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Why do images not appear inverted when looking directly through a pinhole camera? I understand that the way light takes through a pinhole creates an inverted image on a surface behind the pinhole. I remember this effect from school experiments, it's also described in this wikipedia article. I punctured a piece of paper and looked through it (instead of watching the reflection), the image appeared as normal to me. Why is that? Why doesn't the scene appear upside down when looking through the hole?
| Ignoring diffraction, the pinhole can't change the orientation of what you see because it doesn't change the position or direction of any light rays. It just blocks some of them.
When you put a screen in front of an illuminated object, every point on the object emits light in every direction, which hits the screen at every point. You can imagine that the screen contains superimposed images of the object with every possible position and orientation, but you can't see them because they average out to a uniform blur. Putting a punctured sheet between the object and the screen blocks out most of those images, making the remaining one visible. It doesn't create the image, as such. Blocking the light differently would leave different images visible. For example, if you replace the punctured sheet with a laptop privacy filter, you ought to get a blurry upright image (I haven't actually tried this). A filter that creates a sideways image is also possible in principle.
But no matter what filter you put between the object and your eye, you can't change which image orientation your eye sees, because it has its own mechanism for extracting an image from the mess of light rays in the environment, and you are only giving it less to work with.
| {
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Standing wave equation A question about standing wave equation.
We send a harmonic wave that travels down a rope that is fixed at the end with the equation(like in the picture):
$$y = A\sin(kx-\omega t)$$
The wave that travels down a rope gets reflected at the rope’s end and has the equation:
$$ y = A \sin[k(2l-x) - \omega t + π]$$ where $l$ is the length of the rope.
I don't understand this equation. We add $π$ because the waves get inverted when it is reflected, but I don't understand where the $(2l-x)$ part comes from.
| In this problem, fixed end is at $x = l$. Corresponding boundary condition has a form
$$
y(x = l, t) = 0.
$$
Solutions to the wave equation with this boundary condition have the following form
$$
y(x,t) = f(kx-\omega t) - f(k(2l-x)-\omega t). \quad (1)
$$
In this problem, $f(kx-\omega t) = A\sin(kx-\omega t)$. For $t$ large enough, in the $x<l$ area, only the second term in (1) is nonzero and you have reflected wave packet
$$
y(x,t) = -A\sin(k(2l-x)-\omega t) = A\sin(k(2l-x)-\omega t +\pi).
$$
| {
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$\beta$-function of general $\phi^6$ in 3 dimensions In 3-dimensions the $\phi^6$ interaction is renormalizable and the $\beta$-function can be found in many reviews in the $O(n)$ symmetric case, $V(\phi) \sim (\phi_a \phi_a)^3$ where $a=1,\ldots,n$.
What I couldn't find, but I can't believe it's not known, is the $\beta$-function in the general (non-symmetric) case:
$$V(\phi) = \frac{1}{6!} \lambda_{abcdef} \phi_a \phi_b \phi_c \phi_d \phi_e \phi_f$$
where the coupling, $\lambda_{abcdef}$, is totally symmetric under the interchange of indices $abcdef$ and so is the corresponding $\beta$-function, $\beta_{abcdef}$. I imagine the result is known to at least a couple of loops, probably at least 4 or 5.
| The comment by Connor Behan contains the answer:
arxiv.org/abs/1707.06165
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How does a javelin rotate when thrown? As the centre of mass of javelin follows parabolic path, the javelin also rotates about it’s centre of mass while in air. As a result, it hits the ground with it’s nose. Where is this angular momentum coming from? Is there some torque acting on the javelin? Or, Is the angular momentum simply imparted by the athlete when he throws the javelin?
| Please see the related post here. The answer lies in the the wikipedia section on the modern redesigns for javelin throw competitions. It states that
In 1986 the technical committee of the International Association of Athletics Federations introduced new specification for the standard competition javelin. Among other things it was specified that the center of mass should lie forward of the center of air resistance. This new specification gives a nose heavy javelin. The previous standard competition javelin was designed to be balanced. When thrown with perfect balance the javelin would keep climbing for a long time. More and more often the javelin would hit the ground while parallel to the ground, and would continue to slide for a distance. The javelin throw event became rather dangerous; sometimes the sliding javelin would slide all the way onto the running track.
Overall, as the spear moves through the air, aerodynamic friction on the spear affects it's path. Since a completely symmetric spear is unlikely and difficult to realize, besides being unnecessary for the function of a spear, the aerodynamic friction applies a torque on the spear, thus causing it to rotate.
With regards to the new redesign, a balanced javelin is more unforgiving since one has to throw it just right, else the throw would not be curved as desired. On the other hand, the new javelin tends to self-straighten in flight, so the throw no longer needs to be perfect. This redesign therefore shifts the focus from throwing skill to throwing power.
| {
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Does the age of the universe depend upon a specific choice of referential? Does the age of the universe depend upon a specific choice of referential or is there a way to define it in a referential-independent way?
| No. The age of the universe does not depend on any referential system. In order to measure time, you need some physical quantity that's changing to measure time against. In the case of cosmology, it's the time perceived by a typical observer based on the expansion parameter $a$ --see below. In a manner of speaking, it's the time that follows the galaxies in their mutual expansion.
The mathematical basis is the FLRW metric of GR:
$$
-c^{2}\mathrm{d}\tau^{2}=-c^{2}\mathrm{d}t^{2}+a(t)^{2}d\Sigma^{2}
$$
where,
$$
\mathrm{d}\mathbf{\Sigma}^{2}=\frac{\mathrm{d}r^{2}}{1-kr^{2}}+r^{2}\mathrm{d}\mathbf{\Omega}^{2}
$$
and,
$$
\mathrm{d}\mathbf{\Omega}^{2}=\mathrm{d}\theta^{2}+\sin^{2}\theta\,\mathrm{d}\phi^{2}
$$
$k$ is the spatial curvature, $r$ is the actual distance between two galaxies, and $a\left(t\right)$ is called the expansion parameter, and represents the average separation of two standard galaxies measured in this cosmic time. The quantity,
$$
H=\frac{\dot{a}}{a}
$$
is the Hubble parameter, and represents how the scale factor changes with this cosmic time.
The age of the universe is essentially the inverse Hubble parameter:
$$
T=\frac{a}{\dot{a}}
$$
And because the universe is large-scale homogeneous and isotropic, the words "typical observer" are justified, and this time does not depend on which observer you choose. Meaning: on which typical galaxy you sit and watch the universe expand.
Now, we don't live in a FLRW universe. We live in a De Sitter universe (vacuum dominated), but the argument doesn't change (essentially).
| {
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How to inject the maximum acoustic power into a block of granite? I know that we can use transducers that are glued to a surface to achieve this. If I want, for example, to have 200 watts of actual acoustic power in the audible range in the granite, is a transducer the easiest cheapest way? Are there other methods?
| Here is the challenge you have to overcome.
To actually deliver 200 watts of acoustic power into a solid block of granite requires that the transducer be impedance-matched to the load.
If you press on a piece of granite, it presses back really hard and experiences almost no compression in response. This makes it an ultrahigh impedance load, and if your transducer impedance is lower than this, most of the acoustic energy will bounce right off the block and not enter it. Your transducer might be producing 200 watts but 199 of that is reflected off the block; meanwhile you are still pumping 200 watts into the transducer, and it will soon blow up.
This effect rules out the use of any sort of moving-coil loudspeaker as a transducer because they are all designed to drive a very low impedance load: the air in front of the cone.
You are pretty much restricted to piezoelectric-type transducers, as used in very large ultrasonic welding machines, but even then you will need a custom-fabricated horn assembly which will match the impedance of the piezo transducer stack to that of the rock at the frequency of interest. This sort of setup will cost of order ~tens of thousands of dollars.
Note that typical piezoelectric ultrasonic welding machines operate at and above 40 kilohertz. To drive 200 watts into a chunk of rock in the audio range will require many, many transducers.
BTW, what exactly are you wanting to accomplish with such a setup?
| {
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Why is the half-wave dipole the most used antenna design? When producing em waves using a dipole antenna (of length L), you could theoretically use any L and adjust the frequency of the oscillating voltage to get the desired wavelength. Then why are most antennas half a wavelength long? I'd also like to know why it's useful in the context of receiving em waves. Thanks.
| A half-wave dipole is fed with the signal at its mid-point.
Consider the current distribution along the antenna. Some current has to flow in and out at the middle, or power cannot be transferred. But at the ends the current has nowhere to flow. So the current distribution along the dipole takes on a distinct "hump" shape, basically half a sine wave. And that is the clue - half a wave.
Doubling the length of the antenna to a whole wave would de-tune it so that, at that frequency, the current in the middle would be as zero as the current at the ends and no power would be transferred.
In practice, an antenna standing on the ground will be only a quarter-wave long. The ground acts like a mirror, creating a virtual reflection of the antenna so it looks a half-wave long. You can see the effect optically if you stand on the shore of a lake on a still, calm day and the opposite shore is reflected upside down in the water.
| {
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How do I determine the "recoil permeability" and "remanent flux density norm" of a permanent magnet? I am working on modeling a permanent magnet using finite element analysis.* The magnet in question is a neodymium cylinder 1 cm in diameter, and 5 cm in length.
The program uses the equation
$$
\vec{B} = \mu_0 \mu_{rec} \vec{H} + \vec{B}_r
$$
to define the field caused by the permanent magnet, where $\mu_{rec}$ is the recoil permeability, and $\vec{B}_r$ is the remanent flux density.
It is my job to input values for $\mu_{rec}$ as well as the magnitude and direction of $\vec{B}_r$. I have a gaussmeter (i.e. a device that measures the $B$-field) and the magnet. Can I use this to determine $\mu_{rec}$ and $\vec{B}_r$, and if so how?
Toward a solution, I know that the magnetic field may also be expressed as
$$
\vec{B} = \mu_0 \vec{H}+ \mu_0\vec{M}
$$
where $\vec{M}$ is the magnetization vector. Thus,
$$
\mu_0 \vec{H}+ \mu_0\vec{M} = \mu_0 \mu_{rec} \vec{H} + \vec{B}_r.
$$
This makes me think that $\vec{B}_r = \mu_0 \vec{M} $... but if so, I don't understand what the $\mu_{rec}$ is doing hanging out down there. Please help!
*COMSOL Multiphysics 5.6, Magnetic Fields module, if you're curious.
Here is a screenshot
| In general, the relationship between $\vec{B}$ and $\vec{H}$ is nonlinear and history-dependent. The formula
$$
\vec{B} = \mu_0 \mu_{rec} \vec{H} + \vec{B}_r
$$
is an expedient approximation that often works well for rare-earth permanent magnet applications. It refers to the ‘major hysteresis loop’ which is the path followed when $\vec{H}$ cycles between vary large positive and negative values, sufficient to fully saturate the material. By definition, $\vec{B}_r$ is the value of $\vec{B}$ when $\vec{H}=0$. The first term is simply a linear and isotropic approximation for how $\vec{B}$ varies around $\vec{H}=0$.
In principle, the values of $\mu_{rec}$ and $\vec{B}_r$ can be read off of measured hysteresis loops. Vendors of the highest quality NdFeB and SmCo provide such data. (See product literature from Shin Etsu, for example.) When adequate data is not available or it can’t be trusted, the only practical option is to map the exterior field of your magnets and then to fit the data with the help of finite-element calculations. Beware that the magnetic properties may not be spatially homogeneous, especially with poor quality material.
| {
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Anticommutation of variation $\delta$ and differential $d$ In Quantum Fields and Strings: A Course for Mathematicians, it is said that variation $\delta$ and differential $d$ anticommute (this is only classical mechanics), which is very strange to me. This is in page 143-144:
If we deform $x$ we have
$$\delta L = m \langle \dot x, \delta \dot x \rangle dt$$
$$= - m \langle \ddot x, \delta x \rangle dt - d\left(m \langle \dot x,\delta x\rangle \right)\,.$$
Here $\delta$ is the differential on the space $\mathcal F$ of trajectories $x$ of the particle, $d$ is the differential on $\mathbb R$, and the second minus sign arises since $\delta$ and $d$ anticommute on $\mathcal F \times \mathbb R$.
As far as I know, we should have $d(\delta x)=\delta(dx)$. It doesn't make sense to me why a "differential" with respect to the trajectory would anticommute with a differential with respect to time.
| Perhaps the following comment is helpful: If $M=M^{\prime}\times M^{\prime\prime}$ is a product manifold then the exterior differential $\mathrm{d}=\mathrm{d}^{\prime}+\mathrm{d}^{\prime\prime}$ on $M$ is a sum of the exterior differentials on $M^{\prime}$ and $M^{\prime\prime}$. We can assign degree $(1,0)$ to $\mathrm{d}^{\prime}$ and degree $(0,1)$ to $\mathrm{d}^{\prime\prime}$ to form a bicomplex of differential forms. Note that in order for each of the 3 exterior differentials to square to zero, it is important that
$\mathrm{d}^{\prime}$ and $\mathrm{d}^{\prime\prime}$ anticommute.
(A similar situation happens for the Dolbeault complex.)
The variational bicomplex $D=\delta+d$ that OP asks about is a very similar construction. We stress that $\delta$ is an exterior differential; not an infinitesimal variation in the heuristic physics sense.
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Magnetic flux through circular loop due to infinite wire I’m trying to calculate magnetic flux that’s going through circular loop with radius $R$, due to magnetic field of a infinite wire that is in distance $d$ from the center of the loop. $\vec{B}$ vector is parallel to $\vec{dS}$ vector. I know that magnetic field of that wire is equal to $B=\frac{\mu_0 I}{2\pi r}$ and that flux is equal to $$\int_{S} \vec{B}\cdot \vec{dS}=\int_{S} B\cdot dS$$ where $dS$ Is surface of the loop, but what I don’t know is how to change that $dS$ so is possible to solve.
|
fig 1: The geometry of the calculation. Note that the distance of the wire from the center has been taken to be $D$, not $d$.
Let the magnitude of the magnetic field due to an infinite constant current carrying straight wire at a distance $r$ from it be $k/r$.
In the given geometry, the wire is co-planar with the loop. This makes the problem effectively $2$ dimensional. Taking the interior of the circle (in its embedding plane) as the surface of integration of flux calculation, the magnetic field is everywhere (anti)parallel to the surface normal vector, as noted in the question.
Since the magnetic field is only dependent on the distance from the wire, it is constant in the small differential patch of area $2\,y\, dx$, colored gray in fig $1$. The magnitude of the total flux is then given by
$$
\begin{align}
I&=\int_{-R}^{R}\frac{k \,2 y dx}{r}&\\
&=2 k\int_{-R}^{R}\frac{\sqrt{R^2-x^2}dx}{D-x}\tag{1}\\
\end{align}
$$
which equals$^{A.1}$
$$
I=
\begin{cases}
2\pi k (D-\sqrt{D^2-R^2})& D>R\tag{2}\\
\phantom{}\\
2\pi k D & D\leq R\\
\end{cases}
$$
Appendix
*
*The integral in the RHS of equation $1$ is (via Mathematica)
$$
-y+D\tan^{-1}\frac{x}{y}-\sqrt{D^2-R^2}\tan^{-1}\left(\frac{D x -R^2}{y\sqrt{D^2-R^2}}\right)
$$
where $y=\sqrt{R^2-x^2}$ with limits $(x,y)$ going from $(-R,0)$ to $(R,0)$. One has to be careful with the limits here. While the third term is imaginary for $D<R$, it cancels out for the limits.
*First case of Eqn. $2$ has the expected asymptotic form, $ k \pi R^2/D$, for $D\to\infty$. Interestingly, if one considers only its real part, it holds for $R\to\infty$ too.
*The second case of Eqn. $2$ viz. the case of the wire being inside the loop, is remarkably independent of $R$ and depends linearly on $D$. This was indeed verified numerically$(1\sigma)$.
fig 2: Monte-carlo verification of eqn. 2 for $R=1,2\,\,(k=1)$
| {
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What are the weaknesses, if any, of the relational interpretation of quantum mechanics? Carlo Rovelli's 1996 relational interpretation of quantum mechanics (RQM) seems to solve many of the quandaries of traditional theories, including the Copenhagen interpretation (what privileges the observer? Why does he/she instigate wavefunction collapse); the many-worlds interpretation (an infinity of unobservable universes); and spontaneous collapse theories such as that of GRW.
However, there seems to have been limited academic discussion of the theory. Is this due to a shortcoming of the theory, or has there simply not been enough time for it to disseminate?
| I personally consider RQM to be an attempt at the "best of both worlds" between Copenhagen and MWI. By making measurements observer-dependent, it essentially accomplishes the same thing as entanglement of the system and observer in MWI. By refusing to speculate on other worlds, it has the veneer of realism of Copenhagen. (Full disclosure: I am an MWI supporter but am open to all ideas.)
The problem with RQM as far as I can tell is the "sparse ontology" or "flash ontology" problem. It is discussed briefly here. All measurable quantities in RQM are only defined in the exact instant that they are measured; for most instants in time, the state of the system is essentially undefined. The fundamental reason for this is that RQM postulates that the only thing you can talk about in physics is how systems interact rather than how they are.
According to your taste, this may be a positive or negative aspect of the theory.
For me it definitely seems unphysical at best and like borderline solipsism at worst. But I can see obvious counterarguments:
*
*All interpretations of quantum mechanics have something unphysical about them.
*Pretty much all advances in physics since Copernicus have required us to get used to a new counterintuitive idea about reality, and this is probably no greater or worse than the previous ones.
| {
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A Doubt on Hawking Radiation: What Birrell wants to say here? Well, I know basic General Relativity but I still learning the basics of Quantum Field Theory (therefore without mentions on curved spacetimes). But, I'm trying once and a while to become more used to quantum fields and curved spaces.
Now, the notion of Hawking Radiation is paramount but my knowledge is semi-technical. Reading $[1]$, the author said:
At first sight, black hole radiance seems paradoxical, for nothing can
apparently escape from within the event horizon. However, inspection of
$(8.36)$ shows that the average wavelength of the emitted quanta is $\approx M$, i.e.,
comparable with the size of the hole. As it is not possible to localize a
quantum to within one wavelength, it is therefore meaningless to trace the
origin of the particles to any particular region near the horizon.
What worries me is the final consideration when he says "therefore meaningless to trace the
origin of the particles to any particular region near the horizon". So, knowing that phrase from $[1]$, I would like to ask:
Hawking Radiation isn't a phenomena that occurs near the Horizon?
$$ * * * $$
$[1]$ Birrell.N.D. Quantum Fields in Curved Space. page $264$.
| Hawking radiation is observed at future null infinity. One can try to trace back where the radiation observed originated. A common method for this is to use the “geometric optics” approximation, where radiation follows null geodesics. Doing so with the modes of the Hawking radiation suggests that the modes originate arbitrarily close to the event horizon.
However, the geometric optics approximation only works in the limit that the wavelength is small compared to the curvature length scale. As Birrrell points out this isn’t the case for the dominant modes in the Hawking radiation, which have wavelengths comparable to the size of the horizon. The geometric optics results should therefore be taken with a grain of salt. All we can really say is that the modes of the Hawking radiation originate in vicinity of the horizon.
That is the point Birrell is trying to make (I think).
| {
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Active vs passive transformation in right handed particle People often says that active transformation is equivalent to passive transformation.
Suppose that we have a right handed particle that is, the spin and the momentum are pointing in the same direction, call this direction right. Under a passive parity transformation the the spin continues to point to right while the momentum is now pointing to the left.
My question is, in a real world if we inverse the particle momentum , does it turn in to a left handed particle? If not ,does this mean, that passive transformation and active transformation are not equivalent?
Note I am considering here passive transformation as transformation in a measurer apparatus. People often consider passive transformation as coordinate transformation but coordinates are imagination of our mind ,it does has no effect in physics
|
if we inverse the particle momentum , does it turn in to a left handed particle?
Particle is right-handed if spin vector is in the same direction as particle momentum vector and left-handed if these directions are opposite :
So basically answer is that you can inverse particle helicity from right-handed to left-handed (or in reverse) by reversing it's spin OR momentum (but not both). So the answer is YES, it will turn to left-handed from right-handed.
| {
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How do we determine the damping coefficient given the acceleration vs time graph of a 1D mechanical system? Consider a mechanical system about which the only data we have is a graph that shows acceleration vs time. I would like to figure out what the damping coefficient $c$ is.
Instead of displacement as shown in the attached image, the $Y$ axis value would be acceleration. The mass being damped is moving horizontally and does not move at all vertically.
How would I go about this?
| How to Fit
We have data points the oscillate curve $~x(t)~$.
Step 1: obtain the blue points
The period of this curve is $~T=2~$, hence the points are
$$X=T_0+n\,T~,Y=~x(T_0+n\,T)~,n=0,(1),N_p,$$
where $~T_0=\frac T4.$
Step II
Fit the points $[~X~,Y]~$ to the curve $~a\,e^{-\beta\,t}~$ where $~a=x(T_0)$ and you obtain $~\beta~[1/s]$ from the fitting process.
The blue curve is your result.
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Can photons only exist in the state of motion in a medium? Photons are known to travel at a speed of $\rm 299 \ 792 \ 458 \ m / s$ in vacuum. Photons can be absorbed, or absorbed and re-emitted by matter. They slow down to $\rm 225,000,000 \ m/s$ in water with a refractive index of $1.3$, to $\rm 200,000,000 \ m/s$ in glass and to $\rm 125,000,000 \ m/s$ in diamond. So its speed can be varied with variations in the medium.
What happens if the photons can be virtually brought to a stop and is not absorbed by matter? Can the photon exist in a state where it has an energy of $\rm E = h \nu$, but having no velocity?
| It depends how you're defining "motion". As you're talking about photons as opposed to light generally, I'll talk from the quantum perspective rather than just classical waves. See these related questions for more on the different perspectives:
Do photons actually slow down in a medium, or is the speed decrease just apparent?
What is the mechanism behind the slowdown of light/photons in a transparent medium?
Remember that whilst we say that photons/light slows down in a medium, the individual photons still locally propagate at $c$ in between interactions (described by QED). So slowing photons down means increasing the number of interactions with other forms of matter (i.e. absorption and emission). Getting to your question:
What happens if the photons can be virtually brought to a stop and is
not absorbed by matter?
The answer is that either that they're absorbed and not emitted, which doesn't seem to mean what you intended, or the answer to your title question is yes, the massless photons can only exist in a state of motion. "Brought to a stop" doesn't make sense if we're talking about photons themselves.
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Phase difference approximation I'm sitting and trying to solve the equation of the phase difference given by:
$\Delta \phi = k (\sqrt{a^2+d^2} -d) \approx \frac{ka^2}{2d}$
Where $a$ is the size of an aperture and $b$ is the distance of the point at the aperture's center as shown in the figure below.
I'm not a math expert here, so I wondering if anyone can explain the approximation that has been done above here.
Reference:
Applications of Classical Physics by Roger D. Blandford and Kip S. Thorne - Chapter 8 - Diffraction
| You need the binomial expansion
$$ \sqrt{d^2+a^2} \approx d \left(1+\frac{a^2}{2d^2}\right) , $$
which assumes that $d\gg a$.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Energy of Hydrogen Atom (Electron vs Proton) In many textbooks, energy changes of the hydrogen atom are attributed to the electron transitioning between energy levels. However, the energy itself is that of the whole system (proton+electron) so how can we attribute its changes to the electron? what's preventing us from attributing these changes to the proton??
| The energy is indeed that of the whole system, but the electron has a much smaller mass (1/1836) than the proton, so the latter does not contribute much here. But for accurate results you have to consider the proton as well. When you solve the corresponding two-body problem for two masses $m_1$ and $m_2$, you can reduce it to a one-body problem by using the reduced mass
$$\mu=\frac{m_1 m_2}{m_1+m_2}$$
If $m_2$ is much larger than $m_1$ then this equals approximately $m_1$, so $m_2$ doesn't contribute much to the energy and momentum of the whole system.
| {
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Must observables be Hermitian only because we want real eigenvalues, or is more to that? Because (after long university absence) I recently came across field operators again in my QFT lectures (which are not necessarily Hermitian):
What problem is there with observables represented by non-Hermitian operators (by observables, I obviously don't mean the tautological meaning "Hermitian operators")?
One Problem sure is that there are not real eigenvalues. But If I say I want to "measure" some complex quantity, then that alone shouldn't be a problem, I'd be fine with complex eigenvalues then.
This question is answered stating that operators are Hermitian "if and only if it is diagonalizable in an orthonormal basis with real eigenvalues". This doesn't yet seem like a game-stopper to me, as long as I still get an orthonormal basis.
And splitting an arbitrary operator $\hat{O}$ into a Hermitian and an anti-Hermitian part, that would correspond to real and imaginary part of the observable, I could take expectation values just fine.
But maybe I'm forgetting about something, and there are other good reasons why non-Hermitian operators will lead so serious problems.
| Given a separable Hilbert space that underlies a quantum system, the observables on this system are the self-adjoint operators on the Hilbert space. The reason they must be self-adjoint is that the spectrum of an operator is real if, and only if the operator is self-adjoint. The spectrum of an operator is the set of possible measurement values that you can hope to measure in an experiment. The eigenvalues of an operator are a subset of its spectrum.
| {
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"timestamp": "2023-03-29T00:00:00",
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A pedagogical semi-rigorous review of topological phases, topological order, and related subjects I'm looking for a pedagogical review or book about topological phases, topological order, TQFTs, and related subjects.
The ideal thing would be a mix of rigorous definitions and physical examples, with something giving me a physical idea of what's going on.
Right now the only things I can find are either a lot of category theory with no physics at all or a lot of handwaving about the Toric code (after having read about it in a bunch of different sources, I still don't understand what's "topological" about it, or how does it exemplify a more general concept of "topological order").
Does anybody know anything of the sort?
| For topological band theory (single-particle picture) the online course by TU Delft could be a nice starting point. A more thorough approach on topological insulators can be found at the lecture notes by Aboth or at the classic book Topological Insulators and Topological Superconductors by A. Bernevig for instance.
On topological order, TQFTs and the toric code, S. Simon has a nice protobook at his website.
| {
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Analogy for Lentz soliton Analogy for Alcubeirre Warp Drive:
Every explanation of warp drive in layman terms has this sentence in it:
"The Warp Drive will contract space in thier front and expand space behind."
(I am not sure that this is literally what the alcubierre metric describes)
And then show pictures like this:
They also use the analogy of a surf board surfing in the sea.
This analogy of surf board makes it intuitive to understand how warp drives work.
Analogy for Lentz Soliton:
Recently I have been hearing that Eric Lentz published a paper concluding that we can have warp drive without the need of negative energy to expand spacetime.
When I search lentz soliton on google, it shows this picture:
It is absolutely not intuitive how the the lentz soliton works.
In the case of alcubeirre warp drive, it was intuitive that space pushes the drive from behind and pulls from the front.
Question:
I want an intuitive analogy about how the Lentz Soliton works, if there are any.
| Basically, he came up with a way to generate both the push and the pull force without needing to create negative energy, which as far as we know is impossible to do. Since energy and mass are interchangeable due to the well known equation E=mc^2, and mass can warp space time (aka it creates gravity) that in turn means that energy can also warp space time. What warp drive designs do, is contract the space time in front of them, aka they create a gravitational attraction in front of them, and expand the space time behind them, aka ANTI gravity, which since there is no negative mass and there is no way to create any significant amount of negative energy, is impossible. Lentz warp drive design is arranged in such a way that the ship is propelled forwards without needing the negative energy or reverse gravity behind it. Not all scientists are convinced of his designs validity, but he maintains his belief in it’s authenticity.
| {
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Mutual capacitance of a long straight wire and an infinite conducting plate Q. A long straight wire is located parallel to an infinite conducting plate. The wire cross-sectional radius is equal to $a$, the distance between the axis of the wire and the plane equals $b$. Find the mutual capacitance of this system per unit length of the wire under the condition $b>>a$.
Consider a point at a distance $r$ from the axis of the cylinder, since we know that electric field $E(r)$ due to infinite line of charge is
$$E(r) = \frac{\lambda}{{2{\pi}{\epsilon}r}}.$$
So the potenital difference across the surfaces at a distance $a$ and $b$ respectively from the axis of the cylinder comes to be
$$ V= \frac{\lambda \ln (b/a)}{2\pi\epsilon}.$$
So the capcitance is obtained by $\frac{Q}{V}$ . However, this method seems to give incorrect answer. I am not able to understand what I have applied incorrectly.
Note that I have not included the field to induced charges on the plate since the field due to the postive charge developed at the back and the negative charges on the front side of the plates cancel each other. Is this statement correct?
| The field right above a conducting surface should be perpendicular to the surface. Your choice of field does not meet this condition. That is, your choice of field is not a solution of the Laplace's equation with the boundary conditions set by the given system of conductors
To find the right field you can use the method of image charges, that is, you can remove the conducting plate if you imagine to have another long straight wire with opposite charge symmetrically placed with respect to the conducting plate. This ensures that the total field obtained in this way meets the boundary conditions at the conducting plate, because the superposition of the field generated by the original wire with that of its mirror image cancels the component of the field parallel to the plate, at the plate's surface.
The field with the plate will be equal to the one found in the above way, by means of the image charge, thanks to the uniqueness theorem.
Finally, by assuming $b\gg a$, you are actually assuming an infinitely thin wire, for which the surface is orthogonal to any field direction.
| {
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Can an LC circuit be used to produce and receive FM signal? By simply changing the frequency of oscillation of an LC circuit (which you can do with the capacitor alone) you could emit (and receive) FM signal.
Are LC circuits actually used to do that? If not, why not?
| FM signal is modulating the frequency of the signal, i.e. 0 will be encoded to a short signal, and 1 will be encoded to a long signal.
To receive an FM signal, we can first use a bandpass to select the frequency we would like to listen to, and then simply 'read' the signal by parsing the short and long signals to 0s and 1s.
RLC circuit can be used as a bandpass, but to parse the signal, i think you will need another IC to perform this procedure.
| {
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Are soundproof foam shapes reflective of audio frequencies? I came across a previous question regarding how soundproof foam absorbs sound: How does foam "absorb" sound? where the answer explains the properties of the foam itself rather than the shape.
However, in analyzing audio spectrogram images (and messing around with Chrome Music Lab), I noticed a similarity in structure between acoustic foam and 3-dimensional spectrograms. The first two images are spectrograms and the latter two are two different acoustic foam designs.
It appears that, depending on the kinds of soundwaves (music recording, conversation, etc.), the foam would be designed to replicate similar shapes and fit together almost like a puzzle piece.
My question is: Are soundfoam shapes designed to sort of "mirror" the incoming soundwaves?
| Some of the sound is absorbed and some is reflected. A flat surface would reflect more sound than the shaped surfaces which are designed to make more of the reflected wave hit the foam again so more is absorbed.
| {
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Why 2nd Shell can have 8 electrons? I recently watched this video:https://youtu.be/INYZy6_HaQE
and understood why 1st orbital can have only 2 electrons: According to Pauli's exclusion principle, two electrons cannot have the same quantum states.In the first energy level,all other quantum states are same except for the spin. And spin can have two values $+\frac{1}{2}$and $-\frac{1}{2}$.So,there can be two electrons in the first shell.
I need a similar explanation for 2nd shell
In other words, why the 2nd shell can have 8 electrons?
I know that there are other quantum states. I am requesting to explain how many values each quantum state can take.For example spin can have two values.Likewise,how many values can the angular momentum can have?
How all these lead to the 8 electrons in the second shell?
(Don't make it simple,you can use maths.Iam really on to this)
| If you are a high school student or a freshman yet, I will try explaining in terms you find easy. As you must already know, an electron in an atom has a set of 4 numbers known as its quantum numbers. They are $n,l,m$ and $m_s$. Now these 4 can not be the same for 2 electrons. Let us calculate the number of states possible when we talk about the second shell. We have $n=2$.
Now, $l$ can take values $0$ to $n-1$. Thus here $l$ can take the values $0$ and $1$.
For each $l$ there are $2l+1$ values of $m$ possible which go from $-l$ to $+l$.
Thus when $l=0$ $m$ can be only $0$ and when $l=1$, $m$ can be only from the set $\{-1,0,1\}$.
Taking the above things into consideration you have in total 4 cases for $\{n,l,m\}$
*
*$\{2,0,0\}$
*$\{2,1,-1\}$
*$\{2,1,0\}$
*$\{2,1,1\}$
But for each of these the electron can have $2$ values of $m_s$ which are $-1/2$ and $+1/2$. Thus you have in total $8$ combinations of the 4 quantum numbers in the $n=2$ shell. Thus you can't have a 9th electron there because it will have to occupy one of those 8 combinations, which is forbidden by the exclusion principle.
| {
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Why does closing someone's eyes not give them near-sightedness? When we close our eyes aren't we technically looking a really close up piece of skin? That is, our eyelids? If it's so close to our eyes why doesn't it give us bad vision? We know if screens or books or other objects constantly being close to us causes our vision to get worse, why doesn't the eyelid do the same thing? What's the difference?
| The problem comes only when we try to focus on something that is really close to our eyes. We don't try to focus on our eyelid (may be we cannot), and hence its not a problem
| {
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Why electric field is scaled by gamma? Two opposite charges are in a spaceship and are attracted by the electric field $E_s$
But for an observer on earth the Electric force is $$E_e=\gamma E_s$$
Normally the forces are scaled down by $\gamma$ in the earth frame and here also the total force is scaled down.
But why the Electric component of force is scaled up?
Is it because the Electric fields are now closer together because of length contraction?
But i think this is not the answer because if that was the case,then gravitational fields and hence Forces would have scaled up.But this is not the case in nature.
Can you provide a derivation or something which explains how the Electric field is scaled up?
|
In above Figure-01 an inertial system $\:\mathrm S'\:$ is translated with respect to the inertial system $\:\mathrm S\:$ with constant velocity
\begin{align}
\boldsymbol{\upsilon} & \boldsymbol{=}\left(\upsilon_{1},\upsilon_{2},\upsilon_{3}\right)
\tag{02a}\label{02a}\\
\upsilon & \boldsymbol{=}\Vert \boldsymbol{\upsilon} \Vert \boldsymbol{=} \sqrt{ \upsilon^2_{1}\boldsymbol{+}\upsilon^2_{2}\boldsymbol{+}\upsilon^2_{3}}\:\in \left(0,c\right)
\tag{02b}\label{02b}
\end{align}
The Lorentz transformation is
\begin{align}
\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{x}\right)\boldsymbol{\upsilon}\boldsymbol{-}\dfrac{\gamma\boldsymbol{\upsilon}}{c}c\,t
\tag{03a}\label{03a}\\
c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma\left(c\,t\boldsymbol{-} \dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{x}}{c}\right)
\tag{03b}\label{03b}\\
\gamma & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}\frac12}
\tag{03c}\label{03c}
\end{align}
For the Lorentz transformation \eqref{03a}-\eqref{03b}, the vectors $\:\mathbf{E}\:$ and $\:\mathbf{B}\:$ of the electromagnetic field are transformed as follows
\begin{align}
\mathbf{E}' & \boldsymbol{=}\gamma \mathbf{E}\boldsymbol{-}\dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\mathbf{E}\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}\,\boldsymbol{+}\,\gamma\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}\right)
\tag{04a}\label{04a}\\
\mathbf{B}' & \boldsymbol{=} \gamma \mathbf{B}\boldsymbol{-}\dfrac{\gamma^2}{c^2 \left(\gamma\boldsymbol{+}1\right)}\left(\mathbf{B}\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}\boldsymbol{-}\!\dfrac{\gamma}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right)
\tag{04b}\label{04b}
\end{align}
Nothing more, nothing less.
How the Lorentz force 3-vector or the Lorentz force 4-vector are transformed see my answer here Are magnetic fields just modified relativistic electric fields?.
Expressions of the kind $''$...scaled down by $\gamma$...$''$ or $''$...the Electric fields are now closer together because of length contraction...$''$ are misplaced.
| {
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Why don't evanescent waves give rise to electromagnetic waves? I'm reading about evanescent waves for the first time. I understand that even thought no electromagnetic wave is transmitted across the boundary, an electric field is transmitted which decays exponentially into the material.
As far as I understand this is still an oscillating electric field. Can anyone explain, therefore, why it doesn't generate a magnetic field and give rise to an electromagnetic wave?
Or is there an electromagnetic wave moving parallel to the boundary but people just talk about the E field?
| A propagating wave obeys $\omega^2 = k_x^2+k_y^2+k_z^2$ with $k_i^2>=0$. For an evanescent wave at least one wave vector component has $k_i^2<0$. In this direction the wave amplitude decays exponentially, while it propagates in the other direction(s). Such solutions of the wave equation are only possible near an interface. As an example, total internal reflection at a glass-vacuum or glass-air interface is accompanied by an evanescent wave outside the glass.
The existence of such waves are a consequence of the Fresnel equations, which describe what happens at a plane interface between two different effective media, solely characterised by their (complex) indices of refraction. The phenomenon is unrelated to interference.
| {
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Nonlinear extension of Lorentz Group The Lorentz group is defined to be the set of linear transformations that leave $ds^2 = -dt^2 + |d\vec{x}|^2$ invariant. The Poincaré group contains the Lorentz group, but now we allow transformations of the form $t' = t+a$ where $a$ is some constant.
Is there a name for the group of transformations that leave $ds^2 = -dt^2 + |d\vec{x}|^2$ invariant without assuming linearity? The answers to this question here claims that any nonlinear transformation will have some "privileged" point, and hence we disregard them as nonphysical. However, I am curious nonetheless if there is a name for this group and any interest in it.
| It's called the Lorentz group. No, that's not a joke.
*
*Let $A$ be the group of linear transformations that leave
$$
x^2+y^2+z^2-t^2
\tag{1}
$$
invariant. Nonlinear transformations that leave (1) invariant also exist, as shown in the question linked in the OP, but we've defined $A$ to include only the linear ones.
*Now consider the group of all coordinate transformations that leave
$$
dx^2+dy^2+dz^2-dt^2
\tag{2}
$$
invariant, and let $B$ the subgroup that leaves the origin fixed. (To clarify: the prefix $d$ means differential, not finite difference.) The definition of $B$ does not assume linearity.
The group $A$ is the Lorentz group. The group $B$ is also the Lorentz group. They have the same name because they're the same group: $A=B$. To prove this, suppose we have two different coordinate systems such that
$$
dx^2+dy^2+dz^2-dt^2
=
d\hat x^2+d\hat y^2+d\hat z^2-d\hat t^2,
\tag{3}
$$
using hats to distinguish one coordinate system from the other. The form (3) of the line element implies that in both coordinate systems, the set of geodesics through the origin is the same as the set of straight lines — lines whose coordinates are all proportional to each other. The fact that the line element has this form in both coordinate systems implies that if both coordinate systems have the same origin, then the relationship between them is linear, because all coordinate transformations preserve geodesics. (A coordinate transformation is just a smooth re-labeling of the points in spacetime, and the definition of "geodesic" doesn't depend on how we label things.) Therefore, transformations in the group $B$ are automatically linear, and from there we can easily see that it's identical to the group $A$.
| {
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What does it mean for the gravitational force to be "between" two bodies? What is the meaning of the word "between" in the law that the force between two masses at separation $r$ is given by $\frac{GM_1M_2}{r^2}$? I am confused about how can a force be in-between, either it is on body A or on body B, or on both.
Suppose body A exerts force $F$ on Body B, so according to Newton's 3rd law of motion B should also exert a force on A.
Let's consider this case for gravitational force between two bodies. If body A exerts force $g$ on Body B, then B body should also exert a force $g$ on A, but B is also exerting the gravitational force $X$ on A, hence A will also exert force $X$ on B.
So, how are two forces acting?
I have given the representation in this diagram.
| One way to look at this is that forces always come as pairs. For example, you start with a universe with only one object in it, then you add another object and nature will immediately create a pair of forces. It’s not like the Moon feels that the Earth is tugging at it, and retaliates by tugging at Earth itself.
You can’t take such a pair of forces and label one the action and the other the reaction, or one the cause and the other the effect. Rather, both are manifestations of inner workings of nature, and to the best of our current knowledge, those workings aren’t best described in terms of force, but rather as “if things were moving like this before, they’ll be moving like that afterwards”.
Despite not being fundamental, forces are very useful mathematical objects, and the symmetry they exhibit in Newton’s third law is but one among many symmetries the universe has.
| {
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Problem in wedge constraint
My teacher told all this (u can see in picture)
Here both accelerations have component in -j direction, then why do we say that they cancel out?
[X-axis is surface of wedge]
I have problem in photo 2
Nobody answered so I have to ask it again...
| The answer is quite simple. In the constrained motion problems given above, the main constraint is that both surfaces must be in contact with each other. This is possible only if the component of acceleration of the two bodies in the direction perpendicular to that of motion is equal both in magnitude and direction i.e. the relative acceleration of the two bodies in the direction perpendicular to motion must be zero.
Here both accelerations have component in -j direction, then why do we say that they cancel out? [X-axis is surface of wedge]
The discussion here is on the relative acceleration of the two bodies in the direction perpendicular to that of motion.
Relative acceleration is given by:-$$\vec{a_{2/1}}=\vec{a_{2}}-\vec{a_{1}}$$
So, if the component of acceleration of the two bodies in the direction perpendicular to motion is equal both in magnitude and direction, we can expect their difference to be zero which is exactly your relative acceleration of the two bodies in the direction perpendicular to motion. Of course, the two accelerations do not cancel out each other as they are in the same direction!!
Hope it helps you.
| {
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What is the formula to determine the change in pressure when there is a change in flow? [Updated to help clarify my question]
I have a current water flow of 9 GPM (gallons per minute) at 50 PSI through a 1/2 inch diameter pipe pouring out at the end.
I understand that if I reduce the flow at the end (exit) of the pipe, it will increase the pressure of the water exiting the pipe. So, if I add a fixture at the end of the pipe which reduces the out-coming water flow to 2 GPM what will my new exiting water PSI be? What is the formula to determine this?
| According to Bernoulli’s theorem
$$P_1+\frac{1}{2}\rho{v_1}^2+\rho gh_1 = P_2+\frac{1}{2}\rho{v_2}^2+\rho gh_2$$
Since both ends of the pipe is in the same height
$$h_1=h_2$$
Now the equation can be written as
$$P_1+\frac{1}{2}\rho{v_1}^2= P_2+\frac{1}{2}\rho{v_2}^2$$
$\rho$ is the density of liquid
You can use this equation to get the relation between pressure and flow velocity.
Additionally, you can use equation of continuity to get the relation between flow velocity and area of cross section of pipe
$$A_1V_1=A_2V_2$$
| {
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Velocity is relative, which means acceleration is relative, which further implies that forces are relative as well So how would we know whether a force truly exists or not. I can be accelerating a car my 5 meters per second squared but another car accelerating with the same acceleration would think that my car is at rest relative to them. So is there any force on the car? Or are forces just relative and their existence just depends on our reference frame?
| Within different inertial frames, velocities will be different. However, acceleration will always be the same in any inertial frame. Therefore, so will the forces.
Short proof: suppose $v(t)$ is the velocity in one inertial frame, and $v'$ is some constant shift in velocity due to choosing a different inertial reference frame. Then the velocity within that newly-chosen reference frame is $V(t)=v(t)-v'$, and upon differentiating it w.r.t. $t$, you get $A(t)=a(t)$, so the acceleration does not depend on the inertial reference frame, and neither will the force.
Now, in non-inertial reference frames, there exist what we call Fictitious forces (or not real forces), but these do not exist in inertial reference frames.
| {
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Does the Oort cloud act as a kind of shield for the Solar System? Does the Oort cloud act as a kind of shield for the Solar System? When an interstellar object impacts the cloud, does its momentum get absorbed substantially?
| No. The Oort cloud is a cloud of comets and icy asteroids, not a physical barrier. Just like everywhere in the Solar system, it is mostly made up a empty space, and wouldn't absorb the impact of an interstellar object in any meaningful way.
The Sun does protect the Solar System from things like the interstellar wind with it's magnetic field - this is called the heliosphere - but it's separate from the Oort cloud. The Oort cloud actually extends far beyond the heliosphere.
| {
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Can a photon be detected by a "lateral" detector? If I direct a laser pointer north and I put a photodetector eastwards (i.e. at $90^\circ$ ), and I wait for a very very long time (in a perfect vacuum if necessary), will the detector ever be triggered by a photon?
| What the other answers do not address is shielding. Electromagnetism has this phenomenon called shielding and we are able to do it pretty well with our current technologies.
EM waves do spread spherically in space always. The only way to go around this is with shielding, in your case, the laser is using very effective mirrors.
The most common type of laser uses feedback from an optical cavity—a pair of mirrors on either end of the gain medium. Light bounces back and forth between the mirrors, passing through the gain medium and being amplified each time. Typically one of the two mirrors, the output coupler, is partially transparent. Some of the light escapes through this mirror.
https://en.wikipedia.org/wiki/Laser
The reason in your example, why you will only detect photons north, is because the shielding (mirrors) are designed so that photons will only escape in that direction. So the answer to your question is, that if this laser is designed to create a narrow beam, then you will only detect photons north because the mirrors only lets them escape that direction.
That being said, no mirror is perfect, and if you wait long enough, you might detect photons that escape in different directions through the mirrors, knowing that our universe is quantum mechanical.
| {
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Is there power reflected when Fresnel Transmission Coefficient is 1? Suppose a ray does not meet the total internal reflection condition and is transmitted fully through an interface (i.e. Fresnel Transmission Coefficient = 1), can we still expect any power being reflected for that particular ray? Whether it is a ray or wave model, will there be some power coming back to the original medium?
| To conserve energy,
$$1=R+T+A,$$
where $R$, $T$, and $A$ are the power reflection, transmission, and absorption coefficients, respectively (and we assume that scattering is bundled into $R$ and/or $A$). These are all positive, real numbers.
Therefore, if $T=1$, $R=A=0$.
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Why do I feel electric shock even in the presence of wood? I feel mild electric shock when my laptop with aluminum body is kept on a wooden desk is charging, while both my leg rests on the the leg of the table . Why is that ? Why isn't wood behaving like an insulator here ?
| The conductivity of wood depends on its water content. If you're in an arid desert it probably wouldn't conduct electricity, but if it's somewhat humid or the wood is damp for other reasons it's a reasonable capable conductor.
This website says wood can range from rubber to silicon depending on its water content.
| {
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Why doesn't $v_T = \omega r$ involve the direction of its variables? We derived $v_T = \omega r$ by the following procedure, and it's said that $v_T = \omega r$ "is a relation between the magnitudes of the tangential linear velocity and the angular velocity". Why doesn't the formula involve the direction of its variables?
We first define
| The direction is implied to be in the $\hat{\phi}$ direction because it is assumed that (eq 8-9) $r$ is not changing with time. Moreover, it's not really implied since it's the tangential velocity and so is in the tangential ($\hat{\phi}$) direction.
To clarify, let's work with an object spinning in a circle in two dimensions, to keep things simple. A point at distance $r$ in the direction $\hat{r}$ is given by
$$\mathbf{r} = r \hat{r}.$$
where $\hat{r} = \cos\phi \hat{x} + \sin\phi \hat{y}$. The time derivative of this point is given by
$$ \mathbf{v} = \frac{dr}{dt} \hat{r} + r \frac{d\phi}{dt}\hat{\phi} $$
By definition $\frac{d\phi}{dt} := \omega$. Now since it is assumed that $dr/dt = 0$ we are left with
\begin{align}
\mathbf{v} &= r \omega \hat{\phi}\\
&=r\omega (-\sin\phi\hat{x} + \cos\phi \hat{y}).
\end{align}
for the full vector description.
| {
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What is the entropy change of the universe for a rock if it falls from a height into a lake? The rock and the lake are at the same temperature According to my textbook, the entropy change of the universe is $+mgh/T$. I'm confused about why this happens.
after falling (without air resistance), wouldn't the rock possess $K_E = mgh$, which would then be transferred to the lake in the form of heat. Wouldn't this mean that the lake absorbs the same heat energy ($mgh$) from the rock to bring it to a standstill.
Would this not result in change in entropy of universe being $= 0$?
| The first step in determining the entropy change is to apply the first law of thermodynamics to establish the final equilibrium state. In this case, the potential energy of the rock is converted to internal energy of the rock plus surroundings (rest of universe): $$\Delta U+\Delta (PE)=\Delta U-mgh=0$$where U is the internal energy of the system, which, in this case, is the combination of roc plus universe. So, $$\Delta U=mgh$$
This change in internal energy is virtually all in the surroundings, since the final temperatures of the rock and surroundings are the same, and since the mass of the surroundings is so much larger than the rock. Treating the surroundings as an ideal infinite reservoir, we have that $$\Delta S=\frac{Q_{rev}}{T}=\frac{\Delta U}{T}=\frac{mgh}{T}$$where $Q_{rev}$ is the amount of heat that must be transferred to the surroundings in an alternate reversible path (in which it experiences the same internal energy change as the actual irreversible process).
| {
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Does this research paper prove that warp drives are impossible? Does this preprint prove that warp drives are impossible?
J. Santiago, S. Schuster and M. Visser, "Generic warp drives violate the null energy condition"
It states that the NEC (Null Energy Condition) is violated in this paper and many others: E. W. Lentz, "Breaking the Warp Barrier: Hyper-Fast Solitons in Einstein-Maxwell-Plasma Theory"
| No.
Regarding to that specific paper, they analysed only a family of metrics, which apparently does not even cover Lentz ansatz.
In general, be aware of no-go theorems that rely on too many assumptions. Even if you can argue with certainty that in the given mathematical framework something is impossible, you cannot guarantee that such framework, including every single assumption, describe our universe exactly.
| {
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Angular momentum orientation of the orbits of the two stars with respect to the Solar System Happen to see this in Wiki...
The astronomer Karl Schwarzschild observed the motion of pairs of stars orbiting each other. He found that the two orbits of the stars of such a system lie in a plane, and the perihelion of the orbits of the two stars remains pointing in the same direction with respect to the solar system. Schwarzschild pointed out that that was invariably seen: the direction of the angular momentum of all observed double star systems remains fixed with respect to the direction of the angular momentum of the Solar System. These observations allowed him to conclude that inertial frames inside the galaxy do not rotate with respect to one another, and that the space of the Milky Way is approximately Galilean or Minkowskian.[59]
My question is, are there any observations of binary star systems that violate the angular momentum orientation, so far?
| There are lots of observations that show that the line of apsides- the line joining the periapsis points of two stars - changes with time.
The phenomenon is known as apsidal precession.
Apsidal precession can of course be caused by $>2$ bodies in a system, but even for the 2-body systems discussed in the question, there are two mechanisms that lead to apsidal precession - it is predicted by General Relativity and it happens if the bodies are non-spherical and have a quadrupole moment.
In binary star systems, apsidal precession is well established in objects like the Hulse-Taylor binary, where the line of apsides rotates by 4.2 degrees per year.
None of this contradicts conservation of angular momentum - the precession of the apsides is in the orbital plane.
| {
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Elastic collision with one moving object hitting a stationary object In an elastic collision, I understand that momentum is conserved and kinetic energy is conserved. If billiard ball of silver (with velocity $v_{(Ag)}$ impacts a stationary billiard ball of aluminum, I am trying to calculate the velocity of the aluminum ball after the collision, $v_{(Al)}$. After an elastic collision, the impactor is at rest and the impactee has the motion.
Using momentum, $= m \cdot v$
$$m_{(Ag)} \cdot v_{(Ag)} = m_{(Al)} \cdot v_{(Al)}$$
Assuming silver is 4x denser than aluminium, then using momentum, the aluminium ball should have velocity
$$v_{(Al)} = 4\cdot v_{(Ag)}$$
But if we use kinetic energy, $1/2 m \cdot v^2$
$$\frac12m_{(Ag)}\cdot v_{(ag)}^2=\frac12m_{(Al)}\cdot v_{(Al)}^2$$
$$v_{(Al)}^2=\frac{m_{(Ag)}}{m_{(Al)}}\cdot v_{(Ag)}^2$$
$$v_{(Al)}=\left(\frac{m_{(Ag)}}{m_{(Al)}}\right)^{\frac12}\cdot v_{(Ag)}$$
$$v_{(Al)}=2\cdot v_{(Ag)}$$
Somewhere I have lost some neuron connections in my brain because I cannot resolve this conflict. This is a perfectly elastic collision so both momentum and kinetic energy should be conserved.
I have read multiple threads including:
When is energy conserved in a collision and not momentum?
How to calculate velocities after collision?
How can I calculate the final velocities of two spheres after an elastic collision?
Calculating new velocities of $n$-dimensional particles after collision
Velocities in an elastic collision
Summation of the velocities before and after an elastic collision
| If a small mass collides elastically with a larger one which was at rest, the smaller one will bounce back, not stop.
| {
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Why does the intensity of the bright fringes decrease as we move away from the central maxima in Young's Double Slit Experiment? I studied that in Young's Double Slit Experiment the variation of intensity ($I$) of the fringes on the screen with respect to the phase difference ($Φ$) is given by :
$I = 4I_{0} \cos^{2}\frac{Φ}{2}$
$I_{0}$ is the intensity of light coming from each slit.
At maximas or constructive interference, $Φ = nλ$, where $n$ is any whole number and hence we get $I = 4I_{0}$ Below I have given the image of an interference pattern from a laser beam passing through double slit. As you can see as we move away from the central maxima, the intensity decreases and eventually it becomes zero. But how is this possible? According to our equation, the intensity of the centre all the bright fringes should be $4I_{0}$ and hence we should get equal brightness in all the maximas. But why does the intensity decrease and become zero at some point? Shouldn't the interference pattern extend upto infinity and there should be equal brightness at all the maximas? Please explain. I am so confused.
| If you shine a spherical lightwave on the wall, you will observe that away from the line that connects the source with the wall (assuming that the direction of the lightwave is perpendicular to the wall) the intensity of the light will diminish. In the double-slit experiment, you use basically two spherical light waves. The intensities of both waves will drop upon hitting the screen.
| {
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Does Lorentz symmetry breaking always require "stuff"? Following the thread "can Lorentz symmetry be broken?" and the paper "Zoology of condensed matter:
Framids, ordinary stuff, extra-ordinary stuff", it appears that any system with Lorentz symmetry breaking is called a condensed matter system (with some extra details). For a condensed matter system we usually have to have "stuff" (atoms or fermions) to break the symmetry. Question. Can boost symmetry be broken without "stuff"?
Some extra questions. Why is Lorentz symmetry not broken in the Standard model? Is it possible that Lorentz symmetry is broken at higher energies?
| Lorentz symmetry is always broken, as a matter of fact, by implied presence of many reference systems (= stuff), between which we make the recalculation rules for observed results. Markers that break the Lorentz invariance are the relative distances between RFs, their relative orientations, their particular physical properties, etc., absent in the set of Lorentz transformations parameters.
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Book recommendation: Does anybody know a book adopting a more intuitive approach to the topic of Crystal Vibrations (phonons) than the Book by Kittel? I have tried Simon's 'Oxford Solid State basics' and Kittel 8th edition but I am not impressed by both (I mean the content covered through Chapters 4 and 5 in Kittel)
| I don't know what you would or wouldn't find intuitive, so I'll just say which book I like best: J.M. Ziman's Electrons and Phonons: The Theory of Transport Phenomena in Solids https://global.oup.com/academic/product/electrons-and-phonons-9780198507796?cc=us&lang=en&
A lot of the book probably isn't relevant to you (and it's super old-school --- one of the first books on the subject), but the first chapter is my favorite explanation of phonons. FWIW, Ziman's derivation is a little unusual (quantum from the start).
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Why we only need to consider one component of the torque vector when we consider only cases in which the rotational axis is fixed in direction? In my textbook, it is said that "we will consider only cases in which the rotational axis is fixed in direction. As a result, it will be necessary to consider only one component of the torque vector."
I'm confused as to why we only need to consider one component of the torque vector when we consider only cases in which the rotational axis is fixed in direction? Also, what does it even mean by considering one component of the torque vector? Is it saying that we only consider one of the vectors that make up a torque vector (i.e. either $\vec{r}$ or $\vec{F}$)?
| Torque is always given by:- $$\vec{\tau}=\vec{r}\times\vec{F}$$
Since, torque is a vector quantity, it can be broken down into mutually perpendicular components which add up vectorially to give the torque vector just like in the image below.
Though the example is given with acceleration $\vec a$, it is true for torque $\vec \tau$ as well.
This is what is meant with components in the question. It does not mean either $\vec{r}$ or $\vec{F}$.
By one component here, they are probably trying to mean the component of the torque vector along the fixed axis because the other components of the torque vector (which are not along the axis) are automatically nullified as no rotation is allowed about any other axis (due to fixed axis).
Hope it helps.
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Why is it said that antiparticles are a result of combining SR with Quantum theory? I did understand the historical reasons for the discovery of antiparticles in this context. But are antiparticles really a 'consequence' of combining special relativity and quantum theory? Why isn't it better to say that the existence of antiparticles are consistent with QM and SR?
| There are perfectly well-defined non-relativistic theories that contain anti-particles, and perfectly well-defined relativistic theories that contain no anti-particles.
Examples are a dime a dozen. Take a complex scalar field and turn on a Lorentz-violating interaction. Poof: now you have a non-relativistic theory that contains anti-particles. Conversely, take a scalar field and do Lorentz-invariant interactions only. Naturally, everything is its own anti-particle, and the system enjoys full relativistic invariance.
In conclusion, there is no connection between special relativity and anti-particles.
Not even field-theory is related to anti-particles, at least to the extent that you can define a satisfactory notion of "particle" for non-field theories. You can do e.g. a quantum mechanical model a la SKY with complex fermions. This is not a field theory, yet there is a perfectly sensible notion of anti-particle, as inherited from the charge-conjugation symmetry.
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Equipotential as a circle
I just dont understand how with this configuration there could exist a equipotential as a circle. For the assumption that $R>>$ dipole size I think it is there for the approximation of potential due to a dipole $p$ at a distance $r$ $$V=\frac{pcos{\theta}}{4\pi \epsilon_{o} r^{2}}$$
This is all I could make it out.
I'm not asking to solve this question but please explain what I need to look out for and how should be the uniform field be in order for the above to work out?
| Hints:
Do not use the formula for potential. Rather, we know, for a small dipole, the field at an angle $\theta$ and distance $r$ is given by:
$$\vec E=\frac {2kp \cos \theta}{r^3} \hat r+\frac {kp\sin \theta}{r^3}\hat {\theta}$$
Now note that the total electric field at any point on the given circle must be directed towards the normal at that point, otherwise it cannot be an equipotential surface.
Using this information, and the fact that field at $A$ and $B$ must be normal at that point, figure out the direction and magnitude of $E_0$.
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Euler-Lagrange equations in relativity (Goldstein) In order to have a covariant formulation of special relativity, we stop using the time $t$ as a parameter and we choose some invariant parameter.
In Goldstein (third edition), chapter $7.10$, it goes through this derivation making an argument about why proper time $\tau$ can't be the parameter we are looking for because of the constraint on 4-velocities.
\begin{equation}
u_\nu u^\nu=c^2 \tag{1}
\end{equation}
It then chooses another parameter $\theta$ and derives Euler Lagrange equations using this new parameter $\theta$. At the end he chooses
\begin{equation}\theta=\tau\tag{*}\end{equation}
Why does this allow us to ignore the constraint $(1)$?
We're basically using proper time the whole time but ignoring the constraint and using at the end.
I've read something similar on another book (which used $s=\int ds$ where $ds^2=dx^\nu dx_\nu$, thought) which said that to find the variation of action:
\begin{equation}
S=\int Lds
\end{equation}
We should also consider the variation of $ds$ with coordinates. Then it introduces a new parameter that is eventually replaced by $s$. How do you account for that?
| Proper time $\tau$ can be defined as the parametrization of $x^\mu(\tau)$ such that :
$$u_\mu u^\mu = \frac{\text d x_\mu}{\text d\tau}\frac{\text d x^\mu}{\text d\tau} = c^2$$
If you choose an arbitrary parametrization $x^\mu(\theta)$, then you have :
$$\frac{\text d x_\mu}{\text d\theta}\frac{\text d x^\mu}{\text d\theta} = \left(\frac{\text d\tau}{\text d\theta}\right)^2 \frac{\text d x_\mu}{\text d\tau}\frac{\text d x^\mu}{\text d\tau} = \left(\frac{\text d\tau}{\text d\theta}\right)^2 c^2$$
so contraint $(1)$ is indeed lifted.
Since the observables are $x^i$ in terms of $x^0$, the parametrization of the trajectory is irrelevant. The action and the equations of motion should be invariant under reparametrization. It is easier to work out the Euler-Lagrange equation with an arbitrary parametrization, and choose a sensible one (proper time) at the end.
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Does a object gain heat faster the colder it is? Does an object at $-273°{\rm C}$ gain heat faster than an object at $-1°{\rm C}$
| In order to answer your question definitively, one would need to know from what source the object was gaining heat, and how the heat was being transferred from the source to the object.
However, generally heat will be transferred more rapidly where there is a larger temperature difference between the source and the recipient object.
If you poured a cup of water on a steel table at -1C, the water in the cup would take longer to freeze than if you had poured it onto a table at -273C.
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What is the $y$-axis in an electromagnetic wave? Apologies if my question is unclear, any help to clarify it along the way is most welcome.
I'm confused about what we mean when we say electromagnetic 'waves' (say visible light). In the usual mental picture we have of a simple sine wave, what does the y-axis correspond to? In mechanical waves like water or sound, we can plot the vertical displacement of each particle along the x-axis as a value on the y-axis. Alternatively we can fix a specific particle in the water/air and take the x-axis to be time and the y-axis to be its physical displacement.
But for light traveling in a vacuum, there would be no such displacement of particles. So what does the y-axis correspond to? What do we mean by light being/behaving like a wave?
| Of course what is reported onto the $y$-axis is a matter of arbitrary definition. But usually the standard way to depict an electromagnetic wave is to report onto one of the three orthogonal axis the strength of the electric field, on another one the strength of the magnetic field, and the last third axis is taken to be the axe of propagation of the electromagnetic wave.1
Keep in mind that this depiction of an electromagnetic wave is not an abstraction without connection to reality: is not simply a way to represent the wave that we have choosen arbitrarily: in fact it can be proven that the two fields, electric and magnetic, in an electromagnetic wave, are always ortogonal to each other and orthogonal to the direction of propagation of the wave; this follows directly form the Maxwell's laws, that lead to the wave equation.
[1]: With the word "strength" in this context we mean the module of the electric or magnetic field vector in that point in space.
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Wave packet in quantum mechanics? When we talk about light waves or EM waves, we simply say that the wave packet is the superposition of other waves of different wavelengths. In quantum mechanics, we say the same thing; the superposition of many waves associated with electron form a wave packet. I don't understand this, because one wave is associated with one electron. It is only superimposed with other electron waves. Does an electron make superpositions with itself having different wavelengths? How is this possible?
|
Bcs one wave is associated with one electron.
This is not true! You should rather say, There is a unique wave function $|\Psi(t)\rangle$ (or In position basis $\psi(x,t)$) associated with a particular state of system.
There is something very nice about $|\Psi(t)\rangle$ but not very new, If $|\psi\rangle$ and $|\psi'\rangle$ represent the possible wave function of the electron then so does the $\alpha |\psi\rangle+\beta |\psi'\rangle$. That's called the Principle of Superposition.
A wave packet is a simple superposition of different wave forms,
$$\psi(x,t)=\frac{1}{\sqrt{2\pi}}\int \phi(k)e^{i(kx-\omega t)}dk$$
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Why don't electrons accelerate in a circuit? In a circuit, electric field exerts force on electrons, so they must accelerate. Every text book I have read, points that electrons move with a constant drift velocity. How can this happen? Does Newton's law not apply there?
| Yes Newton's law is applied here, and the electrons accelerate in response to the electric field. However, the electrons also undergo collisions with the atoms of the conductor, so on an average, they acquire an initial velocity of zero just after each collision. The electrons then acquire a final velocity $\vec v_f$ before the next collision and the average value of this final velocity is termed as the drift velocity, which is found to be a constant value, simply because the average time interval τ between each collision (called the relaxation time) is very small
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Why light shows its wave-like properties only when it interacts with objects with dimensions close to the wavelength of light? In Young's Double Slit Experiment, we were taught that light behaves as a wave here because the width of the slits are very close to the wavelength of light itself. But why does light behave like a wave only when it interacts with objects that have dimensions close to the wavelength of light? Even in my book no explanation is given as to why this is true. Can someone please explain as to why this is true? I am so confused.
| If you send a water wave (with a fixed wavelength) to an object, a diffraction pattern will form behind the object (assume you send the wave perpendicular to the length of the object). If the wavelength of the waving water is very small compared to the object, the wave will not notably curve around the edges. Almost no diffraction of the wave will be seen (only reflection, approximately). Only at the edges of the object two waves will emerge that can interfere, but because the length of the object is much bigger than the wavelength of the water this effect is small (but it is there). If the wavelength of the water is comparable or larger than the size of the object an interference (diffraction) pattern will emerge. The waves will curve. See here. The same holds for the "inverse": sending a wave to an (infinitely) extended object with an open space in it.
Replace the water waves with electromagnetic radiation, and you will see why waves interfering with objects that have a size comparable with their wavelength will show more wavelike behavior.
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Gauge-fixing conditions in Einstein-Cartan gravity What are the gauge-fixing conditions one needs to impose on the tetrad one-form $e^a$ and the spin-connection one-form $\omega^{ab}$ while working in the Einstein-Cartan formalism where both are independent objects? I am more interested in the case of 4D gravity.
| Analogous to Equation 3.3 of this paper https://arxiv.org/pdf/gr-qc/9406006.pdf, one can propose the following gauge conditions in all dimensions $D > 3$
$$*D* e^a = *[d*\omega^{ab}+(\omega \wedge *\omega)^{ab}] = 0$$
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EMF generated by moving square
I have been studying electromagnetic induction and I have trouble understanding some things.
Say we have a square wire going through a magnetic field like in the picture (height is "h" and total length in a magnetic field is "x"). If we moved that whole conductor to the right because of the magnetic field going into the screen (j) the left-most part of the wire would experience an upwards force and therefore generate movement of electrons in that direction. That's easy enough to calculate F = q * v X B. And generated EMF is now ε = vBh.
Now if we wanted to get the same answer using flux which is Φ = ∫B * da:
Φ = B * h * x so a change in flux is Φ = Bh * dx/dt = Bh * (-v) which makes ε = Bhv, precisely what we got the first time.
My questions is:
If we bring in a new magnetic field that is constant everywhere wouldn't the change in flux be 0 which would mean induced EMF is 0 but if we used the first method to calculate EMF nothing would change and EMF would still be ε = vBh.
| Might as well answer the question in case someone in the future is stuck here as well.
The change in flux is 0 but the EMF in the case of an infinite magnetic field should be 0.
That's because while the force you get using Lorentz force law is indeed still there, now you have to take into account the force that the right-most wire is producing, which is in the same direction leaving you with EMS (around the loop) of 0.
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Why isn't the formation of a black hole stopped by the uncertainty priciple? As long as spacetime is smooth (that is, not quantized), the uncertainty principle can be applied. When a black hole forms the particles that are collapsing get closer and closer (though the metric of space changes). Won't the uncertainty principle prevent them (the particles) from approaching each other too close? That is as long if the spacetime is still smooth. Or will the changing metric prevent this? I'm not asking if the uncertainty principle can be applied to an already formed black hole (if it can be formed), which it obviously can't due to the non-existence of a quantum theory of spacetime. I'm asking if the UP can be applied to the particles that are collapsing to a BH.
| The uncertainty principle does not prevent indistinguishable fermions getting arbitrarily close. All it says is that if they do, they must have an arbitrarily large difference in momentum (or different spin quantum numbers).
In other words, you can cram ideal (i.e. non-interacting, point-like) fermions to as large a density as you like, but the Fermi momentum then becomes commensurately large.
If the fermions become ultra-relativistic then their equation of state becomes $ P \propto \rho$, which is incapable of providing hydrostatic equilibrium (which requires $P \propto \rho^\alpha$ with $\alpha >4/3$, or even higher in a General Relativistic treatment).
| {
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Dirac Current Derivations I am currently self-studying Quantum Field Theory and am using the book An Introduction to Quantum Field Theory by Peskin and Schroeder. I am confused about a derivation presented in section 3.5 (called "Quantization of the Dirac Field"). Equation 3.111 derives a "rotation current density" J which splits up into an orbital angular momentum part and a spin momentum part. This derivation I understand. Next, the book wishes to prove that the Dirac Equation deals with particles of spin $1/2$. To do so, the authors consider the case when the particle is at rest; this allows us to ignore the orbital angular momentum term in equation 3.111. From this it follows (again I understand this):
$J_z = \int d^3x \int\frac{d^3p d^3p'}{(2\pi)^6} \frac{1}{\sqrt{2E_p 2E_p'}}e^{-ip'\cdot x}e^{ip\cdot x}
\sum_{r,r'}\bigg(a_{p'}^{r'\dagger}u^{r'\dagger}(p') + b_{-p}^{r'}v^{r'\dagger}(-p')\bigg)\frac{\Sigma^3}{2}\bigg(a_{p}^{r}u^{r}(p) + b_{-p}^{r\dagger}v^{r}(-p)\bigg)$
However, the next equation says that
$J_z a_0^{s\dagger}|0\rangle = \frac{1}{2m}\sum_{r}\bigg(u^{r\dagger}(0) \frac{\Sigma^3}{2}u^s(0)\bigg)a_0^{r\dagger}|0\rangle$
I am unsure why this result is true. The book gave the commutator relation $[a_p^{r\dagger}a_p^{r'},a_0^{s\dagger}] = (2\pi)^3 \delta^3(p)a_0^{r\dagger}\delta^{r's}$. Note that all $p$'s represent three momentum here and not four momentum. I tried using this commutator relationship when I expanded out the parenthetical terms in $\sum_{r,r'}$ but the $\Sigma^3$ matrix got in between these operators. Can anyone explain mathematically how the second equation follows from the first?
| For the commutator $[J_z, a_{ 0}^{s\dagger}]$ we have non-zero terms only for those that have an $a$ in $J_z^{(s)}$
\begin{align}
[J_z, a_{ 0}^{s\dagger}] = &\, \int d^3 x \, \int \frac{d^3p\, d^3q}{(2\pi)^6} \frac{1}{\sqrt{2 E_{ p} 2 E_{ q}}} \sum_{r, r'} e^{+i ( p - q)\cdot x}
\Big[\Big(a_{ q}^{r \dagger} u^{r\dagger}( q) +b_{- q}^{r } v^{r\dagger}(- q) \Big) \left(\frac{1}{2}\Sigma^3 \right)
a_{ p}^{r'} u^{r'}( p) , a_0^{s\dagger} \Big]
\end{align}
If we act with this on the ground state $| 0 \rangle$ then the $b$ terms vanish and so we get
\begin{align}
[J_z, a_{ 0}^{s\dagger}] | 0 \rangle= &\, \int d^3 x \, \int \frac{d^3p\, d^3q}{(2\pi)^6} \frac{1}{4\sqrt{ E_{ p} E_{ q}}} \sum_{r, r'} e^{+i ( p - q)\cdot x}
u^{r\dagger}( q) \Sigma^3 u^{r'}( p)[a_{ q}^{r \dagger}a_{ p}^{r'} , a_0^{s\dagger} ] | 0 \rangle \nonumber\\
=&\,\int d^3 x \, \int \frac{d^3p\, d^3q}{(2\pi)^6} \frac{1}{4\sqrt{ E_{ p} E_{ q}}} \sum_{r, r'} e^{+i ( p - q)\cdot x}
u^{r\dagger}( q) \Sigma^3 u^{r'}( p)a_{ q}^{r \dagger} (2\pi)^3 \delta^3( p) \delta^{r's} | 0 \rangle \nonumber\\
=&\,\int d^3 x \, \int \frac{d^3q}{(2\pi)^3} \frac{1}{4 E_{ q} } \sum_{r} e^{-i q \cdot x}
u^{r\dagger}( q) \Sigma^3 u^{s}( 0)a_{ q}^{r \dagger} | 0 \rangle
\end{align}
Note that the $a$'s carry no Dirac index, so we can move them across the Dirac spinors without penalty. We can now also perform the $x$ integration, which will give a $\delta^3 ( q)$ which we can then do as well. This gives
\begin{align}
[J_z, a_{ 0}^{s\dagger}] | 0 \rangle
=&\, \frac{1}{4 E_{ 0} } \sum_{r}
u^{r\dagger}( 0) \Sigma^3 u^{s}( 0)a_{ 0}^{r \dagger} | 0 \rangle
\end{align}
| {
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Understanding a Solution to QFT I am self studying QFT from An Introduction to QFT and currently, I am completing problem 3.3(a). Here are some sample solutions that I am using to understand this problem: (https://theoreticalmaximum.files.wordpress.com/2017/07/intro-to-qft-solutions2.pdf). In the middle of page 9, the author writes:
$\frac{1}{\sqrt{2p\cdot k_0}} p_\mu p_\nu \gamma^\mu \gamma^\nu u_{R/L}(0) = \frac{1}{\sqrt{2p \cdot k_0}} p_\mu p_\nu\eta^{\mu \nu} u_{R/L}(0)$.
Isn't $\gamma^\mu \gamma^\nu = \eta^{\mu \nu} - i\sigma^{\mu \nu}$ (from the identity $\sigma^{\mu \nu} = \frac{i}{2}[\gamma^\mu, \gamma^\nu]$)? This seems to suggest that $p_\mu p_\nu\sigma^{\mu\nu} = 0$, but I do not completely understand why this is true. Can anyone explain?
| As @Cosmos Zachos comments, the contraction of an antisymmetric tensor with a symmetric one vanishes. By antisymmetry of $\sigma$,
$$p_\mu p_\nu \sigma^{\mu \nu} = -p_\nu p_\mu \sigma^{\nu \mu},$$
But as the indices are summed over we can relabel $\mu \to \nu$ on the RHS and we have
$$p_\mu p_\nu \sigma^{\mu \nu} = -p_\mu p_\nu \sigma^{\mu \nu},$$
which can only be true if both sides are equal to zero.
| {
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Where is energy in energy density? I was learning about energy density and it seemed to be defined as the potential energy per unit volume in an electric field
$\frac{dU}{dV} = \frac{1}{2}\epsilon E^2$
But how can just the electric field have a potential energy on its own without presence of any charge? What is causing this energy to be present in an electric field?
|
But how can just the electric field have a potential energy on its own without presence of any charge? What is causing this energy to be present in an electric field?
Remember the mechanical definition of energy. Energy is the capacity to do work. The electric field can exert a force on a moving charge, so it does have energy. Furthermore, the field itself must be able to carry the energy since energy can leave all of the charges. Finally, the field energy also represents other definitions of energy, like the Noether theorem based definition.
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Geodesic incompleteness of static spherically symmetric solution Static spherically symmetric solution of Einstein equations is given by the metric
$$
ds^2=f(r)dt^2-\frac{dr^2}{f(r)}-r^2d\Omega^2,
$$
where $f(r)=1-(kr)^2$, $d\Omega^2$ is the metric of unit sphere.
I struggle with the notion of geodesic incompleteness. It is said that the space-time with metric given above is geodesic incomplete, but I can't understand how to show it.
| Geodesic incompleteness is defined by the existence of a geodesic that, under the evolution of time from $-\infty$ to $\infty$, terminates at some finite affine parameter. Hence, it is enough to construct a geodesic (by computing the Christoffel symbols and solving the geodesic equation with respect to the given metric) to obtain $s(t)$, inverting it into $t(s)$, and show that $t\to\infty$ or $t\to -\infty$ for some finite $s$.
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Name for critical lines in parameter space and plots thereof Suppose I study a dynamical system as a function of some control parameters, and I find that the nature of the attractors changes discontinuously (or non-analytically) at certain critical values (or along critical lines) in the space of control parameters. These lines separate different basins of attraction. What do you call a plot of these basins in the space of control parameters? A phase diagram of the dynamical system? What is a proper terminology for the transition lines? (Any standard references for this kind of thing?)
|
What do you call a plot of these basins in the space of control parameters?
Such a plot is called "parameter space" (for example), but the specific regions are not usually called "basins" (a term more associated with the system's phase space), but simply "region" or "set" (e.g., here).
A phase diagram of the dynamical system?
When appropriately defined, it can of course be understood, but it might not be the best choice, since "phase" is most often understood as a synonym for "state", as in "phase space", "phase portrait", "phase plot" and even (though seldom) "phase diagram". That said, some dynamical systems papers (for instance this) do use the term "phase diagram" in the statistical sense, so this does remain an option.
What is a proper terminology for the transition lines? (Any standard references for this kind of thing?)
I'm not aware of any established nomenclature and usually generic descriptive terms such as "boundaries" or "lines" are used.
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Why is power transmission carried out at low current high voltage? My textbook states that power is transmitted at high voltage and low current since $P=I^2R$ and as the current has a small magnitude, the heat dissipated across the transmission lines is less than when we carry it out at high current and low voltage. But $P=I^2R$ can also be written as $P=V^2/R$ and hence a discrepancy would occur. Where am I going wrong?
| As we have $P = I\Delta V $, the same amount of power can be delivered either at high currents and low potential differences or at low currents and high potential differences. In either case, the same amount of internal energy is transferred to the surroundings by heat as you correctly pointed out, so the decision is made primarily based on economic reasons. From the definition of resistance, we have: $$ R = \frac{\Delta V}{I} $$ Hence, if low potential differences and high currents were used, the resistance of the materials would have to be lower in order to deliver the same power. Materials that have low resistances include rare metals like silver and gold that are too expensive to be used commercially. Hence the choice of low currents and high potential differences is made, so that inexpensive copper wires can be used that generally have a high resistance. This high potential can also be conviniently reduced or increased by a transformer.
Hope this helps.
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How is light emitted by an incandescent lamp? I am looking for better understanding of how light is produced in an incadescent lamp. More specifically: how is the kinetic energy of electrons converted to light?
*
*Are we dealing with interband transitions or with intraband relaxation involving photons? Is this Bremsstrahlung (electrons lose their energy as light when colliding with crystal impurities/defects)? Or is this a thetmal radiation resulting from Joule heating?
*How is the emission affected by presence of impurities and imperfections of the crystal lattice? Do phonons play a role?
*What properties make a material more suitable for use as a filament: should it be a metal? Should it have a crystalline structure? Will any metal produce light, if a high current is passed through it in vacuum?
Update
The term describing the processes in the incadescent lamp is thermal bremsstrahlung, see the posts on this subject here and here.
| The only requirement for radiation to occur, in insulators or conductors, is acceleration of charges or magnetic fields. Bound electrons surrounding a nucleus can be stimulated to radiate by thermal agitation of the nucleus. Rotations, vibrations etc. All atoms have either dipole or multipole magnetic moments, these will also radiate when agitated by heat. This is thermal spectrum radiation, under certain conditions it can have a "black body spectrum" Obviously an electron transitioning from one, non radiating stable atomic state, to another non radiating stable atomic state, will also radiate briefly, this form of radiation is the source of line spectra.
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Is there a special quantity for direction? As in, since vectors have magnitude and direction. And scalars have only magnitude, is there a special quantity for direction?
| You can represent a vector as
$$\mathbf{v}=v\ \hat{v}$$
where $v$ is the magnitude of the vector (that is the length of the arrow if you like) and $\hat{v}$ is the unit along the vector $\mathbf{v}$. You can refer to the former as magnitude and later as the direction of a vector.
For example: In
$$\mathbf{v}=2\hat{i}$$
$2$ is the magnitude of the vector while $\hat{i}$ is the direction of vector.
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Does mass have an effect on Centripetal Acceleration? I am using an online simulation for a lab concerning Centripetal Acceleration. When I change the mass the graph indicates that the magnitude of the acceleration is constant. According to the Centripetal Acceleration formula: $a=v^2/r$, this is true because no mass is present in the relationship. However, when I use Newton's Second Law of Motion, $a=f/m$, I can see that the mass and the acceleration are inversely proportional. Both of these ideas are found when I look them up online, now I am a bit confused on which one might be more valid.
| It depends on what you are looking at.
If you are applying a constant centripetal force to objects of different masses, then they will each experience a different centripetal acceleration.
If a bunch of different masses are under going circular motion around a circle of radius $r$ with speed $v$, then they will all be experiencing the same centripetal acceleration (but different centripetal forces).
So "does mass effect centripetal acceleration?" is not specific enough to have an answer. You need to add in what else you are considering, i.e. what you are holding constant and what you are allowing to change as you change the mass.
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Why doesn't current flow through an open branch? I know that current doesn't flow through open branch because current can't flow through air due to its high resistance .But i was thinking , what's the problem if current flows through an open wire (assumed 0 resistance for matter of circuit solving). I mean, isn't there a possibility that the charge flowing through the wire keeps accumulating at its end as it can't flow through air as battery is incapable of flowing it through the air. Why do we have to entirely abandon that wire?
| The answer is current actually does flow and charges do accumulate at ends. That's why arcs form in open circuits at high voltage. And as you mentioned the flow would be a momentary one and not continuous.
When we connect a battery to an open wire, the potential of the battery does push charges around. But as soon as the charge reaches the end and accumulate, they create an opposing potential and attain an equilibrium when the voltage induced at the open end is equal to the battery voltage.
Now, the voltage generated when $Q$ charge is put on an object is given by its Capacitance. So how much charge flows depends on the capacitance of the lose wire. Generally a bigger piece of wire can hold more charge at a given voltage. So if the lose wire is longer, more momentary currents will flow. ( apart from many other ways of explaining, this also explains why longer antennas capture better signals )
Also, the current flow will be momentary if the voltage is a constant voltage. If it is a varying voltage, the charge stored would also vary creating small
persistent currents. example : line testers work even if we isolate ourselves from ground. This is because charge enters and exits our body due to varying voltage.
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In LED's do the number of charge carriers (electrons and holes) decrease with time? According to page-1268-69 of Halliday, Walker & Resnick's Fundamentals of Physics (10th edition),
To emit enough light to be useful as an LED, the material must have a
suitably large number of electron-hole transitions....What we need is
a semiconductor material with a very large number of electrons in the
conduction band and a correspondingly large number of holes in the
valence band. A device with this property can be fabricated by placing
a strong forward bias on a heavily doped p-n junction, as in Fig.
41-16. In such an arrangement the current I through the device serves
to inject electrons into the n-type material and to inject holes into
the p-type material. If the doping is heavy enough and the current is
great enough, the depletion zone can become very narrow, perhaps only
a few micrometers wide. The result is a great number density of
electrons in the n-type material facing a correspondingly great number
density of holes in the p-type material, across the narrow depletion
zone. With such great number of densities so near each other, many
electron-hole combinations occur, causing light to be emitted from
that zone.
Now, if electron-hole pairs are ceasing to exist due to recombination and resulting in a greater number of "gridlocked" electrons and light, who will continue to conduct electricity? Won't current flow stop after a while due to the absence of electron-hole pairs?
| No, the current won't stop flowing as the battery maintains constant electromotive force. Some free electrons will lose their kinetic energy and become gridlocked. You can view this phenomenon, that is, free electrons losing their kinetic energy, as free electrons losing kinetic energy to a resistance, say a light bulb.
| {
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Why is entropy of gas reamains constant during adiabatic expansion? In adiabatic compression, we add energy to the gas via piston and decrease volume. Now the increase in energy and decrease in volume will cause increase and decrease in entropy respectively, but it turns out(according to the books) that the entropy of gas remains unchanged. How to prove it mathematically? Although I know that energy given in adiabatic process is zero but the energy is given indirectly via piston and that confuses me.
| We measure change in entropy according to the following integral from state 1 to 2:
$$dS = \int_1^2 \frac{dQ}{T} $$
For adiabatic change $dQ = 0$ from 1 to 2
Therefore the integral becomes
$$dS = \int_1^2 \frac{0}{T} $$
Since $dS=0$, no entropy change from 1 to 2.
What might be confusing you is whether the Piston itself is able to conduct heat. When we define an adiabatic system, by definition the walls and the piston itself are not able to transfer energy in the form of thermal energy (heat). But only in terms of mechanical energy (work).
We can expect the piston to be built of a thermally insulating material although this doesn't happen practically.
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Can information travel faster than speed of light in this situation? I know the answer is no but I have a thought experiment that seem to be violating that. Imagine two persons living on two different planets namely A and C which are 10 light years apart. There is a planet in between, B, which is located exactly at the same distance from A and C. If two persons from A and C get on their hypothetical spaceships at the same time, that can travel around .99C, and move towards B, they can talk to eachother in around 5 years. Didn't information just convey between two persons at the speed of ~2C in this case?
| The limit here is about how far the information can travel between two points. If people who originate at those points move while the message is propagating, that can reduce the time it takes for the people to get the message, of course, because it cuts the distance that the message has to travel to reach them. (But it does not reduce the minimum time necessary for the message to get all they way to where the people started the scenario.)
So in your case, the travelers from A and from C arrive at B in just over 5 years. No violation there as you specified they were traveling at $0.99c$. We're good to this point.
Upon arrival at B, the travelers can speak to each other essentially instantly. No problem here either because they are in the same location, so the information is traveling (approximately) 0 distance. Speed is not a factor.
Now if the traveler who started from A wants to relay a message back to his home world, he can do that. But his message will have to travel the 5 light-years between B and A, so that will take time. Exactly 5 years if he sends the message on a carrier that propagates at speed $c$. So a message relayed all the way from C to A by this process takes a bit more than 10 years, computed as a the bit more than 5 years it took the travelers to reach B from their respective home worlds and the 5 years it took for the relayed message to travel from B to A. That's consistent with the limit of 10 years that it would have taken if the message had been sent directly from C to A at the speed of light from the start. Still no violation.
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Can atom be excited above ionisation level? I know this sounds dumb, but it puzzles me when I search for ArI levels catalogue in NIST. It shows argon's first ionisation level as $15.7596119 \; \text{eV}$ in row 427, with more rows after with energy above that.
This is puzzling. How can atomic energy levels go beyond ionisation level? Shouldn't the atom be stripped off an electron and becoming an ion on those levels?
NIST documentation put some words on this, but it doesn't explain the question I want to know.
Chances are that this is an idiotic question out of poor knowledge, please forgive my ignorance if that's true.
| The atom is a whole quantum mechanical entity, the electrons occupying orbitals at definite energy levels. Scattering photons of the appropriae frequency an electron has a probability to be removed, it does not need to be in the shallowest energy level.
If one could radiates the atom with the frequency/energy of a lower level , there will be a quantum mechanical probability that the electron from that level would be extracted, and the atom would be ionized because it would have excess positive charge corresponding to an inner electron leaving. See this enter link description here
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Can it be shown in experiment that the momentum (or position) states of the electron and proton in hydrogen are entangled? The states of the electron and proton in hydrogen are entangled. Which means that the momentum and position of both are entangled. Can this be shown in an experiment, so if you measure the momentum (or a small range of them) of the proton you will know in advance what the momentum of the electron will be?
If you first have to separate them, to do a measurement on them, will the information of the entanglement show itself in the measurement? As the information of entanglement of two spin states can show itself in measurement?
| Indeed, the states of the electron and the proton in a hydrogen atom are entangled. The wave function of a hydrigen atom has form
$$
\psi(\mathbf{x}_e, \mathbf{x}_p) = e^{i\mathbf{k}\mathbf{R}}\phi(\mathbf{x}_e- \mathbf{x}_p),
$$
where
$$
\mathbf{R}=\frac{m_e\mathbf{x}_e+ m_p\mathbf{x}_p}{m_e+m_p}
$$
is the position of the center-of-mass of the atom, whereas $\phi(\mathbf{x}_e- \mathbf{x}_p)$ is one of the hydrogen eigenfunctions discussed in any quantum mechanics text. Function $\psi(\mathbf{x}_e, \mathbf{x}_p)$ can be expanded in terms of momentum eigenstates of electron and proton
$$
\psi(\mathbf{x}_e, \mathbf{x}_p) = \int d^3\mathbf{k}_e \int d^3\mathbf{k}_pe^{i\mathbf{k}_e\mathbf{x}_e}e^{i\mathbf{k}_p\mathbf{x}_p} \varphi(\mathbf{k}_e, \mathbf{k}_p),
$$
which is an entangled state. We could further calculation the coefficients $\varphi(\mathbf{k}_e, \mathbf{k}_p)$ using the form of the wave function given above as:
$$
\varphi(\mathbf{k}_e, \mathbf{k}_p) = \int d^3\mathbf{x}_e \int d^3\mathbf{x}_pe^{-i\mathbf{k}_e\mathbf{x}_e}e^{-i\mathbf{k}_p\mathbf{x}_p}\psi(\mathbf{x}_e, \mathbf{x}_p) =\\
\int d^3\mathbf{q}\tilde{\phi}(\mathbf{q}) \delta^3\left(\mathbf{q}+\mathbf{k}\frac{m_e}{m_e+m_p}-\mathbf{k}_e\right)
\delta^3\left(\mathbf{q}-\mathbf{k}\frac{m_p}{m_e+m_p}-\mathbf{k}_p\right)=\\
\delta^3\left(\mathbf{k}-\mathbf{k}_e-\mathbf{k}_p\right)
\tilde{\phi}\left(\frac{m_p\mathbf{k}_e + m_e\mathbf{k}_p}{m_e+m_p}\right),
$$
where $\tilde(\mathbf{q})$ is the Fourier transform of $\phi(\mathbf{x})$.
Separating the proton and the electron would mean changing the state by ionizing the atom - whether the two will remain entangled depends on the process of separation. Measuring a momentum of one of the two particles is hard due to their strong binding, but this probably could be done in accelerator experiments. We then see from the equation above that the momentum state of one particle is related to the momentum state of the other - if we do manage to measure the momentum of the electron, we woudl obtain the momentum of proton as $\mathbf{k}_p=\mathbf{k}-\mathbf{k}_e$. It could be much easier to emasure their angular momenta.
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Are the accelerations the same at either end of a moveable pully? Given a moveable pulley with a fixed pulley on either side,
Is the acceleration of the left weight (m1) the same as that of the right weight (m2)?
Intuitively, I would imagine it to be, since if m1 drops by 10 metres, then m3 would rise by 5 metres, and thus m2 would drop by 10 metres. But mathematically the accelerations don't appear to be the same... Perhaps there is a mistake in my reasoning.
Assume the system is released from rest, an inextensible and massless string is used for the pulleys, and there is no friction forces acting on the system.
| While it is certainly possible that $x_1 = - \frac{1}{2} x_3$ and that $x_2 = - \frac{1}{2} x_3$, as your intuition suggests, it is not necessarily the case. (We define $x_1$, $x_2$, and $x_3$ to be the displacements from the original positions of the masses.) If both of the equations were automatically true, then rope would never move relative to $m_3$, since the same amount of rope would be going over the side pulleys at all times. But hopefully it seems intuitive that you could come up with some highly unbalanced set of masses (with $m_1 \gg m_2$, for example) where the rope should slide over $m_3$.
Instead, you should look to find one equation that relates $x_1$, $x_2$, and $x_3$ all together. This equation will be a mathematical statement that the overall length of the rope does not change. Once you have this equation, you can differentiate it to get a parallel statement concerning the accelerations of each mass.
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Time derivatives of the unit vectors in cylindrical and spherical In cylindrical and spherical coordinates, the position vectors are given by $\mathbf{r}=\rho \widehat{\boldsymbol{\rho}}+z \hat{\mathbf{k}}$ and $\mathbf{r}=r \hat{\mathbf{r}}$, next to next, and their derivatives with respect to time are
$$
\dot{\mathbf{r}}=\dot{\rho} \hat{\boldsymbol{\rho}}+\rho \dot{\varphi} \widehat{\boldsymbol{\varphi}}+\dot{z} \hat{\mathbf{k}}
$$
$$
\dot{\mathbf{r}}=\dot{r} \hat{\mathbf{r}}+r \dot{\theta} \widehat{\boldsymbol{\theta}}+r \sin \theta \dot{\varphi} \widehat{\boldsymbol{\varphi}}
$$
I wonder what would be the time derivatives of the unit vectors of the basis themselves. Is there any straightforward way to deduce their expressions?
| It doesn't make a lot of sense to talk about the "time derivatives of the unit vectors of the basis themselves", because the unit vectors themselves are constant with respect to time. Instead, in curvilinear coordinates, the basis vectors are vector fields, with different values from point to point in space. The unexpected terms that arise in the expressions you've written are because the unit vectors are not constant with respect to space, and any trajectory that moves through space will see these unit vectors vary because of their motion through space.
To make this more concrete, think about $\hat{r}$ as a vector field: $\hat{r}(r,\theta,\phi)$. If we then have a particle that has a trajectory given by $\mathbf{r}(t) = r(t) \; \hat{r}(r(t),\theta(t),\phi(t))$, then we can derive the velocity vector as follows:
\begin{align*}
\frac{d\mathbf{r}}{dt} &= \frac{dr}{dt} \hat{r} + r \frac{ d \hat{r}}{dt} \\
&= \frac{dr}{dt} \hat{r} + r \left( \frac{\partial \hat{r}}{\partial r} \frac{dr}{dt} + \frac{\partial \hat{r}}{\partial \theta} \frac{d\theta}{dt} + \frac{\partial \hat{r}}{\partial \phi} \frac{d\phi}{dt} \right)
\end{align*}
where we have used the product rule and the multi-variable chain rule.
If you carefully calculate $\frac{\partial \hat{r}}{\partial \theta}$ and $\frac{\partial \hat{r}}{\partial \phi}$, you can show that they're equal to $\hat{\theta}$ and $\sin \theta \hat{\phi}$ respectively, and so the expression you have for spherical coordinates is obtained. The key point to note here, though, is that the $\hat{r}$ vector field does not depend inherently on $t$; it's only a function of your position in space.
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Motivation of Keeping Supersymmetry in String Compactification In Candelas, Horowitz, Strominger, and Witten's famous paper [1] about string compactification, they ask that supersymmetry should not be broken in the resulting 4d theory. Then combined with other reasonable requirements, they concluded that the 6d compactified space has to be a Calabi-Yau manifold. By "supersymmetry should not be broken", we mean it is unbroken at the Planck (or compactification) scale.
I can understand that keeping supersymmetry can solve the hierarchy problem. But I want more arguments/motivation. Could anyone help me? The more the better.
[1] Candelas, P., et al. “Vacuum configurations for superstrings.” Nuc. Phys. B258, 46–74
(1985).
| The motivation is that supersymmetry is actually a consistency condition for string theory. Non-supersymmetric string theories exist but all of them have problems such as tachyons or anomalies.
Other interesting point is that we don't know any single example of a stable non-supersymmetric vacua constructed within string theory. This does not imply in any way that superstring theory should be unable to describe non-supersymmetric physics, it's just that maybe string theory is predicting that the only universes that will last forever are the supersymmetric ones; the rest of them belong to the swampland or are meta-satble, as it seems to be the case of our universe. You can hear this point of view in this talk of Cumrun Vafa (min3 7:42).
Reference:
Why string theory implies supersymmetry.
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Indistinguishability of Quantum States and its Consequences In the book Quantum Computation and Quantum Information, there is a discussion about how if states are not orthonormal then there is no quantum measurement capable of distinguishing the states.
I am interested in the consequences of this. What does this mean physically? How is this applied when manipulating quantum states?
I am sure there are other questions that stem from this that I am not even thinking of, but that would be interesting to explore. So any help in understanding the consequences of this would be greatly appreciated.
| Let assume two states $|x\rangle$ and $|y\rangle$ and their inner product $\langle x|y\rangle$. If $\langle x|y\rangle = 0$, then both state can be perfecly distinguished and there is no uncertainty which is which. Examples of such states are $|0\rangle$ and $|1\rangle$ or $|+\rangle$ or $|-\rangle$. As $\langle x|y\rangle$ tends to $1$, the states are more and more similar and less distinguishable. In extereme case when $\langle x|y\rangle$ is 1, it holds that $|y\rangle = \mathrm{e}^{i\theta}|x\rangle$. Such states differ in global phase only and they are absolutely indistinguishable. This all means that not only two orthogonal states are distinguishable but others as well. However, as $\langle x|y\rangle$ goes to one, the states are more and more similar.
Note that you can employ so-called swap test to calculate value $|\langle x|y\rangle|^2$ and decide how two state are similar to each other.
| {
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When I walk down the stairs where does my potential energy go? When I leave my room I walk down three flights of stairs releasing about 7kJ of potential energy. Where does it go? Is it all getting dispersed into heat and sound? Is that heat being generated at the point of impact between my feet and the ground, or is it within my muscles?
Related question, how much energy do I consume by walking? Obviously there's the work I'm doing against air resistance, but I feel like that doesn't account for all the energy I use when walking.
| Into the motion of your body.
When you go down the stairs, your muscles take the vertical motion down the stairs and add a horizontal component to it. As you travel down the stairs, you will be accelerated vertically by gravity, as your potential energy is converted into kinetic energy. The faster you go vertically, the faster you go horizontally, so unless you deliberately slow yourself down, you're going a lot faster horizontally at the bottom of the stairs than you were at the top of them. As a result, in addition to the conversions into heat and sound energy covered by other answers, a portion of the potential energy you possessed at the top of the stairs has been converted into kinetic energy.
| {
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Mechanical energy in a body moving upwards Why is it that mechanical energy is always conserved, I mean when an object is thrown in air, why does the kinetic energy convert to potential energy and not any other form of energy?
| Mechanical energy is not always conserved. It is only conserved if the system is not subject to a net external force.
In your example the object and the earth are the system.
If you throw an object in air mechanical energy of the system will not be conserved because the air exerts an external force due to air resistance dissipating part of the kinetic energy of the system as heat due to air friction. This results in a gain of gravitational potential energy less than that if there was no air.
Hope this helps.
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Graph Interpretation of Gravitational Waves In the image is the data recorded by the LIGO's 2 observatories in USA. What is its interpretation? I mean what does the zig zag lines represent? Similarly, what does the blond red and blue lines (that seem like exponentially increasing up) represent? Could you please clarify? (Note: I am a highschool student so please make sure a high schooler like me too understands it.)
| This is showing what is called a black hole merger. It is two black holes that are orbiting each other, and after time, they get closer to each other and then finally combine.
These graphs show the frequency spectrum versus time of the gravitational wave signal (each graph is from each detector of the LIGO). As time passes, the frequency of the signal increases, corresponding to what is described as a "chirping" noise.
This black hole merger happened because orbiting black holes rotated around each other at increasing frequency and energy (corresponding to the oscillatory pattern on the graphs), coming closer together after each revolution, where finally at the end they merge forming one black hole.
As all of this is happening, the binary black hole system releases a massive burst of gravitational waves in its final instants before merging. Also as this happens, the binary system radiates gravitational waves, which are waves in spacetime, and this is what the LIGO is detecting.
| {
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Why do I feel centrifugal force If I move with constant speed on a turn? Title sums it up pretty much, I'm studying for my physics exam right now and I just can't wrap my head around this
| Because as the car turns around a corner it is forcing you away from your straight inertial path so your path has to curve with the car. As the seat, seat belts, or whatever part of the car you are touching curves away from your straight inertial path it applies pressure on you forcing you away from the straight line you would continue on if no forces were applied.
| {
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Scalar field displacement from the minimum of the potential gives rise to particles/dark matter, why? In This paper (Kobayashi et al -- Lyman-alpha Constraints on Ultralight Scalar Dark Matter: Implications for the Early and Late Universe) it says, at the beginning of Section 3.1:
A light scalar field stays frozen at its initial field value in the early Universe. Hence, any initial displacement from the potential minimum gives rise to a scalar dark matter density in the later universe.
I don't understand this statement. Can someone explain its meaning? Why would such a configuration give rise to matter later in the universe? Is it due to the fact that later in the universe the scalar field would oscillate and oscillations can be seen as particles?
Sorry if the question is not clear, I studied physics quite a long time ago and study these things in my free time so there are many gaps in my understanding of fundamental physics and Cosmology. Feel free to be as technical as you wish but please remember I'm not expert or anything
| The answer seems to be much less deep than what OP expects. The field $\phi$ is to be thought of as a purely classical field, its magnitude literally representing the density of matter. So if its value is large, it means there is a lot of matter, and if low, very little.
As the field is locked to its initial value, if this is non-zero, it will represent a non-zero density at all times. Hence the claim: if there is a displacement it leads to matter density at later times.
Note: $\phi$ is not quite the matter density. The density is actually $\rho\propto\phi^2$ but this is irrelevant for this discussion. If $\phi$ is non-zero so is $\rho$, and viceversa.
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Why is difference of points not a valid definition for a vector in curved space? In page-49 of MTW (1973 edtn), the following picture is shown:
After seeing this picture, the question which arose in my head is why exactly can we not define a vector as difference of points in curved space?
| Because in curved spacetime you can't take the difference of a vector at point P and point P' unlike in Minkowski spacetime.
Imagine abovementioned two points separated by $dx^{\mu}$ and let's call the vector at point P $a_{\mu}(P)$ and the other $a_{\mu}(P')$. You can infinitesimally transport the vector at P by $dx^{\mu}$. Thus $A_{\mu} = a_{\mu} + \delta a_{\mu}$ which is a vector at point P'. Now you can take the difference, hence it becomes
$$a_{\mu}(P') - A_{\mu}(P') = (a_{\mu, \nu} - \Gamma^{\lambda}_{\mu \nu} a_{\lambda})dx^{\nu}$$
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Why is the work done by internal conservative forces equal to negative of the change in potential energy of the system? In my Physics class I studied that work done by an internal conservative force is equal to the negative of the change in potential energy of the system. That is,
$$dW_{internal conservative force} = dU_{system}.$$
But I have a question. Why is the work done by only 'internal' conservative force accounted for? There may also be external conservative forces which behave no differently from internal conservative forces. I mean they also have the capability of storing energy in the system in the form of potential energy, right? Then why don't we account for external conservative forces. Shouldn't the equation actually be :
$$dW_{internal conservative force} + dW_{external conservative force} = dU_{system}.$$
Why work done by only 'internal' conservative force is equal to the negative of the change in potential energy of the system? Shouldn't it be : Work done by all the conservative forces including internal conservative forces and external conservative forces is equal to the negative of the change in potential energy of the system. Why do we consider only internal conservative forces and not external conservative forces? Can someone please explain?
| In short, because:
Conservation of energy theorem =>
$T + V = Constant$
So: $dT + dV = 0$
So: $dT = - dV$
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Unexpected different results from Newton's second law Please don't ban me. I read through Homework-like questions and I know they should ask about a specific physics concept and show some effort to work through the problem. I hope the question is ok.
I recently came across a mechanic basic problem which I wanted to solve following two different approaches. The odd thing is that the approaches bring different results, so probably there might be some wrong reasoning which I cannot explain. Can anyone explain why results are different?
Here's the figure of the system:
The question is about getting the friction static coefficient between the plane and the bar.
We know the bar weight $P=980N$, and the angle $\theta=60°$, the system is in equilibrium, and the static friction is at its maximum value.
Solution 1
Definition of static friction:
$f_{max} =\mu _{s} N\ \rightarrow \ \mu _{s} =\frac{f}{N}$
Newton's second law, forces perpendicular to the plane
$y:\ N=P\cdot \cos 60°$
Newton's second law, torques relative to the right edge of the bar:
$f\cdot L\sin 60°=P\frac{L}{2} \ \rightarrow \ f=\frac{P}{2\sin 60°}$
Static friction coeff:
$\boxed{\mu _{s} =\frac{1}{2\sin 60° \cos 60°}=1.2}$
Solution 2
Definition of static friction:
$f_{max} =\mu _{s} N\ \rightarrow \ \mu _{s} =\frac{f}{N}$
Newton's second law, forces perpendicular to the plane
$y:\ N=P\cdot \cos 60°$
Newton's second law, torques relative to the left edge of the bar:
$T\cdot L\sin 60°=P\cdot \frac{L}{2}\rightarrow T=\frac{P}{2\sin 60°}$
Newton's second law, forces parallel to the plane:
$x:\ f+T=P\sin 60°\rightarrow f=P\sin 60°-T=P\sin60°-\frac{P}{2\sin 60°}$
Static friction coeff:
$\boxed{\mu _{s} =\frac{\sin 60° -\frac{1}{2\sin 60°}}{\cos 60°}=0.6}$
The solutions for the static friction coeff. are different, in particular solution two is $\frac{1}{2}$ solution 1!!
| The first solution is wrong. When you calculate the torques relative of the right edge of the bar you forgot about the torque of force $N$. The correct equation should be
$$ f\cdot L \sin\theta + N \cdot L\cos\theta = P \frac{L}{2} $$
and then you'd get the result that agrees with the second solution.
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If $E=mc^2$, then why do different substances have different calorific values? Today during a classroom discussion, I realised that if we consider the equation $E=mc^2$, then we are establishing a relation between energy and mass but we often observe that different substances produce different amount of energy when they are burned. For example: Burning a kilogram of wood will not produce same amount of energy as 1 kilogram of petrol.
| Different $E$. Burning is a chemical process that releases some amount of chemical energy, which is vastly less that the total rest energy contained in the initial substance. The $E$ in $E = mc^2$ is how much energy you'd get if you converted all of the mass into pure radiation and left nothing left, which is a totally different process from burning.
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Does light travel in a straight line from a third observer's view? We are always taught that light travels in straight path in a homogeneous medium, we even do experiments in primary school with cardboards with holes, "proofing" that light does not bend.
But on a wider perspective, let's say sending a pulse of light from our Sun to another star across the galaxy (and assuming it is not blocked by anything in the way nor goes near any massive source of gravity), and you observe far away 'above' the galaxy, would that pulse of light travel in a curved path (sort of like how a spacecraft would travel)? Or would it ignore the whirl of the galaxy and travel completely straight when observed in that perspective?
Edit: Let me rephrase my question. If light is massless, it would not carry the inertia from the light source (e.g. my flash light). Then it means if I shine a laser beam into the sky, I should be able to see a tiny curve in the laser's path (maybe not with my naked eye, but some special camera), because the laser pointer in my hand is moving with the Earth, right?
| According to the GTR, there is no such thing as a classic straight line in a universe with Gravity - they are replaced by the concept of a geodesic.
The weaker the gravity, the closer the geometry approximates Euclidean geometry with straight lines. Since gravity "curves" space-time, and light travels through space-time, light cannot ignore gravity
| {
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Could metal rods conduct geothermal energy to the surface? Currently geothermal heat pumps circulate a working fluid through a loop running in either a deep well or a long series of more shallow trenches. Boring a well is expensive and digging trenches chews up a large chunk of land.
Could an array of metal rods (rebar connected to a sacrificial anode?) driven into the ground effectively conduct heat to a heat-exchanger at the surface? Would this cost less than the other methods?
| Since the characteristic time of diffusion is L²/D, where L is the characteristic length and D is the diffusivity (here, the thermal diffusivity), the characteristic speed is D/L, or 10-7 m/s for a 1 km deep probe and very thermally conductive copper. We can pump liquid far faster than this.
Looking at the conductive power delivery, we have kAΔT/L for the probe, where k is the thermal conductivity, A is the cross-sectional area and ΔT is the temperature difference. Compare this to ρCVΔT/t for advective transfer, where ρ is the material density, C is its specific heat capacity, V is its volume, and t is the turnaround time. Dividing by AΔT, we’re comparing the magnitudes of k/L and ρCv, where v is the liquid pumping speed. We find again that v need only exceed 10-7 m/s for water to best a strongly conductive rod.
The conclusion is that when L is large (or even longer than millimeters), pumping liquid is generally a far more effective way to transfer thermal energy than to rely on conduction.
(I’ll leave it to you to incorporate the relevant cost comparison, which takes this topic into engineering.)
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How can the Wigner function of squeezed states be non-negative? It is always said that when the Wigner function of quantum states takes a negative value, then it is a clear signature of non-classicality of this particular state. It is also well-known that the Wigner function of squeezed states is completely non-negative, yet, we still call squeezed states quantum. How does one explain this supposed contradiction here? is the Wigner function just not a "strong enough" measure of non-classicality? if so, what is? Also, what defines non-classicality for that matter (specifically for squeezed states)?
Thanks!
| There is no contradiction because positivity of the Wigner is not enough to guarantee classicality: squeezed states are precisely examples of this. What is known for pure states (Hudson’s theorem) is that only Gaussian states have non-negative WFs (at least in the $xp$ plane), but the theorem is silent on the classicality of the Gaussian states.
Most non-classical states have singular $P$-functions, although it is recognized that the criteria is not so simple and there are recent results in
Damanet, F., Kübler, J., Martin, J. and Braun, D., 2018. Nonclassical states of light with a smooth P function. Physical Review A, 97(2), p.023832
that suggest a way of engineering non-classical states so that their P-function behaves more smoothly.
I do not know of a necessary and sufficient condition for a state to be classical, although negativity in the Wigner function guarantees it is NOT.
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Double Slit Interference pattern - horizontal or vertical? What determines whether in a double slit experiment the interference pattern will be horizontal or vertical for example? Is it mostly seen horizontal because the slits are small enough horizontally for interference pattern to be seen? Also, I sometimes see a "rectangular" interference pattern when using a lens to focus the light. How does that happen? Shouldn't we expect a circular pattern?
| TL;DR
*
*If the slits are sufficiently long the diffraction pattern will be spaced in the same direction as the slits.
*If the length of the slits is of the order of the wavelength then the diffraction pattern will appear "rectangular".
*One should only expect a circular interference pattern if the aperture has a circular symmetry for example a circular hole.
Explanation
The interference pattern observed from the double-slit experiment produced by particles is the same as that produced by light due to wave-particle duality. However, the scales of the interference patterns will differ as the scale depends on the wavelength.
The interference pattern can be calculated using the Fresnel-Kirchhoff diffraction integral. There are two commonly used limits of this integral, the Fresnel limit (screen or point source close to aperture) and the Fraunhofer limit (screen and source far from aperture).
For this question, we will consider the Fraunhofer limit as the results are easier to understand intuitively and also easier to calculate as in the Fraunhofer limit the amplitude of the interference pattern on a parallel screen is proportional to the Fourier transform of the aperture function (a function that is $1$ where light (or particles) can pass through and $0$ where they cannot). The intensity is then just the square magnitude of the amplitude.
Thus, for a single slit (a rectangle) then the aperture function $h\left(x,y\right)=\Pi\left(x/w\right)\Pi\left(x/h\right)$ the intensity of the interference pattern is:
$$I\propto\operatorname{sinc}^2\left(\frac{1}{2}wk\sin\left(\theta\right)\right)\operatorname{sinc}^2\left(\frac{1}{2}hk\sin\left(\phi\right)\right)$$
where $\Pi\left(x\right)$ are Rectangular functions ("top hat functions"), $w$ and $h$ are the width and height of the slit, while $k\equiv\frac{2\pi}{\lambda}$ where $\lambda$ is the wavelength and finally $\theta$ and $\phi$ are the angles in the $x$ and $y$ directions to the normal. We can approximate $\sin\left(\theta\right)=\frac{x}{L}$ and $\sin\left(\phi\right)=\frac{y}{L}$ when close to the centre of the interference pattern.
For a double-slit we get an extra factor of $\cos^2\left(\frac{1}{2}Dk\sin\left(\theta\right)\right)$ where $D$ is the slit separation so:
$$I\propto\operatorname{sinc}^2\left(\frac{1}{2}wk\sin\left(\theta\right)\right)\operatorname{sinc}^2\left(\frac{1}{2}hk\sin\left(\phi\right)\right)\cos^2\left(\frac{1}{2}Dk\sin\left(\theta\right)\right)$$
I recommend plotting these equations (replacing $\sin\left(\theta\right)=\frac{x}{L}$ and $\sin\left(\phi\right)=\frac{y}{L}$) using plotting software (for example Desmos) and tweaking the parameters to see how this affects the interference pattern. You should find as the slits get longer there appears to be no variation vertically, as the slits get narrower then in the double-slit experiment the intensity drops off less slowly.
Finally, for a circular hole, we will get Airy disks (the Wikipedia page linked has some nice images of diffraction patterns).
| {
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Why is acoustic intensity inversely proportional to density of the medium? The definition of sound (acoustic) intensity is given by
$$ I = {p^2 \over {\rho c}} \;\;\;\; \text{or} \;\;\;\; I = {p^2 \over {2\rho c}}$$
I've seen both definitions in different textbooks and am not sure which equation is more accurate. But in either case, the relation of intensity and density of air is:
$$ I \propto {1 \over \rho} $$
This contradicts my intuition.
I would hear no sound in a vacuum because there is no molecules in the medium to scatter the vibration from the source. If I start adding some molecules in the medium (low density), the source will vibrate these molecules in the medium, acoustic energy can be transferred and sound may be heard if there are just enough molecules hitting by eardrum (and within the audible frequency range). If I add a whole lot more molecules in the medium (high density), many more molecules will vibrate and hit my eardrum, and I would therefore hear a louder sound (or detect a higher acoustic intensity).
So why the equation suggests the opposite relation? Could it be that the instantaneous pressure is also a function of density, or $ p(t, \rho) $, in a way that pressure is proportional to density of higher order to offset the drop of $ 1 \over \rho $ have on intensity?
| A candid way to visualise that $density / amplitude$ relationship is the following image:
*
*Imagine the medium as a 1 or 2 or 3 dimensional lattice of marbles linked to their nearest neighbors by springs (a N-dimensional array of harmonic oscillators, if you wish).
*The density (of the simplistic medium) is related to the mass of the marbles, consequently, for a given (fixed) amount of energy transferred to the medium, the amplitude (of the oscillations) will diminish for different (increasing) values of the mass, to keep kinetic energy constant.
| {
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Do $2s$ and $2p$ orbitals have same energy? While solving the Schrodinger equation for the H atom, we get $E_n$ depending exclusively on $n$ (actually on $\frac{1}{n^2}$). Then I thought 2s and 2p orbitals must have the same energy.
But while reading Molecular orbital theory in Atkin's Physical chemistry book, I found written that "$2s$ and $2p_z$ orbitals have distinctly different energies".
Can anyone please explain this?
| In a multielectron system by solving the Schrodinger equation you will see that the energy of the orbital doesnt only depend on n ,it depends on l as well.
| {
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What are the symmetries of circular billiards that makes it integrable? I have often heard that integrability in is equivalent to extensively many conserved quantities $A_i$, i.e. the Poisson bracket $\{H,A_i\}=0$ or in quantum mechanics $[H,A_i]=0$.
*
*What are the conserved quantities $A_i$ for classical circular billiard, which is integrable?
*What are the conserved quantities in the quantum circular billiard?
My guess is that since momentum $P$ and angular momentum $J_3$ commute with $H$, we could just say that the $A_i$ are the set of all functions $f(P,J_3)$, but I'm not sure.
| 2D circular billiard is Liouville integrable:
*
*The Hamiltonian $H$ itself is always an integral of motion for an autonomous system.
*The boundary force is a central force, so the angular position variable $\theta$ is cyclic, and hence the angular momentum $J_3$ is an integral of motion.
Linear 2-momentum $\vec{p}$ is not an integral of motion at the boundary; its square $\vec{p}^2$ (or any function thereof) is an integral of motion, but not independent of the Hamiltonian $H$.
(Interestingly, elliptic billiard is also Liouville integrable.)
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Entanglement in 2D Harmonic Oscillator system Let's assume a 2 dimensional harmonic oscillator system with the Hamiltonian $\hat{H} = \frac{1}{2} p_x^2 + \frac{1}{2} p_y^2 + \frac{1}{2} \omega_x^2 x^2 + \frac{1}{2} \omega_y^2 y^2$ with the solution of the ground state being simply the product of the the ground state of each independent mode.
$\psi_{gs}(x,y) = \psi_{x}(x) \psi_{y}(y) \propto e^{-w_xx^2} e^{-w_yy^2} $ which is clearly a separable solution.
But when we rotate our coordinate system by $\pi/4$ we, get the new coordinates to be
$ x' = \frac{1}{\sqrt{2}}(x + y)$ and $y' = \frac{1}{\sqrt{2}}(x - y) $
and expressing the solution in those coordinates will lead to the solution
$\psi_{gs}(x',y') \neq \psi_{x'}(x') \psi_{y'}(y') $ and hence a signature of entanglement.
How can rotating a physical system (or just someone choosing a different coordinate system) lead to it being entangled?
| Entanglement means the system not separable in any basis (or coordinate system). It’s not always easy to show there is no transformation (such as the inverse of the one you propose) that would bring a system to an explicitly separable form.
The development of practical entanglement witnesses to show entanglement irrespective of the coordinate system or basis is still an active area of research.
| {
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Experimental result of the atomic nucleus volume by scattering alpha particles from the atomic nucleus. Investigation of the electron volume by what? Rutherford's alpha particle experiments marked the beginning of the determination of the volume of the atomic nucleus.
How were the experiments conducted that led to the statement of the point-like electron?
|
How were the experiments conducted that led to the statement of the point-like electron?
The particle physics experiments are mainly scattering experiments, and the theory that fits the existing data and predicts future data is the quantum field theory, QFT, of the standard model, SM.
The axiomatic table in the SM has all the particles point like in the QFT, i.e. for each individual vertex in the Feynman diagrams the particles are considered as points in the integration, no extent or complexity. The fit of theory to data is very good, this means that it is not necessary to change the axioms.
Now the complete calculation is different than the first order diagram, each addition of higher order diagrams brings corrections and if one looks at the "shape" of the electron it will not be a point but a locus in probability space
But the experimental evidence for assuming the vertex particles pointlike comes from the successful fit of the SM to data, up to now. Anyway, experiments are running in order to find disagreements with the SM, who knows what the future will bring?
| {
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Uniqueness of the definition of Noether current On page 28 of Pierre Ramond Field theory - A modern primer the following is written:
"we remark that a conserved current does not have a unique definition since we can always add to it the four-divergence of an antisymmetric tensor [...] Also since $j$ [the Noether current] is conserved only after use of the equations of motion we have the freedom to add to it any quantity which vanishes by virtue of the equations of motion".
I do not understand what he means by saying, any quantity which vanishes by virtue of the equations of motion.
| In Noether's first theorem, the continuity equation$^1$
$$ d_{\mu} J^{\mu}~\approx~0 \tag{*}$$
is an on-shell equation, i.e. it holds if the EOMs [= Euler-Lagrange (EL) equations] are satisfied. It does not necessarily hold off-shell.
Hence we can modify the Noether current $J^{\mu}$ with
*
*terms that vanish on-shell, and/or
*terms of the form $d_{\nu}A^{\nu\mu}$, where $A^{\nu\mu}=-A^{\mu\nu}$ is an antisymmetric tensor,
without spoiling the continuity eq. (*).
--
$^1$ The $\approx$ symbol means equality modulo EOMs.
| {
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Does real life have "update lag" for mirrors? This may sound like a ridiculous question, but it struck me as something that might be the case.
Suppose that you have a gigantic mirror mounted at a huge stadium. In front, there's a bunch of people facing the mirror, with a long distance between them and the mirror.
Behind them, there is a man making moves for them to follow by looking at him through the mirror.
Will they see his movements exactly when he makes them, just as if they had been simply facing him, or will there be some amount of "optical lag"?
| If there were such a lag, you could use a moving mirror to measure the speed of light.
This was first done in the 1920s by Michelson. (See e.g. this link which hosts papers from 1924 and 1927.) Michelson built an octagonal solid with mirrored sides which spun at several hundred times per second, and used this spinning mirror to measure that light took about 230 microseconds to make the 44-mile round trip between Mt. Wilson and another mountain, whose location is not uncertain but whose name is different in different reports.
Human persistence of vision is measured in milliseconds, not microseconds, and the unaided human eye can’t resolve images in mirrors from such distances. So this isn’t an effect that you could observe naked-eye in a mirror that fit in a stadium, for any reasonable definition of “stadium.” The lag is real, but it takes experimental finesse to observe it.
An astronomical version of “mirror lag” is a supernova light echo.
| {
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Smallest proposition given a state $\psi$ Today, in my lecture the following was mentioned:
Given a state $| \psi \rangle$ in a Hilbert Space $H$, the smallest proposition which is true given this state is given by the projection operator $| \psi \rangle \langle \psi |$, which is the smallest projection operator that projects onto the one-dimensional subspace $| \psi \rangle$.
There are few things I don't understand here. What is meant by smallest projection operator? Also why is the outer product $| \psi \rangle \langle \psi |$ the smallest?
| This terminology is not familiar to me, but I can think of only one possibility :
For Hermitian operators, there is a notion of positivity (and therefore an order). In this case we would say that a hermitian projector $P$ is smaller than another $Q$ if :
$$\forall |\varphi\rangle, \langle \varphi|P|\varphi\rangle\leq \langle \varphi|Q|\varphi\rangle$$
We see that this is equivalent to the range of $P$ being included in that of $Q$. As any closed subspace containing $|\psi\rangle$ will contain the range of $P$, we see that $P$ is the smallest orthogonal projector with $P|\psi\rangle = |\psi\rangle$.
| {
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What does it mean for a nucleus to be more stable? When we say that a nucleus is more stable than another nucleus, are we saying that the nucleus is at a lower energy level than the other nucleus? In chemistry, a more stable compound is one that is at a lower energy level, is this the same when we talk about a nucleus?
If yes, wouldn't this make all nuclei with a larger mass more unstable since they will be at a higher rest mass energy.
But clearly, this is not the case as the binding energy per nucleon graph increases initially before decreasing, suggesting that a heavier nuclei can be more stable than a lighter one.
| Your reasoning based on the ‘rest mass energy’ is not really relevant to the analysis. On earth, a nucleus at higher altitude would hold more gravitational potential energy. But I don’t think you would intuitively imagine that this would make it less stable. My point is, you could define many different kind of energies/potentials in which you put your nucleus, that doesn’t make it relevant to its stability.
Instead, just recall a very basic idea of the physics of what you are studying to guide your choice of a relevant quantity . The protons and neutrons are held together by the nuclear force. So what you want to minimize is the potential due to that force. Gravitational fields would be irrelevant to describe how strongly the nuclear force holds the nucleus together, right? Similarly, the mass energy doesn’t really help to describe it.
So the energy that needs to be smallest is the one due to all interactions between nucleons. If this energy of interaction is smaller when the nucleons are packed together vs when they are apart, then they wanna stick together. The more negative that difference is, the more stable the nucleus (i.e. the more energy needed to take it apart)
| {
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Why is centrifugal force called fictitious? When an object undergoes rotation, from the object's reference frame, which is a non-inertial reference frame, the object feels there is a radially outward force, a centrifugal force, acting on it. However, from an inertial reference frame, this force doesn't exist at all. That's why it is called a fictitious force.
My argument is, who are we to say what is fictitious or not. The object at the non-inertial frame really feels the centrifugal force! So, it is a real force for the object.
Suppose, there are two inertial reference frames $S$ & $S'$ and $S'$ is moving with a velocity v that is a significant fraction of the speed of light. From $S$ it would seem that time is going slower for $S'$. Surprisingly, it would seem from $S'$ that time is going slower for $S$ as well. Now, who is right? Answer: Both of them are right.
So, is it really right to call centrifugal force fictitious just because it doesn't exist in an inertial reference frame?
| I disagree that you feel centrifugal force. A person in a centrifuge actually feels their reaction to the centripetal force. If you sit in a car that is subject to harsh acceleration, you 'feel' as if you are being pushed back in your seat. There is no force pushing you back- it is simply the result of your inertia.
| {
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Can we cool Earth by shooting powerful lasers into space? In a sense, the climate change discussion revolves around the unwanted warming of the earth's atmosphere as a whole.
It seems a bit too obvious to be true, but could we cool the atmosphere by simply shooting that unwanted energy somewhere else?
Energy might be collected from remote expanses where it would otherwise be somewhat pointless to harvest it due to lack of habitability and resulting anticipated losses due to transmission (ocean surface, ???)
If so, what would be a good place to shoot it?
| Cheap and effective method: We should paint roofs and streets white.
Besides, a small part of the Sahara would currently be enough to supply humans with electricity from solar cells. On the rest of the Sahara there could be solar cells to power your lasers. But this is only possible if the power loss of cells and lasers that escapes into the environment is less than the part of the sunlight that does not leave the earth as reflected radiation nad heats the earth.
| {
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What's the difference between inelastic X-rays scattering and Raman scattering? In solids, inelastic scattering of X-rays can produce or absorb a phonon, which is equivalent to saying that solid ends up in excited vibrational level ( or if it was in a vibrational level to begin with, it ends up in ground state). But isn't that exactly what happens in Raman scattering (stokes and antistokes). As Raman spectroscopy uses light from X-ray to Infra-red in all regions, what is the difference between X-ray scattering and Raman and IR spectroscopy? And what is X-ray raman spectroscopy?
| Raman Scattering is indeed an inelastic scattering process. Raman scattering is synonymous to inelastic scattering of photons.
Raman scattering is usually done in IR region, the reason being vibrational spectra of most molecules are in this region. That doesn't mean Raman spectroscopy can't be done with high frequency waves though, in fact enhanced X-ray Raman spectroscopy (called Resonant Raman Spectroscopy) is also pretty common nowadays, especially in crystalline solids which have excitation spectra in that range.
| {
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