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How would the Aurora light on Earth look like if there wasn't a magnetic field? Here are some pictures of the aurora light.
The beautiful phenomenon of Aurora is a well-known one, seen in the northern (Aurora Borealis) and southern parts (Aurora Australis) of the globe. Here is another short but easy readable article about it.
Solar winds send charged particles towards the Earth. They are part of the Van Allen radiation belt (together with cosmic rays) and thanks to the Earth's magnetic field we are shielded off from them.
Depending on the kind of the in storming particles and their energies in the solar wind different colors of lights can be seen in the northern and southern parts of our planet. The Earth's magnetic field lets the particles spiral to the north or the south.
But suppose the Earth's magnetic field was absent. Probably life wouldn't exist, but what would there be to see in the sky?
Would we see a sky that would be flickering (when a reaction takes place between an incident solar particle and the Earth's atmosphere) in different colors? Would we see the same kind of one color band (caused by the solar flares)? Would it be a combination of the two? Would the atmosphere in the night light up the same everywhere on Earth (though the closer you get to the equator the brighter)? Or what? let's assume the atmosphere won't be blown away because of the solar winds.
| We wouldn't have a magnetosphere if we didn't have a magnetic field. Therefore the solar wind wouldn't be able to interact with it and cause an aurora.
Once the solar wind strips away our atmosphere the sky would probably look a lot like the moon's sky (black with stars), or mars. https://en.wikipedia.org/wiki/Extraterrestrial_skies
| {
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Evaluating Potential Energy Integral in Quantum Chemical Calculations My question is what are the steps for taking an integral of the following form?
$$\int e^{-\alpha|\mathbf r- \mathbf R_a|^2} {1\over|\mathbf r- \mathbf R_b|} e^{-\beta|\mathbf r- \mathbf R_b|^2} dV$$
This integral is commonly seen when attempting to do Quantum Chemistry calculations with a Gaussian type basis set. I have tried to use wolfram alpha to solve this problem but it fails to give a solution.
| Since the integral is over all space, it makes sense to shift the origin of your coordinate system to $\mathbf{R}_b$. Then your integral becomes $$I = \int e^{-\alpha|\mathbf{r}'-\Delta\mathbf{R}|^2}\frac{1}{|\mathbf{r}'|}e^{-\beta|\mathbf{r}'|^2}dV',$$ where $\Delta\mathbf{R}\equiv\mathbf{R}_a-\mathbf{R}_b$.
The next trick to use would be to choose to align the $z'$ axis along the direction of $\Delta\mathbf{R}$. Then you'll have that
$$I = 2\pi\int_0^\infty r'^2dr'\int_0^\pi\sin\theta' d\theta' \, e^{-\alpha(r'^2-2r'\Delta R\cos\theta'+\Delta R^2)}\frac{1}{r'}e^{-\beta r'^2},$$
where I've already performed the integration over $d\phi'$. The above integral should be do-able by hand (or by Wolfram Alpha).
| {
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If atoms were held by gravitational (instead of electrical) forces
"If atoms were held together by gravitational (instead of electrical) forces, a single hydrogen atom would be much larger than the known universe."
- from Grifffith's Introduction to Electrodynamics
I don't understand this. If the atoms were held together by gravitational forces, then the atoms will have to be extremely close together, but why extremely big?
| At first sight, you would indeed expect that the electron has to be much closer to the nucleus, so the force needed to keep the electron in a circular orbit will be the same as that in the case of the electric force.
But for the centripetal force (directed inwards and provided by the gravitational force or the electric force), proportional to $\frac 1 {r^2}$ (for conservative forces), to be in balance with the centrifugal force, proportional to $\frac 1 r$ (which is the condition for a circular orbit), the distance where this balance will occur will be the bigger the smaller the force which binds the electron and the nucleus.
So essentially it's because of this difference in the $\frac 1 {r^2}$-dependence of the force which delivers the centripetal force and the $\frac 1 r$-dependence of the centrifugal force which will increase the radius of a circular orbit if the centripetal force gets smaller (as is the case for the gravitational force).
| {
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Fourier series analysis of string vibration In case of a plucked string,the amplitudes of successive frequencies fall by 1/n^2.
In case of a string which is struck so that say at x=a only the string has a velocity,say v,initially,then the amplitudes of successive frequencies fall by 1/n which implies that it is more enriched with harmonics than plucked one.
Again,for a string struck at midpoint such a way that the initial velocity at each point varies linearly with its distance from nearest end from zero to v,then the amplitudes fall by 1/n^3.
My question is that,they all can be showed by fourier analysis applying proper boundary conditions.
But what is the reason behind it physically?why a struck string has more enriched vibration with different frequencies,than a plucked one,and why the amplitudes fall at such different rates for different cases?
| Sine waves are absolutely continuous, so they are good at approximating functions that are also absolutely continuous. By this I mean that a Fourier series for an absolutely continuous function will generally converge fast.
If the function is continuous but has discontinuities in the gradient, like a triangle wave, the convergence will be slower because it's hard to get the discontinuity in the first derivative using sine waves. If the function is discontinuous, like a square wave, then the convergence will be even slower because it's even harder to get discontinuities in the function using sine waves.
So for example the Fourier coefficients of a triangle wave (discontinuities in $f'$) fall as $1/n^2$ while the coefficients of a square wave (discontinuities in $f$) fall as $1/n$.
The point of all this is that the instant after the string has been displaced (struck, plucked or whatever) it will have some profile and the shape of that profile will determine the convergence of the Fourier series used to describe it. The rate at which the higher harmonics fall away will depend on how jagged the initial shape of the string is.
At this point I have to resort to arm waving because I don't know anything about playing stringed instruments. However plucking the string involves smoothly deforming it sideways then releasing it, while striking the string delivers an impulse at one point on the string. It seems a reasonable supposition that the initial profile of a struck string is more jagged than that of a plucked string so the Fourier coefficients will fall off more slowly.
You indicated you were after a physical rather than a mathematical discussion, which is what I have attempted. If you're interested in the maths then Wikipedia has a nice article on the convergence of Fourier series.
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Why do different letters sound different? If one sings the letter "A" and "M" at the same volume and pitch, the two letters are still differentiable. If both pitch and volume are the same however, shouldn't the sound be the exact same?
| Here's something different to note. Let's say I would tell a band to play the musical note "C3". The bass, the guitar, the piano, the voice, the banjo, all of them sound different and yet we perceive them as the same note that has been played.
Similar, think of a sung "A" and a sung "B" (as in "bee") as an instrument respectively. They have their own unique "sound" to them, and yet they can both be used to create the same "musical note" of a certain given pitch and volume.
What makes a C3 note of a sung "A" different from a C3 note from a sung "B" then?
(Or what makes a C3 of a piano different from a C3 of a guitar?)
Note what "same pitch and volume" actually means. I'll keep it simple.
Pitch: perceived frequency
Volume: air pressure or amplitude
Here are two pictures to illustrate what I mean:
Both of them have the same amplitude, or volume.
Both of them have the same perceived frequency, or pitch.
Thus both of them play the same musical note we perceive.
But looking at the wave form, you could probably tell that they will sound differently, even though we would perceive them as the same note.
This difference is similar to a piano C3 vs a guitar C3.
Essentially: The same perceived frequency and air pressure creates the illusion of the same musical note perceived by a listener. Completely different wave forms (sounds) can be perceived as the same musical note, as long as their wave form "look the same" (the two pictures above illustrated what I mean with that).
So a sung "A" and a sung "B" are actually quite different from each other. But if sang with the same pitch, they will produce the same musical sound (as perceived by humans).
Source of the images used
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Schwarzschild coordinates beyond the event horizon We can write down the metric of the Schwarzschild black hole in Schwarzschild coordinates.
On page 6 of the notes by Leonard Susskind of a course given at the Perimeter Institute titled 'Black Holes and Holography.' we find the following:
However, note that the Schwarzschild coordinates are only formally valid for $r > R_{s}$, and must be analytically continued within the event horizon.
Which aspect of the metric in Schwarzschild coordinates indicates that the coordinates are only valid outside the event horizon?
| I think the coordinates are valid inside the horizon as well, albeit inside the horizon $r$ is timelike and $t$ is spacelike. They are not valid at the horizon since both $r$ and $t$ have zero coefficients in the metric there. That means you need some other coordinate system there to connect the exterior and interior solutions.
| {
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Find the speed of the block in this question
The figure shows a block (assumed to be a point) being pulled by an ideal string across an elevated pulley. At any time $t$, let the horizontal distance of the block be $x$ from the pulley. The length of the string is $l$ and the height of the pulley is $h$ from the ground. The string is pulled with a speed $u$ as shown. Using calculus I found the speed of the block to be $u \sec \theta$ (where $\theta$ is the inclination of the string with the horizontal) which is given as the correct answer.
So my doubt is: why can't we simply resolve $u$ along the horizontal and say the block's speed is $u \cos \theta$? It doesn't seem wrong to me but isn't correct for some reason. Could someone please explain why?
| Cosθ=(length of X/length of L). You need to find (delta length X/delta length L) or dx/dl. One is a relationship between lengths, the other is a relationship between changes in length. The two are just different.
| {
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What am I failing to understand about a light in a material? In a material, a photon's velocity becomes slower, photon's wavelength becomes shorter, but photon's frequency doesn't change.
If there is a material that makes low frequency photons have very short wavelength photons, and if I stick my finger in the material, do I get a radiation burn? What kind of effects can I have from the short wavelength photons in a material except for the refraction?
| The energy of a photon is related to its frequency. In vacuum you can convert freely between frequency and wavelength - but once you have a material with a high refractive index, it's not so simple.
Your question about "radiation burn" is assuming that the energy of a short wavelength photon is always higher. That's not the case. The thing about the photon that doesn't change with medium is the frequency - and that is the thing you need to look at.
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Why are springs shaped the way they are? Why are springs coiled the way they are? Why not some other shape? Is the shape due to its elasticity or something?
| Well, metals can bend easily. If a length of metal is in the shape of a coil, then this bending can occur over the entire spring at once. Perhaps minimal temporary deformation at each point prevents permanent deformation. In any case, it maximizes lengthening ability while minimizing "effort" by the spring. I don't know if there are other shapes that would work better, but a cylindrical coil is certainly a simple and effective solution.
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Simultaneity in Newtonian mechanics How would Newtonian mechanics answer the train and moving light question?
The setup is:
A train is moving in the positive x_axis with speed c/2. A person stands in the middle of the train. There are two light bulbs at both ends of the train. The light goes off at the same time (absolute time in Newtonian physics). The person standing in the middle of the train would perceive both lights independently.
Outside the train there is a stationary observer. Let's assume the train is already to the "right" of the observer (in x_axis) when the lights go off. Would the stationary observer observe the rear light before the front light?
The reason why I am asking this is that the relativity of simultaneity is often attributed ONLY to special relativity. Here, would Newtonian mechanics also predict that the stationary observer observes different simultaneity than the moving observer in the train?
| Under C19th Newtonian theory, light was usually treated using the ballistic emission theory of light, which would say that the speed of light generated by bulbs inside the train would by default be cTRAIN and not cPLATFORM.
So if the central observer sends a trigger signal to cut the lights, that signal travels at cTRAIN, both lights cut out, the final wavetrain signal travels at cTRAIN, and both lights appear to the central train observer to go off at the same moment.
If by this time, the train observer (with the train moving across our page from left to right) is already to the right of the platform observer, then the platform observer will see the rear light go off first for two reasons:
*
* Because it's closer, and
* Because the signals from the approaching rear of the train will (for platform observers) be travelling faster than those from the receding front end of the train.
| {
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Dirac's quantization of electric charge I was following Nakahara's book Geometry, Topology and Physics, specifically subsection 10.5.2. I do understand how he obtains the equality,
$$ \Delta \varphi = \int \mathrm d\varphi = \int \limits^{2\pi}_{0} 2g \, \mathrm d\phi = 4\pi g$$
and that therefore this has to be a multiple of $2\pi$ for $t_{NS}$ to be uniquely defined. But I don't see directly how,
$$\Delta \varphi / 2\pi = 2g \in \mathbb{Z},$$
is equivalent to
$$ \frac{eg}{\hbar} = 2\pi n ~(n\in \mathbb{Z}).$$
Can somebody tell me what I am missing?
| It is just a matter of convention. Nakahara uses the gauge covariant derivative $D_\mu \phi = ( \partial_\mu + \mathcal{A}_\mu ) \phi$ in the chapter. This is equivalent to the common definition $D_\mu \phi = ( \partial_\mu + i e/\hbar A_\mu ) \phi$ if $$\mathcal{A}_\mu = i e/\hbar A_\mu \,.$$ Nakahara's definition of the magnetic charge $g$ is
$$4 \pi i g = \int \mathcal{F}\,.$$
If you define your magnetic charge as
$$\tilde g = \int F\,,$$
you find that $g = \frac{e \tilde g}{4 \pi \hbar}$, and
$$2g = \frac{e \tilde g}{2 \pi \hbar} \in \mathbb{Z} \,.$$
| {
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Flat metric induced from Schwarzschild The task is to find a function $f(r)$ such that the induced metric from the Schwarzschild metric
$$ds^2 = -\left(1-\frac{2m}{r}\right) dt^2 + \frac{1}{1-\frac{2m}{r}} dr^2 + r^2 d\Omega^2 $$
on the level set $\{t=f(r)\}$ is flat. My first attempt was to guess
$$ dt=0 = f'(r)dr$$
but it led me nowhere. My other idea was to introduce advanced and retarded coordinates $v=t-r$ and $u=t+r$ but there also I'm stuck. Maybe someone could give a guidline, a hint or a direction.
| You have $dt=f'dr$ so $ds^2=(\varphi^{-1}-f'^2\varphi)dr^2+r^2d\Omega^2$ with $\varphi:=1-\frac{2m}{r}$. We want the $dr^2$ coefficient to be $1$, so $f'^2=\frac{1-\varphi}{\varphi^2}$. You can take the rest from there.
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Biot-Savart law and magnetic field of a ring I have to calculate the magnetic field along the axis of a ring of radius $R$ on which circulates a current $I$ using the Biot-Savart law. The Biot-Savart law as given in my (really bad) course states
$$\mathbf{B}=\frac{\mu_0}{4\pi}\oint_{C}\frac{\mathbf{I}\times \mathbf{r}}{|\mathbf{r}|^3}\mathrm{d}l $$
where in this case
$$C=\{(\rho, \phi, z): \rho=R, \, z=0,\,\phi\in[0,2\pi[ \}$$
and $\mathbf{I}=I\hat{e}_{\phi}$
we have $\mathbf{r}=\rho\hat{e}_{\rho}+z\hat{e}_z$, thus
$$\hat{e}_\phi\times \mathbf{r}=\hat{e}_\phi\times(\rho\hat{e}_{\rho}+z\hat{e}_z)=-\rho\hat{e}_z+z\hat{e}_\rho$$
and thus
$$\mathbf{B}=\frac{\mu_0}{4\pi}\int_{0}^{2\pi}\left[\frac{-\rho\hat{e}_z+z\hat{e}_\rho}{(\rho^2+z^2)^\frac{3}{2}}\right]_{\rho=R, \,z=0}\mathrm{d}\phi$$
which of course gives something which is constant and wrong. I really don't understand how this formula could give anything which makes sense, since every spatial variable will disappear with the integration. I found other versions of the law around which include notations such as $\vec{dl}\times\vec{r}$ which I just don't understand, I don't know what it means to take the cross product of a differential with something. Do I have a wrong Biot-Savart law? If not, what am I doing wrong?
Thank you.
| I usually approach these problems less mathematical. Griffiths uses the following figure:
The part $dl'$ yields a piece of the magnetic field $dB$. All the horizontal components cancel so we only have to account for the vertical components. The integral becomes:
$$ B(z) = \frac{\mu_0}{4\pi} I \int \frac{d\textbf{l'}\times \textbf{r}}{r^2} = \frac{\mu_0}{4\pi} I \int \frac{dl'}{r^2} \cos\theta $$
The cross product between dl and I is more clearly visible in the figure. You take the cross product of a small part of I with the vector r. The vector r is a unit vector so it's length is 1. The angle between the 2 vectors is 90 degrees so you only have dl left. The cosine picks out the vertical component. Solving this integral is trivial.
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Why is light bent but not accelerated? Light is bent near a mass (for example when passing close to the sun as demonstrated in the famous sun eclipse of 1919). I interpret this as an effect of gravity on the light.
However, it seems (to me, at least) that light is not accelerated when it travels directly toward the (bary-)center of the sun. The same gravitational force applies yet the speed of light remains constant (viz. $c$).
What am I missing?
| One thing that the previous answers are missing -- the light is accelerated; it just is accelerated according to the rules of special relativity, which says that it cannot pick up speed when already travelling at the speed of light.
Instead, it gains kinetic energy the way a photon gains kinetic energy -- by being blueshifted to a higher frequency, which does translate to more energy, according to the Planck relation $E = h\nu$.
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Charging current of capacitor In one of my books there is a figure
where G is a neon lamp. Basically the capacitor gets charged once the switch is closed up to a certain spark-current $U_Z$ where the neon lamp gets switched on so the capacitor can discharge to a certain charge-current $U_L$. Further it says that from the charging current
$U(t)=U_0(1-\exp(-t/RC))$
of the capacitor it follows that the periodicity is
$\displaystyle T=RC\cdot\log\frac{U_0-U_L}{U_0-U_Z}$.
How exactly does this equation follow?
I am not familiar with the proper english terms in electrical engineering so I might have mixed up voltage, current, etc. I hope, it's still clear what I mean.
| The time to get from zero to $U_L$ is obtained by solving (putting $RC=\tau$)
$$U_L=U_0\left(1-e^{-t/\tau}\right)\\
U_0-U_L = U_0 e^{-t/\tau}\\
\log(U_0-U_L) = \log(U_0)-t/\tau\\
t = \tau \log\frac{U_0}{U_0-U_L}$$
The time to get from $U_0$ to $U_H$ is similarly obtained. When you take the difference between these numbers and rearrange, you get the expression from your book.
Do you think you can do the rest of the derivation yourself now?
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Equation of motion for a falling slender bar
I have a few question about equations of the uniform slender bar motion.
The bar is released from rest on a horizontal surface as shown in the figure and falls to the left. The surface is rough so friction forces are applied at bottom end of the bar. The coefficient of friction is μ and the mass of the bar is 'm'
First, I want to know the angle β when the end of the bottom end of the bar begins to slip. I know that if the x-axis direction force gets bigger than the static friction force it will start to slip. But I can't figure out the equation about the x-axis direction force as function of θ.
Second, I want to know the angle α when the bottom end of the bar starts to lift off the ground. This situation will happen when the normal force is zero. But I can't figure out the equation of the normal force as a function of θ.
I want to know the equation to calculate the alpha and beta.
The problems are not related, it's a different problems.
Any hints or opinion would be appreciated.
| Take this diagram as a hint:
Take torques by considering the center of mass as the pivot. The torque due to friction must equal the torque due to the normal reaction for the rod to not slip.
I think you'll arrive then at the answer for the coefficient of friction.
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Canonical Momenta and gauge invariance Suppose there is a charged particle moving in an Electromagnetic field for which the vector potential is A.
Now we change A by gradient of some scalar(which is time independent) Hence, the scalar potential remains unchanged.
Now, I want to investigate the effects of such a gauge transformation in Quantum Mechanics.
*
*The expectation value of $\hat{x}$ (position operator) and Mechanical Momenta would remain unchanged under gauge transformation, right? ( 1st question)
But the expectation value of Canonical Momenta is not gauge invariant despite the physical situation being the same.
*Does that imply that we cannot associate an observable with the Canonical Momenta i.e we cannot directly measure it by equipments.
Is it not an observable? (2nd question)
| You are correct on both counts. Mechanical momentum is a gauge invariant physical quantity, but canonical momentum isn't. That means that canonical momentum is as measurable a quantity as the vector potential - which is to say, not at all unless you further specify what gauge you want it in.
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Non-constant tension in rope Could somebody explain to me intuitively how tension is not the same in a rope with mass? My physics teacher (when regarding a massless string) told me that the tension is always equal because if you pull at one side more then the other side has to pull just as much to keep it in equilibrium resulting in the tensions being equal. This fits really well intuitively with me, but if the string has a non-negligible mass how can I adapt this idea? Or replace it if necessary?
| If you pull on a massless rope from both ends, the tension at every point must be the same. Why?
Because the net force on any segment of the string must be zero, because it's a massless object and hence, tension must be same at every point.
In a string with mass, we can apply two different forces at the two ends of the string because of which it has a net acceleration.
Now, the Tension at any point must be such that it can account for the acceleration of the mass of that segment of the string.
Suppose string has length L and constant mass density g. Now, you apply a force F on right. The acceleration of the entire string is
(F/gL).
If you want to find the tension at a distance L/4 from the right. What will it be?
F - T = m.a
i. e T = F - (gL/4)*(F/gL) = 3F/4.
You take a different distance and the Tension would be different because it is dragging with it a different length of the string.
Think of the string as many many blocks with small links attached in between.
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Insulated box of air at temperature $T$, is dropped from height $H$, what is the temperature of the air inside box on the ground? In an effort to understand more about the first law I was wondering about the potential energy term and how it would influence the gas inside the box I made this question and tried to answer it.
I will assume that the air follows the ideal gas law, there is no air resistance and the box stops moving immediately when it reaches the rigid ground.
The first law:
$$
0 = \Delta PE + \Delta U\\
gH = c_v (T' - T)\\
T' = gH/c_v + T
$$
would this be correct or missing something? (i.e. invalid assumptions..)
I am also wondering if the energy is fully transferred to the ground instead..
| Exactly, $dU=\delta Q -\delta W$, and here $\delta W=mgh$ and $\delta Q=0$. You can run some numbers to see that the value is very small for realistic settings.
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Energy contributions of Hamiltonian density In Lancaster and Blundell, Quantum Field Theory for the Gifted Amateur, p.99, the Hamiltonian density is
\begin{equation}
\mathcal{H}=\frac{1}{2}[\partial_0\phi(x)]^2+\frac{1}{2}[\nabla\phi(x)]^2+\frac{1}{2}m^2[\phi(x)]^2,\tag{11.5}
\end{equation}
and it tells us that the energy has contributions from
*
*a kinetic energy term reflecting changes in the configuration in time,
*a 'shear term' giving an energy cost for spatial changes in the field, and
*a 'mass' term reflecting the potential energy cost of there being a field in space at all.
In the equation above, i think the first term is the same as the classical mechanics. But i don't understand why second (shear) and third (mass) term are represent potential energy.
| Actually the first two terms are kinetic part, and the last term is the potential energy.
To get a correspondence to the classical mechanics, just consider the vibration of a string. When the string is displaced from equilibrium, a segment associated with the interval dx has a length:
$$dl = \sqrt{dx^2+d\psi^2}=dx\sqrt{1+(\frac{d\psi}{dx})^2}\approx dx\left[1+\frac{1}{2}(\frac{d\psi}{dx})^2\right]$$.
($\psi$ here is the displacement)
The potential is proportional to the string deformation:
$$dU\propto dl - dx = dx(\frac{d\psi}{dx})^2$$
This potential is due to the deformation of the string. The second term in the Hamiltonian is corresponding to this, which is related to the spatial changes in the field.
I am not sure about the correspondence for the last term (potential term), but it's maybe corresponding to the potential of a oscillating spring (correct me if it's wrong):
$$T=\frac{1}{2}m\omega^2 x^2$$
| {
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Does current run forever in water? (assuming the supply voltage is there forever) Suppose pH of water is $6$, I think this means there is one $\text{H}^{+}$ ion for every $10^6$ water molecules.
When we plug in the battery, I believe we see a current as the $\text{H}^{+}$ ions drift to the $-ve$ side of the battery and suck the electrons injected by the negative plate of the battery. Similarly, $\text{OH}^{-}$ ions drift to the $+ve$ plate of the battery and give an electron away to the positive plate of the battery. This way $\text{H}^{+}$ and $\text{OH}^{-}$ ions neutralize themselves as they contribute to the current. Since the ions were neutralizing themselves, would the current cease to exist after some time when all the $\text{H}^+$ ions in the water were used up ?
| The ions are converted into gases $H_2$ and $O_2$ at the electrodes, so water is gradually being removed from the container - but it requires a very very large amount of charge to flow in order to convert all of the water into gases.
The $H^+$ ions exist as hydronium ions $H_3 O^+$. There is a reversible equilibrium in the water :
$2H_2O <=> H_3O^+ + HO^-$
The product of concentrations of the ions is fixed at a certain value which depends on the temperature of the water. If one or both types of ions are removed by electrolysis, more water molecules dissociate to keep the product of concentrations constant :
$K_w = [H_3O^+] [HO^-]$.
At 25$^{\circ}$C the product $K_w \approx 1.0 \times 10^{-14}$ when concentrations are measured in moles per litre.
In ideal conditions there will always be ions available to conduct electricity through the water. It will keep flowing until either the battery is fully depleted or the water is completely converted to gases.
In practice, an insulating layer may build up at the electrodes which prevents electrolysis from taking place. This layer is caused by other reactions at the electrodes due to impurities in the electrodes or the water.
See wikipedia articles Electrolysis of Water and Self-ionization of Water.
| {
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Why large excavated holes can create air currents? My friend shared this post on my wall, but I got doubt: How can a hole like this create air currents which suck helicopterss also? I didn't really understand.
| The text seems to be very sensationalist. If this was the case then there would be a no fly zone over the grand canyon.
The one explanation I can think of is a draft due to the temperature difference at the top and bottom of the hole. The temperature at a great depth is much higher than that at the surface, and this causes a draft of cold air from the surface to the depths of the hole.
A naive helicopter might end up hitting the sides (and consecutively fall in) or something (if it happens to hover over the hole while not having enough lift) but calling it "sucking" is too narrative-ey.
| {
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If the universe is constantly expanding, shouldn't we go away from other planets? I also saw that every point in the universe is the center. So what does it mean that the universe is expanding? It depends on the repository?
| I think, as far as I understand it, the expansion of the universe is overshadowed by other factors at smaller scales.
The gravity between the earth and the sun is strong enough that even as the universe expands, the distance between the earth and the sun is not affected.
Gravity continues to affect structure more than the expansion of the universe until you get to the largest scales, the distances between galactic clusters. At that scale, on average, everything is moving away from everything else because the universe is expanding and the distances involved are so great that gravity is not strong enough to keep it all together.
| {
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Help in an integration step in QFT by Lewis H. Ryder There is an integration step I can not figure out and is frustrating.
We start from the equality
$$ \dfrac{\partial^2\phi}{\partial x^2} = \dfrac{\partial V}{\partial \phi} $$
and by integration process we are supposed to get (eq. 10.8 in the textbook):
$$ \dfrac{1}{2}\left(\dfrac{\partial\phi}{\partial x}\right) ^2 = V(\phi)
$$
Maybe I'm just overcomplicating it but I can not understand how this is done.
$\phi =\phi(x,t)$ but for this case $\dfrac{\partial\phi}{\partial t}=0$ and $\phi$ approaches zeroes of $V(\phi)$ when $x\rightarrow\pm\infty$.
Then my idea was to integrate by $d\phi$ both sides to get the RHS of eq. 10.8 and for the LHS I tried integrating by parts using $$d\phi=\dfrac{\partial\phi}{\partial x}dx$$
but got no success yet, and also the fact is that I don't even think what I'm doing is correct since is $V(\phi)$ and not $\phi$ what tends to zero when $x\rightarrow \pm \infty$.
Any help would be appreciated
| Just think about it as classical mechanics. Relabel $\phi$ to $x$ and $x$ to $t$ for
$$\frac{d^2 x}{dt^2} = \frac{dV}{dx}.$$
This is Newton's second law with $m = 1$ and an extra minus sign. Then the result is energy conservation with an extra minus sign,
$$\frac12 v^2 - V(x) = E$$
and is proved the same way. Presumably, boundary conditions set $E = 0$, giving the desired result.
| {
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Stationary waves and transfer of energy I have seen a question posted similar to this, but I am not sure if it answers what I was wondering about.
Essentially, we are taught that there is no average energy transfer of a wave to its surroundings. I understand that it intuitive for a mehanical wave on a rope, or even an electron that is bound to an atom. However, stationary waves can also form in microwaves, which are responsible for heating up food. This is also fairly obvious, as the electric and magnetic fields interact with the dipoles of the water molecules.
However, there is a contradiction, and I am not sure exactly where my misunderstanding is. Could someone point it out? Thanks!
| In a empty microwave oven, you create standing wave. And if it's empty, there is no transfer of energy to anything.
Now let's say you put a glass of water right in the middle of your MW oven, The molecules of the water (H20) have resonancies due to their internal structure : vibration, rotation etc... When the frequency of the oven is one of those frequency, their is a transfert of energy from the standing wave to the liquid (heat for the liquid).
That's why you should place you plate in the middle of the oven, that's where the electromagnetic field si the most import and it's the place where you can transfert the more enrgy from the standing wave to you plate. And of course, the oven supply more power to keep the standing wave at about the same amplitude. But you need to inject power into your standing wave
| {
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Why does a massless, frictionless piston move from high pressure to low pressure? Consider an ideal gas kept in a rigid cylinder with a movable massless, frictionless piston at the top. Let the pressure inside the cylinder be $P$ at pressure exerted by the surrounding on the cylinder be $p$. Let $$ P>p $$
Now since the piston is massless net force on it must be 0 (since $m=0$). This implies that the piston can move with any acceleration. But it moves in only one direction that is outwards. Why doesn't it moves inwards?
| If you had a massless, frictionless piston, it would oscillate infinitely fast. There will be an outward force initially, so it will accelerate outward (acceleration = infinity).
I'm not entirely sure how you have deduced that the force is 0. I'm quite sure it is not, if the pressures are different.
| {
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How is the Falcon 9 aerodynamically stable while ascending? How is the Falcon 9 aerodynamically stable while ascending? The rocket doesn't look like it should be able to fly straight. With the large, largely empty fairing at the top and the heavy engines at the bottom and no fins, it looks like it would want to flip around. Is it simply really well controlled by computers to not stray from the prograde vector, or are the engines really light, or what?
| It isn't.
Stabilisation is achieved by gyros which command the rocket nozzles to swivel so a sideways component of thrust can be used to balance the rocket. This has been true since the days of the A-4, which used moveable graphite fins sticking into the rocket exhaust to stabilise the device.
Before, rockets used much larger fins and a long ramp for stabilisation, and the rocket had to gain enough speed for the fins to become effective. Only by adding artificial stabilisation did bigger rockets become possible.
| {
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Magnetic force on open circuit? Let's say we have a straight horizontal wire and we let it drop inside a magnetic field which will be parallel to the ground(coming out of the screen). Charges inside the wire feel a force due to their movement inside the magnetic field .
Let's say the field is coming out of the screen. Positive charges gather on the left and negative on the right. If we had a loop I would have no trouble with this but during the charges' movement do we consider that we have a current? Therefore leading to a magnetic force opposing the bar's drop or not?
|
So I'll have to ask again, does this ever stop? Is equilibrium ever
reached?
Prompted by the comments to reexamine the question, I now better understand the question.
Essentially, this is the canonical rod moving in a perpendicular magnetic field extended to the case that there is a constant external applied force.
Clearly, unless there is an oppositely directed magnetic force that cancels the applied force, the rod will accelerate.
But an accelerated rod implies an increasing magnetic force (on charge) parallel to the rod.
Further, if the mobile charge is moving parallel to the rod (an electric current), there is a magnetic force against the external applied force.
What then is equilibrium (the stationary case)?
As user Farcher points out in the comments, the work done by the external can go into the potential energy of the separated charge but it can also go into the kinetic energy of the rod.
(to be continued... it's late now)
You're correct that there is a current but it is a transient current.
As the mobile charge redistributes due to the magnetic force, another force develops - the electric force due to the field from the redistributed charge.
Charge moves just until the electric force and magnetic force on mobile charge cancel out.
| {
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Bounds of Integration (with respect to something that is not time) I have been reading Richard Feynman's lectures and came across an interesting proof regarding the Earth's gravitational force. At one point in the proof, Feynman uses the following the integral:
$\int_{R+a}^{R-a} dr$
(13.18 on http://www.feynmanlectures.caltech.edu/I_13.html)
In this integral, r is the distance between a point in space and the surface of the Earth, R is the distance between that point and the center of the Earth, and a is the radius of the Earth. I interpret this integral as summing up all of the dr's going around the Earth. The proof itself makes sense to me, I am just confused about the bounds of integration.
As $\int_{R+a}^{R-a} dr$, I interpret the integral as summing up the dr's starting on the right side of the Earth and going to the left side. However, in this sense, $\int_{R-a}^{R+a} dr$ should be the sum of all the dr's starting from the left side and going to the right side. Conceptually, I feel as if these should be the same, but mathematically $\int_{R+a}^{R-a} dr = -\int_{R-a}^{R+a} dr$. My question is, how did Feynman choose the ordering of his bounds of integration? It does not appear arbitrary, but I am not sure how the decision was made. Thank you!
| Reading through the proof, it seems like he actually starts by working in $x$, then proves the relationship between $x$ and $r$ (which includes the negative sign): $2r~dr = -2R~dx$.
Keeping the direction of the integration the same, you can see that the natural order for $r$ is from larger ($r=R+a$) to smaller ($r=R-a$):
The sign change (going from x positive to r negative) is absorbed by removing the negative sign that was in front of the original expression for W.
Image source: http://www.feynmanlectures.caltech.edu/I_13.html modified to include direction of integration
| {
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Neutron stars - only neutrons? I was at a museum recently, and there was a display on neutron stars. It said that neutron stars are made only of neutrons, which honestly didn't make much sense to me - neutrons decay very quickly on their own, so how do neutron stars "last", so to speak?
So naturally, I checked wikipedia, which provides this diagram:
Electrons and protons do seem to be present. Nowhere is there a "layer" that's only neutrons.
This leads to two questions:
*
*Is Wikipedia wrong, or the museum (normally I'd trust wikipedia, but this is a not-insignificant museum that I went to)?
*Is there anything that could explain the museum display if wikipedia is right?
| Neutrons are stable in the neutron star because there is such high pressures that neutrons are unable to decay. The states for protons and electrons make sure that the neutrons do not decay. Neutrons do decay but not enough to make a dent on the neutron star. Also any neutrons that do manage to decay to protons most likely convert back to neutrons. There is no complete layer with neutrons. Just more neutrons the further down you go in a neutron star. Both the museum and Wikipedia are right. When they say composed of "all neutrons" they do not literally mean made of completely neutrons. Instead they mean neutron stars have tremendous amount of neutrons but still have a small amount of other stuff.
| {
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Time Constant of a circuit - capacitors in parallel or series?
When I charge the circuit as shown where the switch is at A, the p.d. across C1 is 10V, which is correct.
When I discharge the circuit, however, by changing the switch S from A to B, the total capacitance of the circuit is 200μF as C1 and C2 are in parallel.
This must mean the time constant of the discharging circuit is
$$200 \times 10^{-6} \times 100 \times 10^3 = 20s$$
In the solution, however, the time constant of the discharging circuit is quoted as 5 seconds, where it states C1 and C2 are in series when the switch is at B.
Who is right, the solutions or me?
|
the total capacitance of the circuit is 200μF as C1 and C2 are in
parallel.
They're not in parallel for either switch position.
When the switch is in position A, C2 and the 100k resistor are in series but one end of the resistor is 'dangling' so there is no path for current through the series combination.
When the switch is in position B, C1 is placed in series with the C2 + 100k series RC combination. This should be obvious since there is only one path for current, all three circuit elements have identical current.
If the capacitors were parallel connected, C1 and C2 would 'split' the current through the 100k resistor but clearly, all of the current through the resistor is through either capacitor and so, the capacitors are series connected.
As you already know, the equivalent capacitance of 2 identical series connected capacitors is 1/2 the individual capacitance and thus
$$C_{eq} = 50\mu\mathrm{F}$$
and the time constant is
$$\tau = 100\mathrm{k\Omega}\cdot 50\mu\mathrm{F} = 5\mathrm{s}$$
| {
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Velocity of a Mechanical Wave on a String I recently read a derivation for an equation which governs how quickly a wave is transmitted along a string, $v = \sqrt{\frac{T}{\mu}} $, where T is the tension in the string, and $\mu$ is the mass per unit length along the string. The derivation makes sense but gives a more mathematical and geometrical account as to why this is the case.
Could someone please explain more qualitatively why an increase in density of the string would reduce the velocity of transmission, and why an increase in tension would increase the velocity?
I can see, in very vague terms why an increase in tension would cause neighbouring elements of the string to more quickly follow the motion of preceding elements when for instance, a pulse is sent down the wave as follows:
though I cannot visualise the effects of density. I can imagine each element of the string having more mass and their movements becoming more 'sluggish,' but would this not affect the frequency of their oscillations?
| If you assume that tension and density are the only factors then dimensional analysis will lead you straight to that formula except for an undetermined constant. If you want an answer without such an assumption, then maybe you have to accept that the math is needed after all.
| {
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Derivative with respect to the radius of a moving particle on a non-frictional surface Basically I was just wondering about the following framing of question:
How can one determine the derivative of the velocity of the particle with respect to the radius when decreasing the force F on the string - (the particle is attached to a string that goes through a hole, as in figure). The surface on which the particle moves, has no friction.
FIGURE (not an inclined plane):
My attempt: So far, I've realized that when decreasing the force F, the radius R will be bigger. I also know that the angular momentum will be preserved. Although, now I'm stuck - so I'm very thankful for every responses/hints to help me solve this problem.
PS. Sorry for my bad english.
| How much of a quantitative answer do you expect? Because it kind of an abstract question.
First of all, you say "I've realized that when decreasing the force F, the radius R will be bigger." This is not true in a case where the force pulling on the string is bigger than the centrifugal force. The only thing you can be sure of is that $\frac{d^2R}{dt^2} > 0$, but for $\frac{dR}{dt}$, it depends on the initial conditions.
Secondly, you don't have a function for $F(t)$ so it's tough to come up with something numerical. What you know is that there are two forces in the problem, $\vec{F} = -F\hat{r}$ from the string and $\vec{F}_{cen} = \frac{m u_{\theta}^2}{r} \vec{r}$, both radial. Writing the sum of forces, you have
$$
\sum \vec{F} = \left( \frac{mu_{\theta}^2}{r} - F \right) \hat{r} = m\frac{d\vec{u}}{dt} \equiv m \frac{du_r}{dt} \hat{r}
$$
because there are no angular force. So $\frac{du_{\theta}}{dt} = 0$ which as you stated corresponds to conservation of angular momentum. Then you have
$\frac{du_r}{dt} = \left( \frac{mu^2_{\theta}}{r} - F(t) \right)$ which you could try to integrate but $r$ depends on time as well and it because complicated at this point.
I guess that still gives you a clue what to do... I Hope it helps you!
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How to choose the best Gaussian surface? Are there some basic rules so that I could choose a Gaussian surface according to my physics problem? Are there any guidelines for how to choose an appropriate Gaussian surface for the system of charges you are analysing?
| Choose the Gaussian surface in such that the electric field at every point on it is constant. Ultimately, you should be looking for symmetries, since it would simplify calculations a lot.
For example, consider finding the magnitude of the electric field due to an infinite thin sheet of charge, having a uniform positive charge density $\sigma$. We choose the Gaussian surface to be a cylinder going into the sheet and out of it. Finding the flux coming out of the Gaussian cylinder and splitting the integral for the flux coming out of the end faces, we have
$$
\oint_S \vec{E} \cdot \vec{dS} = \oint_{end face}\vec{E} \cdot \vec{dS} + \oint_{end face} \vec{E} \cdot \vec{dS}
$$
$$
\oint_S \vec{E} \cdot \vec{dS} = 2\oint_{end face}E dS\tag{1} = 2ES
$$
Since $\vec{E}$ and $\vec{dS}$ are in the same direction.
Equating $(1)$ to $\dfrac{q}{\epsilon_0}$ by Gauss's Law, we have,
$
E = \dfrac{q}{2S\epsilon_0} = \dfrac{\sigma}{2\epsilon_0}$, since $\sigma = \dfrac{dE}{dS} = \dfrac{E}{S}$
You could write it in vector form as $\vec{E} = \dfrac{\sigma}{2\epsilon_0}\hat{k}$, depending on the direction chosen.
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Why does a surface always exert force normal to it? In whichever angle an object is thrown at a surface, the surface always exert force normal to it. But why? According to Newton's third law, if an object hits a surface at an angle, the reaction force provided by the surface must be equal and opposite to the applied force by the object. But why does the surface always exert force normal to it?
| The force is normal to the surface only if no friction is present. If we have some sort of sticking between the thrown object (e.g. ball) and the surface (e.g. earth), there will be a non-normal component of the force.
Just think of an ball which slides (w/o friction) on the surface of the earth. As there is no friction, there can not be a force.
Now go back to the original question. Let's consider the motion of the ball as a superposition of two motions:
*
*The vertical motion, which is in normal direction (say y-direction), and
*the horizontal motion (say in x-direction).
If we consider the horizontal motion to be frictionless, there can not be a force in x-direction. Hence, we are left with a normal force, which reflects the ball.
Once you include friction to the problem, you will get a superposition of two forces. From that you can derive the direction of the total force. In general it will not be along the moving direction of the ball.
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What produces an electromagnetic field? A charge that moves, or a charge that changes over time? I've learnt an electromagnetic field is produced by moving charges, i.e. a current.
Is it the case, or is it actually the fact that the charge is changing at a given location?
I mean: imagine I have a charge $q_1$ located at a given point of the space. Then imagine the charge changes and becomes $q_2$, but without moving (this is probably impossible? but let's imagine). Does it produce an electromagnetic field?
Or, without "imagining" a charge changes over time:
Is the EM field produced by the movement of the charge, of by the fact the charge vanishes from its location?
| Regarding a change in the charge, when you say
this is probably impossible? but let's imagine
─ let's not. Electric charge is conserved, both globally and locally, i.e. it obeys the continuity equation
$$
\frac{\partial \rho}{\partial t}+\nabla\cdot\mathbf j=0,
$$
and this is absolutely critical for Maxwell's equations to be internally consistent. All changes in the charge distribution can ultimately be traced down to the physical transport of particles moving from one place to another.
If you want to "imagine" a world where the laws of physics are so different that the total charge enclosed inside a given surface can change without there being a corresponding flux of charge through the surface, then you're no longer describing electromagnetism as we know it, and it is completely pointless to speculate about how radiation might look like in such a theory.
| {
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Why is the spring force equal to the external force? So we are studying about springs, as of now, the assumption is that they are massless.
My teacher told me that when we extend a spring, or compress it, there is a force called the spring force which tries to reform the spring back it its original form. This force acts inwards and outwards respectively.
He also told us that the magnitude of this force is equal to the force we apply.
What I want to ask is why it this force always equal to the force we compress the spring with?
Wouldn't the equality suggest that the spring will still remain deformed? If equal amount of force is coming from both sides, and if the spring is deformed then wouldnt it remain deformed?
Shouldnt the spring force be slightly greater than the external force to reform the spring?
| Well, think of it this way: as you push into a spring the more it gets deformed and the more it gets deformed the more it pushes back. Eventually, you reach a position when the applied force is just equal to the force of the spring and at that moment you cannot push further and the spring is at rest now. The spring will remain at this deformed state until the external force is removed.
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First law of thermodynamics as conservation of energy We have that $\Delta U = Q + W$. What I don't see is how this formula relates to the law of conservation of energy. Can someone please clarify?
Does this mean that $\frac{dU}{dt}=\frac{dQ}{dt}+\frac{dW}{dt}=0$, so that $\frac{dQ}{dt}=-\frac{dW}{dt}$?
| Since your second question has been answered in the comments, I will answer your first question.
Let's examine in words what $\Delta U=Q+W$ means: "Any change in internal energy arises from a flow of heat into/out of the system and/or work done by/on the system." Put differently: "The only two ways internal energy can change are if heat flow occurs or if work is done." Both heat flow and work are examples of energy transfers (indeed, both have units of energy). Also, note that any energy transfer to the system that is not classified as heat flow is automatically classified as work done on the system.
Putting this all together, we can restate the equation as follows: "In order to change the internal energy of a system, you must add or subtract energy from the system." This is an indirect way of stating that energy is conserved.
It might be easier to see this in the case of an isolated system, where no external heat flow or work occurs, so $Q=W=0$. In that case, the law reads: "In an isolated system, total internal energy does not change." This is a direct statement of conservation of energy.
| {
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Scalar Field Theory for Gravity While reading the book Gravitation Foundation and Frontiers by Padmanabhan, I came across the Lagrangian for a scalar theory of gravity. But the coupling term consist of trace of the Energy Momentum Tensor. If we change the coupling term to $F(\phi)T_{ab}T^{ab}$, what prevents it from being a viable theory?
| If we linearize the equations of motion about a $\phi = 0$ background (and assume the usual sort of kinetic term), we will find a non-relativistic limit of something like $\nabla^2 \phi \propto \rho^2$, which is inconsistent with Newtonian gravity.
| {
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How to shift AGN x-ray spectrum to rest frame I have limited information to shift a spectrum (in the x-ray 0.5-10 keV bandpass) at redshift z=2 to rest frame. I have a plot of normalized (photon) counts s$^{-1}$ keV$^{-1}$ by energy (keV), and so I have no information to compute the flux density. Is there a way to shift the spectrum I have to rest frame? Or do I really need luminosity distance, etc.?
| The frequency is related to the redshift by $$\frac{\nu_{\rm obs}}{\nu_{\rm emit}} = \frac{1}{1+z}$$
Another useful relation is the fact that, for any redshift,
$$\frac{I_{\nu}}{\nu^3} = {\rm constant}$$
where $I_{\nu}$ is the specific radiative intensity.
The specific count rate (counts per area per time per energy interval) is proportional to $I_{\nu}/\nu$. (Do you see why?)
That should give you enough information to derive how the measured count rate changes with $z.$ Of course, to know the actual value, not just the function of $z$, you'd need to know the actual count rate measured, not just the normalized value.
| {
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Understanding the units of number density I understand that we can measure any general number density $n$ as,
$$ n = \frac{N}{V}$$
for total number $N$ and volume $V$. This puts the units of number density as $\text{length}^{-3}$ e.g. $\text{cm}^{-3}$
Now suppose the number density varies with radius as e.g.,
$$ n(r) = N_0 r^{-2}$$
Clearly the units at any particular radius are still $\text{cm}^{-3}$, but then what are the units of $N_0$ and $r$?. It seems to me that for different distributions e.g. $N_0 r^{-3}$ the units would be different. But then how can the units of $n$ always be $\text{cm}^{-3}$
| So $r$ has units of length, for example $\mathrm{m}$, so $r^{-2}$ has units $\mathrm{m}^{-2}$. to remain dimensionally consistent $N_{0}$ would have to have units $\mathrm{m}^{-1}$. Note that here, we have number density in 3 dimensions, giving a number per unit volume. We might imagine number density on the plane, or on a line. In this circumstance, the dimension of $N_0$ would change. It might be of interest to you to note that $N_0$ in this case is only a dimensionless number when the power law matches the dimension of the space we are considering.
| {
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Time dilation and Transverse Doppler Effect – where does energy disappear? Kinematic effects of Special Relativity like time dilation and length
contraction are well known.
Article in Wikipedia makes it clear:
https://en.wikipedia.org/wiki/Time_dilation
“In the special theory of relativity, a moving clock is found to be ticking slowly with respect to the observer's clock. If Sam and Abigail are on different trains in near-lightspeed relative motion, Sam measures (by all methods of measurement) clocks on Abigail's train to be running slowly and similarly, Abigail measures clocks on Sam's train to be running slowly.”
One practical way to measure the amount of time dilation is measuring frequency of relatively moving source of radiation at points of closest approach – the Transverse Doppler Effect. That means, if Sam radiates with proper frequency f, Abigail measures frequency $f_s/\gamma$. If Sam measures frequency of Abigail, he measures $f_a/\gamma$. That means, the photon redshifts and redshifts again.
According to this logic, if Sam and Abigail exchange a photon, the photon will finally vanish. Or will Sam and Abigail will vanish themselves? Where does the energy disappear?
Please don't confuse it with relativistic Doppler redshift when emitter and absorber recede.
| Let's say Sam has some light trapped in a box. According to Agibail that light in the box has momentum into the direction of motion of the box.
Transverse light according to Agibail is such that it has only transverse momentum.
If Sam opens the lid on the top of the box then according to Sam the out flowing light has no longitudinal momentum. According to Agibail it has longitudinal momentum. (Top is the side that points towards Agibail)
So Agibail must instruct Sam to point the box in the backwards direction. And now we can see where the energy "disappears".
It becomes kinetic energy of Sam, according to Agibail. It becomes kinetic energy of Agibail according to Sam.
| {
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Two-point correlation function in Peskin's book I am reading Peskin's book on QFT and I reached a part (in chapter 4) where he is analyzing the two-point correlation function for $\phi^4$ theory. At a point he wants to find the evolution in time of $\phi$, under this Hamiltonian (which is basically the Klein-Gordon - $H_0$ - one plus the interaction one). Anyway, when he begins his derivation he says that for a fixed time $t_0$ we can still expand $\phi$ in terms of ladder operators in the same way as we did in the free (non-interaction) case (this is on page 83), i.e. $$\phi(t,\mathbf{x})=\int{\frac{d^3p}{(2\pi)^3}\frac{1}{E_\mathbf{p}}(a_\mathbf{p}e^{i\mathbf{x}\mathbf{p}}+a_\mathbf{p}^\dagger e^{-i\mathbf{x}\mathbf{p}})}.$$ I am not sure I understand why can we do this, for a fixed time. When we wrote this in term of ladder operators for the free case, we used the KG equation in the free case, which resembled to a harmonic oscillator in the momentum space, and hence we got the ladder operators. But now, the equation of motion is different (it has $\phi^3$ term, instead of 0, as before), so can someone explain to me why we can still use the same formula as before, even if the equation of motion is different?
| I am now very sure that Peskin made a mistake here. Please check the lecture Notes by Weigand
on page 43, where it is said clearly that The crucial difference to the free theory is, though,
that $\phi(x)$ cannot simply be written as a superposition of its Fourier amplitudes $a(\vec{p})$ and $a^{\dagger}(\vec{p})$ because
it does not obey the free equation of motion...
| {
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Alternative form of scalar particle dynamics in the gravitational field I am interested in the dynamics of a single (classical) spin-0 particle in the General Relativistic spacetime of arbitrary signature and dimensionality. In the 1-st order (Palatini) formalism, this can be written as
$$ S[X] = -m \intop_{X} d\tau \sqrt{\eta_{IJ} e_{\mu}^{I} e_{\nu}^{J} \dot{X}^{\mu} \dot{X}^{\nu}}, $$
where $X(\tau)$ is (an arbitrarily parametrized) particle worldline.
However, I have recently seen another action in the literature (e.g. in this action about 3d Euclidean quantum gravity on page 6):
$$ S[X] = \frac{m}{2} \intop_{X} e^0 = \frac{m}{2} \intop_{X} d\tau \, e_{\mu}^0 \dot{X}^{\mu}. $$
It appears to violate gauge invariance, but as far as I understand, authors are claiming that its fine (not exactly sure why though). (Actually, in the mentioned article a more general form of the action is considered which includes the classical angular momentum $s$, but for now I am only interested in the $s = 0$ case).
Question: from a brief look I can tell that the dynamics given by this action is drastically different from the dynamics given by the geometrical "length of the worldline" action above. Which one is physically more suited to describe interactions of (classical) particles with gravity?
| I think I just realized how to answer my own question.
The linear action is just a special case of the "length of the worldline" action in the
$$ e^0 \parallel dt;\quad e^{1,2,3}\perp dt $$
case. So this second action is obtained from the first one after gauge-fixing.
The whole point of the paragraph, though, is that introducing particle degrees of freedom is equivalent to loosing gauge invariance along the worldline. So by making this choice of gauge-fixing we introduce just as much information as we would by introducing a dynamical particle trajectory.
| {
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On the average gas pressure over a pipeline This question stems from gas flow through pipelines, quoting this paper (link):
For the calculation of the $z$ factor one needs the average pressure
$p_{av}$. An obvious choice is the arithmetic average $$\tag{1}
p_{ava}=\frac{p_b+p_e}{2}$$
However the pressure over the pipeline is not linear but quadratic.
If the influence of the $z$-factor is ignored, the pressure over the
pipeline is $$\tag{2} p(x)=\sqrt{p_b^2-cx}$$ with $$\tag{3}
cL=p_b^2-p_e^2$$
The average pressure over the piple line $p_{avl}=$
$$\tag{4} \frac{1}{L}\int_0^L \sqrt{p_b^2-cx}\ dx=\frac{-2}{3cL}(p_b^2-cx)^{\frac{3}{2}}|_0^L=\frac{2}{3cL}(p_b^3-p_e^3)$$
Elimination of $cL$ gives
$$\tag{5} p_{avl}=\frac{2}{3}\frac{(p_b^3-p_e^3)}{(p_b^2-p_e^2)}$$
$p_b$=upstream pressure; $p_e$=downstream pressure; $L$=length of pipe; $x$ is distance along length of pipe; $c$=?(author did not state)
Questions:
*
*How does one derive Eqn. (2)? (Since the $z$-factor is ignored, I guess one would start with the ideal gas law,$pV=nRT$? I guess this derivation also assumes isothermal flow.)
*I guess to answer the first question, one needs to know what the $c$ variable is...what is it?
*If one were to not ignore the influence of the $z$-factor, how would one derive the average pressure over the length of pipe? (assuming isothermal flow)
| 1 and 2. To get the constant c, you treat the flow as fully developed locally, calculate the Reynolds number, represent the density in terms of the pressure (using the ideal gas law), determine the friction factor, and determine the local shear stress at the wall. This all then gives you the gradient of $ p^2$. You then integrate along the pipe.
You are correct, that the derivation assumes isothermal flow.
*Without ignoring the z factor, you would use a similar procedure, but the problem would have to be done iteratively. You could guess the inlet pressure, then integrate the local pressure gradient numerically to the outlet end of the pipe. Then you would check to make sure that the outlet pressure matched the known outlet pressure. If not, you would adjust the inlet pressure and iterate until it did.
| {
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Is this an electromagnetic wave without the magnetic part? Jefimenko's Equations are:
$$
\begin{align}
&\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{\rho(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') - \frac{1}{|\mathbf{r}-\mathbf{r}'| c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \mathrm{d}^3 \mathbf{r} \\
&\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r} \\
& \mbox{where the retarded time is: }t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}
\end{align}
$$
I noticed the following: In the E-field, the second and third terms fall at $1/r$. Meaning, they are radiative terms. That is, electromagnetic radiation. Analogously, In the B-field, the second term falls at $1/r$, thus radiative. However, in the B-Field, there is no radiative term depending on $\partial\rho / \partial t$.
With that in mind, consider the following situation:
$$
\frac{\partial\mathbf J}{\partial t} = 0
\quad\quad\mbox{and}\quad\quad
\frac{\partial\rho}{\partial t} \neq 0
$$
Then, the magnetic field will not be time varying, and won't be radiative. In this situation, the only radiative fields would be electric. Meaning, an electromagnetic radiation without the magnetic part! How is this possible? Is there a mistake somewhere? What prevents it from happening?
An approach was to allow time-varying $\rho$ with charge conservation by means of constant not-null-everywhere $\mathbf J(\mathbf r)$. By continuity equation, one can have a time varying $\rho$ one wants, by simply choosing $\mathbf J$ cleverly. I haven't made further progress with this.
$$
\nabla\cdot\mathbf J + \frac{\partial\rho}{\partial t} = 0
\quad\implies\quad
\frac{\partial\rho}{\partial t} = -\nabla\cdot\mathbf J
$$
| To add to Emilio Pisanty's answer: for the situation you are considering, Jefimenko's equations simplify to
$$
\begin{align}
\mathbf{E}(\mathbf{r}, t) &= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{\rho(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\
\mathbf{B}(\mathbf{r}, t) &= \frac{\mu_0}{4 \pi} \int \frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r} \\
\end{align}
$$
As Emilio points out, if ${\bf J}$ does not depend on time then the continuity equation requires that $\rho$ depends linearly on time. Writing it as $\rho({\bf r'}, t) = a({\bf r'}) + b({\bf r'})\, t$ gives
$$
\begin{align*}
\mathbf{E}(\mathbf{r}, t) &= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{a({\bf r'}) + b({\bf r'})\, t_r}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{b({\bf r'})}{|\mathbf{r}-\mathbf{r}'|^2 c}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\
&= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{a({\bf r'}) + b({\bf r'})\, \left(t - \frac{|{\bf r} - {\bf r'}|}{c} \right)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{b({\bf r'})}{|\mathbf{r}-\mathbf{r}'|^2 c}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\
&= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{a({\bf r'}) + b({\bf r'})\, t}{|\mathbf{r}-\mathbf{r}'|^3}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\
&= \frac{1}{4 \pi \epsilon_0} \int \left[ \frac{\rho({\bf r', t})}{|\mathbf{r}-\mathbf{r}'|^3} (\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r}.
\end{align*}
$$
All retardation effects cancel out, and the electromagnetic fields are given by the usual Coulomb and Biot-Savart laws applied to the instantaneous rather than retarded sources! In particular, they fall of like $1/r^2$, and are not radiative.
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How to evaluate the Lorentz force at a surface where the field is discontinuous? I'll take a simple case as an example. You have a constant and uniform magnetic field inside an ideal infinitely long solenoid, with currents circulating all around the thickless coils (so there's a surface current with a field discontinuity there).
\begin{equation}\tag{1}
\vec{B}_{\text{inside}} = \mu_0 \, n \, I \: \vec{z},
\end{equation}
where $\vec{z}$ is the unit vector oriented along the solenoid's main axis. The field is 0 outside the solenoid (ideal case) : $\vec{B}_{\text{outside}} = 0$.
The magnetic field exerts a Lorentz force density $\vec{f} = \vec{J}_{\text{sol}} \times \vec{B}$ on the currents that are creating that same field, so there is magnetic pressure acting on the solenoid. That pressure should be proportional to the field's energy density.
Now, how do you evaluate the magnetic field, at the solenoid's surface, that should act on the current density $\vec{J}_{\text{sol}}$ ? $\vec{B}_{\text{inside}}$ given above ? $\vec{B}_{\text{outside}} = 0$ ? The average defined as this :
\begin{equation}\tag{2}
\vec{B}_{\text{average}} = \frac{\vec{B}_{\text{inside}} + \vec{B}_{\text{outside}}}{2} = \frac{1}{2} \, \mu_0 \, n \, I \: \vec{z} \quad ?
\end{equation}
or what else ? If it's the average (2), how can you justify it ?
| It is indeed the average, as you suspect. To understand why, witness that the current is spread over a nonzero thickness. Begin with an azimuthal directed current density $J(r)$ (sketch this) and you get two equations:
$$\frac{\mathrm{d} \,B(r)}{\mathrm{d}\,r} = -\mu_0\,J(r)\quad\text{(}\vec{B}\text{ axially directed)}$$
$$\frac{\mathrm{d} \,F(r)}{\mathrm{d}\,r} = r\,\theta\,L\,J(r)\,B(r)\quad\text{(Lorentz force on sector of azimuthal subtense}\,\theta,\,\text{length}L)$$
Eliminate $J$ to find that $B$ varies with $F$ according to $\frac{\mathrm{d} \,F}{\mathrm{d}\,B} = -r/\mu_0$; this simple equation shows you very simply that the force on an infinitely thin current sheet is calculated with the average of the magnetic fields inside and outside the coil.
The same principle works in calculating the force exerted by a plane wave on a perfect conductor when the former is reflected or absorbed by the latter. You get the same answer whether you use half the magnetic field at the on the outer surface acting on the current sheet, or whether you do the full calculation with exponentially dwindling fields with depth owing to the skin effect.
| {
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Deformation of a self-gravitating sphere from two forces I have a fluid sphere (say a gas or a liquid of uniform density, under its own gravity) on which forces is applied to its surface. I would like to find its approximate shape (most probably an oblate ellipsoid), from the forces applied on its (initialy) spherical surface. Using spherical coordinates, the radial (pressure) and tangential (shear) forces are these :
\begin{align}\tag{1}
F_r(\vartheta) &= C \sin^3 \vartheta, \\[12pt]
F_{\vartheta}(\vartheta) &= 4C \, \sin^2 \vartheta \, \cos{\vartheta}, \tag{2}
\end{align}
where $C$ is an arbitrary positive constant. In vectorial form :
\begin{equation}\tag{3}
\vec{\boldsymbol{\mathrm{F}}} = C \sin^3 \vartheta \, \vec{\boldsymbol{\mathrm{u}}}_r + 4C \, \sin^2 \vartheta \, \cos{\vartheta} \, \vec{\boldsymbol{\mathrm{u}}}_{\vartheta}.
\end{equation}
There's an axial symetry around the $z$ axis. The deformation may be considered "weak", as a first approximation ($C$ may be "small", compared to the gravitationnal force on the surface : $C \ll G M^2/R^2$).
Note that the pressure force is 0 at the poles, and maximal at the equator, so it tends to "squash" the sphere to an oblate ellipsoid (of unknown ellipticity). The shear force is 0 at the poles and at the equator.
Any idea about how to find the deformation's ellipticity?
| The problem is to find $r(\theta, \phi) = r(\theta)$ with r the radius of the ellipsoide.
I think we can assume that the gravitationnal field created by the fluid is the same as the one from a sphere (weak deformation).
$$\mathbf{G}(r) = - \frac{\mathcal{G}M}{R^3}{r} \mathbf{e}_r$$
Where M is the mass of fluid, and R the radius of the sphere
Fluid statics gives
$$ \boldsymbol{\nabla} P = \mathbf{f}_v$$
With P the pressure and $\mathbf{f}_v$ the gravitationnal force.
So we reach
$$ P(r, \theta, \phi) = P(r) = P(0) - \frac{\rho GM}{2}\frac{r²}{R^2}$$
And we should try to make it equal the tangential part of the pressure F.
With $\alpha$ the angle between the normal vector to the surface for the sphere and the ellipsoid, we have
$$ \mathbf{F}_N = \mathbf{F}_R \cos(\alpha) + \mathbf{F}_{\theta} \sin(\alpha)$$
As we can assume $\alpha <<1$,
$$ \mathbf{F}_N = \mathbf{F}_R+ \alpha\mathbf{F}_{\theta} $$
And I think we can find
$$\alpha = \frac{1}{r} \frac{\textrm{d}r}{\textrm{d}\theta}$$
So we have a differential equation on $r(\theta)$! The inital condition would be that the volume must remain the same.
| {
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Attenuation of radio frequency transmissions in space In a science fiction movie some aliens come to earth and when asked why, they say they received our radio transmission and came to investigate, the radio transmission being the first television broadcast (the Berlin Olympiad in 1936). So, the idea is that it took decades for the signal to reach them.
I am wondering how attenuated such a signal would be after traveling for 50 years in space, assuming the transmitter was about 100,000 Watts?
| What you are looking for is called free-space path loss.
Let's assume the signal is broadcast from a spherical source (e.g. the signal isn't sent with a directional antenna). Neglecting all other sources of loss (e.g. diffraction, reflection), the simplest way to calculate this loss is $\frac{1}{4\pi r^2}$.
The signal must be received by another antenna as well; therefore we must also take into account the aperture efficiency. This describes the ability of an antenna to pick up a signal, which is dependent on wavelength, and given by the equation $A_{eff} = \frac{\lambda^2}{4\pi}$. A wonderful derivation can be found elsewhere on this site.
All together, this gives us the free-space path loss (FSPL):
$$ FSPL = \left(\dfrac{4\pi r}{\lambda}\right)^2 $$
Where $r$ is the distance between transmitter and receiver and $\lambda$ is the wavelength of the signal.
Useful resource: http://www.radio-electronics.com/info/propagation/path-loss/free-space-formula-equation.php
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Why would a steel tube vibrate under high voltage? Can anyone explain why a vertical steel tube, standing with one end in contact with the ground under a $230 \space kV$ electric transmission tower, would physically vibrate at high frequency? I'm not well-grounded in electricity laws, but I suspect it may have to do with eddy currents.
| The high-voltage line is inducing eddy currents in the steel tube, and these are reacting against the cyclic magnetic field that surrounds the wire- and the steel tube becomes "microphonic". If the excitation frequency is close to one of the tube's fundamental (mechanical) harmonics, then the tube will "sing". This occurs in power transformers that have poorly-clamped steel laminations; They hum with varying degrees of loudness when power is applied to them.
| {
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Are all correlation functions in a CFT non-zero? I am particulary interested in the Ising CFT. Is it clear/true that for any field $\phi$ and a large enough number $r \in \mathbb R^+$, we have that $\langle \phi(x) \phi(y) \rangle \neq 0$ if $|x-y| > r$?
In case such fields exist, are they expected to also be realized in lattice models? Or are they figments of the continuum description?
| For CFTs we actually know the precise functional form of the 2-point functions
\begin{equation}
\left\langle \phi(x)\phi(y) \right\rangle = \frac{C_\phi}{|x-y|^{2\Delta_\phi}}
\end{equation}
where $\Delta_\phi$ is the scaling dimension of your field and $C_\phi>0$ for non-trivial fields. You can look up the formula in DiFrancesco's Conformal Field Theory Eq. (4.55). This means that, yes, it is true that $\left\langle \phi(x)\phi(y) \right\rangle \neq 0$ even for $r=0$.
As for the second part of your questions I don't know.
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"source": "stackexchange",
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When short circuits are not exactly short circuits?
Here $R1 = 2 \Omega$, $R2= 4 \Omega$ and $R3= 4 \Omega$
Though there looks to be a short circuit in this diagram, my teachers say that this circuit can easily be redrawn into simple parallel circuit. As far as I know a short circuit is an alternate way for current to move where there is no resistance whatsoever. So all of the current will flow through a short circuit if found.
Here if the current flows through that part of the wire ACB then after that R2 and R3 being the same it will get confused which way to go and even if it goes both sides then some part of the current is bouncing back and moving from the negative side of the cell to the positive. This is just a conjecture.
So can anyone please describe why there is no short circuit in the circuit though it seems to be? And also how to understand by looking at any circuit that there is actually no short circuit though there seems to be one.
| It is best to label all the nodes.
$D$ and $F$ are in fact one node and are on one side of each resistor.
$A, \, B$ and $C$ are in fact one node and connected to the other side of each resistor so the three resistors are in parallel.
All the wires are doing is ensuring certain parts of the circuit are at the same potential.
If wire $C$ was not there you would only have current flowing through resistor $R1$ and likewise if wire $DF$ was absent current would only flow through resistor $R3$.
In both cases the remaining wire would be a short circuit across two of the resistors.
As drawn currents flow through all the resistors.
| {
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Why do opposite dipoles of same charge and equal magnitude cancel each other out?
In carbon dioxide, there are 2 dipoles of equal magnitude and charge pointing in opposite directions.
Why can't both oxygen atoms obtain a partially negative charge while the carbon atom contains a partially positive charge twice in magnitude? Shouldn't two oxygen atoms pulling on a single carbon atom result in more positivity?
Why is carbon dioxide truly nonpolar? Why does this make sense?
| Sean's answer is definitive, but I think it's worth adding a few comments.
Why can't both oxygen atoms obtain a partially negative charge while the carbon atom contains a partially positive charge twice in magnitude? Shouldn't two oxygen atoms pulling on a single carbon atom result in more positivity?
The oxygen atoms do get a partial negative charge and the carbon atom does get a partial positive charge. The partial positive charge on the carbon is twice the magnitude of the charges on the oxygen atoms - your diagram is a bit misleading in this respect. I would draw it as:
However this does not produce a dipole moment for the reasons Sean explains. The first non-zero moment is the quadrupole moment. Your title asks:
Why do opposite dipoles of same charge and equal magnitude cancel each other out?
and the answer is simply that the dipole moment is a vector, and the dipoles of the two halves of the molecule add in the usual way vectors add. Since the two dipoles are equal and opposite they add up to zero.
Why is carbon dioxide truly nonpolar? Why does this make sense?
Be careful with the word nonpolar. It does not mean has no dipole moment. Actually I'm not sure that polar has a precise meaning but it generally means interacts with electric fields. Carbon dioxide interacts with electric fields because it has a quadrupole moment, and also because it is polarisable.
| {
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Why can a regular infrared camera not show temperature (thermography)? There are a lot of questions here dealing with infrared cameras and thermographic cameras. I think I understand the reason why a thermographic camera is able to retrieve the temperature values from any object and convert them to a falsecolor representation, but why is a "regular" infrared camera not able to retrieve this information? What are the differences between these cameras? Is it just the sensor within the camera?
| Partly a definition, thermographic means shows temperature - so any camera that shows temperature is thermographic and any that don't aren't !
To measure temperature a camera needs a couple of features. It needs to be sensitive to a wavelength that the object is emitting. A room temperature object has a peak emission around 10um so an infrared security camera using a Silicon sensor that is only sensitive to 1um isn't going to see much thermal emission from a room temperature object.
To obtain anything like an accurate temperature you also need to measure the amount of infrared energy (the brightness) at more than one wavelength, then by comparing the relative amount of infrared you can estimate the peak of the blackbody curve and so the temperature.
| {
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Converting 1 Kelvin into eV I am using $E=\frac{3}{2} kT$ to find the equivalent of 1 K in eV using $k=1.38\times 10^{-23} \,\mathrm{\frac{J}{K}}$. I get $E= 1.29 \times 10^{-4} \,\mathrm{eV}$, but Wikipedia says it's $8.6\times 10^{-5} \,\mathrm{eV}$. Could anyone tell me what Wikipedia did different then me?
| It is easier to see when you use k in eV per K which is
$$
8.6173303(50)×10^{-5} \ \mathrm{\frac{eV}{K}}
$$
Wikipedia is simply using E = kT
You should always care about your units. Numbers have no physical meaning without them.
| {
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Why does sugar dissolve faster in hot water compared to cold water? Why does sugar dissolve faster in hot water compared to cold water?
| I will explain it with the help of Le Chatelier principle. It states that the rate of reaction depends on the temperature. If the reaction is exothermic, then decrease of temperature will increase the rate to compensate the effect caused by decrease of temperature. However the rate will decrease if the temperature is increased, so that the equilibrium is again obtained. Similarly it happens with endothermic reactions but inversely.
I hope I was able to answer your question.
| {
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Buoyancy Force and Density I am doing a practice problem about buoyancy force. If there are two objects, (one made of wood and one made of metal), and we hold them underwater, will the buoyancy force of both objects stay the same? That's what I believe at least.
| Yes the buoyant force is the same. The buoyant force is defined as the weight of the water displaced or $\rho V g$, where $\rho$ is the fluid density and $V$ is the volume displaced. Because the amount of water displaced is the same for each object (I assume you are stating that they have the same volume), the buoyant force is the same for both.
| {
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Why does Newton's Third Law actually work? My father explained to me how rockets work and he told me that Newton's Third Law of motion worked here. I asked him why it works and he didn't answer. I have wasted over a week thinking about this problem and now I am giving up.
Can anyone explain why Newton's Third Law works?
For reference, Newton's third law:
To every action there is always opposed an equal reaction: or the
mutual actions of two bodies upon each other are always equal, and
directed to contrary parts.
| Newton's Third Law works because the universe tries to be fair. If you push against something it makes no sense not for it to push back against you. Your hand pushes on the table, and the table pushes back just as hard against your hand. If it didn't push back, your hand would go straight through the table. The world would literally fall apart without that law.
Things do go through each other. A swimmer goes through water; you walk through the air all the time. But in both cases, things are literally bouncing off you. Air molecules bounce off your body as you move, and water molecules bounce off your body as you swim. Both the air and the water push back against you just as hard as you push them. That's why you feel resistance when trying to walk against the wind, or why it's a lot harder to run in water the deeper it is.
| {
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Does phononic "Superconductivity" exist? Can solids be cooled to a sufficiently low temperature such that sound can travel through them without acoustic impedance?
It seems that if I had a solid in the shape of a torus, sitting in ambient vacuum, cooled to absolute zero, and then decided to introduce a compression along a circular slice (rest of the solid still at 0), then this wave should propagate indefintely looping around the torus.
Can this behavior occur at $T_c > 0$ if one takes into account Bose-Einstein condensation of phonons?
| I think there is an important difference between electrons and phonons that has to be considered here. The superconducting transition entails the formation of a many-particle state, where electrons attract each other and bind together through phonon-mediated interactions. In effect what is happening is that the motion of the electrons and lattice vibrations interfere coherently to form a quasiparticle that can move freely through the many-body vacuum. At finite temperature the main source of resistance in a metal is electron-phonon scattering, but this source is eliminated when you form a coherent electron-phonon state, so resistance drops basically to zero (although scattering of Cooper pairs by impurities and lattice defects is still possible, and will give a small residual resistivity).
However, there is no equivalent process for phonons. There is no mechanism through which they will interfere coherently with their scatterers (mainly, lattice defects, anharmonic effects, and electrons) to create quasiparticles that propagate freely. As mentioned by user157979, what phonons can do at low temperatures is form a Bose-Einstein condensate, which is a completely different type of object, with different properties.
| {
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Does gravity have anything to do with Van Der Waals forces? Does gravity have anything to do with Van Der Waals forces? Just throwing this out there, I was wondering if they do because gravity is such a weak force and the VdW forces at the molecular level could seem to be a good intermediary force between gravity and the forces acting within atoms. Given that there are so many atoms and molecules within objects like the earth doesn't it seem possible that an extrapolation of the VdW forces could make a good candidate for a theory of gravity?
| No. Van der Waals forces are electromagnetic, not gravitational. "Van der Waals forces" are actually a family of different effects. Dipole-dipole Van der Waals falls off as $1/r^7$ (non-retarded) or $1/r^8$ (retarded). These forces can be understood using only electromagnetism and quantum mechanics. Gravity is unnecessary in their explanation, and gravitational forces fall off as $1/r^2$.
Gravity accelerates all objects equally, whereas the magnitude of Van der Waals forces depends on the polarizability, which varies from material to material. They are not similar and neither can explain the other.
| {
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Why does battery generate less terminal voltage difference when current flows? The terminal voltage difference of a battery means the difference of voltage between the two termials of a battery. Now, a battery has a voltage at the positive terminal and a voltage at the negative terminal. Voltage means the work needed to be done to bring a poistive one coulomb charge from infinity to that terminal. Now, how does the value of this work change when current flows? (The value of this work must change for both of the terminals for the terminal voltage difference to be changed, isn't it?)
I want an intuitive explanation.
| Some reasons the voltage of a battery goes down under load:
*
*Internal resistance. A simplistic first-order model of a battery is a voltage source in series with a resistance. As you draw current, the current causes a voltage across the resistance. That voltage subtracts from the internal voltage source to give you what you get on the terminals.
*Chemical depletion. Starting with the simple model above, add the fact that the actual internal voltage source doesn't put out a constant voltage either. As a battery is used, its chemical energy is used up. This changes the chemistry so that the open-circuit voltage goes down.
*Ion migration bottleneck. At high currents, there is also the effect that the ions can't migrate fast enough to the other electrode. This sortof clogs up the works, also decreasing open-circuit voltage. This is why you can see batteries sometimes have a low open-circuit voltage immediately after sourcing a lot of current, but have that voltage recover at least somewhat over time.
| {
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What is the Planck scale magnetic field strength? Using the constants $\mu_0$ (or $\varepsilon_0$), $c$, $\hbar$, $e$ and $G$, it is possible to define two quantities with units of magnetic field :
\begin{align}
B_1 &= \sqrt{\frac{\mu_0 c^7}{\hbar G^2}} \equiv \sqrt{\frac{c^5}{\varepsilon_0 \hbar G^2}} \approx 8 \times 10^{53} \, \mathrm{T}, \tag{1} \\[12pt]
B_2 &= \frac{c^3}{G e} \approx 3 \times 10^{54} \, \mathrm{T}. \tag{2}
\end{align}
Which one is really the Planck magnetic field?
While $B_2$ is simpler, I suspect it should be $B_1$, because it doesn't use the electric charge unit. $e$ is not exactly as universal as $\mu_0$. $B_1$ uses the Planck constant, so it's consistent to call it a Planck "unit", while $B_2$ doesn't use that constant. Also, because of the square root, $B_1$ is a bit more of the same shape as the Planck length :
\begin{equation}\tag{3}
L_{P} \equiv \sqrt{\frac{\hbar G}{c^3}}.
\end{equation}
The Planck units are presented on wikipedia: https://en.wikipedia.org/wiki/Planck_units but it doesn't tell anything about the magnetic field.
We could also argue that $B_1$ is the answer because we can find it by equating the magnetic field energy density with the Planck density (dropping all the dimensionless constants) :
\begin{equation}
\frac{B_1^2}{2 \mu_0} = \frac{M_P c^2}{L_P^3}.
\end{equation}
But then, we could also find $B_2$ by equating the Planck cyclotron angular frequency with the Planck energy :
\begin{equation}
\hbar \omega_{\text{cyclotron}} \equiv \hbar \, \frac{e B_2}{2 M_P} = M_P c^2.
\end{equation}
Both methods are arbitrary.
So what is the Planck magnetic strength?
| A Planck magnetic field strength is not uniquely defined.
Actually you can construct many of them:
$$B= \frac{c^3}{Ge}\left(\frac{e^2}{4\pi\epsilon_0\hbar c} \right)^n$$
where $n$ is an arbitrary number, and the factor in brackets
is the dimension-less fine-structure constant $\alpha \approx \frac{1}{137}$.
| {
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How can a monochromatic X-Ray tube produce a spectrum in XPS? I thought that if we use monochromatic source, we can only get one peak, if it exists.
Because Photoelectric Effect allows only the electron that have the corresponding frequency(energy level) that can be excited.
appreciate your help to rectify my understanding.
| Monochromatic X-rays produce electrons from a multiplicity of orbitals,
producing a range of electron kinetic energies. One can analyze the outgoing electron
energy to produce a spectrum. It's a spectrum of Xray photon energy
minus binding energy of the electron, so for a fixed Xray energy, it's
the electron binding energy that is represented.
Over a small range of energies, you can give monochromatic excitation and analyze the output electrons' individual energy. Over a larger range, the
'monochromatic excitation' from a synchrotron source can be swept through
a range of frequencies. There is a significant improvement in Xray
absorption efficiency for electron binding energies at or slightly above the
incident Xray energy, so sweeping the Xray source energy is useful.
| {
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Expression for angular friction Consider say a door rotating about its axis. Is there, in general, any expression for the frictional hindrance to its motion? I was thinking in line with the coefficient of friction for linear motion on a surface.
| Not in terms of dry (coulomb) friction. But in terms of damping, relating angular speed to torque $$\tau = c_\theta \,\omega$$
where $c_\theta : [\rm \frac{N m}{rad/s}]$
The source of this damping is a thin lubrication layer and the corresponding shearing of the lubricant. It is easy to estimate the damping coefficient from the geometry and the viscosity of the fluid.
| {
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If someone were to put really small objects 10x spaced on a background, would they see the objects or the background? If someone were to put a sheet full of 1 micron x 1 micron black squares as a grid on a white piece of paper, and spaced them 10 microns apart (up-down, left-right, obviously diagonal doesn't count), would someone looking at it from a normal distance see black or white? If white, would they be able to see the black at all?
| What you describe is identical to the process called halftoning, which is used in print to give the illusion of greys with only black and white, or the illusion of full color with only 4 pigments. You describe a pattern which is 1% covered with black cubes, and 99% open space, so it would appear to be the "1% black" color of grey.
It's pretty hard to distinguish 1% black from 0% black, unless you have some reference, so people most likely would not see it at all. If you coated half of the paper this way, and left half uncovered, most people could distinguish the covered half as a faint grey.
| {
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Definition of symmetrically ordered operator for multi-mode case? As I know, Wigner function is useful for evaluating the expectation value of an operator. But first you have to write it in a symmetrically ordered form. For example:
$$a^\dagger a = \frac{a^\dagger a + a a^\dagger -1}{2}$$
For single mode case where there is only one pair of creation and destroy operator the symmetrically ordered operator is defined. But for multi-mode case,how is it defined? For example, how would we write
$$a_1^\dagger a_1 a_2^\dagger a_2$$
in a symmetrically ordered form (such that we could easily evaluate its expectation value using Wigner function)?
| A quasiprobability distribution depends on the symbol/operator ordering prescription. E.g.:
*
*For Weyl/symmetric ordering of Hermitian operators, one uses the Wigner quasiprobability distribution.
*For Wick/normal ordering of creation & annihilation operators, one uses the Glauber–Sudarshan $P$ representation.
*For anti-normal ordering of creation & annihilation operators, one uses the Husimi $Q$ representation.
| {
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How does a man inside bubble ball accelerate without an external force? As Newton's Laws states an object should be in rest or in constant velocity if no external force is applied. A man inside a stopped car cannot push the car as he is not giving any external force.But a man inside a bubble ball can make it move. What is the theory behind this?
https://www.holleyweb.com/images/human_sized_hamster_ball_free_walking.jpg
| The simple answer is that there is an external force, that of friction towards the ground. The man and the ball can thus change their velocity by using the friction provided by the ground, much in the same way we can walk by using the frictions on the ground.
An important follow up question, that the OP touches upon, is how the ball is actually made to move. Here again friction plays an important role by applying different amount of force in different direction a net movement can be achieved in any particular direction.
As an example, if the man decides to jump forward inside the ball, that will make the ball move forward. The reason is that when jumping forward the friction on the ground stops the ball from moving backwards, but when he lands, he transfers his momentum to the ball, which begins to roll.
| {
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Conservation of energy problem with water hose? When we squeeze the end of the pipe while watering plants , the water speeds up , but the amount of water being released is the same ? From where does this excess kinetic energy come from ?
| Since mass=density×volume, m=dAx
Since the kinetic energy remains conserved
K.E=1/2mv^2=1/2(DAx)v^2.
As K.E is constant, as Area decreases velocity must increase.
A is area, v is velocity and x is length of water column.
| {
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How could I measure the colour spectrum of a light bulb and investigate how closely it matches a black body radiation curve? Here is my research question:
What is the colour/spectrum produced by each globe type? What is the temperature equivalence? How closely does a globe match a black body radiation curve?
I will be testing this on halogen, compact fluorescent, and LED bulbs. How could I measure the colour specturm of the bulb? I have been suggested to take photos of the glowing bulb then use photoshop to analyse the colour. Is that a possible solution?
How could I match it to a black body radiation curve? Would I have to plot it out? From research it doesn't seem like a curve that could be hand-drawn, especially CFL and LED:
| You will need to use a spectrometer to measure the irradiance at various wavelengths. A camera will only measure three wavebands (R/G/B) so this will not give you the necessary resolution to make a meaningful comparison of spectral composition.
| {
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Observing Two-Beam Interference at Home I want to know how difficult it would be for me to observe two-beam interference at home.
I have:
*
*A laser pointer.
*A non-polarizing beam-splitter.
*A mirror.
*Two concave lenses.
*An uneven shaky floor, some chairs, and tape.
*Patience that spans an entire day.
This is the sketch of the setup that I have in my mind:
where the laser, lenses, and so on are taped to the chairs.
How close do the two paths lengths have to be to each other? In my case, the two path lengths will differ by dozens of centimeters.
The laser spot from the laser pointer is not even uniform. Will that be an issue?
Are there any other ways this could go wrong? Is there any advice on how I could observe two-beam interference?
| The setup you describe is similar to the one used in creating holograms. All you would need to add would be the beam spreaders.
Unfortunately, this demands a level of stillness which is hard to achieve without a properly isolated optical bench. Red lasers are roughly 650nm, and if you want to see effects, you're going to want the path length to be stable down to 1/10th of that or better (1/10th being a reasonable heuristic). That kind of precision is not found with a few chairs. To do this, holographers set up an optical bench that is properly isolated. A DIY solution which has been recommended to novice holographers is to get an innertube and put a heavy piece of granite on top (sometimes the places which cut granite counters will have leftovers that you can get!). The heavier the better! Anchor everything to that piece of granite. Then, turn off your air conditioner and wait a bit -- the differences in air densities will affect holograms, so it will affect your measurements as well.
| {
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Different values for same integral calculated via contour integration Perhaps this question should be asked on the Math community, but I think that, because of the specific example that I will cite, I might get a better insight on my doubt here.
So, why can we get two different results (mathematically) according to whether we choose to circumvent or to contour a pole when calculating an integral? The most obvious example being the Feynman prescription on QFT and related calculations.
I know that most integrals to which contour integration techniques are useful doesn't exist and, in these cases, what we do is 'just' a way of giving them a quantitative meaning (call it the Cauchy's Principal Value, if you want), however still can't understand intuitively or formally what causes the difference on the values calculated from different choices of contour.
| This is simply a consequence of the Cauchy Goursat Theorem and Residue Theorem: the first tells you that an integral is invariant under an homotopy of a contour unless the the region between a contour and its homotopic image includes a new singularity; the second tells you that the integral changes by $2\,\pi\,i$ times the residue of any new pole that becomes included under an homotopy as long as the contour does not contain branch points or cross brach cuts.
The intermediate case, that leads to the Cauchy principal value, is where the contour goes exactly through the pole; this behavior can be understood by "dinting" the contour inwards by a semicircular dint of radius $\epsilon$ to avoid the pole and then evaluating the limit of the explicit calculation of the contour on this semicircle as $\epsilon\to0$.
As to the physical meaning of all these variations, one has to look in detail at the physical assumptions underlying the operations on a case-by-case basis to determine this meaning.
| {
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Shouldn't work be the same in all coordinates? We know that the work done by a force $\mathbf{F}$, along a path $\mathbf{x}$, is given by:
\begin{equation} W = \mathbf{F}^T \cdot \mathbf{x}
\end{equation}
However, suppose that i apply some change of basis, given by a matrix $A$. So, $\mathbf{F}$ will become $A\mathbf{F}$ and $\mathbf{x}$ will become $A\mathbf{x}$. And so
$$W = (A\mathbf{F})^T \cdot A\mathbf{x} = \mathbf{F}^TA^TA\mathbf{x}$$
Which may not be equal to $\mathbf{F} \cdot \mathbf{x}$. What am i missing? If the path and the force are both the same, shouldn't the work be the same in both cases, no matter what basis am i using for $\Bbb{R}^3$? I am supposing that $A$ is not necessarily orthonormal.
| Let us write the dot product like this:
$$\vec F^T \vec x$$
where the $T$ means transpose. If we now apply the change of coordinates we get:
$$\vec F'^T \vec x'=(A\vec F)^T A\vec x$$
$$=\vec F^T A^T A \vec x$$
Now a change of coordinates matrix must be orthogonal (in this case) so
$$A^TA=I$$
Hence we get:
$$\vec F'^T \vec x'=\vec F^TI \vec x=\vec F^T \vec x$$
which is coordinate independent.
EDIT
Sorry I missed the statement about orthogonal matrices in the question. The point is that we actually only expect the work done to remain the same under orthogonal transformations. Orthogonal transformations correspond to rotations (and reflections) under which we do not on physical grounds expect the work to change. If the matrix is not orthogonal we then are doing things like stretches - these changes units and with the same scalar product we do not expect to get the same answer.
As a side note as ACuriousMind stated in the answers to one of my questions a proper calculation could be done but this would involve a change in the scalar product.
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How does one find the wavefunction of a particle in its rest frame? In classical mechanics, the orbital angular momentum of a particle is defined as $\textbf{L}=\textbf{r}\times\textbf{p}$. This is zero in the rest frame of the particle where $\textbf{p}=0$.
Quantum mechanically, $\textbf{p}$ is an operator. So putting $\hat{\textbf{p}}=0$ in $\hat{\textbf{L}}=\hat{\textbf{r}} \times\hat{\textbf{p}}$ and claiming that the orbital angular momentum of a quantum particle is zero in its rest frame does not make sense. One must look at the value of $\hat{\textbf{L}}^2$ on the "wavefunction in the rest frame" of the particle.
How does one find the wavefunction of a particle in its rest frame?
| The rest-frame wavefunction $\psi(\boldsymbol x,t)$ is the one such that
$$
\boldsymbol 0\equiv\langle \boldsymbol p\rangle=\int_{\mathbb R^3}\psi^*(\boldsymbol x,t)(-i\boldsymbol \nabla)\psi(\boldsymbol x,t)\ \mathrm d\boldsymbol x
$$
If $\boldsymbol k\equiv\langle \boldsymbol p\rangle$ is non-zero, we just need to redefine the wave-function:
$$
\psi(\boldsymbol x,t)\to\mathrm e^{-i\boldsymbol k\cdot\boldsymbol x}\psi(\boldsymbol x,t)
$$
which satisfies $\langle\boldsymbol p\rangle\equiv \boldsymbol 0$ by construction. This is just a translation in momentum space,
$$
\tilde\psi(\boldsymbol p,t)\to \tilde\psi(\boldsymbol p-\boldsymbol k,t)
$$
which obviously has zero mean.
More generally, if you have a system of many particles, the rest-frame of the system is, by definition, the one where $\langle\boldsymbol p\rangle\equiv\boldsymbol 0$, where $\boldsymbol p$ denotes the total linear momentum:
$$
\boldsymbol p=\sum_i \boldsymbol p_i
$$
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Can transverse sound waves be polarized? I know that polarization only occurs in transverse waves and polarization of light occurs as EM wave is a transverse wave. But sound waves are both transverse and longitudinal in solids. So can polarization occur for the transverse part? But we cannot stop the sound wave from propagating by any medium except vacuum. Because it will propagate through the stopping medium(like an analyzer but for sound).
Even if it gets polarized somehow(I don't think it can get polarized) then how can we observe it, since any sound reaching our eardrums will be longitudinal as the medium in front of our eardrums will be air, and so no polarization will occur in longitudinal waves.
See the 7th and 8th line in this image(source:- https://en.wikipedia.org/wiki/Polarization_(waves)).
I am a little confused now.
P.S. This may seem as a possible duplicate but all other answers didn't clarify my doubt.
EDIT:- Based on the answers, it seems that shear waves can be polarized. So my question is how to polarize these shear waves?
| These transverse waves are known as S-waves or shear waves and yes they can be polarized. Waves that are polarized in the horizontal direction are known as SH-waves and those in the vertical direction are known as SV-waves. These waves exhibit the property that when they meet a boundary between two mediums (say solid -> air) they can turn into P-waves, which are the more well know compression waves.
When they are polarized as SH-waves it means that the particle motion in the solid is contained in the horizontal plane, and the same goes for SV-waves except in the vertical plane.
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Traveling Wave Equation $\sin(kx-wt)$ vs $\sin (wt-kx)$ In my textbook most if the times it uses $A\sin(wt-kx)$, but occasionally there is a problem using $A\sin(kx-wt)$
So i just changed it from $A\sin(kx-wt) \to -A\sin(wt-kx)$
but does the amplitude change to $-A$?
Is the wave going downwards first?(as negative amplitude would imply)
|
Is the wave going downwards first?
That all depends an what you mean by downwards and what type of change you're interested in.
Let's investigate "downwards": The wave solution describes the disturbance of some quantity from its equilibrium solution. It may be a displacement of a section of string from its rest position (say, in a violin or cello or bass). What is "downwards" for the bass? The string is almost vertical.
Or the wave could be a pressure wave in some medium. In that case, molecules are being condensed and rarified along the path of transmission. How is downward defined? Actually, downward is irrelevant.
I believe I understand why you say "downward" and that is because of a routine association between the variable $y$ and the local gravitational pull direction. That is an association that you need to break. The variable $y$ can represent anything, and for the standard wave solution it is merely the disturbance about an equilibrium position.
Now, on to the type of change. You could investigate how the wave disturbance from equilibrium, $y$, changes as you move along the direction of travel of the wave (the $x$ direction) for a given instant of time, $t$, OR you could investigate how $y$ changes for a particular point, $x$, as the time moves forward ("time keeps on slippin' into the future" according to Steve Miller Band).
Let's consider $y=A\sin(kx-\omega t)$ as our starting convention. If we take a time snapshot, start at $x=0$ and slide along the $+x$ direction, the disturbance initially becomes larger in the (pre)defined positive $y$ direction, then it turns around and goes back in the negative $y$ direction. On the other hand, if we choose a particular $x$ location and monitor what happens with time something different will happen. Consider watching $x=0$, our previous starting point. As time progresses, that point will move in the $-y$ direction, reach a maximum negative and turn around and move back toward equilibrium. Again, downward is irrelevant. And the choice of positive $y$ direction is arbitrary.
The importance of the sign of $y$ only shows up when one makes an association with a physical system.
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Does amplitude affect time period for spring mass system? I know that with the formula $T=2\pi\sqrt{\frac{m}{k}}$ the time period is not related to the amplitude. However, would amplitude matter if i do this experiment in real life. Would a greater amplitude result in more friction of some sort?
| No, the amplitud does not affect to the period.
For a spring-mass system:
if you solve the differential equation $$m\ddot x +kx=0$$ you get a solution that looks like this
$$x(t)=A_0 \cos (\omega t -\delta)$$
Where both the amplitude ($A_0$) and the phase angle ($\delta$) are in fact arbitrary constants which could be any real number, but the frequency $\omega $ is determined by the mass and the stiffness constant of the spring (or its equivalent in other mechanic system).
And similarly to the damped oscillator.
At the moment this model works well enough to predict the results of all mechanics systems which performs small oscillations (like a pendulum, or a string), so an experiment in real life should behave as the theory says.
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How many dimensions are there in the electric field? I am not a physicist. I am buying some polariser for my camera. Circular polariser intrigues me. Basically you pass light through a linear polariser, then through a waveplate, you get circular polarisation.
Wikipedia says the following:
By adjusting the thickness of the wave plate one can control how much the horizontal component is delayed relative to vertical component before the light leaves the wave plate and they begin again to travel at the same speed.
Does that mean the electric field in an electromagnetic wave is a 2D vector field? I am a bit confused. I thought in 3D space, the electric field should be 3D vector field.
| Electromagnetic waves such as visible light are transverse waves. This means that the direction of oscillations of the fields is perpendicular to the direction of propagation of the wave. In general we can decompose the electrical field into two orthogonal components along the oscillation plane and the relative phase between these components gives rise to different polarizations. For example, a linearly polarized electromagnetic wave propagating along the $z$ direction has the following form
Note how the electric field (as well as the magnetic field) is fully contained to a plane.
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Does a glass of water at room temperature emit (infrared?) radiation While reading the introduction to Feynman's lectures, it's mentioned how a glass of water cools down through evaporation, when some molecules get a bit extra energy and break free. If it's not a closed system, energy will be gradually taken away from the cup, hence blowing at the soup helps move those molecules away so that they don't reenter the surface.
But I thought that all bodies also radiate heat? Does a cup of water also emit low frequency radiation, or is my understanding incorrect?
| Yes, all matter above absolute zero emits radiation. To quote wiki:
When the temperature of a body is greater than absolute zero,
inter-atomic collisions cause the kinetic energy of the atoms or
molecules to change. This results in charge-acceleration and/or dipole
oscillation which produces electromagnetic radiation, and the wide
spectrum of radiation reflects the wide spectrum of energies and
accelerations that occur even at a single temperature.
This continuous release of energy would eventually cool the source to a lower and lower temperature except your glass of water is in contact with a heat reservoir (the room) which compensates for the energy loss.
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Why do objects feel heavier when held with an extended arm than with a bent arm? I do realise that this is due to torque and that torque is at maximum when the angle between the direction of the torque and the force acting on the object is 90 degrees. I would like to know if it is the angle between the forearm and the elbow that causes the situation I described in the question or whether if it's due to something else.
| You should always choose to calculate the torque with respect to the axis of rotation in each case.
In one of the cases, that point is your elbow and so the distance is elbow-hand, while in the other case it is the distance shoulder-hand=whoele arm.
Evidently a longer distance makes greater torque, and so more propensity to turn around.
Evidently, the rest of your body would turn around in the opposite direction if there weren't other forces. The thing is that your body makes forces in the appropiate points so that you stay stood up, but the nature of these forces is not relevant here. Just care about the torque produced by the weight in your hand.
| {
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Total eclipses becoming rare soon? Given that the moon is receding from the earth and that the average angular size of the moon is approximately equal to the angular size of the sun at the moment, what is the current rate at which total eclipses of the sun are decreasing? ( let's say frequency rate in total eclipses / 100 years )
| In about 620Myr
The moon is receding at around 3,8 cm/yr and eclipses will stop (ie the shadow won't reach the Earth) when it has moved another 23500km
| {
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What causes the phenomena of a ring appearing around the shadow of an airplane So I was sitting in an airplane and I saw a ring appearing around the shadow of the plane on the clouds. What causes this phenomena? I've added an edited image, so the effect is more pronounced.
| This phenomenon is called a Glory. It is caused by back scattering from individual water droplets. There is a lot more detail to be found on this site: http://www.atoptics.co.uk/droplets/gloab.htm
Wikipedia also has an article on it, which claims that the physics is not fully understood.
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How do magnetic cores (ferrites) guide magnetic field? As far as I know, magnetic materials have magnetic dipole moments which align when they are under influence of the outside magnetic field.
Basically they increase the magnetic field strength. But people also say that they guide magnetic field. Does the magnetic field strength decrease around them compared to the state before they were introduced near a coil? Do they really guide magnetic field, or just increase the magnetic field inside them so that magnetic field around them seems small, even though it hasn't decreased?
| A ferrous metal will attract the magnetic field lines around it to pass through it. Thus, it will decrease the density of magnetic field lines outside the magnet and increase the same inside itself.
So, yes the magnetic field strength decreases in area where a ferrous metal is introduced.
Ferrous metals attract the magnetic lines and cause the ones near it to pass through it. this is what some people refer to as guiding magnetic field.
Overall the magnetic field remains the same, but it is in the precise environment of the ferrous metal introduced, that one can see the changes.
The behavior will depend on permeability of the ferrous metal and the intensity of the external magnetic field.
| {
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Does work done depend on the frame of reference? Suppose I am sitting on a bench and looking at a moving car. Force is applied on the car by its engine, and it makes it displace, hence some work is done on the car. But what if I am sitting in the car and looking at the bench? The bench covers some displacement, but who has applied force to it? Is any work done on it?
| Let me address your scenario first. What you suggested is bad physics because it breaks the conservation of energy. It doesn't work because the car is not an inertial frame. An inertial frame is arguably defined as one in which newton's laws work. We know that the bench frame is inertial because momentum is conserved as the car accelerates. The car frame is not inertial so it is meaningless to apply newton's laws and subsequently, energy conservation (unless you introduce fictitious forces, but those are limited to only a few scenarios). If this makes the definitions and statements in newton's laws seem circular, it's because they are. They empirically work relative to the fixed stars, (and any galilean shift of that) and one can show that they are true for frames purely accelerating due to gravity, as in the earth, so they are very useful anyway.
Now to answer the question you literally asked. Momentum change of a particle is conserved between inertial reference frames. However, energy change of that particle across a chosen path is not preserved, once you apply the appropriate transformations. The work done on a closed system however, is zero in any inertial reference frame, this is the conservation of energy and it is all that is needed to allow the frames to agree with each other.
For instance, think of an elastic collision between two particles with equal and opposite velocities magnitude $v$, in frame A, the work done on each particle is $mv^2$. Now choose a frame with a particle at rest, the work done is now $2mv^2$, the energy change of the system in both frames is 0.
| {
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How much power in a lightning strike? The Biblical story of 1 Kings 18:20-40 relates how a lightning strike ignited a bulls carcass, as well as a wood pyre, both of which had been thoroughly wetted with abundant water. Is there enough power in a lightning strike to do this? Does the power of a mountain top lightning strike differ from the power of a lightning strike elsewhere?
| In this link the following estimates are given
From articles in Windpower Engineering & Development, we learn that lightning bolts carry from 5 kA to 200 kA and voltages vary from 40 kV to 120 kV.
Here we find an average estimate :
An average bolt of lightning, striking from cloud to ground, contains roughly one billion (1,000,000,000) joules of energy. This is no small amount, enough to power a 60-watt lightbulb for six months plus a forgotten open door refrigerator for a day. In the forms of electricity, light, heat and thunder, this energy is all released by the flash in a matter of milli- or even microseconds.
| {
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Is a lightening strike deterministic? This relates to a previous question: Is a dice roll deterministic?
Essentially, I'm trying to get a better understanding of how quantum indeterminacy interacts with the macro world.
It seems to me that lightening involves charge, and might be subject to quantum indeterminacy in terms of how the lightening expresses itself as a bolt, but that determination is well beyond my knowledge and expertise.
| Quantum effects appear at the microscopic level, or at highly organized solids and fluids (superconductivity, superfluidity for example). This is due to the very small value of the Planck constant in the Heisenberg uncertainty, which is obeyed automatically macroscopically down to micron levels.
Lightning strikes follow classical electromagnetism and thermodynamics. It is potential differences and the phases of matter as energy is released and the air heated to high temperatures. Lightning has temperatures of plasma as discussed in the answer here .
There is an indeterminacy coming from the boundary conditions for the classical electromagnetic + thermodynamic solution, coming from the classical statistical mechanics of the ensemble of atoms on the way between the high potential of the cloud and the earth, but it is classical indeterminacy, not quantum.
| {
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With respect to what we are saying space is homogeneous or space-time is isotropic? I don't really understand what we are talking about when we say space is homogeneous. What we are measuring? My notion is: it should depend on the entity and with respect to that entity one can decide space is homogeneous or not!
May be I'm asking stupid question, actually the fact is I don't understand the phrase "space is homogeneous" or "space-time is isotropic"
| You are correct in the sense that there are physical quantities, observables, like scalar and vector fields that are, in general, varying in space.
However, that's not what the phrase "spacetime is isotropic" is meant to convey. What this means is that, in the absence of anything to break the symmetry, each point in spacetime is no different than any other point in spacetime.
Without reference to anything else, spacetime is considered as homogenous everywhere.
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Do charges move to the outer surface of a conductor to minimize the potential energy? We can think the charges go to the outer surface of a conductor to minimize the electrostatic potential energy of the system. We can check this using a simple calculation using a charged sphere.
A uniformly charged sphere would have $20\,$% more energy than that of a very thin spherical shell with the same radius and the same charge as in the former case. As the potential energy would be less if we distribute charges uniformly over the surface rather than distributing the charges uniformly throughout, we can say, the conductor prefers having charges on the outer surface. Here's my question then:
Suppose I have a conducting sphere on which a charge $Q$ is to be distributed. Now, suppose I divide the charge into two equal parts, $Q/2$ each, and place them at two diametrically opposite ends of the sphere. Now the potential energy of the configuration becomes $75\,$% less than that for charges distributed uniformly over the surface. Then why doesn't the conductor prefer this kind of distribution which can minimize the potential energy further?
| If you put two equal point charges, say two nuclei with $+|e|Z$, at the opposite points of the conducting sphere, they will stay there. The free electrons of the conductor will start to move due to attraction to these charges until the friction losses (=resistance) dissipate the extra energy into heat. Otherwise there will be current oscillations dissipating due to radiative losses.
In case of negatively charged nuclei the situation will be similar - the electrostatic forces will move free charges until the minimum of the potential energy is reached, the extra energy being lost to heating or to something else.
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How does one obtain observables from a wave function? I'm beginning to study the quantum chemistry (my background is computer science and computational mathematics) and I'm not sure if I understand well the basic concepts, like wave function and operators.
I've read questions
What is a wave function in simple language?
Use of Operators in Quantum Mechanics,
and I understood that the wave function describes all the observable quantities of a system, e.g. an electron.
But what should I do when I want to obtain some of these quantities? Let's say I want to obtain spin and momentum. If I use the linearity property of quantum operators, $\hat{O}$ being the operator, $o$ its eigenvalue and $\psi$ the wave function, like this
$$\hat{O}\psi = o\psi$$
does $o$ always equal the observable value? In other words, is it enough to use the above equation and express the $o$ variable to get an arbitrary observable value?
| $\hat{O}$ will have a set of eigenvectors $\phi_i$ and eigenvalues $o_i$.
You can expand your wave-function $\psi$ in terms of the basis set $\phi_i$, i.e. $\psi=\sum_i c_i\phi_i$.
If you make a single measurement then your wave-function $\psi$ will collapse to one of the eigenvectors $\phi_i$ with probability $|c_i|^2$ in which case your measurement is the corresponding eigenvalue $o_i$.
In many cases we are interested in the average rather than a single measurement, in which case you evaluate
$$ \langle\psi|\hat{O}|\psi\rangle = \sum_i |c_i|^2o_i $$
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Thermodynamics : Doubt in First Law I just started my study of thermodynamics. I am having problem in uderstanding a concept.
Suppose I drop an ice-cube and a rock of cube shape. Which one will reach the ground first? I know ratio of Work to Heat equals Joules Constant. Therefore any change in potential energy will result in heat which in return will melt ice cube thus mass is changing. Will internal energy of ice cube change till it reaches the ground? If mass of ice cube is changing what will be the ratio of kinectic energy of ice cube to that of rock.
| The falling process can be quite complex when we come to answer your with one word: stone or water.
When an object falls, it will be subjected to the following two forces: gravity force and air drag force. If the air drag force can be ignored, the gravity force acts on the object converting the potential energy to a kinetic energy. The answer can easily be that they reach the ground at the same time.
When there is viscous force, the object shape and its surface viscous force play an important role (e.g. a feature vs. a stone of the same mass). The viscous force is a function of velocity. If they have the same shape, then they reaches the ground at the same time.
Viscous force does work that converts kinetic energy to heat. This heat will heat up air in vicinity and ice. It can melt the ice cube. The water can change shape easily. With the shape change, the ice will have less drag and reach the ground first.
Viscous force can also shear off some of the water, whereas it would not tear off any material of the stone. This will make the ice smaller and less drag. Ice will hit the ground first.
The ice may completely evaporate before it reaches ground. In this case, a rock will hit the ground first.
Now, back to the first law, there are components of potential energy, kinetic energy and internal energy. But there is also an important component: work. Nothing can be ignored and all stone should be turned.
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Closed container buoyant force A closed container of water contains a Styrofoam block attached to the bottom of the container by a massless string. When the system is accelerated upwards, what happens to the tension in the string?
Background: I am an MCAT Physics teacher and this question was on an MCAT. Assuming the water is compressible, Newton's first law would indicate that a higher than normal pressure would result at the bottom of the container and a lower than normal pressure would result at the top of the container. The net result would be an increase in buoyant force (force due to a pressure gradient); thus, increasing the tension. The question I have is, "If the water were considered to be incompressible, would the tension still increase?"
| Buoyancy does not depend on compressibility. The variation of pressure with depth is linked to the weight of a cylinder of liquid having cross-section $A$ extending from depth $y=-D$ to the surface $y=0$. The density is $\rho(y)$ and the pressure is $p(y)$.This cylinder has a weight $$W=gA\int_{-D}^0\rho(y)dy$$ that must be balanced by the difference between the pressure forces on its two ends $A(p(-D)-p(0))$. this gives
$$p(-D)=p(0)+g\int_{-D}^0\rho(y)dy$$
and hence $$\frac{dp}{dy}=-g\rho(y)$$
For any real fluid, we can take $\rho$ to be a known function of $p$ and integrate this equation to find $p(y), \rho(y)$. A common model for the atmosphere follows from assuming an isothermal relationship.
However, to investigate a hypothetical fluid that is totally incompressible, we can just set $\rho$ to an arbitrary constant $\rho_0$
so that $dp/dy=-g\rho_0$ and the buoyancy of any object having volume $V$ and totally submerged at any depth is $\rho_0gV$.
You can check this question for information on the actual changes in ocean density from surface to deepest deeps. The change in pressure is a factor of about 1,000. The change in density is less than 5%. If the density actually changed as much as the pressure does, it would not be possible for ocean exploration vehicles to dive as they do.
They could not overcome the buoyancy force.
ADDITION An answer that has now been deleted stated correctly that the effect of accelerating the system is merely to increase the effective value of gravitational acceleration. Therefore both mass and buoyancy forces increase in the same ratio.
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Nature of Work done by Forces I am confused about the possible nature of work that a conservative and non conservative force can do.
*
*Do Non conservative forces like friction only do negative or zero work and not positive?
*Can conservative force also do negative work, I know Gravitational Force is conservative force and it can do negative work, but I read somewhere that net work done by conservative forces is positive.
Can someone please clarify?
| Yes force of friction can do positive work
Work done by conservative forces may be positive.negative or zero
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Time dilation for non-physicists Apologies in advance, as I'm not a physicist, and may use terms incorrectly.
In the movie Interstellar, the planet Miller has a time dilation of one hour to seven Earth years. This has brought up several questions for me:
*
*At what point would someone (outside the gravitational force surrounding Miller) begin experiencing the time dilation?
*How long would it take to send data from the surface to said someone in space outside Miller?
*Would a one hour audio clip recorded on Miller take seven years to play once transmitted?
| It was a fictional story. It was not well researched. According to Newtonian physics, it would take about a year to accelerate to the speed of light at 1G. If the movie were more realistic, then once you're that deep in a gravitational potential well, it would probably take many years of apparent time on the ship to accelerate to a spot where the gravitational time dilation factor is only 2. I think neutron degenerate matter has the highest ratio of bulk modulus to density of any material. I think that a neutron star at its tipping point just barely large enough to begin the runaway effect towards collapsing into a black hole doesn't have enough gravitational time dilation at it surface for 23 years to be perceived as 2 hours on the surface.
| {
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How can the presence of a solvent be included in a polymer bead-spring model? I have a bead-spring model of a polymer melt and I would like to simulate what happens when a solvent is added to the system.
The most direct way to do this would be to add solvent particles into the system. But my simulations already take enough time so I would like to avoid this approach.
I think the best approach would be to change the interaction energies of the system. However, I'm not sure how I should change the interaction parameters to correctly represent the presence of a solvent.
Is this the correct approach? How can I determine appropriate parameters to mimic my solvent?
My system uses a 6-12 LJ potential and a FENE bonding potential.
| I assume you are currently running a deterministic simulation which explicitly computes the time evolution of all degrees of freedom using Newton's equations. In order to implicitly simulate a solvent, what you want instead is a simulation that computes the time evolution of only a subset of relevant degrees of freedom (i.e. those of your polymers). It turns out that the evolution of this subset can be described using the stochastic Langevin equation. Instead of explicitly simulating solvent particles, it applies noise to your polymers that mimics random collisions with solvent particles.
For a very detailed description of why this is the case, see section 2 of Roux et al..
| {
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Why is unit of pressure (psi) used to determine things like bite force of an animal? Whether in tv documentaries or journals, whenever they talk about an animal's bite force, it's measured in PSI anytime imperial units are used (ex: National Geographic, NIH Journal). Many even seem to highlight the fact that it's pounds per square inch.
But when it's used in SI unit context (ex: BBC Earth), it's written in Newtons, a measurement of force.
It seems to me that a force should be measured in a unit of force and not pressure. Additionally, pressure alone would seem to tell us nothing since area is never specified, thus unknown force.
So why is a unit of pressure so dominantly used to indicate an animal's force?
For example, this science daily shows the croc having a bite force of 3700lbs. But then other sources (along with every other web search) will say 3700psi. This National Geographic article even writes psi side by side with newtons as if it was pound-force.
| Thank you for asking this question! You are absolutely correct. Bite force (any force) should be reported in units of pounds-force or Newtons. Better yet, I would model the jaw as a hinge and report the torque around the hinge since the force would likely be higher closer to the back of the jaw. I suspect when bite forces are measured, the analyst is reporting the result from a pressure gage directly without multiplying by the area of the sensor. I see this all the time in mechanical testing, especially if a hydraulic-based sensor is used, but it sure would be nice to understand this!
| {
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What do they mean when they say that it does not require any work to move a charge from one point to another in an equipotential surface? In the textbook it says that no work is required to move a charge from one point to another on an equipotential surface. Do they mean work by the electric field or work by anything? Because clearly the object cant just magically move sideways with nothing.
| "it takes no work" in the same sense it takes no work to move an object on a perfectly frictionless, flat surface. It is true and theory, but moving an object requires accelerating it at least a little bit, which requires some work, as you point out.
| {
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Direction of friction in complicated physics problems I want to ask how to determine direction of friction in complicated mechanical problems (for example 5 masses each over other and 3 forces on these masses in 3 different directions and friction between every masses)?
Is there a absolute way to find direction of friction or I have to calculate it each time for each different case ?
| If the block is moving, the direction of the frictional force is in the opposite of the relative motion of the block compared to the frictional surface.
If on the other hand it is a static situation the frictional force is in the direction opposite to the direction that the blocks would move if it were a friction-less system.
| {
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Why does a calculation to count objects covering a certain area seem to give nonsensical units? Suppose you want to estimate the number of atoms in a rectangular sheet of graphene. You might estimate the sheet to have $10^{7}$ atoms along one edge and $2*10^{7}$ atoms along the other edge. Multiplying while keeping track of units, we get
$$10^{7}\text{atoms} * 2*10^{7} \text{atoms} = 2*10^{14} \text{atoms}^2$$
But obviously, there are $2*10^{14}$ atoms, not $2*10^{14} \text{atoms}^2$. What is wrong with this calculation?
| What is wrong with your calculation, is that you substituted a "count" for a dimension.
The correct way to do it, would be to determine the length of $10^7$ atoms. Assuming they take 1cm, then $1cm^2$ would have $1x10^{14}$ atoms, and the sheet (1 cm by 2 cm, or 2 $cm^2$) would contain ($1x10^{14}$ atoms/$cm^2$ x 2 $cm^2$ =) $2x10^{14}$ atoms.
| {
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Lorentz Transformations Vs Coordinate Transformations I'm really confused about Lorentz transformations at the moment. In most books on QFT, Special Relativity or Electrodynamics, people talk about Lorentz transformations as some kind of special coordinate transformation that leaves the metric invariant and then they define what they call the Lorentz scalars. But from my point of view (which is somehow grounded in a background from differential geometry), scalars and the metric, which is a tensor, are invariant under any "good" coordinate transformation and that's a lot more than just Lorentz transformations, so I don't see why there's a special role for the Lorentz transformations in special relativity. Saying that the metric is invariant under Lorentz transformations is non-sense to me, because indeed it should be under any type of coordinate transformation if it's a well defined metric on a Minkowski manifold.
It seems to me that Lorentz transformations should be relating observers (frames) and not coordinate systems - that would make more sense to me, but usually people mix both concepts as if they were exactly the same. I'd like to understand what it means when one says that some scalar is Lorentz invariant. If someone could clarify me this conceptual confusion, I would be really grateful.
| Looking at your question i guess i have a simple answer.if any two observers are in a specific type of coordinate frame (cartesian polar.....e.t.c)and they want to know the energy momentum position and velocity then they will use lorentz transform to find out each others position velocity energy momentum.but if one observer is in cartesian frame and the other is in polar then they must take also the coordinate transform too from polar to cartesian or vice versa.we often confuse between coordinate frame and reference frame.there is a subtle difference.coordinate frame such as cartesian polar cylindical system.but reference frame is observers frame.we can quantify reference frame by using any type of coordinate frame
| {
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What exactly happens in the basic QED Feynman diagram? When a photon is absorbed by an electron, I think that the following things happen:
*
*The electron changes in momentum, angular momentum and energy.
*The phase of the electron wave function changes by a fixed angle given by the coupling constant.
Is this correct? Is this complete?
| As for the first point, any of the quantities by which the electron is characterized are changed due to conservation laws, which are the 4-momentum conservation law, angular momentum conservation law, charge conservation law. The photon doesn't carry electric charge but it definitely carries the 4-momentum and the angular momentum.
As for the second point, you should note that there are no such meaning as the electron's wave function during the scattering. Within the Feynman perturbation theory, instead of the wave functions we're dealing with the Fock multi-particles states, and Feynman diagram just tells you how to write the amplitude corresponding to transition from two different Fock states. Therefore it isn't completely correct to talk about the wave function's phase. You can, however, talk about it when considering the scattering of the particle in the external field, which is not so strong that particles creation occurs.
Let me also clarify the SuperCiocia statement about the Feynman diagrams. In order to talk about the physical intermediate states during the scatterings (including also your example) you should use the so-called old-fashioned perturbation theory (you can find information about it by googling), in which the time (and therefore the energy) are initially "dedicated". The sum of all of the possible intermediate states is possible to be rewritten in Lorentz covariant form, which is known as Feynman perturbation theory. In this form the time is of course not "dedicated", and therefore there is no meaning of intermediate states. We introduce the propagator instead, and it corresponds to something like the particle travelling forward in time "plus" particle travelling backward in time.
| {
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Sufficient conditions for a mapping to be canonical in Hamiltonian Mechanics My professor mentioned:
A simple way of testing whether a mapping $(q,p)$ to $(Q,P)$ is canonical is by examining:
$$P · dQ − p · dq$$
and if it equals to $dA$ (a differential) then it is canonical.
However, I'm wondering why is this the case, since the requirements for canonical map is that at first is $$P ·dQ − Kdt = p·dq − Hdt + dS$$ (so that the closed contour integral of $P ·dQ − Kdt$ to equal that of $p·dq − Hdt$. Then what about the $Kdt$ and $Hdt$?
| Be advised that there are different definitions of a canonical transformation (CT), cf. e.g. this Phys.SE post.
Your last definition of a CT agrees with the definition in e.g. Landau & Lifshitz and Goldstein, while your professor is listing a sufficient condition for a symplectomorphism, which is called a CT by e.g. Arnold.
| {
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