that? Is it equivalent to showing that the minimal surface area for a simply connected volume in $\mathbb{R}^3$ with a sufficiently smooth boundary is that of a sphere, i.e., the result of the isoperimetric inequality?
| Another way to look at it is the following.
The main force on the molecules will come from other water molecules and be due to cohesion. The system will try to minimize it's energy and bond the molecules together as much as possible. This means minimizing the surface which results in a sphere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/153840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
What does amplitude in wavelength of light physically mean?What oscilates with time in photon? Like amplitude in wavelength of water waves signify the displacement of water particles about their mean position.
| Although you can (as you obviously know) think of electromagnetic radiation as either a particle or a wave, it's easier in this case to think of it as a wave.
As a thought experiment, if you wave a magnet near a piece of wire, an electric potential will be induced in the wire. Likewise, if you pass current through a wire, a magnetic field will be produced around the wire. You can restate those two observations as: "a changing magnetic field produces an electric field; and a changing electric field produces a magnetic field."
If you work out the math to describe those interactions, you essentially get Maxwell's field equations. In a nutshell, if you want to produce an electromagnetic wave, you start by creating a time-changing electric field (say, by running electrons back and forth in an antenna). The changing electric field produces a changing magnetic field. The changing magnetic field in turn produces a changing electric field, et cetera. So an EM wave is just an electric field and a magnetic field leapfrogging their way through space.
To finally answer you question, the "amplitude" of the wave is the strength of the electric and magnetic fields involved. They aren't in units of distance, as are sound waves or water waves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/153935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can velocity be an undefined quantity? We have the image below displaying the uniform velocity by time-distance graph. At every point velocity is constant but what if distance and time both become zero as at origin in the graph is? The velocity must become undefined as $\frac{0}{0}$ is undefined in mathematics or shall we call it zero? If zero then why?
| Actually you have plotted the graph of displacement $x(t)$,
And here the $x(t)=vt$ ,
$v$ is some constant.
now lets take the slope of the graph i.e
$ dx/dt =v $
The slope is constant (equal to $v$ and we physically call it velocity)
now what is the slope at (0,0)?
since the slope is constant it will be still $v$,right?
hence at the origin the velocity is $v$ and clearly it is defined.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/154067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 0
} |
How is energy transferred in Joules law of heating? Joule's law of heating states that an accelerated electron loses its energy, which is then converted into heat energy, by colliding with vibrating atom i.e ions in their lattice site. but we know atom consist of electrons and a nucleus. Where does it collide? How does energy get transferred?
| That's a very hard question to answer with the appropriate level of detail! Very broadly speaking in an ideal metal all atoms are forming a perfectly regular crystal lattice. Conduction band electrons can move freely around these atoms, which makes it easy to pass a current trough the metal.
In a (theoretical) metal with perfect crystal lattice the electrons wouldn't be losing any energy and the metal would behave similar to a superconductor. However, in reality metals are never ideal, they have so called defects. A defect is a place in the lattice where an atom is missing or where it's sitting in the wrong position, or there could be an atom of a different element replacing one of the metal's own.
When electrons pass such a defect, they encounter a discontinuity and get deflected from their ideal path. This can only happen if the momentum of the electrons change and because of momentum conservation that has to change the momentum of the atoms around the defect. The momentum change also transfers kinetic energy from the electrons to the metal lattice, which is the heat that is generated when a current flows trough a conductor.
In reality these processes have to be described with quantum mechanics and that's so complicated that we are still researching many of these phenomena (although simple resistive Joule heating is understood fairly well).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/154242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can electric field representation be obtained from Enge representation using Maxwell's equations? Suppose we have a long electric capacitor. Let $L$ be its length ($z$ coordinate), $W$ its width ($y$ coordinate), and $D$ its full height (full aperture; $x$ coordinate). Let $L\gg W\gg D$.
The shape of it is similar to that of parallel plate capacitor. Actually, it is an electrostatic deflector and is slightly curved (general concept shown in figure), which can apparently be ignored to some extent.
The fringe field falloff is described by the Enge function
$$
F\left(z\right)=\frac{1}{1+\exp\left(a_{1}+a_{2}\left(\frac{z}{D}\right)^{2}+a_{3}\left(\frac{z}{D}\right)^{3}+a_{4}\left(\frac{z}{D}\right)^{4}+a_{5}\left(\frac{z}{D}\right)^{5}\right)}
$$
How can the field near the reference orbit, which runs through the centerline of the device and onwards, be found from the Enge function using Maxwell's equations?
The Enge function coefficients were obtained from analysis of the degenerate case of a parallel plate capacitor composed of infinite semi-planes.
| Let $E_{0}$ be the $x$ component of the electric field at $z\rightarrow-\infty$. Then the $x$ component of the electric field $E\left(0,0,z\right)
$ on the reference orbit can be approximated by $E_{x}\left(0,0,z\right)
=E_{0}F\left(z\right)$.
By Gauss's law $\nabla E\left(x,y,z\right)=0$ near the reference orbit. Assuming the absence of magnetic fields, the Maxwell-Faraday's equation gives $\nabla\times E\left(x,y,z\right)=0$. Also, due to midplane symmetry and $W\gg D$ we can take $\frac{\partial}{\partial y}E\left(x,y,z\right)=0$ near the reference orbit, as well as $E_{y}\left(x,y,z\right)=0$.
The relevant Maxwell's equations are:
*
*$\frac{\partial}{\partial x}E_{x}+\frac{\partial}{\partial z}E_{z}=0$; and
*$\frac{\partial}{\partial x}E_{z}+\frac{\partial}{\partial z}E_{x}=0$.
We apply the second equation at points $\left(0,0,z\right)$ to obtain the Taylor's expansion of $E_{z}$ by $x$ and $z$. Then we apply the first equation at points $\left(x,0,z\right)$ to obtain the Taylor's expansion of $E_{x}$ by $x$ and $z$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/154323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why are position and velocity enough for prediction and acceleration is unnecessary? In classical mechanics, if you take a snapshot and get the momentary positions and velocities of all particles in a system, you can derive all past and future paths of the particles. It doesn't seem obvious why the position and its first derivative are enough and no further derivatives are needed.
For some reason the accelerations (forces) can be expressed by formulas that only mention the position and velocity of particles. For example, the gravitational force only requires knowing positions but some electromagnetic things need velocities as well. Why doesn't anything need the second derivative (acceleration)?
Does this say something about the universe or rather about our way of analysis?
Could we come up with a theory that only requires a snapshot of the positions? Could we devise a set of concepts and formulas where the second derivative is also required for prediction and instead of forces we'd be talking about stuff that induces third derivatives of motion?
Does modern physics (e.g. relativity) have something to say about this curious thing?
|
Why doesn't anything need the second derivative (acceleration)?
Only Newton's gravity law does not use acceleration in the expression for force. In electromagnetic theory with retarded fields, forces are functions of past positions, velocities and accelerations of the charged particles.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/154456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
A precise definition of macroscopic and microscopic objects What are the formal definitions of "macroscopic object" and "microscopic object". How can one differentiate between them? I mean, is there any fixed condition by which we can distinguish between them?
| I suspect this will be closed as "opinion based". I don't believe there is a canonical answer. Usually microscopic scale relates to phenomena that occur on a level much smaller than the system under consideration (atoms in a crystal when you are thinking about the crystal, for example).
There is an analogy with micro- and macro-economics. Micro-economics describes how individuals make economic decisions (spend, save, etc). Macro-economics describes how the system ("the economy" of a city, country, world) evolves as a result of the micro behavior.
They are almost always related - with one being the aggregate of the other. And they may require a different level of effort to observe - although I would not say it's necessarily "naked eye vs aided eye".
For example, when you consider the evolution of a galaxy, the motion of the moons around the planets in the solar systems may be considered "microscopic" - "small compared to the scale of the system under consideration". This means it may not be possible to set an absolute limit on the size where "microscopic" becomes "macroscopic".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/154529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Different frequencies working together How do the different waves of EM spectrum present in the environment not interfere with each other? If they do, how does everything work properly?
The radio waves of mobile phones and wi-fi work together. Why don't they collide with each other, since they are physically present?
| It's a good question, but the answer isn't particularly simple. The wikipedia page on the Superposition Principle is a decent starting point. If you feel like you understand the basics, there's a nice simulation of superimposing different frequency components that might illustrate the idea for you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/154688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the energy of a Gaussian wave packet? Suppose we have a potential barrier situation, that is $V(x)$ is zero everywhere except on the interval $[-a,a]$, where it is equal to some $V_0 > 0$. Introduce some Gaussian shaped wave packet to the left of the barrier, moving right.
What is the energy of the packet (i.e. of the system described by this wavefunction) at each instant of time?
Well, the wavefunction $\psi(x,t)$ is not an energy eigenstate, so the question is asking about the expected value of the energy, I suppose. Does that just mean carrying out the calculation
$$\langle E(t) \rangle = \int_{-\infty}^\infty dx \ \langle \psi_t | x \rangle \langle x | H \psi_t \rangle \quad ?$$
| As Nemis L. pointed out, the expectation value $\langle H\rangle$ is constant, because of Ehrenfest's theorem:
$$\frac{d}{dt} \langle H \rangle = \frac{1}{i \hbar} \langle [ H,H ] \rangle = 0.$$
The other way of seeing this is that the state can be written as a superposition of orthogonal energy eigenstates.
Obligatory image:
Goldberg, Schey, and Schwartz, Computer Generated Motion Pictures of One Dimensional Quantum Mechanical Transmission and Reflection Phenomena", Am. J. Phys., 35, 177 (1967).)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/154752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What's the true reason behind thermal expansion? Thermal expansion is a normal concept everyday. There are 2 explanations:
1, thermal expansion result in stress, then result in deformation
2, thermal expansion result in deformation, then result in stress
I am confused about it. Could you explain the thermal expansion?
| Roughly speaking solid matter is on a lattice form,
A three-dimensional lattice filled with two molecules A and B, here shown as black and white spheres.
The molecules fit like LEGO , the forces tying them together are mainly the spill over electric field forces , attractive and repulsive forming the patterns of the lattice.
In a single crystal one quantum mechanical solution applies and the atomic distances are at their lowest/ground energy state, which has vibrational and rotational degrees of freedom of the molecules and atoms.
Thermal input increases the energy transferred to the lattice and this means that the atoms/molecules transit to higher energy levels by absorbing thermal photons. Higher energy levels for each atom mean higher average distance in the solution of the potential well for each of them. This necessarily means expansion, which will be transferred by electromagnetic interactions, from atom/molecule to atom/molecule. This will induce stress macroscopically, the addition of the impulses will have macroscopic consequences, expansion.
Correspondingly in cooling the atoms/molecules return to the ground state emitting thermal photons and that is the noise heard as the lattice contracts.
So it is the change in energy levels at the microscopic framework, leading to change in average distances which will manifest as stress.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/154821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Resource cost and noise effects in quantum teleportation of multible (entangled) qbits Suppose you have n qubits that are in an unknown state (may be entangled, etc).
Can you teleport this state by teleporting each qubit individually (using a Bell state and a classical channel)?
If not, how many classical bits and what kind of Bell states are needed? Do we suffer an exponential blowup? Does adding small amount of noise (i.e. imperfect hardware) have a large impact on these costs?
| Yes, a many-qubit state (even if it entangled) can be teleported by teleporting each qubit separately using one (perfect) Bell pair and two bits of classical communication. (This is the idea of quantum teleportation: The qubit, including all of its quantum correlations, is teleported.)
This can be seen by observing that teleportation of a qubit implements the identity channel on the qubit, i.e., it maps any state $|\alpha\rangle$ to itself (while preserving the phase!). More precisely, teleportation (like any physical map) is a completely positive trace preserving (CPTP) map which is of the form $\mathcal E(\rho)=\rho$, i.e., it also preserves coherences. Several independent teleportations thus implement the identity on $N$ qubits, i.e., they preserve any states (including entangled ones).
On the other hand, if there is noise (i.e., the entangled states are not perfect) you will need to use an encoding/decoding scheme. For general noise (however small it is), this will only work asymptotically, i.e., you will need to teleport a large number $N$ of qubits using $cN$ imperfect Bell pairs (with $c>1$ some constant), and in the limit $N\rightarrow\infty$, the teleportation will work perfectly given $c$ is not larger than the quantum capacity of the channel.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/154885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
London equation and Maxwell equation I am reading Tinkham's Introduction to Superconductivity (2nd ed) (Amazon link).
On pages 4-5, they state that:"
The second London equation 1.4, when combined with the Maxwell equation ${\rm curl}\ h = \frac{4\pi J}{c}$ leads to:
$$\nabla^2 h = \frac{h}{\lambda^2}$$
where equation (1.4) is:
$$\ h=-c\ {\rm curl} \ (\Lambda J_s )\tag{1.4}$$
$$\Lambda = \frac{4\pi \lambda^2}{c^2}$$
If I take a curl on (1.4) and equate it with Maxwell equation I get:
$${\rm curl} \ h = - c \ {\rm curl}\ {\rm curl}\ (\Lambda J_s) = -c(\nabla(\nabla \cdot) -\nabla^2)\Lambda J_s=(\dagger);$$
I don't see how do they get $\nabla^2h =h/\lambda^2$. I do get: $\nabla^2 J =J/\lambda^2$, if the first term in $(dagger)$ is zero, and $J_s=J$.
So is this a misprint in the book, or am I mistaken?
| Hints:
*
*Given the Maxwell equation:
$$
\nabla\times\mathbf h=\frac{4\pi}{c}\mathbf J
$$
Take the curl of both sides, what do you get? How can (1.4) be used here?
*Vector calculus tells you
$$
\nabla\times\nabla\times\mathbf a=\nabla\left(\nabla\cdot\mathbf a\right)-\nabla^2\mathbf a
$$
What assumptions must you make to get the London equation?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/155084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
planewave Ansatz for modelling phonon dispersion in crystals From Ashcroft's "Solid State Physics", for one-dimensional monatomic Bravais lattice, the equations of motion of ions are:
\begin{equation}
M\ddot u(na)=-K[2u(na)-u([n-1]a)-u([n+1]a)]
\end{equation}
And we seek solutions of the form:
\begin{equation}
u(na,t)\propto e^{i(kna-\omega t)}
\end{equation}
My question is:
Why do we know the solutions to above equations can be expressed in this form?
| This is the classical treatment to model vibrations in solids, using the analogy with vibrations of a one-or-two dimensional monatomic or diatomic chains. Which basically boils down to writing Newton's equation of motion to find out the force on each mass when the whole system constitutes of masses attached by Hookean springs, i.e. for our purposes the potential holding the atoms together is taken as quadratic.
The first differential equation you wrote corresponds to the monatomic 1D case, which in a more compact notation would be (only considering its 2 closest neighbours and $u_n$ being the relative displacement of atom $n$): $$M\ddot{u_n}=F_n=K(u_{\rm n+1}+u_{\rm n-1}-2u_n)$$
As an attempt to find a solution to Newton's equations here one usually makes use of the planewave Ansatz in order to describe the normal modes of vibration as waves:
$$u=Ae^{ikna-i\omega t}$$
Where $A$ is just an amplitude of oscillation and $k$ and $\omega$ are the wavenumber and frequency of the proposed wave. Now bear in mind that all we're trying to reach at the end is the phonon dispersion relation $\omega(k)$ in a given crystal. The educated guess of planewaves here stems from the fact that we're dealing with perfectly periodic systems, in which the normal vibrations, or in other words phonons are modelled as infinitely extended plane waves. Of course for a real crystal, filled with imperfection, things will not be so simple anymore. This is why we need models like PCM: Phonon Confinement Model, where due to imperfections, the perfect periodicity is perturbed and the waves describing phonons become localized, i.e. not simple planewaves anymore. But that's another story.
For a more general understanding on planewave Ansatz, you will find more useful information on this post. At the end of the day, the planewave choice is nothing but an educated guess which works out perfectly for our purposes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/155242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is Boltzmann distribution contradicting with the fundamental assumption of statistical thermodynamics? In equilibrium statistical physics the fundamental assumption of statistical thermodynamics states that the occupation of any microstate is equally probable (i.e. $p_i=1/\Omega, S=-k_B\sum p_i\,{\rm ln}\,p_i=k_B{\rm ln}\,\Omega$). But for isolated system in equilibrium we also have Boltzmann distribution which states $p_i=e^{-\beta E_i}/Z$, where $E_i$ are the allowed energy levels. So the two $p_i$ matches if and only if there is one single allowed energy level. How can we resolve this conflict?
| To clarify, the full assumption is that all states are equally probable, in the absence of any knowledge about the state of the system.
With the Boltzmann distribution (AKA canonical ensemble) this assumption doesn't apply since we have knowledge about the system. In particular we know that if the system is put into contact with a thermodynamic heat bath of temperature $T$ (the same $T$ as in the $e^{-E/(kT)}$ distribution), the system will remain in statistical equilibrium (the distribution will not change). This property, of being in equilibrium with other systems of the same temperature, is special to the Boltzmann distribution and is what makes it so useful.
It's worth noting that many texts give a deceptive statement of the fundamental assumption, that "for isolated systems all states are equally probable". This is not true - isolated systems can be in any distribution. For example if a system is in the Boltzmann distribution because of contact with a heat bath, and that contact is removed to make the system isolated, then its state continues to be described by the Boltzmann distribution. Only after new information about system's state is obtained, does its probability distribution change to something else.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/155351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Cosmic Background Radiation and redshift vs. temperature I get that the CMB has gone from high energy photons, to low microwave photons today. And that is probably due to the redshift from the expanding Universe. But, since CMB is a black body it is temperature dependent, i.e. it has a high peak at high temperatures, but then gradually gets less "peaky" and goes towards longer wavelengths.
So, is it the decreasing temperature of the Universe that is increasing the wavelength of the CMB radiation, or is it the expansion, and therefore redshift, or are they somewhat linked ?
| The blackbody radiation due to the cosmic microwave background was in thermodynamic equilibrium with the rest of the Universe at a temperature of about 3000K when it "decoupled".
The decoupling event (I never understand why it is often call recombination, since the protons and electrons were never combined previously), where electrons were captured by protons, made the universe very transparent to the radiation in it. Hence the blackbody spectrum was fixed at this point, at a temperature of 3000K, and would be the same today if it were not for the expansion of space.
As space expands, all wavelengths get longer by the same factor of $1+z$, where $z$ is the current observed redshift of a cosmic radiation source. The epoch at which the CMB was emitted corresponds to $z \sim 1100$, so all wavelengths have been stretched by a factor of $\sim 1101$.
Wien's law relates the wavelength the peak of the blackbody distribution to the recirocal of its characteristic temperature. Hence this peak wavelength is also shifted and so the current characteristic temperature of the CMB is $\sim 3000/1+z$, which gives the 2.7 K observed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/155433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
General question about the potential barrier problem: Why does $\exp( kx)$ diverge when $x>0$ in the case when $E < V(x)$? For the two images below, the first potential barrier has particles approaching it where $E > V_o$ & the second has a particle that has $E < V_o$, where $E$ is the energy of the particles and $V_o$ is the potential of the barrier:
For the first case, when $x < 0$, the particles have a part that is reflected from the barrier (marked by coefficient $B$) and a part that is transmitted (has a coefficient $A$). When $x > 0$, the particles do not have any part that is reflected from the barrier and only a part that is transmitted (has a coefficient $C$).
For the second case where $E < V_o$, the particles again can be reflected or transmitted when $x < 0$, but for the solution when $x > 0$, the book says:
(Images from "Quantum Mechanics Concepts And Applications" by Nouredine Zettili)
My question is:
I am confused about why there is this difference here? What about $E$ being less than $V_o$ that we need to look if $\exp(kx)$ diverges or not? Why was this not taken into account in the first case when $E > Vo$, i.e doesn't $C \times \exp(ikx)$ also diverge in the first case when $E > V_o$?
| Just try to solve the Schroedinger's equation. You'll find that for $x>0$, if you take the solution as $e^{ikx}$, then
$$k=\frac{\sqrt{2m(E-V_0)}}{\hbar}.$$
Now if $E<V_0$, then $k$ is imaginary, i.e. $k_2=ik$ is real, and thus $e^{ikx}=e^{k_2x}$ is real. Thus in case $E<V_0$ one of the solutions vanishes at infinity, and another one diverges due to the properties of exponential function.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/155518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Why does a pitcher with lemon juice have foam, while one with pure water does not? Whenever I pour water into lemon juice (pouring directly from the tap into the pitcher, not quietly along its edge) I get a foam on top:
The same pitcher with water (same water tap, pitcher, time between the water poured and the picture, temperature):
Another answer discusses the formation of foam when pouring water but I did not find any mentions of lemon juice being particularly surface active (for instance Kitchen Mysteries: Revealing the Science of Cooking only mentions that lemon juice adds water to various sauces and modifies the surface active elements like oil, but does not mention lemon juice's own surface active properties).
Which substances in lemon juice could help to form such a persistent foam (it lasts at least 10 minutes)?
| Manufactured 'lemon juices' such as cordials typically contain emulsifiers which can act as surfactants to lower the surface tension of the liquid. This would result in a foam forming more easily when air becomes 'trapped' during pouring water into the jug.
Even homemade lemon squash which often has honey added as a sweetener will produce a foam because the honey can have a similar effect (as an emulsifying agent).
If it is pure lemon juice producing the foam, my hypothesis would be that naturally occurring amphiphilic molecules in lemon juice act as the emulsifying agent.
In fact, orange, lemon and grapefruit juices naturally contain a class of amphiphilic molecules called phospholipids, including phosphatidylethanolamine and phosphatidylchlorine, which have surfactant properties and help give them a 'cloudy' appearance. This would have something to do with the foaming observed when you pour water into a jug containing the juice as air becomes trapped beneath the surface.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/155663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Does the diffracted electron radiate photons? When electron is diffracted after the slit it might follow different direction, than before the slit. That means, that going through the slit it gains some acceleration. And accelerated charge emits photons.
Thus - does the diffracted electron radiate photons?
| I believe the answers given so far are off the mark and do not directly answer the OP's question: do diffracted electrons emit bremsstrahlung radiation/soft photons?
So I give the answer here: The answer is a resounding yes!
However since the diffracted electrons are scattered in the far forward direction (tiny scattering angles), the momentum/energy transfer is minute and therefore the emitted photons have extremely long wavelengths. Therefore, in any practical diffractive experiment, the radiation is undetected. That the radiation is not detected is very important in order to maintain coherence of the system, and therefore leave the diffraction pattern unspoiled.
(I think CuriousOne was alluding to this kind of answer in the comments section of others' answers)
An amusing thought just crossed my mind: it may be possible to observe the radiation for components of the electron trajectories deflected at sufficiently large scattering angles, in which case it is possible to deduce, in principle, the direction of the scattered electron, leading to the loss of the interference pattern at larger angles!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/155743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Tensor Product vs. Direct Product for three spin-1/2 particles Let us consider three spin-1/2 particles and only focusing on their intrinsic spin $S$. The Hilbert space has then to be $\mathcal H = ℂ^2 ⊗ ℂ^2 ⊗ ℂ^2$. The spin can be described by $V ∈ \text{SU(2)}$ and the fundamental representation $\mathcal D_{1/2}$ with
$$\vec{S} = \hbar\vec{M} = \frac{1}{2}\hbar\vec{\sigma}.$$
Let us choose for the base of $ℂ^2$ (1 particle):
$$\left\lvert\frac{1}{2},\frac{1}{2}\right\rangle = \left(\begin{array}{cc}(1)\\(0)\end{array}\right)≡\lvert\vec{e}_3\rangle, \quad \left\lvert\frac{1}{2},-\frac{1}{2}\right\rangle = \left(\begin{array}{cc}(0)\\(1)\end{array}\right)≡\lvert-\vec{e}_3\rangle. $$
Furthermore according to the Clebsch-Gordan series one gets:
\begin{align}
\mathcal D_{1/2}⊗\mathcal D_{1/2}⊗\mathcal D_{1/2} & = \mathcal D_{1/2} ⊗ (\mathcal D_1 ⊕ \mathcal D_0) \\
& = (\mathcal D_{1/2}⊗\mathcal D_1) ⊕ (\mathcal D_{1/2}⊗\mathcal D_0) \\
& = \mathcal D_{3/2} ⊕ \mathcal D_{1/2} ⊕ \mathcal D_{1/2}.
\end{align}
So we are left with 8 states in the combined 3-particle system.
Questions:
*
*If one would simply consider the direct sum of the three particles, i.e. $ℂ^2 ⊕ ℂ^2 ⊕ ℂ^2$ we would only have 6 states, correct?
*What is the simplest picture to see the consequences of doing this instead of taking the tensor product?
*Maybe one could also give me a good (physical) example for the difference of $ℝ^3 ⊗ℝ^2$ versus $ℝ^3 ⊕ℝ^2$ (phase space?).
I have searched for similar questions and found some stuff. However I am not in particular interested in the look of the outcoming states (i.e. their spin momentum) but more in the differences if one were not considering the tensor product.
| This is mainly about your question 2.
A decomposition of the Hilbert space into a direct sum $\mathcal H = \mathcal H_1 \oplus \mathcal H_2$ represents that the system can be in a state in $\mathcal H_1$, or a state in $\mathcal H_2$ (or a superposition, of course). In your example, $\mathcal H = \mathcal D_{3/2} \oplus \mathcal D_{1/2} \oplus \mathcal D_{1/2}$, the system can be in states of spin $\frac 3 2$, or spin $\frac 1 2$ (and superpositions thereof).
A decomposition of the Hilbert space into a tensor product $\mathcal H = \mathcal H_1 \otimes \mathcal H_2$ represents that the system is described by a state from $\mathcal H_1$ and a state from $\mathcal H_2$ (and superpositions of such states). This is naturally the case for multi-particle systems: a description of the system means giving a state for each particle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/155847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How confident can we be that the speed of light in a medium is constant? I have recently found this article http://news.harvard.edu/gazette/1999/02.18/light.html it tells that physicists have been able to slow the speed of light. Is this hokum? If not how is it possible to use light as a measure of distance? Surely doing this would effect the waveform of light rendering it impossible to tell how far it has travelled or how long it has been travelling.
| That article is about slowing the speed of light in a physical medium (note the mention of the 'exotic medium' in the third paragraph), a common and well-known phenomenon (even ordinary transparent mediums like water or glass will slow light waves traveling through them to some extent), the newsworthy aspect was just the huge degree of slowing they were able to achieve. From what I understand, in quantum theory that basic explanation is that the photons are being repeatedly absorbed and re-emitted as they travel through the medium (not by individual particles of the medium but rather by collective vibrational modes known as 'phonons', see the discussion here), which slows them down even though they still travel at the speed of c between absorptions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/155950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to calculate U-spin for baryons I am trying to calculate U-spin for sigma baryons. I don't know why U-spin for $\Sigma^{+}$ and $\Sigma^{-}$ and $\Sigma^{*+}$ is 1/2, but for $\Sigma^{*-}$ is 3/2?
I know that $\Sigma^{+}$ and $\Sigma^{-}$ are octet, and $\Sigma^{*+}$ and $\Sigma^{*-}$ are decuplet, but I don't understand their U-spins.
| U-spin is very similar to isospin which interchanges u and d quarks, except, now, U-spin interchanges d and s quarks, so, in the Eightfold-way weight multiplets, it transitions to lower right from upper left---unlike for left to right for isospin.
So, behold!, in the baryon octet, the $\Sigma^-$ is part of a U-doublet, the $\Sigma^0$ mixing with the $\Lambda^0$ to provide the center of a U-triplet, and the $\Sigma^+$ of a doublet.
In the decuplet, the $\Sigma^{*+}$ is in a doublet, the $\Sigma^{*0}$ a triplet, and the $\Sigma^{*-}$ a quartet. These are exactly the values u=1/2 and 3/2 you are puzzled by, above. Two pictures are worth a thousand words.
The underlying reason is the asymmetry of the triangular decuplet weight diagram. The top apex of the increasing-V decuplet triangle is the $\Delta^{++}$, so, as you jack up the V-spin, you transition from a U-quartet ultimately to a U-singlet in it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/156021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Electron scattering to measure the nuclear radius I have been taught that you can find out the size of a nucleus of an atom by firing electrons at high velocities at the atom. This causing scattering due the positive charge of the nucleus and diffraction due to the small de Broglie wavelength.
Part of this relies on the first minimum of the diffraction pattern occurring at $\sin \theta = \frac{1.22 \lambda}{d}$, where $d$ is the diameter of the nucleus. This is a variant on the same equation that I came across in light diffraction where $d$ is the width of the slight, or spacing between elements of a grating. Why should $d$ here now become the diameter of a nucleus rather than the distance between nuclei?
Thank you for all your help
| If we evaluate the DeBroglie wavelength of electrons (e.g. with this online calculator), then we find that an electron energy of 1 eV leads us to $1.2\times10^{-9}\,\mathrm m$ or about 1 nm (that's the size of small organic molecules with 2-3 benzene rings), 15 eV for the size of a hydrogen atom. If we use 1 GeV electron energy we finally get down to $1.2\times10^{-15}\,\mathrm m$, which is roughly nuclear size. So if we want to make atomic measurements, a little tabletop experiment will do, to measure the size of nuclei with scattering directly takes a medium-sized accelerator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/156282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Given a wave function $\psi(x)$, is there always a potential $V(x)$ such that $\psi(x)$ is an eigenstate? Given any unit norm wave function $\psi(x)$ which is in the Hilbert space, can we always find a $V(x)$ such that the $\psi(x)e^{-i\omega t}$ is a solution of the corresponding Schrödinger equation? (I mean the Hamiltonian which uses the potential $V(x)$.)
| As the question is about physics and not mathematics the point is that the Schrödinger equation is only defined if a real valued potential exists.
So any wave function - defined as a solution of the given equation - has necessarily at least one corresponding potential. If someone wants to solve your intended problem mathematically she should take in account that a wave function in most cases is not different from zero on the whole real axis and that it is usually defined on a three dimensional space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/156373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Raising and Lowering indices of tensor Why we use metric tensors $g$ to raise or lower indices of tensors, why not using other (invertible) order-2 tensors to do the job?
| Given a finite-dimensional vector space $V$ over some field, there is no natural isomorphism with $V^*$, its dual vector space. If $V$ is equipped with an inner product, this can be used to define an isomorphism between $V$ and $V^*$ by sending any linear form $\omega\in V^*$ to the unique vector $v_\omega \in V$ that is such that
$$\omega(u) = (v_\omega,u),\qquad\forall u\in V.$$
Given a basis $\{e_k\}_{k=1,\ldots,n}$ of $V$, this isomorphisms allows to define the so-called dual basis of $V^*$ by taking vectors $e^k$ that satisfy to
$$(e^i,e_k)=\delta^i_k$$
and mapping them back to $V^*$.
These ideas can be generalised "fibre-wise" to manifolds. Hence on a Riemannian manifold $(M,g)$, the metric $g$ provides an isomorphism between the cotangent bundle $T^*M$ and the tangent bundle $TM$. So given a basis of $T_pM$, there is a dual basis on $T^*_pM$, and the way a vector of $T_pM$ is mapped to the corresponding 1-form in $T^*_pM$ in terms of component w.r.t. these basis is through the components of the metric tensor. In a coordinate-free approach you would have that, given a 1-form $\omega\in T^*M$, there is a vector field $v_\omega\in TM$ such that
$$\omega(u) = g(v_\omega,u),\qquad\forall u\in TM.$$
Locally, when referring to a chart $(U,\phi)$, you can stick to the coordinate basis $\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^n}$ of $TM$ on $U$, whose dual basis is denoted by $\text dx^1,\ldots,\text dx^n$, and with the property that
$$\text dx^i\left(\frac{\partial}{\partial x^k}\right) = \delta^i_k.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/156525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do electromagnetic waves travel in a vacuum? This is perhaps a total newbie question, and I will try to formulate it the best I can, so here it goes. How does an electromagnetic wave travel through for example, the vacuum of space?
I usually see that waves are explained using analogies with water, pieces of rope, the strings of a guitar, etc, but it seems to me that all those waves need a medium to propagate. In fact, from my point of view, in those examples the wave as a "thing" does not exist, it's just the medium that moves (involuntary reference to The Matrix, sorry).
But in space there is no medium, so how does a wave travel? Are there free particles of some sort in this "vacuum" or something? I believe the existence of "ether" was discarded by Michelson and Morley, so supposedly there isn't a medium for the wave to travel through.
Moreover, I've seen other answers that describe light as a perturbation of the electromagnetic field, but isn't the existence of the field, potential until disturbed? How can it travel through something it does not exist until it's disturbed by the traveling light in the first place? (this last sentence is probably a big misconception by me).
| Photons are released in packets. So,when the source is generating the electromagnetic wave it's actually being released in packet forms. These packets are self entities that travel on it's own and doesn't need any medium to sustain it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/156606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 4
} |
Why aren't (domestic) kettles insulated? In my experience of buying and using kettles, I have come across none which are insulated.
The obvious reasons as to why it would be beneficial being that heating time would be reduced, similarly, less power hence money would be required to heat an arbitrary volume of water. Some kettles become very hot on the outside so safety is also a factor!
Is there a reason why this is so, apart from the costs involved? I.e. cost of manufacture vs. operating cost over the product lifetime.
| Most kettles are silver to minimize heat loss through radiation. (They also have small exit holes at the top to minimize heat loss of steam because conversion of liquid water to steam requires latent heat)
I expect the reason that there is usually no thermal insulation is that kettles heat water very quickly and because the air outside the kettle is a poor conductor of heat the ammount of heat lost by conduction/convection is probably minimal compared to the ammount of enrgy that goes into the heating of the liquid to make it boil.
By contrast, a hot water tank in a central heating system stores hot water for long periods of time, so it makes sense to carefully insulate hot water tanks.
In summary, I think one way to think about this is that the air outside the kettle is good enough thermal insulation for the short time that the kettle boils the water.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/156922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Do particles always flow from high to low pressure? In a recent question, it was stated that
particles in high pressure air always flow to lower pressure.
In a pipe with a constriction, fluid flows from from low to high pressure after the constriction. (From here.)
How are these concepts related?
| It doesn't necessarily flow from higher pressure to lower, but from higher energy to lower energy, as per Bernoulli's theorem. That's the answer you want here when you are pertaining to the venturimeter.
Bernoulli's theorem states that total energy remains same between any two points. Total energy includes pressure energy plus kinetic energy and potential energy (datum head). So to maintain equilibrium if pressure reduces at any point, then kinetic energy has to increase, which means velocity increases.
$$\text{total energy}=\text{pressure energy}+\text{kinetic energy}+\text{datum energy}$$
Note that fluid flows from region of higher energy to lower energy. And not from higher pressure to lower pressure, which is a very common myth. Let's see the proof.
Let's look at venturimeter. Let section 1 be normal area section at starting point. Let section 2 be throat section. Let section 3 be the normal section after throat.
First half has converging section. Area reduces and therefore velocity increases to maintain equal discharge between two section to obey continuity ($\text{discharge}=\text{vel}\times\text{area}$). Now as the velocity increases its pressure has to decrease to maintain total energy equilibrium by Bernoulli equation.
$$\text{result: velocity 2 > velocity 1 and pressure 1 > pressure 2.}$$
So yes: fluid just flowed from higher pressure region to lower. The next half is more important than first half. Let's see what does it have.
In the second half the area increases which implies the velocity reduces in diverging section to satisfy continuity equation. Which means pressure energy has to increase to satisfy Bernoulli equation has to increase.
$$\text{result : velocity 2 > velocity 3 and therefore pressure 2 < pressure at 3.}$$
So fluid still flowed from lower to higher pressure region, which is against the myth of fluid only flows from higher pressure to lower pressure.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/157038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
} |
Was everything in the Universe "created" from light? Is the following true?
The only matter existing directly after the big bang was electromagnetic radiation.
| No.
We don't know what happened in the very early stages of the Big Bang because we have no experimentally tested theory that takes us back that far. However, courtesy of the LHC we have an experimentally tested theory that takes us back to a time called the electroweak epoch, and we can use this theory to answer your question in the negative.
Electromagnetism is a low energy effective theory. It works below energies of somewhere around a teraelectron volt, but above that energy it has to be replaced by a unified theory of the electromagnetic and weak forces called (somewhat obviously) the electroweak theory. The discovery that proved this (not that anyone seriously doubted it) was the discovery of the Higgs boson at the LHC in 2013.
Anyhow, the electroweak theory tells us that during the electroweak epoch there were four massless vector bosons. At low energies these become the Z, W$^+$, W$^-$ and the photon, but above the electroweak transition the four bosons were indistinguishable.
I confess I don't know how to calculate the particle content during the electroweak epoch, so I shall say nothing further on the subject. Nevertheless we can be confident that whatever was floating around at those energies was not just photons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/158132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
What's the closed-form of the sum relating to the DOS of simple harmonic motion? In order to calculate the density of states of single particle in the simple harmonic potential, we would calculate that
$$
D(\epsilon)=\sum_{n}\delta(\epsilon-\epsilon_n)
$$
where $\epsilon_n=(n+1/2)\hbar\omega$. In the limit $\hbar\omega\ll1$,we find that
$$
D(\epsilon)\approx\frac{1}{\hbar\omega}\theta(\epsilon).
$$
But I want to know how to calculate the $D(\epsilon)$ exactly by means of some special function for example.
| There is a formal manipulation that can answer your question using well-known formulas. So, let me write
$$
\epsilon_n=(2n+1)\epsilon_0
$$
being $\epsilon_0=\hbar\omega/2$ and
$$
\delta(\epsilon-\epsilon_n)=\int_{-\infty}^{+\infty}\frac{dt}{2\pi}e^{-i(\epsilon-\epsilon_n)t}.
$$
Then,
$$
D(\epsilon)=\int_{-\infty}^{+\infty}\frac{dt}{2\pi}e^{-i\epsilon t}
\sum_{n=0}^\infty e^{i(2n+1)\epsilon_0 t}
$$
where I have formally exchanged the sum with the integral (just one of my yet-to-be-justified mathematical steps). The sum is normally not converging. E.g. if we truncate the spectrum at $n=k$ one gets
$$
\sum_{n=0}^k e^{i(2n+1)\epsilon_0 t}=
\frac{e^{i(3+2k)\epsilon_0 t}-e^{i\epsilon_0 t}}{e^{i2\epsilon_0 t}-1}
$$
that for the exponential becoming $1$ yields that the sum goes to infinity just like $k$. But physicists have a lot of resources to cope with these situations. We can resort to one of the techniques in the Hardy's book and introduce a converging factor into the series as
$$
\sum_{n=0}^\infty e^{i(2n+1)\epsilon_0 t-\delta n} = \frac{e^{\delta+i\epsilon_0 t}}{-e^{2i\epsilon_0 t}+e^{\delta}}
$$
and the limit $\delta\rightarrow 0$ yields the result we wanted. So, finally
$$
D(\epsilon)=-\int_{-\infty}^{+\infty}\frac{dt}{4\pi i}e^{-i\epsilon t}\frac{1}{\sin\epsilon_0 t}
$$
that is the final result provided we add a rule to circumvent all the poles arising due to the sine function at the denominator (see below). Just notice that for $\epsilon_0 t\ll 1$ one has $\sin\epsilon_0 t\approx \epsilon_0 t$. Now, you have to add a rule to circumvent the pole at $t=0$ and this is done in the standard way by adding a $i\epsilon$ at the denominator yielding
$$
D(\epsilon)=-\frac{1}{\hbar\omega}\int_{-\infty}^{+\infty}\frac{dt}{2\pi i}e^{-i\epsilon t}\frac{1}{t+i\epsilon}.
$$
This is exactly the definition of the $\theta$ function and so
$$
D(\epsilon)\approx \frac{1}{\hbar\omega}\theta(\epsilon).
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/158263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Pressure equation working to get units kg m/s$^2$, but stuck? Why does $kg/m^3 \cdot m/s^2 \cdot m$ become $kg\,m/s^2$ (i.e. one newton)
When I do the working, I get $kg/(s^2 m)$ (i.e. metres is on the bottom rather than the top)
My working is:
\begin{align}
\frac{kg}{m^3} \cdot \frac{m}{s^2} \cdot m&= \frac{kg}{m\cdot m\cdot m} \cdot \frac{m}{s\cdot s} \cdot m \\
&= kg \frac{m\cdot m}{m\cdot m\cdot m\cdot (s\cdot s)} \\
&= kg \frac{m^{-1}}{s^2} \\
&= \frac{kg}{s^2 m}
\end{align}
My working for the metre indices: $m^2/m^3 = m^{-1} = 1/m`$
My problem is the metre and its indices. The metre is the opposite way round, but I can't see why. I know the first solution is correct and I am wrong, but I can't see where my working is wrong?
| By multiplying the units you gave in the question, it looks like you're referring to $P=\rho g h$, where $P$ is pressure, $\rho$ is mass density, $g$ is gravitational acceleration, and $h$ is the depth in some fluid (of density $\rho$) at which the pressure is being measured/calculated. Recall that the definition of pressure is force per unit area (as mentioned in the comments above). So, pressure is not the same thing as force. To get a force from a pressure, you have to multiply it (the pressure) by an area (again, as mentioned in the comments above).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/158351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is differential geometry used in solid state? I'm an undergraduate in physics interested in a career in solid state. While I know that any additional math is helpful--I am on time constraints, and can only take a few supplemental classes.
That said, is differential geometry used much in solid state physics? I'm aware of things like Fermi Surfaces, but wonder if much diff. geo. techniques are actually used.
| There is theory of dispersion in crystals. One can say that the differential geometry is used there. I think it is Group theory + differential geometry.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/158416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Dumbbell rotation
What if apply impulse to dumbbell consisting of two masses and massless rod? It is clear for me that left dumbbell will move straight line. But it is not clear for me what kind of motion will have right dumbbell. It will certainly have rotational component, but will its center of mass move?
Let's define distance between masses as $2r$ and assume that we start from rest. Then after the impulse $F dt$ we have $v_1 = F dt / m.$ Also i can find velocity of mass to which force was applied as $v_1=\frac{\int r\times F dt}{mr},$ because $Iw = \int r\times F dt.$
Now i can find velocity of the center of mass, $F dt = 2mV_{cm},$ hence $V_{cm} = \frac{F dt}{2m}.$ From this i can see that dumbbell will have not only rotational component of speed which breaks my intuition:) Furthermore, $w = (v_1 - V_{cm})/r.$
As conclusion, in both cases dumbbells will have the same linear component of speed, but in right case it also will have rotational component.
You may think that it is a stupid question, but i just want to understand well how it works, and i have nobody to ask because i'm learning physics on my own. Thanks.
| It looks like you've already done all the work. It is correct. Yes the center of mass will move.
Any unbalanced force on the object will cause an acceleration of the center of mass. $F=ma$
Any unbalanced force on the object that does not go through the center of mass will provide a torque and a rotational acceleration. $Fd = I\alpha$
Both of these happen simultaneously. It's not the case that the linear acceleration is reduced because the force is off-center.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/158566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Centripetal force in this example? I have a general question regarding the centripetal force. In the example of a ferris wheel where there is a normal force pushing up against the person and the gravitational force pulling the person down, which force is centripetal? I know that the centripetal force counters the linear velocity, tangent to the circle of motion, which allows the object or person to stay in circular motion but which force is actually pulling it towards the center, the gravitational force or the normal force? Also, would it be correct to say that the net force equals zero (since the person is neither moving towards or away from the center) in this example or does the net force equal the centripetal force (since the centripetal force has to counter the linear velocity --- if this is correct, how would I compare the two since linear velocity is not a force)?
I know that if a car is moving around a banked curve, the horizontal normal force will be centripetal but what about in other examples such as the ferris wheel? Also would the net force of a car moving around a bank curved be zero since it is neither moving towards or away the center?
tl;dr - is the net force in a centripetal force example zero or is the net force equal to the centripetal force? Also, how would I relate this to the linear velocity that cancels it out?
Thanks for the help!
| The component of the net force pointing to the center is the centripetal force. In your Ferris wheel example, we can choose three points: top (12 o'clock), bottom (6 o'clock) and halfway (3 o'clock). At 12, centripetal is normal force plus gravitational force; at bottom they are on opposite direction ($F_c=F_n-F_g$) and at 3 only the normal force points inwards while gravity speeds up or slows down the wheel'a rotation (depending on direction of motion).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/158676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is there a physical interpretation of a tensor as a vector with additional qualities? What is a tensor? has been asked before, with the most highly up-voted answer defining a tensor of rank $k$ as a vector of a tensor of rank $k-1$.
But if a scalar is defined as a physical quantity with a magnitude, a vector as a physical quantity with both a magnitude and direction, can a tensor be defined in a similar way as a vector with additional qualities?
| If you want antisymmetric tensors, there are well known geometric building blocks. For instance if you take two orthogonal vectors you can multiply them to get the oriented plane they span (with the orientation determined by the order you multiplied them). Similarly for three mutually orthogonal vectors, note there are 6 ways to multiply them, but because they pairwise anticommute, there are only two orientations. In an $n$-dimensional space you might have as many as $n$ mutually orthogonal vectors so have a rank $n$ object.
Now if you want to consider adding these higher rank antisymmetric things you can insist that addition distribute across multiplication. And now you need to make a choice, either assume all vectors anticommute, (not just the orthogonal ones) and use $\wedge$ for the multiplication symbol and get a grassmann algebra. Or you assume that a vector multiplied by itself gives it's squared length and you get a clifford algebra.
But actually, I lied about having to make a choice since we used the $\wedge$ to denote the antisymmetric product on vectors, we can have both products sinc ethey are defined on the same thing (linear combinations of products of orthogonal vectors). So we can have the grassmann algebra product and the clifford algebra product, and call it a geometric algebra. The basic building blocks are the products of orthogonal vectors, which are little oriented 1-volumes, 2-volumes, 3-volumes, ... or $n$-volumes. Everything else is a linear combination of those.
However, tensors do include symmetric tensors, so this doesn't give you those tensors as geometric objects. So you might have to think of the full space of tensors as multilinear functions on vectors.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/158890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
What position of the center of gravity can make the front wheels of the car lift off the ground? I have a question regarding the position of center of gravity required to just lift off the front wheel of a vehicle
Consider a vehicle of mass $m$ having a center of gravity at height $h = 0.5m$ from the ground. The coefficient of friction between the tire and the ground is 1. Assume that the engine supplies just enough torque to utilize all the friction force without causing the wheels to spin.
My question is where should the center of gravity of the vehicle should be located in relation to the rear wheels to make the wheels just lift off the ground. I have solved the question as shown, and I get a 'NEGATIVE' value for L2 meaning that COG should be 0.5m behind the rear wheel, but the solution says that it should be 0.5 m infront of the rear wheels.
Can someone help me out on this!
Since, my handwriting is not clear, I am writing the equations here too
Equilibrium in vertical direction
(1) $N_1 + N_2 = mg$
Equation of motion in horizontal direction
(2) $F_{tr} = ma$
Here $F_{tr}$ is the traction force on the rear tire which propels the vehicle. Also,
(3) $F_{tr} =\mu N_2$
The balance of the torque on the rear tire $ \to $ the net moment on the rear tire about the contact point is zero
(4) $mgl_2 + 0.5 \ ma = 0$.
Now, since the vehicle must just lift off the ground
(5) $N_1 = 0; \ N_2 = mg$.
Using the equations (2), (3), and (5),
(6) $a = \mu g$.
Now using the equation (6) in the equation (4),
(7) $l_2 = -0.5 \ \mu = -0.5$
Now, the final value of $l_2$ is negative, which means that it is opposite to the assumed direction. So the center of mass should be 0.5 m behind the rear wheels.
The only difference between my method and the solution manual which I am referring to is that they have considered inertia force(pseudo force) on the vehicle, and thus, they get the answers $l_2 = 0.5$, which means 0.5 m in-front of the rear wheels
| It's just a simple sign-error.
You wrote
$$ mgl_2+~ma0.5=0~.$$
But when you accelerate the car to the right then you get a reaction force $ma$ which points in the other direction, so your $m \vec a$-arrow should point to the left. The equation then reads
$$ mgl_2-~ma0.5=0~.$$
Then you have your torque balance about the rear wheel and $l_2=0.5$ at the end.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How much additional light does Earth receive from the Sun due to Earth's gravitational field? I was reading about how gravity affects light, and that got me wondering how much additional light is collected by the Sun due to the Earth's gravitational field.
Is it a significant amount of light (>1% of total light)? Is it significant enough to be considered when estimating the surface temperature of a planet?
| The amount of deflection of light (the bending of the null geodesic) passing a star is a deflection angle of 4m/R where m is the mass of the star and R is the radius of the star. The mass of the sun is $2 \times 10^{30}\rm\,kg$, where as the mass of the Earth is $6 \times 10^{24}\rm\,kg$. The radius of the earth is $6 \times 10^6\rm\,m$, and the radius of the sun is $7 \times 10^8\rm\,m$. Very roughly, this seems to indicate that the extra sunlight attracted to the Earth is very much less than 10,000 times less than the effect of the sun itself, which is barely detectable.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
If two photons collide, does the resulting particle have zero velocity? If two photons traveling in opposite directions along the same line collide, will the resulting particle have a velocity of zero relative to the rest of time space in the instant of the collision?
| Generally no, because velocity is not a conserved quantity. It is momentum that is conserved in all interactions. For photons, the magnitude of momentum is simply
$$ p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}, $$
so photons with different energies/frequencies/wavelengths will have different momenta. If the total momentum is nonzero before the collision, it will be nonzero after.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
What is the effect of torque steering on a motorcycle with a long wheel base? From experience, it appears motorcycles with a large wheelbase coupled with a long handlebar (say a modified Harley Ape Hangar) shows a noticeable tendency to veer to a side during acceleration vs a motorcycle with a short wheelbase and short handlebars. (eg. Triumph Speed Triple); is this an example of torque steering (as explained here)?
Or is there more to it? From a design standpoint, what would be the ideal ratio between wheelbase and handlebar length?
| Although your question is simple, the answer involves some rather complex modeling of dynamic systems to analyze and explain the stability of the bicycle (as well as motorcycles) and what parameters affect it. Here is an on-line article that doesn't go too deeply into the mathematics, but gives you an idea of the things one needs to consider:
http://www.phys.lsu.edu/faculty/gonzalez/Teaching/Phys7221/vol59no9p51_56.pdf
I'm a control systems scientist, but have not studied or worked with non-holonomic systems of which, I believe, the bicycle system fits into. Non-holonomic means that you cannot integrate some or all of the states, and in the case of the bicycle I believe that arises from multiple dimensions of space that have to be considered in the model.
If you are seriously interested in studying the problem, there are many papers in the IEEE Xplore Digital Libarary that have addressed such analyses mostly contributed by members of the Control Systems Society. I recall one issue of Control Systems Magazine that dealt with this subject and even analyzed more exotic configurations including back-wheel steering. Supposedly back wheel steering is very unstable and can be dangerous at high speed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Ampere's law of circular path when "bulging" out
The picture shows a use of Ampere's law. A circular path is chosen.
$$\oint \vec B \bullet \mathrm{d}\vec l=\mu_0 I_{encl}$$
When using Ampere's law we are talking about the current enclosed. That means, the current through the surface enclosed by the path.
The text to the picture says:
For the plane circular area bounded by the circle, $I_{encl}$ is just the current $i_C$ in the left conductor. But the surface that bulges out to the right is bounded by the same circle, and the current through that surface is zero. So $\oint \vec B \bullet \mathrm{d} \vec l$ is equal to $\mu_0 I_{encl}$, and at the same time it is equal to zero! This is a clear contradiction.
(The topic is displacement current.)
I am confused of why they can simply let the area (the plane surface enclosed) bulge out like this. This means that the total current through any surfaces bounded by the same path must always be the same.
In the derivation earlier in the book of Ampere's law, only a plane surface was considered. What is the explanation that Ampere's law still applies when finding the line integral, nomatter the enclosed surface's shape in space?
| The explanation of this is enclosed in Stoke's theorem. In the most general formulation, Stokes theorem asserts that
$$\oint_{\partial\Omega}\alpha = \int_\Omega\text d\alpha$$
where $\partial\Omega$ is the boundary of the cycle $\Omega$ and $\text d\alpha$ is the exterior derivative of the form $\alpha$.
In the case of electromagnetism this specialises to
$$\oint_{\partial S} \mathbf B\cdot\text d\mathbf l = \int_S\nabla\times\mathbf B\cdot\hat{\mathbf n}\ \text dS,$$
where $S$ is any surface with boundary and $\partial S$ is its boundary, which is a closed path. Now by Maxwell's equations, $\nabla\times\mathbf B = \mathbf J$, and the RHS then becomes the total current flowing across $S$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Mass in special relativity? Is the mass of a object at rest defined by $$E=mc^2$$
where $m$ is the rest mass. I.e. does the rest mass include every thing from thermal to gravitational potential energy and every other possible energy that it could have at rest. And thus if we write the following:
$$total\ energy=mc^2+potential\ energy+thermal\ energy $$
are we double counting the potential energy and the thermal energy?
| No, $E=mc^2$ covers only the mass energy of the object. Requirement of gravitational potential or thermal energy correctly require additional terms to the energy equation.
It's worth noting they are usually neglected because they are so small compared with the mass energy. Consider a rock of mass 1kg a few metres above the surface of the Earth:
*
*Mass energy = $mc^2$ $\approx 10^{17}$J
*GPE $\approx$ $mgh$ $\approx$ $50$J
*Thermal energy = $mTq$ $\approx$ $300$kJ .
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Why does a plane look like it's slower than a car? Recently, I was in a car and noticed a plane in the sky. What was interesting was that the plane seemed to go slower than my car because I passed the plane. Is there a physics reason to this?
| There can be only two possibilities:1. In case the plane and the car were moving in the same direction the plane was indeed flying with slower speed than the car.2. In case the plane was flying with greater speed than the car the plane was going in some other direction. The component of velocity of the plane in the direction of motion of the car matters here which is lesser than the velocity of the car in your case.Another way of seeing is (as the OP said the plane and the car were pretty much going in the same direction):
The car C can only see the background object B as coming out from the mouth of the plane P if the distance traveled by the plane PQ is smaller than the distance traveled by the car CD and the car will pass the plane also if B,P and C are always in straight line then the plane is traveling slower than the car and the car will pass the plane at the place where BPC is perpendicular to the ground. If the plane travels equal to or faster than the car then the car would see the background object B as coming out from the tail of the plane.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Apparent depth and virtual image position
(source: bbci.co.uk)
Why does the virtual image appear right on top of the object and not a little to the right, for example? Is it explained by some formula or just symmetry of the geometry?
| The answer is that, when looking at an object, in order to determine where it is, you need several rays coming at slightly different angles and from all over the object's surface. Using Snell's Law and tracing back the rays to the point where they converge you get the full image and its apparent position. The next image represents this idea:
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Symplectic geometry in thermodynamics There seems to be analogues between Hamiltonian dynamics and thermodynamics given the Legendre transforms between Lagrangian and Hamiltonian functions and all of Maxwell's relations. Poincarè tried to generalise classical mechanics to the realm of statistical thermodynamics with ergodic theory yet I believe his model is not complete(?)
Therefore as my main question, does symplectic geometry underpin thermodynamics? I am currently reading about KAM theory (please see my other question regarding this) and was wondering can indeterminism in perturbation theory and chaos lead to entropy and the second law?
| No answers yet? So let's take a shot at a (partial) answer:
Therefore as my main question, does symplectic geometry underpin thermodynamics?
No. In thermodynamics, we're dealing with a Legendrian submanifold of a contact manifold (cf Wikipedia). Thermodynamic variables are canonical coordinates on that manifold.
Morally speaking, in case of symplectic geometry, canonical coordinates map any symplectic manifold to the cotangent bundle $T^*\mathbb R^n$ with symplectic form $\omega = d\theta$.
In case of contact geometry, canonical coordinates map any contact manifold to the first jet bundle $J^1\mathbb R^n$ (essentially $\mathbb R\times T^*\mathbb R^n$) with contact form $\alpha = dz + \theta$ (in both cases, $\theta$ denotes the canonical 1-form of the cotangent bundle; $z$ is the coordinate of the first factor).
On the jet bundle, the submanifold in question is given by the prolongation of some state function - a thermodynamic potential expressed in its natural variables. Eg for $U = f(S, V)$, we end up with a coordinate expression
$$
(S, V, U, T, p) = \left(S, V, f(S, V), \frac{\partial f}{\partial S}(S, V), \frac{\partial f}{\partial V}(S, V) \right)
$$
Please insert minus signs as appropriate ;)
[I] was wondering can indeterminism in perturbation theory and chaos lead to entropy and the second law?
As far as geometry is concerned, there isn't really anything special about entropy, ie this question has to be answered at the lower level of statistical mechanics; I'm happy to leave that part of the question to someone else...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/159980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.