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GR Verification for a Charged Black Hole For a charged ($Q$) rotating ($L$) mass ($M$), the Kerr-Newman equations give the angular deflection of light. Has there been observational verification (I would prefer to use only the angular deflection of light) that all three $QLM$ parameters are needed?
I am wondering if all observed light deflection could be fit just as well with the simpler Schwarzschild model (or, if not, with the Kerr model). I have some doubt about the way that electromagnetic field energy bends spacetime in General Relativity.
| I don't think there has even been observational evidence of Kerr-specific effects. While gravitational lensing is well known these days I don't think any of the objects studied have been rotating fast enough for the difference between the Kerr and Schwarzschild metrics to be apparent. Well, not in lensing anyway - Gravity Probe B did measure frame dragging.
The chances of us ever finding an astronomical object with a net charge big enough to have any measurable effect on the geometry are essentially zero. There is no known way for a charge imbalance that big to develop and too many ways for it to neutralise. I doubt there will be observational evidence for the effect of charge on spacetime any day soon.
Observational evidence of black holes remains rather indirect even these days.
| {
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What is meant by rest in rest-mass? As far as I know only photons are considered to have no rest-mass. In common words when it doesn't move it 'disappears'.
*
*Electrons and quarks should have a rest-mass. But are they really at rest?
In atoms and molecules is always a kind of zero-point energy left which implies that there is still some 'movement' in the particles.
*
*So when an electron or quark is really at rest does it still exist?
*Or does it 'disappear' just like a photon?
*And hence is there really a difference in mass-property of particles or is the equation $E=mc^2$ already suggesting that there is no real property difference in mass between a photon and electron? Or there is?
| The term rest mass is a poor one because it implies it's the mass measured in the rest frame. But photons have no rest frame, and indeed any particle subject to some form of confinement has a $\Delta p\gt 0$ so its rest frame is somewhat poorly defined.
The modern term is invariant mass, which is simply the mass in the equation for the total energy:
$$ E^2 = p^2c^2 + m^2c^4 $$
Using this definition the electron has an invariant mass of $\approx 511$keV while the photon has an invariant mass of zero.
While, as you suggest, it can be difficult to get an electron into a state where it is completely at rest you can approach this state to any desired degree of accuracy. And if you do then you'll find the rest mass approaches the invariant mass.
| {
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What are some resources for learning about x-ray powder diffraction? I am looking for up-to-date, applied resources to learn about x-ray powder diffraction. There is a lot of jargon with which I am unfamiliar. I am trying to refine theoretical curves to collected data using the program Maud using the Rietveld Method but I am unsure about which parameters to adjust.
| As with any new thing one the best ways to do it is to look for publications. If you don't understand the jargon/notation, go to their references and look at those papers and keep recursively applying that method until you go back far enough to where you understand, then you can read back up to the first one.
Edit: I understand this is not a resource recommendation per se, but it is a general rule of thumb for researching when you find yourself immersed in notation/jargon/terminology with which you are not familiar.
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Is it possible to have solid light? Is it possible to have solid light?
If so, what would it be like?
| This depends upon how you want light to solid. If you want light to be solid in the way the "Star Wars" movies have light sabers, I would say no. There are however materials that trap photons so they have zero velocity. Photons are in a sense trapped, and these are sometimes called artificial black holes. The energy of the photons then contribute a tiny amount of mass to the material trapping them.
| {
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What is the color of a group of trillions of electrons, protons, and neutrons Since an electron is smaller than visible light, then what what color would a group of electrons (trillions of electrons) be if there were enough of them to be seen by the eye? What color would a group of trillions of protons be? Color of trillions of neutrons? I don't mean a group of electrons, protons, and neutrons mixed together into atoms, I mean a group of each of them separately. Would they be an actual color (red, black, green, etc), clear but visible (the color of water, glasses's lenses), or invisible?
| The 'color' would be an ultra-bright burst of gamma rays as the trillions of electrons rush apart, frying both you and your eyes to a crisp.
More seriously, if you confined the cloud of electrons, it wouldn't emit any particular color on its own -- for instance, there could be no optical transitions since there are no nuclei.
If you shined light on it, it might behave like the electrons in the 'free electron model' of metals. Since the electron response determines the color, I guess it'd be a shiny grey, like typical metals. For more detail, see section 5 here.
| {
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Uncertainty in the distance between Sun and other planets I have read about the orbit distances between Sun and the planets and have come to know for example: Earth is around 150 million km away from the Sun.
However I have seen that tht value is only an average radius of the orbit.
I however cannot find the uncertainty of this measurement.
I have considered the eccentricity of the planets and thought of using it to determine uncertainty.
So then I would have: Earth distance from Sun = $(150\times10^6\pm 2.5\times 10^6)km$. As
Earth's eccentricity is around 0.017
So I am wondering if this would be the right way to determine the uncertainty of the radius of the orbit distance.
| The variation in the orbital distance between aphelion and perihelion is not an "uncertainty" - it is certainly variable.
The ratio between aphelion distance and perihelion distance is $(1+e)/(1-e)$.
The NASA fact sheet http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html gives the eccentricity $e$ to 8 decimal places, so I assume (though I may be being pessimistic) that the orbital separation range is known to this sort of precision. However, it will also change from orbit to orbit depending on the positions of all the other planets, though this will be a second order effect compared with the overall eccentricity of the orbit. The bottom line is - how accurately do you need to know the number?
| {
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Does the charge leakage of capacitors lead to photon emission? Capacitors will leak charge over time. This charge is basically electrical current leaking through insulating layer of the capacitor. I am wondering, if there is some emission of photons as there is is a potential acceleration of the electrons that are part of the leakage?
Sorry if there are major logical mistakes in my assumptions as I am not a physicist.
| In principle any acceleration of an electron causes some radiation, and an electron has to accelerate in order to leak from one plate to the other. However:
*
*the velocities, and therefore the accelerations, of electrons in electrical circuits are small. Calculating the electron drift velocity is an exercise routinely given to students and the results tend to be fractions of a millimetre per second.
*the radiation from an accelerating charge is negligible unless the acceleration is very high. For example see my calculation here.
So, in the situation you describe, to say that the radiation is utterly negligible would be a considerable understatement.
| {
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Growth and Decay, Law or not? The differential equation for decay that applies to radioactive decay is:
$$dN/dt=-kN$$
for a positive constant k and number of particles N. My question is: is this, strictly speaking, a "Law"? I have seen this differential equation refered to as a "Law" and sometimes not, so what is the deal?
Thanks in advance for your help.
| Physics is about quantified observations of nature, modeled mathematically. The mathematical models are rigorous and self consistent but in order to connect to measurements extra postulates are needed, which define the connection of the mathematical formulae to data. Laws are parts of these postulates. In the same way that axioms are not provable within a mathematical theory, physics laws are not provable through the mathematical model, they are part of the physical axioms which pick up the solutions that are relevant to the data.
Your example may be called a law by some, but it is not crucial to the model, it is not an axiom or postulate, it is part of the output of the mathematics, in my opinion.
| {
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Fluid mechanics paradox: The force needed to 'balance' a fluid using a piston We know that the pressure in a fluid (like water) is dependent on the depth. Consider this example:
For the first setup, solving the pressure at the bottom yields
$P_1 = \rho gh = 1000kg/m^3 \bullet 9.8m/s \bullet 1m = 9800Pa$
And from the definition of pressure, $P = F/A$, the fluid exerts a force on the piston equivalent to:
$F_1 = P_1A = 9800Pa\bullet 1m^2 = 9800N$
so the piston will exert a force of $9800N$ to maintain equilibrium.
Which should be reasonable since this is the actual weight of the water: $W_{water}= 1000kg \bullet 9.8m/s^2 = 9800N$
For the second setup, solving the pressure at the bottom yields
$P_2 = \rho gh = 1000kg/m^3 \bullet 9.8m/s \bullet 1.5m = 14700Pa$
And
$F_2 = P_2A = 14700Pa\bullet 1m^2 = 14700N$
so the piston will exert a force of $14700N$ to maintain equilibrium.
Note that the water at each setup occupy the same amount of volume.
The question is: Is this assumption correct? If so, then how do the forces and pressures in the water arrange in such a way that it now requires more force to balance the water than its original weight? And how is setup 2 different from setup 3, where you have a solid mass of the same shape and weight?
| The quick answer is that a long as the surface area of the piston, and the volume/mass of the remain unchanged, the pressure stayed the same.
So the erreanous step is to assume that $ P_2 = \rho g h $. This invalid as the cross section area changes with height.
| {
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Why are ceramics good electric insulators? I know it depends a lot on the composition, so not all are great electric insulators. So what makes it good or bad? And is it different from what makes them good thermal insulators?
Power line insulators are ceramic and they have to stave off huge voltages. That's what got me wondering.
| Most ceramics are ionic compounds, in which electrons are immobile. This is different to metal, in which the atoms are in a "sea of electrons" that are free to move.
Note that ceramics have some kind of conductivity, it's just extremely low. The conductivity of copper, for example, is ~6×107 S/m. Most ceramics have conductivities in the range of 10-3 S/m to 10-5 S/m. That's around 10 orders of magnitude less conductive than copper, but it is 10 orders of magnitude more conductive than air, for example (10-15 S/m).
| {
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The analytical result for free massless fermion propagator For massless fermion, the free propagator in quantum field theory
is
\begin{eqnarray*}
& & \langle0|T\psi(x)\bar{\psi}(y)|0\rangle=\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}.
\end{eqnarray*}
In Peskin & Schroeder's book, An introduction to quantum field theory
(edition 1995, page 660, formula 19.40), they obtained the analytical
result for this propagator,
\begin{eqnarray*}
& & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}=-\frac{i}{2\pi^{2}}\frac{\gamma\cdot(x-y)}{(x-y)^{4}} .\tag{19.40}
\end{eqnarray*}
Question: Is this analytical result right? Actually I don't know
how to obtain it.
| Yes it is correct. The derivation in P&S is straightforward but I will expand on it a bit. The key observation is that
\begin{equation}
\int\frac{d^4k}{(2\pi)^4}e^{-ik\cdot(y-z)}\frac{i\gamma^{\mu}k_{\mu}}{k^2+i\epsilon}
=-\gamma^{\mu}\partial_{\mu}\int\frac{d^4k}{(2\pi)^4}\frac{1}{k^2+i\epsilon}e^{-ik\cdot(y-z)},
\end{equation}
where the integral on the right hand side is the Feynman propagator for a
massless scalar. Performing the $k$ integrals to get to position space yields
\begin{equation}
\int\frac{d^4k}{(2\pi)^4}\frac{1}{k^2+i\epsilon}e^{-ik\cdot(y-z)}=\frac{i}{4\pi^2}\frac{1}{(y-z)^2-i\epsilon}.
\end{equation}
If you aren't sure about this last step, it is easier to consider the massive case and then take the limit as $m\rightarrow 0$ at the end. Schwinger parameters are also helpful for proving this identity.
Once you have transformed to position space you simply act with $-\gamma^{\mu}\partial_{\mu}$ to arrive at the final expression.
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Can energy conservation be derived from Newton's laws? Can the idea of energy conservation be derived from Newtons's laws? From inspection of his laws you can vaguely discern a relationship but I want to know if you can manipulate his laws to prove it.
If not, what else did it take in history? What other assumptions about the world were needed to take that next step? Or is it only a fact of nature that can be proven experimentally, much like Newton's laws.
| The answer is not. Newton's laws are equivalent to linear momentum conservation, but it does not implies mathematically energy conservation.
The elementary proof that is usually given assumes that involved forces derive from a potential energy. In that case the answer is yes, taking into account potential energy.
But consider for example two bodies of the same mass $m$ in rest at time $t=0$, which experiment a time dependent repulsive force $f(t)=mt, t \geq 0$ ($f_1(t)=mt$,$f_2(t)=-mt$). Energy at $t=0$ is $E=0$. The force verifies Newton's third law. Momentum is conserved because $p(t)=mt^2/2-mt^2/2=0$ but the kinetic energy is $E(t)=1/2mt^4/4+1/2mt^4/4$, which increases over time.
| {
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Why do thin films need to be thin? No matter what thickness a piece of glass is wouldn't its optical thickness be close to an integer multiple of a wavelength such that it could create interference effects? I feel like I am missing something here.
|
Ignoring for simplicity's sake the usual refraction that takes place at the interfaces of the media, if:
$$|OA|+|AB|=D\big(\frac{1}{\cos \theta}+\tan \theta\big)=n\lambda$$
with $n$ an integer and $\lambda$ the wave length, then we have positive interference.
But as we increase $D$, the distance $\Delta$ also increases, so for high values of $D$ these rays can no longer interfere.
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Is there any effect on mechanical waves by electromagnetic waves (and vise versa)? Do electromagnetic waves like light and gravitational waves (due to moon for instance) affect on mechanical waves like sound?
Can sound change the path of light?
| Any physical phenomenon is potentially capable to cause some change to any other phenomenon, more or less directly. If it was not the case, the physical world could be divided into completely independent realms; there would not be the one single world we call Nature.
Practically though, many if not most of the actually existing interactions between systems can be ignored, or just treated as perturbations in models taking into account only the most important ones. This is because interactions happen in a wide range of order of magnitudes. For example you would not usually include electromagnetic interactions between Moon and Earth when modelling their respective motion, although it certainly does play some part in the actual interplay of the two bodies (both having a magnetic field). If you do not ignore negligible effects, well even nocturnal urban lighting does play a part by sending photons to the Moon, pushing it away from Earth!
As a side note, the fact that some interactions are so much less intense than others is very useful: it allows us to use them as measuring devices. As shown in another answer, we can use Schlieren photography as a straightforward way to display air density because indeed the path of light is altered by compression waves, but only marginally so. If the dependence of electromagnetic waves on air density was more intense, it would be more complicated to decorrelate both effects.
| {
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Projectile motion with air friction force/resistance
We have to find the x (the distance, if you didn't know that then I'm not sure if you should be doing this problem) that the projectile travel during the time in the air until the time it hits the ground. I can do this no problem without air resistance, but I have no idea what to do if there is. My teacher hinted to me to do it using energy formulas but even after doing that, I got stumped. Can someone please tell me or show me how to do this problem using either motion formulas or energy formulas?
| I assume that air resistance force is parallel to the velocity vector but in opposite direction.
We have:
$$a_x=\large{\frac{10\cos\theta}m}$$
$$a_y=\large{\frac{10\sin\theta}m}-g\;\Longrightarrow\;a_y+g=\large{\frac{10\sin\theta}m}$$
$$\tan\theta=\large{\frac{v_y}{v_x}}$$
Then,
$$\large{\frac{v_y}{v_x}}=\large{\frac{a_y+g}{a_x}}\;\Longrightarrow\;\large{\frac{a_y+g}{v_y}}=\large{\frac{a_x}{v_x}}=C\;\textrm{(constant)}$$
Left side of equation above is a function of $y$ and the right side is a function of $x$. So, it must be equal to a constant like $C$. Hence, we will have:
$$a_y-Cv_y+g=0$$
$$a_x-Cv_x=0$$
Or
$$\large{\frac{\mathrm d^2y}{\mathrm dt^2}}-C\large{\frac{\mathrm dy}{\mathrm dt}}+g=0$$
$$\large{\frac{\mathrm d^2x}{\mathrm dt^2}}-C\large{\frac{\mathrm dx}{\mathrm dt}}=0$$
You should solve these differential equations by considering to the initial conditions. Then, you can find the time of falling by substitution $y=y_0-1000$ say $t_f$. Finally, the range that you want is determined $R=x(t_f)-x_0$.
| {
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Third Brillouin zone for a quadratic 2D lattice As far as I understand, the construction of Brillouin zones stems from the relation$$
2 \vec{k}\cdot \vec{G} +G^2 = 0
\,,$$where $\vec{k}$ is the wave vector and and $\vec{G}$ is the reciprocal lattice vector. This condition is supposed to be fulfilled when $\vec{k}$ terminates on a line normal to $\vec{G}$ at half of the length of $\vec{G}.$
If so, why does the third Brillouin zone take the form of Figure 1, rather than Figure 2, for a quadratic lattice? The second figure shows the area enclosed by lines situated half of $2b\hat{x}$ and $2b\hat{y}$ from the origin, perpendicular to the $x$- and $y$- directions respectively.
$\hspace{50px}
{\def\place#1#2#3{\smash{\rlap{\hskip{#1px}\raise{#2px}{#3}}}}}
\place{0}{220}{\begin{array}{c}\textbf{Figure 1} \\ \hspace{250px} \end{array}}
$
$\hspace{50px}{\def\place#1#2#3{\smash{\rlap{\hskip{#1px}\raise{#2px}{#3}}}}}
\place{0}{220}{\begin{array}{c}\textbf{Figure 2} \\ \hspace{250px} \end{array}}$
| One reason is that the volume of your third BZ is twice the volume of your first BZ.
To see it another way, notice that the black dot on the far right (the middle of the three vertical dots) is a BZ center, a $\Gamma$ point. The black dot directly above that one (far upper right corner) is also a $\Gamma$ point, but it must be in the next BZ. Half-way between these two $\Gamma$ points must be a zone boundary, exactly as shown in your Figure 1.
| {
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Is the energy needed for a current through a straight and a coiled wire different? When you add current to a straight piece of wire does it use less electricity than if it was coiled? The power wire on telephone pole's are curved while buried cables are pretty strait in comparison. Does the curve of the wire and proximity to itself make a difference in energy consumption?
| Okay I'll have a go at answering this, although it may be a make belief scenario.
Looking at where you got your inspiration from, he stated using a magnet as a core for an electromagnet, so your curiosity must have piqued from the idea that the wire carrying electric could itself be magnetic. Because of polarization and magnetization, all the magnetic dipoles in the wire would align with the potential of the wire, which would send you on a kind of wild goose chase, but hey lets do some maths anyway.
The steps to find the efficiency would be this:
*
*Find the general force of the electrons in the wire through the
magnetism of the wire.
*Find the force of the electron through the magnetic field generated
by their current density
*Take the ratio of these forces, multiply by the distance the
electrons have traveled and find its difference from one.
1
The force on the electron from the wire itself
$$ F_{wire} = q(\hat{v_{e^{-}}} \times \hat{B_{wire}}) $$
2
The force of the electron generated by the rotation of it's own current density (lol)
$$ \hat{j} = nq\hat{v_{e}}A $$
$$ \nabla \times \hat{j} = \hat{M} $$
$$ \frac{\hat{M}}{\mu} = \hat{B_{j}} $$
It would be important to maintain the unit vector of the magnetization such that you know the direction of the magnetic field.
$$ F_{e^{-}} = q(\hat{v_{e^{-}}} \times \hat{B_{j}}) $$
3
The work:
$$ W = F_{wire} \cdot \hat{r} $$
$$ W = F_{e^{-}} \cdot \hat{r} $$
Where I guess r would be the integral of the electron's velocity vector.
The efficiency would then be, with:
$$ W_{1} \leqslant W_{2} $$
$$ \epsilon = 1 - \frac{W_{1}}{W_{2}} $$
| {
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Prove that an electron in a hydrogen atom doesn't emit radiation According to electrodynamics, accelerating charged particles emit electromagnetic radiation.
I'm asking myself if the electron in an hydrogen atom emits such radiation. In How can one describe electron motion around hydrogen atom?, Murod Abdukhakimov says the total momentum of the electron is zero, hence it does not emit radiation. Could someone prove this statement ?
It may be an obvious question, but I can't figure out why the total momentum of the electron should always be zero, in any energy state.
| I believe some of the answer in the links are correct, others are less obvious and might even be confusing. I am not gonna repeat the arguments there, but to stress the following idea. You cannot demonstrate that using classical electrodynamics. The theory as is does not apply to quantum objects and thus it was modified. The equations are the same, they are now incorporated into a different algorithm, the Schroedinger equation (if we limit the expansion to semi-classical quantum mechanics), and the measurement formalism of quantum mechanics. As with many theories that have been generalized, they still work well in many realms, in this case you do not need to use quantum mechanics when dealing with the electromagnetic properties of "most" macroscopic objects, but the reverse is not true, classical electrodynamic predictions fail often when you deal with quantum objects.
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The expression of the density in terms of molecular mass and the distribution function I am reading a book about the boltzmann equation, the author gives the expression of the fluid density $\rho$ as follows:
$$\rho(\mathbf r,t) = \int {M\,f(\mathbf r,\mathbf c,t) \, \mathrm d\mathbf c}$$
where:
*
*$\rho$ is the fluid density.
*$\mathbf r$ is the position of particles.
*$\mathbf c$ is the velocity of particles.
*$t$ is time.
*$M$ is the molecular mass.
*$f$ is the Boltzmann distribution function which gives the number of molecules at time $t$ positioned between $\mathbf r $
and $ \mathbf r + \mathrm d \mathbf r$ which have velocities between $\mathbf c$ and $ \mathbf c + \mathrm d \mathbf c$
Can you, please, help me to derive this expression of fluid density?
Thank you
| Density is the average amount of mass per unit volume $\rho(\vec{r},t) = \frac{M}{V_r}$.
Distribution function is defined as a number of particles per unit phase space volume $ f(\vec{r}, \vec{c},t) = \frac{N}{V_r V_c} $ (which is space volume times velocity volume). Each particle has mass $M$, so to get the total mass density, we need to sum distribution function multiplied by particle mass over all velocities in the given space volume.
Then, summing over small unit volumes in the velocity space $d^3 c$
$$\rho(\vec{r},t) = \sum_\vec{c} M f(\vec{r}, \vec{c},t) \Delta V_c \approx M \int f(\vec{r}, \vec{c},t) d\vec{c} $$
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Physical meaning of enthalpy I've been reading about thermodynamics and reached the topic about enthalpy . I've understood its derivation but I don't understand its physical meaning ... Also I don't understand why they have divided by the mass of gas to get to the specific enthalpy equation . what's the use of it? I know the meaning of all state variables the enthalpy contains but I can't see the benefit of combining them together to have the enthalpy ..
| In physics one of the most fundamental concepts is the conservation of energy and in thermodynamics we systematize, in an ideal manner how to account for the energy and changes in energy in systems. So basically a means of categorical naming, bookkeeping.
The units of enthalpy are energy units such as Joules. And for a homogeneous system, the enthalpy is the sum of the system internal energy and the pressure energy.
As energy, enthalpy is potential in a system in the form of chemical bonds; the making and breaking of these bonds. The direction in which enthalpy changes tells us which way heat is flowing: if $\Delta H < 0$ heat flows out of the system (exothermic), and if $\Delta H > 0$ heat flows into the system (endothermic).
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Are the partial derivatives of Lagrangian in the varied action functional derivatives? In particle mechanics Lagrangian $L$ depends upon position, velocity (and may be explicitly on time), whereas in field theory the Lagrangian density ${\cal L}$ similarly (or analogously) depends upon the field and its derivatives. When we derive Euler-Lagrange equation of motion we vary the action,
In particle mechanics, $$\delta S=\int_{t_1}^{t_2}~\mathrm dt\left(\frac{\partial L}{\partial q}~\delta q+\frac{\partial L}{\partial \dot{q}}~\delta \dot{q}\right)\tag{1}$$
In field theory, $$\delta S=\int_\sigma~\mathrm d^4x\left(\frac{\partial \mathcal{L}}{\partial \phi}~\delta \phi+\frac{\partial \mathcal{L}}{\partial (\partial _\mu \phi)}~\delta \partial_\mu \phi\right).\tag{2}$$
Now, Lagrangian is a functional, it maps functions (position, velocity or fields and their derivatives) into another function (or a real number). Like, $$L:F×F→F;(q(t),\dot{q}(t))↦L[q(t),\dot{q}(t)]. \tag{3}$$ So my question is that are the partial derivatives of the Lagrangian w.r.t. position and velocity functions or fields and their derivatives functional derivatives?
| Yes, OP is right. In the field-theoretic case, the partial derivatives in OP's first formula (1) should be replaced with functional derivatives
$$ \delta S~=~\int_{t_1}^{t_2}\!\mathrm{d}t\left(\frac{\delta L}{\delta q}~\delta q+\left. \frac{\delta L}{\delta v}\right|_{v=\dot{q}}~\delta \dot{q}\right),\tag{1'}$$
where the Lagrangian
$$L[q(\cdot,t),v(\cdot,t);t]~=\int \! \mathrm d^3x~ {\cal L}(q(x,t),v(x,t), ~\partial_x q(x,t), \partial_x v(x,t),~\ldots , t) $$
is a functional. The ellipsis $\ldots$ indicates dependence of possible higher-order derivatives. See my Phys.SE answers here and here for further details.
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Multipole expansion of the electromagnetic field In Jackson's Classical Electrodynamics, section 9.7, he develops the multipole expansion of the electromagnetic fields in terms of the vector spherical harmonics and the spherical Bessel and Hankel functions. His expansion is somewhat confusing, and I was wondering any other reference doing the same expansion, in some other manner?
| It is better to avoid vector spherical harmonics by expanding the electric and magnetic scalar potentials instead of the vector fields. The electric potential
is expanded in every EM textbook. The magnetic scalar potential is treated in Section 7.10 of "Classical Electromagnetism, 2nd Edition" by Jerrold Franklin
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Electric field dependence on distance How can it be proved that for a point charge, $E$ is proportional to $$1/r^2$$ using the concept of Electric field lines (or lines of force)? I tried to show that if field lines are close, then magnitude of Electric field is higher. But, I couldn't show the given dependence.
| As such there is no real theoretical proof to the inverse square dependence of the electric field in classical electrodynamics. It is an experimental fact famously known as the Coulomb's law. When combined with the superposition principle, it gives us the Gauss's law of classical electrodynamics:
$$\nabla \cdot\mathbf E = \frac{\rho}{\epsilon_0}.$$
But, one can also think of the Gauss' law as an experimental fact and from it, he/she can derive (with suitable physical assumptions) the inverse square dependence of the electric field of a charged sphere in the following manner:
Let's take a spherical charge whose charge $Q$ is distributed spherically symmetrically within some radius $a$. Consider a surface centered at the center of the sphere and having a radius $R>a$. Now, one can argue that at each point of the spherical shell, the only direction, an electric field can have is either radially outward or radially inward. Also, if the electric field points radially inward at one of the points on the spherical shell then it should point radially inward at every other point on the spherical shell. Also, the magnitude of the electric field must be the same at each and every point of the considered spherical shell.
Thus, the integral $\displaystyle\iint_{S} \mathbf E\cdot \mathrm d{\mathbf A}$ (where $S$ denote the integration over the spherical surface) can also be written as $4\pi R^2E$ where $E$ is the magnitude of the electric field - taken positive if it points radially outward and negative if it points radially inward. (This is just a convention - you could alter it and still get the right physical direction of the electric field provided you use the vector calculus properly.) Now from the Gauss' law, this integral must be equal to the total charge inside the spherical surface divided by $\epsilon_0$. i.e. $Q/\epsilon_0$. Therefore,
$$4\pi R^2E = \dfrac{Q}{\epsilon_0}$$
Or, $$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{R^2}\;.$$
Since, there was nothing special about the radius $R$ except for $R>a$, we can consider this formula to be true for any $R>a$.
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Can an inhabitant of a spacetime region measure its curvature tensor? So, lets say that I am an ant living on a 2-D spherical surface that is stretching to the equator...like half a sphere. I can not describe this surface in terms of the outside coordinates only someone living in the outside world can do this. So, can I really determine that I am in fact living on a spherical and not flat surface? If I try to measure the phi coordinate change as I move over the R coordinate (phi is the angle that runs from 0 to 2pi and R runs from 0 to rpi where r is radius of a sphere) at two separate places as I move to the north I can measure that this distance is getting smaller. I hope that I am clear enough about what I mean. But, would not the measuring stick also get smaller? Making it imposible for the inhabitant of this world to determine its geometry. So how can someone living in this world determine its geometry while his measuring equpment distorts acording to this?
| You could construct a circle of radius $r$ and measure the circumference $c$. In a flat space, $c=2\pi r$. However, in a positively (negatively) curved space, $c<2\pi r$ ($c>2\pi r$).
Alternatively, you could pick three non-collinear points and measure the three angles that they form. In a flat space, the sum of angles will be 180$^\circ$. However, in a positively (negatively) curved space it will be $>180^\circ$ ($<180^\circ$).
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How does isolating transformer protect from geting shocked From what i know, isolating transformer makes it that you would have to touch both ground and phase cables to get shocked, but why exactly is that.
There should be some potential difference between phase (transformer output) and ground (that we stand on), and as long as there is this difference, the charge is going to flow, and conduct a shock.
Or do I miss something here?
| If you had one sweaty hand on the isolated circuit, and one bare foot on the wet Earth, there would be no path by which current could flow through your body, and return to the isolated circuit. A (probably) trivial amount of current could flow because of the AC voltage, and (probably) very weak capacitive coupling between the Earth and the circuit. But, I'm guessing that the capacitive effect would not be enough for you to feel unless the voltage was very high.
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Why can we see the moon when it is between the Earth and the Sun? A rather stupid question, why can we see the moon when it is between the Earth and the Sun?
| The diagram you drew is flat, but the solar system is not. The Moon's orbit is not in the same plane as the Earth's orbit. Wikipedia has a nice diagram:
Because of this, when the Moon is "in between" the Earth and the Sun, it is usually a little "above" or "below" the Sun as well. You can observe this for yourself: one or two days after the new moon, look for the Moon in the sky at sunset. It will be above the Sun, and the bright arc will point toward the Sun. But it will also be a little off to one side or the other. That angle is a combination of the Moon's orbital angle with the ecliptic and the Earth's axial tilt.
(They do sometimes line up perfectly, and then we have a solar eclipse, as mentioned in other answers.)
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Is potential difference the difference in electric potential energy or electric potential? Referencing the book Physics for Scientist and Engineers, Ninth Edition, the book says that "Potential Difference should not be confused with Difference in Potential Energy."
I also reviewed several internet sources that say "Potential Difference is the difference in Electric Potential Energy between two points."
What is the difference between potential difference and a difference in potential energy?
| Technically "potential difference" is the difference in electrical potential, i.e. $\Delta V$, not the difference in electrical potential energy, $\Delta U$. Potential difference ($\Delta V$) is also called voltage, in certain contexts.
However, many people and sources are sloppy about their terminology, and they will say just "potential" when they really mean potential energy. An expert could tell which is meant based on context - or, in some cases, that it doesn't matter. Since potential energy is related to potential by $U = qV$, if the charge $q$ is known and constant, you can usually say the same things about either quantity $U$ or $V$.
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Interactions preserving separability Consider the interaction (described by a unitary matrix U) of two qubits initially in a separable state |ab⟩ = |a⟩ ⊗ |b⟩, such that after interaction the composite system is in state U|ab⟩.
Are there unitary matrices U for which U|ab⟩ is also separable, independent of |ab⟩?
If so: How can these matrices be characterized?
If not so: How can the separable states |ab⟩ for which also U|ab⟩ is separable, be characterized?
| Any linear operation $U$ which maps all product states $|a\rangle_A|b\rangle_B$ onto product states must be of the form
$$
U \equiv (U_A\otimes U_B)\mathbb F\ ,
$$
where $U_A$ and $U_B$ act independently on the two parts of the system, and $\mathbb F$ swaps the two parts, i.e.,
$$
\mathbb F:|a\rangle_A|b\rangle_B \mapsto |b\rangle_A|a\rangle_B\ .
$$
This is proven in https://arxiv.org/abs/quant-ph/0407118 (Sec. II C, Result 1).
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Is there a prohibited region in $P-V$ plane? Polytropic process generalize the particular thermodynamic processes with
$$P V^{n}= \mathrm{constant}$$
Where, if $n$ changes, the curve on $P-V$ plane changes, as shown in the diagram.
The orange region is not touched by any curve, so there is no value of $n$ for which the gas goes directly in to the orange region.
Why is that? I do not see any particular reason why there should not exist a process to make the gas go into the orange part.
| It's simply inherent to the definition of polytropic processes that they don't allow the system to increase both its pressure and volume at the same time. That doesn't mean you can't increase a system's pressure and volume. You just need a non-polytropic process to do so. For example, it could be a compound process consisting of two polytropic processes with different values of $n$, with one being run in reverse - like half of a heat engine cycle.
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How did physicists know that only negative charges move? I have phrased similarly another question about how physicists knew that two charges exist, positive and negative. The purpose of the question is not necessarily to educate me historically. It's just that I wish to know about classical subjects without making the atomic assumption.
I know that electrons (elementary negatively charged particles) move in contrast with protons (elementary positively charged particles) because electrons have small mass and orbit the nucleus while protons are stuck in the nucleus of atoms. Roughly at least!
Charging by induction works because of the transfer of electrons, negative charge between a conductive sphere and the ground. But it could be very well explained by the transfer of either positive or negative charge. But only the later happens.
Was there any (thought) experiment to show that only negative charges happen to move/transfer?
| Physics's don't know that only negatively-charged particles move. We can create ion currents on demand in many environments. We do know that the current flowing in a metal wire is negatively charged particles in motion.
As for how to determine that, you do a Hall effect measurement.
The measurement works by subjecting a current in a relatively wide bar to a magnetic field perpendicular to both the current and the width of the bar and then measuring the potential difference across the width. In this era of turn-key precision voltage measurements is easy enough to do in a high school laboratory if the students can follow the underlying arguments.
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Why don't high pressure gases stored in containers lose energy? Containers holding gas at a high pressure don't slowly lose the internal energy of the gas. It seems like the high speed particles would collide with the metal walls and slowly transfer their energy to the slower particles outside the container.
Even if the pressure is from more particles in the container, they can do work when released so they have energy. Shouldn't that energy dissipate over time?
| There are two ways you can change the internal energy of a gas, one is macroscopic, that is, performing work on or by the gas, if the gas either expands or contracts. The other is microscopically through heat. If the compressed gas is at the same temperature than the outside gas, these microscopic collisions will not result in an exchange of energy, because the speed distribution of the particles is a function of the temperature, not of the pressure, so it is not correct to assume that the particles inside the container are faster than those outside, and thus there will be no net transfer of energy.
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What Postulate States Entropy must be Positive? I've been reading the Postulates of Classical Thermodynamics, and I haven't found anywhere to be said that the Absolute Entropy of a system has to be a positive number.
The third one states that the Temperature, $dU/dS$, must be Positive, since $S$ is monotone increasing with $U$, and that as $U(S)$ approaches flatness, $S$ approaches zero. But this doesn't mean that $S$ can't be negative. Consider for instance the function $U = kS^3$, which is monotone increasing, and where $S$ approaches zero as $dU/dS$ approaches zero from both sides, meaning $S$ could be negative.
Im pretty sure $S$ can't be negative, but I'd like to see the actual proof or Postulate that says so in CLASSICAL (not Statistical) Thermodynamics. Maybe it's included in the definition of entropy and I missed it?
| A better definition of Entropy is given by the statistical mechanics definition of the Gibbs Entropy:
$ S = -k_B \sum{p_i \ln{p_i}}$
Where $S$ is the entropy, $k_B$ is Boltzman's constant and $p_i$ is the probability of the $\mathrm{i^{\mathrm{th}}}$ microstate to be occupied. For any probabilities less than 1, the quantity is clearly always positive. It can be shown that this definition is consistent with the classical thermodynamic entropy
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Finding the coefficient of static friction on slope My Problem
1.0kg block is sitting on a horizontal incline plane. The plane is tilted into the block begins to slide. You notice that the plane is inclined at an angle of 32 degrees above the horizontal when the block begins to slide. Calculate the coefficient of static friction that acts between the block and the plane.
My Solution
I tried solving this by doing FgSinθ=μCosθ and manipulating it into Sinθ/Cosθ=μ which is Tanθ=μ. This comes out to Tan(32)=0.62
My Question for You
But my question is that it's already moving at 32 degrees, so am I finding the coefficient of KINETIC friction the way I did it. If so then how do I find the coefficient of STATIC friction given the information I have.
| You have found the critical angle $\theta_c$ at which the block begins to slide. That gives you the coefficient of static friction $\mu_s = \tan\theta_c$. Kinetic friction comes into consideration when the block is actually moving.
Before the block can move, the force $mg\sin\theta$ acting down the incline must be at least equal to the maximum possible value of the static friction force, which is $\mu_s mg\cos\theta$, acting up the plane. When the block just begins to move at angle $\theta_c$ these two forces are equal :
$mg\sin\theta_c = \mu_s mg\cos\theta_c$
$\mu_s = \tan\theta_c$.
When it moves the sliding block might - and usually does - accelerate down the slope because kinetic friction $\mu_k$ is often less than static friction $\mu_s = \tan\theta_c$. The rate of acceleration $a$ is given by
$a = g\sin\theta_c - \mu_k g\cos\theta_c = (\tan\theta_c - \mu_k)g\cos\theta_c = (\mu_s - \mu_k)g\cos\theta_c.$
If the block does not accelerate down the slope ($a=0$) but moves at constant velocity then $\mu_k = \mu_s = \tan\theta_c$. If the block does accelerate ($a\ne 0$) then you can rearrange this relation to find $\mu_k$.
| {
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If the universe stopped expanding, when would we realize it? Suppose in this moment the universe stopped expanding, how long would if take and how would we register the change?
Since the bodies are still and it is space that is expanding in between, we wouldn't notice the change even if the expansion would completely stop in a short time?
| A few million to about 30 million years from now. I.e., we'd be able to measure a change in the previously observed redshifts for galaxies a few to about 10 Mega parsecs away. A parsec is a little more than 3 light years.
The reason is that closer in galaxies are in our local group or cluster, and we are to a great extent gravitationally bounded to them. Our relative velocities then depend more on each other than the cosmological expansion. If we go further then the cosmological expansion starts becoming the dominant effect. We would have to see more than one Galaxy doing that, but it certainly would be deemed as totally unexpected to see even one radically having changed its velocity.
Of course, since it might have happened a few to 30 million years ago, we could start seeing that tomorrow.
Now, something huge cosmologically had to happen for the expansion to stop. Maybe some huge increase in the matter density. If so, we might see some of that in more local galaxies, maybe even ours, as we would see the local gravitational effects. So we actually then would discover it pretty quickly, say a year or two for astronomers to realize they are seeing something and not simply errors in everything they calculate from more local observations. If not an increase in mass or energy density, then some strange field would have settled to some value other than its current one, and again it could affect things everywhere. If it happened 30 million light years away we'd wait the same time as in my first answer.
But none of that makes much sense.
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Problem on Emissivity and absorptivity I have read the radiation chapter, where I have been introduced with the terms emissivity and absorptivity. emissivity tells about the ability to emit heat energy as thermal radiation compare to a black body. and absorptivity is the amount of heat absobed by body devided by the incident heat energy .but my question is ,are these emissivity and absorptivity constant or depends on temperature?
Actually I had a misconception that these quantities are intrinsic property of body
If I have really wrong concept and they depends on temperature then tell me and
Tell me how they depends on temperature
I mean I need The mathematical expression
| In a way, these are intrinsic properties of a material that depend on temperature. The temperature of the object dictates what a black body curve looks like and which wavelengths are emitted in what amounts. The emissivity of a material is not always uniform across all wavelengths, which means at different temperatures, you might get different emission profiles depending on the emissivity of the material at each wavelength along the curve. Similarly, the absorptivity can differ with wavelength. This means the absorptivity and emissivity values for a material might be not just different, but variably different. So the material might absorb well at the wavelength of incident radiation and emit poorly at its current temperature, but as a result, it may then heat up into a blackbody range where it is a good emitter. Furthermore, it also means the amount of energy the material absorbs (when pointed at a blackbody object) could depend on the temperature of the object it is pointed at.
There is no mathematical relation to provide you. Absorptivity and emissivity are intrinsic properties of a material. We test materials and empirically find the wavelength dependence of those values. It varies greatly from material to material, which means any time you want to use a different material, you should look up the $\alpha$ and $\epsilon$ values for the wavelengths you expect to be relevant. Usually, however, it is sufficient to look up wavelength ranges, such as visible light, near IR, UV, etc.
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What would be the optimal diameter of gold grains if you want to fill a pool with them and swim in it? I was wondering, if I were to fill a pool with gold/gold dust, what would be the optimal diameter of the grains to minimalize the friction to a point at which you would be able to swim in it ((and if swimming is not possible, what size would pose the least resistance))?
This is assuming the grains are spherical and the normal gold-on-gold friction coefficient applies.
Thanks!
| It's not so much about a grainsize. As long as the grains are big enough to aloud air(gas) flow through them, you can make them liquid by"Fluidization";
"Fluidization (or fluidisation) is a process similar to liquefaction whereby a granular material is converted from a static solid-like state to a dynamic fluid-like state. This process occurs when a fluid (liquid or gas) is passed up through the granular material."
Basically You can easily swim in Gold, sand or any solid granular material if you fill a whirlpool with it, and blow air through it like shown in this video.
Edit: The given comment forces me to expand the answer. This is about Buoyanycy. It has nothing to do with density. (Explained in video at 6 min->) Another aspect is terminal velocity. If the gas speed is higher than the terminal velocity of the object floating in fluidized granulars, then the object will raise to the surface.
The terminal velocity of Human is 53 m/s, which means that it can be written;
$$\frac{v^2}{2}=\frac{p}{\rho}$$
Where we know the $\rho=19.32$ (Gold, $kg/m^3$) and $v=53 m/s$, so we can solve;
$$\frac{v^2\rho}{2}=p=27134 Pa$$
Which alouds us to solve the maximum granulat size.
The area of the granulat is;
$$A=\pi r^2$$
And the weight of the granulat is;
$$m= \rho \frac{4}{3} \pi r^3$$
Which produces a simplifed model to calculate the maximum $r$, as $p=\frac{F}{A}$ and $F=ma$ and $a=g$, so
$$ p=\frac{\rho \frac{4}{3} \pi r^3 g}{\pi r^2}=\rho \frac{4}{3} \pi r g$$
So the r can be solved to be;
$$ \frac{3p}{4 \pi g}=r$$
Which with $p=27134 Pa$ gives r= 660 m.
(Somehow I don't believe this answer's magnitude, but I leave it here, and get back to it later. Good night)
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Does the speed of light in vacuum define the universal speed limit?
*
*Is light the thing causing the universal speed limit to be $299\,792\,458\,\mathrm{m/s}$? So the universal speed limit would be different if light travelled faster or slower?
*Or, is $299\,792\,458\,\mathrm{m/s}$ the universal speed limit anyway and light just goes that fast? Light is just something we commonly associate with it because it goes super fast.
| Above all, speed of light is the speed of propagation of fields through space. While light may be slowed down when crossing matter, fields (electromagnetic fields, gravity) are always propagated at c. One of the consequences is the "speed limit for causality" mentioned by DavidZ and the speed limit for transmission of information.
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How / why do infrared thermometers work? How can reading the intensity of infrared radiation coming from an object let you calculate its temperature? Could any wavelength be used to measure temperature - why infrared, and not, for instance, green, or low frequency radio waves? And would an infrared thermometer give an accurate temperature of, say, an infrared LED?
| *
*Every object with nonzero temperature (so, every object) radiates EM radiation
*The spectrum of a black body (that is, a body that absorbs all light at all wavelengths) is given by Planck's law.
*Generally, nothing is completely black (except for empty space, which is why cosmic microwave background has a very nice black body radiation spectrum). In general, emissivity varies with wavelength: everything that reflects some light (has color in any part of the spectrum) doesn't follow Planck's law anymore, it multiplies it by some factor for each wavelength.
Conclusion: you can measure the temperature using the Planck's law by measuring emitted light at some wavelength (or many wavelengths for an accurate reading), only if the object is black at that wavelength. It's handy to do this in that portion of the infrared spectrum, where most materials have no special behaviour. In green wavelengths, it would only work for black, red and blue objects (but not for green/yellow/cyan/anything that reflects green). In radio wavelengths, the signals would be way too weak to detect with such a small sensor, and also, due to large wavelength, there's no way to target the measurement to a small object (you'd basically get an average of half the sky due to diffraction), and if an object itself is smaller than the wavelength, it's emission footprint is small (it's basically transparent for that wavelength, so the assumptions required for the Planck's law to work don't hold.
| {
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Simple analytic examples of Multi-scale Entanglement Renormalization Ansatz (MERA) I want to understand Multi-scale Entanglement Renormalization Ansatz (MERA) with very elementary examples. So far I could find references which are mostly based on numerics. It would be a great help if someone could give a very simple example explaining the basic concepts of MERA or give some references (papers/theses/reviews) that implement the operations (i.e, isometry and disentangler) in analytic fashion (even for a simplest possible system).
Actually, I do not have any 'practical' knowledge about tensor networks in general as I am from theoretical HEP background. So references on (analytic) MPS or PEPS with simple examples will be really useful.
| You may want to start from the paper Entanglement Renormalization: an introduction. In this work author describes basic consepts of MERA, things like isometries and disentanglers and why one needs to use them in order to produce MERA.
Article Algorithms for entanglement renormalization is more advanced and is focused mainly on description of algorithm itself. It will be a good exercise to build MERA for simple Hamiltonian based on information provided in aforementioned article.
| {
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Large-scale antimatter production From what I can find, presently the only known means of producing antimatter consist of directing particle accelerators at various targets, and only infrequently getting positrons or anti-protons as byproducts of particle interactions.
Assuming a sufficiently large source of energy: Have more efficient means of producing antimatter been conceived?
And based on these is there a known upper bound on antimatter production efficiency? For example:
*
*If all known theory requires that an anti-particle be produced with its particle pair, then efficiency will always be under 50%.
*Are there known processes for "transmuting" matter into anti-matter? If so, do these allow for a theoretical upper-bound on production greater than 50% of input energy?
| This Centauri Dreams article claims that a 0.01% efficiency is possible with current technology if there was a dedicated production facility.
Having searched for information on this problem myself, I have found nothing more of significance.
If you can get past the google scholar paywalls, you may want to look through the articles by Robert Forward who is referred to as the source of the information.
| {
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What happens to Pauli's argument (that says that there is no time operator) when applied to $X$ operator for some simple systems? An argument by Pauli is usually referred to in the literature when it is stated that there cannot be a time operator in quantum mechanics. This argument can be found as a footnote to P63 of W. Pauli, The general principles of quantum mechanics, p. 63. Springer, Berlin 1980. The corresponding page is available from google books (look for P.63):
https://books.google.co.in/books?id=hVjsCAAAQBAJ&printsec=frontcover&dq=pauli+general+principles+of+quantum+mechanics&hl=en&sa=X&ved=0ahUKEwiOwKX1vvXNAhWJMI8KHSAODC0Q6AEIHDAA#v=onepage&q=discrete%20eigen%20values&f=false
But I do know cases, for example a particle on a circle or other cases with periodic boundary conditions, where momentum operator $P$ has only discrete eigen values. But then shouldn't Pauli's argument also say that there is no $X$ operator for such systems, since the commutation relation between $X$ and $P$ is the same as the commutation relation between $T$ and $E$ assumed by Pauli?
| Commutation relations between position and momentum operators, in quantum mechanics, are valid only when the actions of either of the two operators is contained in the domain of the remaining one. In particular we have:
$$
[x,p]\psi = (xp)\psi-(px)\psi.
$$
In order for the above to be defined $\psi$ must be in the domain of definition of both operators and, also, $x\psi$ must be in the domain of $p$ (and viceversa). For the particle on the circle it is just not the case.
Also, commutation rules between time and energy are not actual commutation rules, rather they are indicative statements that only hold in particular cases.
| {
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GR says that time and space are aspects of the same thing, yet there is no observable for time in QM I understand that the topic of a time operator in quantum mechanics has come up more than a few times so forgive me if this is a repeat question but I couldn't find anything specific to my question.
My question is more to do with the relationship between general relativity and quantum mechanics.
My understanding is that according to GR, time and space are elements of the same thing, the spacetime manifold. So I would have thought that (up to a point at least) the mathematics of time would be the same as for position. Yet in QM, there is no operator for time... Why is this? I am prepared to accept that time is a parameter in QM, not a variable, but then why is the same not true for space? General relativity allows me to calculate a spacetime metric but QM tells me time is not a variable...
To be clear, my question is: Why does general relativity tell me that space and time are very similar, but quantum mechanics tells me they are very different?
Furthermore, is this in any way related to the incompatibility between QM and GR?
| Because time itself is the driving force of motion, whereas space is not, space is expanding as is time, but space doesn't measure motion, it's a measure of the area we occupy. If I have 80 billion units of space but no measurable time then the energy of motion would not exist, therefore nothing could move and space would remain a fixed location. Since time and space are relatively infinite, in each aspect a measurable distance, they couldn't get around how time and space operate so they made it the same entity.
| {
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What is the lowest energy atomic transition ever detected and identified? In this concise and insightful answer to Which is larger, all known, bound nuclear energy levels, or known, bound atomic energy levels? it was pointed out that atoms can have an extremely large (possibly infinite) number of bound excited states.
The Rydberg formula represents a simple model for the energy of transitions in a hydrogen-like (one-electron) atom. Originally written for the measured vacuum wavelength of the photon released by the transition:
$${1\over\lambda_{vac}}=\text{R}_{\infty}Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$
because the original work was done with grating spectrographs. Z is the atomic number of the nucleus. It is often written in terms of energy as well:
$$E=h\nu = \text{Ry}Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$
where the Rydberg energy Ry is roughly 13.6 eV.
As was pointed out, as $n$ increases, the number of energy levels for this simple model increases (mathematically, at least) without limit. So possible transitions can have arbitrarily small energy, and therefore long wavelength.
It's often the low energy transitions measured in radio astronomy that allow mapping gas temperatures and distribution, rather than the optical transitions.
My Question: What is the lowest energy atomic transition ever detected and clearly identified? I've added the discussion of the Rydberg model just to help explain why there can be extremely low energy transition. If there is a really compelling case for a lowest energy record that was achieved in - for example - a diatomic molecule, that would be OK here - as long as it is a well defined and unambiguously identified transition.
Naturally occurring atomic hyperfine transitions are fine (no pun intended) but I want to rule out things like NMR, where the transitions and their energy are caused and defined by (respectively) an externally applied fields. This should be about transitions of the atom itself.
| The 21cm line from atomic hydrogen is pretty low. Caused by the transition from parallel to anti-parallel spins between the atom and the electron.
| {
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Principle of Least Action Question Let's say we have a particle with no forces on it. The path that this classical particle takes is the one that minimizes the integral
$$\frac{1}{2}m\int_{t_i}^{t_f}v^2dt.$$
So if we graph this for the actual path a particle takes it is a straight, horizontal line on the $(t,v^2)$ plane. But couldn't we lessen the integral if we first slow down and then speed up near the end to create a sort of parabolic line that has less area under the $(t,v^2)$ plane? So why doesn't the particle take this path? What am I missing in my thinking?
| You're missing that Dirichlet boundary conditions
$$ x(t_i)~=~x_i \quad\text{and} \quad x(t_f)~=~x_i $$
are implicitly implied. The stationary action principle is not well-posed without boundary conditions.
| {
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How many photons are needed to make a light wave? What is the smallest number of photons needed to make a "light wave"?
In other words, how many (coherent?) photons start to exhibit classical behavior?
For example, how many photons are needed to get linear polarization? (Single photon has circular polarization.)
| Even though there is a single photon in a volume of your choice the light is still a wave.
An experiment was performed which proved this. In this experiment a Michelson interferometer was set up and the incident light is so weak that only one photon was in the whole setup at a time. A photographic plate was used to detect the interference pattern. Now just imagine one photon is being split up by the beam splitter and combined on the detector to give the interference pattern.
After several hours of the exposure people have recovered the classical interference pattern is generated (as if one photon has interfered with itself).
Hence the interference pattern (the classical proof that light is a wave) is just our perception, It remain wave all the time whether it is one photon or one million.
| {
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$(1/2, 0)$ representation of the Lorentz Group $SO(1,3)$ Let us consider the $(j, j') = \left(\frac{1}{2}, 0\right)$ representation of $SO(1, 3)\cong SU(2) \otimes SU(2)$.
$j = \frac{1}{2}$ corresponds to $SU(2)$ generated by
$$ \tag{1} N_i^+ = \frac{1}{2} \left(J_i + i K_i\right); \quad i =1, 2, 3. $$
$j' = 0$ corresponds to $SU(2)$ generated by
$$ N_i^- = \frac{1}{2} \left(J_i - i K_i\right); \quad i =1, 2, 3. $$
For $j' = 0$ representation of $SU(2)$, the generators
$$ \tag {2}N_i^- = [0] = 0 \Rightarrow J_i = iK_i$$
Eq.(1) then implies that
$$ \tag{3} N_i^+ = \frac{1}{2}(iK_i + iK_i) = iK_i = \frac{1}{2} \sigma_i;$$
where the Pauli matrices $\sigma_i$ are the generators of $SU(2)$ for $j = \frac{1}{2}$.
Therefore $K_i = \frac{-i}{2} \sigma_i$, $J_i = i K_i = \frac{1}{2} \sigma_i$ and an element of $\left(\frac{1}{2} , 0\right) = \exp\left(i \vec{\theta} \cdot \vec{J} + \vec{\phi} \cdot \vec{K} \right)$.
My Question:
In Eq. (2), $ \quad N_i^-$, $J_i$ and $K_i$ all are $1 \times 1$ matrices. Then how can we substitute $J_i = iK_i$ in Eq. (3), where $N_i^+$ is a $2 \times 2$ matrix? Addition of a number with a $2 \times 2$ matrix is not possible.
Inspiration: This question is inspired by the derivation provided in the book named "Symmetry and the Standard Model" by Matthew Robinson (page: 122).
| This is what happens when physicists try to do group theory but don't bother introducing the proper mathematical notions.
*
*There is no isomorphism between $\mathrm{SO}(1,3)$ and $\mathrm{SU}(2)\times\mathrm{SU}(2)$, the former is non-compact, the latter is compact. More around this apparently confusing topic can be found in this answer. Furthermore, using the tensor symbol $\otimes$ is wrong, the product is a direct product, not a tensor product, of groups, see also this question.
*What is true is that there is an equivalence of finite-dimensional representations of the algebras $\mathfrak{so}(1,3)$ and $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, the latter is the compact real form of the complexification of the former. Indeed, the map between them is given by using $N_i^\pm = \frac{1}{2}(J_i \pm\mathrm{i}K_i)$ as the basis of the latter in terms of the basis $J_i,K_i$ of the former. This is also not an isomorphism of Lie algebras, it is just the case that the finite-dimensional representations of these algebras are equivalent.
*The argument in the bit that confuses you is supposed to go as follows: You are given the $(1/2,0)$ representation $\rho : \mathfrak{su}(2)\oplus\mathfrak{su}(2)\to\mathfrak{gl}(\mathbb{C}^2)$. Since $\rho$, as a representation, is a Lie algebra homomorphism, you know that $\rho(N_i^-) = 0$ implies $\rho(J_i) = \mathrm{i}\rho(K_i)$. Here, all matrices $N_i^-,J_i,K_i,0$ matrices are two-dimensional matrices on $\mathbb{C}^2$. You know that $\rho(N_i^-) = 0$ as two-dimensional matrices because of how the $(s_1,s_2)$ representation is defined: Take the individual representations $\rho^+ : \mathfrak{su}(2)\to\mathfrak{gl}(\mathbb{C}^{2s_1+1})$ and $\rho^- : \mathfrak{su}(2)\to\mathfrak{gl}(\mathbb{C}^{2s_2+1})$ and define the total representation map by
$$ \rho : \mathfrak{su}(2)\oplus\mathfrak{su}(2)\to\mathfrak{gl}(\mathbb{C}^{2s_1+1}\otimes\mathbb{C}^{2s_2+1}), h\mapsto \rho^+(h)\otimes 1 + 1 \otimes \rho^-(h)$$
where I really mean the tensor product of vector spaces with $\otimes$. For $s_1 = 1/2,s_2 = 0$, this is a two-dimensional representation where $\rho^-$ is identically zero - and the zero is the two-dimensional zero matrix in the two-by-two matrices $\mathfrak{gl}(\mathbb{C}^2)$.
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Spring behaviour under high atmospheric pressure We have a spring inside a chamber. We compress the spring and then let it decompress freely. Will its decompression (its speed and displacement) be the same if the air pressure of the chamber is $1\;\mathrm{atm}$ or $3000\;\mathrm{atm}$? If not, how will it be affected?
| In 3000atm it's speed of decompression will be slower because it is facing greater air density and the expanding spring has to move it. There will also be less "resonance" as the denser air damps the spring movement.
| {
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Why is the resistance in a parallel circuit smaller than the resistance in a series circuit? So I was told in the physics class that the resistance in a parallel circuit is smaller than the resistance in a series circuit. Why does that happen?
Is this statement also true for circuits which have no resistors or resistance-offering devices connected to them?
And also from my calculations the total resistance in a parallel circuit is smaller than the resistances of any of the devices connected in the circuit. Now I really don't understand how this can happen.
| From the wording of the question I would assume the OP didn't want formulas or a very technical answer, so I'll attempt answering in layman's terms.
What does resistance do? It resists the flowing of current. Given the same voltage, the bigger the resistance, the smaller amount of current can flow.
Now, imagine that there is a resistor. You put another one in parallel. Now, current can flow through both ways, so it's easier for current to flow. Imagine that previously there was only one lane on the highway, now there are two. More cars can get through during the same time. So, parallel resistors allow more current. This means that the system as a whole can let more current through, so its resistance is lower.
Now, imagine two resistors in series, after each other. Current has to pass through both of them, and both will resist the flow of current. So less current can pass through, therefore the system as a whole "resits the current more".
| {
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Why is current the same in a series circuit? So I am a 10th grade student and my teacher told me that the current is the same at every point in a series circuit. It does split up in parallel circuit but it then recombines and the current flowing out of the battery is the same as the current flowing back into it.
My question is - Why does the current remain the same?
I mean let's say that there is a light bulb somewhere in a series circuit. Now, current(or electrical energy) will flow into it and then convert into light energy.
But if the amount of current flowing into the filament of the bulb = the amount of current flowing out of the filament and at the same time it is producing photons(light energy) [and some heat energy too] then aren't we creating energy ? Which is not possible.
I am really confused and can't seem to grasp this idea. Please help.
| Current is a measure of how many electrons go past a particular point in the circuit every second. So there are electrons rushing into one side of the light bulb, and rushing out of the other side. The number rushing IN each second is equal to the number rushing OUT each second. If that wasn't the case, then there'd be a build-up of electrons inside the light bulb; which would hinder the flow of further current.
Although the number of electrons rushing into the light bulb each second is equal to the number rushing out of it, each electron loses a little bit of energy on its way through the light bulb; and that's where the light comes from.
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Does the charge of a black hole affect its space-time geometry? Does a black-hole of a given mass and angular momentum have the same space-time geometry regardless of its charge? I'm pretty sure that an electric field can accelerate a charged particle but doesn't curve space-time so the only way an electric field can affect space-time is by accelerating a mass which then produces a different gravitational field because it has a different position and velocity because of its acceleration. Does that mean a charged black hole will have the same gravitational field outside its event horizon because the charge in gravitational field inside the event horizon produced by accelerating charges can't escape the black hole and therefore the electric field of a charged black hole will not accelerate another charged black hole?
| So the answer is no.
An electric field has energy and energy generates a gravitational field, just like any mass.
See the charged black hole solution is the Wikipedia article https://en.m.wikipedia.org/wiki/Charged_black_hole
The charge of a black hole, if nonzero, changes the metric and solution to account for the charge and electric field. That Wikipedia article has reference to the solution, called Reissner Nordstrom, for spherically symmetric, non rotating black holes. If they rotate it is the Kerr Newman solution.
Both exist because charges have electric fields, and those have energy. And because charge is a conserved quantity charge is not radiated away in a black hole (the no hair theorem came about because conserved quantities can not be radiated away, and that is mass, angular momentum and charge).
So charged black holes have a different, but similar, gravitational field as uncharged ones. See the Wikipedia articles.
Since black holes have charge, and a static electric field that is manifest outside the black hole (just like the static gravitational field is), they definitely can interact with charged particles or bodies, including othe charged black holes.
| {
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Confusion in a trick in solving an energy eigenfunction Given a non-relativistic energy eigenfunction for a central potential $\left|\Phi \right>$
In solving relativistic hydrogen atom, one of the terms is
$$
\left<\Phi\middle|\frac{e^2}{r}\middle|\Phi\right> \tag{1}
$$
I have read a trick to solve it is:
$$
\left<\Phi\middle|\frac{e^2}{r}\middle|\Phi\right> =\left<\Phi\middle|\frac{-e}{2}\frac{\partial}{\partial e}\left[\frac{\hat {P}^2}{2m}-\frac{e^2}{r}\right]\middle|\Phi\right>
=
\left<\Phi\middle|\frac{-e}{2}\frac{\partial}{\partial e}\middle[\hat {H}\middle|\Phi\middle>\right]\tag{2}
$$
and it gives the correct value $$-\frac{me^4}{\hbar^2n^2} .\tag{3}$$
In order to understand this trick, I tried this in another term:
$$
\left<\Phi\middle|\frac{e^4}{r^2}\middle|\Phi\right>
=\left<\Phi\middle|\frac{e^2}{r} \frac{e^2}{r}\middle|\Phi\right>
=\left<\Phi\middle|\frac{e^2}{4}\left[\frac{\partial}{\partial e}\left(\frac{P^2}{2m}-\frac{e^2}{r}\right)\right]^2\middle|\Phi\right>
,\tag{4}
$$
assuming $(\frac{\partial}{\partial e}\hat{H})^\dagger = \frac{\partial}{\partial e}\hat{H}$.
then this equals
$$
\left<\frac{e}{2}\frac{\partial}{\partial e} \left(\frac{\hat{P}^2}{2m}-\frac{e^2}{r}\right)\Phi\right|\left|\frac{e}{2}\frac{\partial}{\partial e}\left(\frac{\hat{P}^2}{2m}-\frac{e^2}{r}\right)\Phi\right>
= \left(\frac{me^4}{\hbar^2n^2}\right)^2
.\tag{5}
$$
However, since this very term is supposed to tell us how the relativistic effect destroy the symmetry, while this result gives us no degeneracies breaking. As it turns out, the correct answer for this term is $$\frac{m^2e^4}{(\ell+1/2)\hbar^4n^3}.\tag{6}$$
It takes me so long that I still can't figure it out where is wrong, I suspect $\frac{\partial}{\partial e}\hat{H}$ is not Hermitian.
| The theory behind the trick is based on the Hellmann-Feynman (HF) theorem
$$ \frac{dE_{\lambda}}{d\lambda}~=~\langle \psi_{\lambda} | \frac{d\hat{H}_{\lambda}}{d\lambda}| \psi_{\lambda} \rangle,\tag{A}$$
which works with a single derivative, but not with a square of a derivative, cf. OP's failed calculation (5) for the expectation value $\langle\frac{1}{r^2}\rangle$. Incidentally on the Wikipedia page, the correct result (6) for $\langle\frac{1}{r^2}\rangle$ is obtained via the HF theorem by varying wrt. the azimuthal quantum number $\ell$ rather than the electric charge $e$. (Concerning a subtlety in the variation wrt. $\ell$, see also this related Phys.SE post.)
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Does the force on the bottom piece of a hummingbird feeder depend on the height of the water in the reservoir? Intuitively I would think that the more nectar you put into the reservoir the more weight it will put on the bottom piece (the tray that holds the open pool of nectar that the birds can feed out of). While I was trying to think about this I thought about doing some free body diagrams. The reservoir would have a force acting upwards on it from the hanger, and a downwards force from the sheer from the stored nectar. The bottom piece would have an upwards force from where it threads onto the reservoir and I can't really figure out how to do the downwards force. I'm thinking if I could calculate the pressure all along the interface between the pan and the nectar that would do it..
| The force the fluid does on the bottom piece does not depend on the height of the water column of the reservoir. It does depend on the height $h$ of the water column in the plate.
This can be easily seen by the fact the water is static so the pressure at any horizontal plane is the same. The pressures in $a$ and $b$ are the same. The force of the water on the plate is $F=pA=\rho g h A$, where $\rho$ is the water density and $A$ is the area of the plate.
It is definitely counterintuitive so I had to go get an empty bottle of wine and a plate to check it. As long as the bottle does not touch the plate, the latter weights the same no matter if the bottle is full or almost empty.
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Quick question: Curvature form of a connection on the trivial bundle Let $L=\mathbb{R}^2\times U(1)$ be the trivial $U(1)$-bundle over $\mathbb{R}^2$. Define a connection $\nabla=d+A$ where $A=fdx+gdy$ is an $\mathbb{R}$ valued $1$-form on $L$. That is, $\nabla$ gives a distribution $\mathcal{H}$ on $L$ - the horizontal distribution.
The distribution $\mathcal{H}$ is obtained as the graph of $-A$ as a linear map from $\mathbb{R}^2\rightarrow\mathbb{R}$. A horizontal lift $\tilde{X}$ of a vector field $X$ on $\mathbb{R}^2$ is given by $\tilde{X}=(X,-A(X))$.
Let $\alpha$ be the projection onto the vertical direction on $L$ i.e. $\ker(\alpha)=\mathcal{H}$, and define the curvature $2$-form $\Omega_{\nabla}$ of the connection $\nabla$ by
$$\Omega_{\nabla}(X,Y)=\alpha([\tilde{X},\tilde{Y}])$$
The following is expected to be true
$$\Omega_{\nabla}=-dA?$$
Here is my confusion:
Let $z$ be the local vertical coordinate, $X=\partial_x$ and $Y=\partial_y$. Then $\tilde{X}=-f\partial_z+\partial_x$ and $\tilde{Y}=-g\partial_z+\partial_y$. And
$$-dA(X,Y)=-X(A(Y)+Y(A(X))+A([X,Y])=-\partial_xg+\partial_yf$$
while
$$[\tilde{X},\tilde{Y}]=(f(\partial_zg)-g(\partial_zf)-\partial_xg+\partial_yf)\partial_z$$
therefore
$$\Omega_{\nabla}(X,Y)=\alpha([\tilde{X},\tilde{Y}])=f(\partial_zg)-g(\partial_zf)-\partial_xg+\partial_yf.$$
Where is the mistake?
| $f$ and $g$ are defined on the base space to begin with. You would need to extend to functions on the total space in order to define taking $\partial_z$.
| {
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Could a radar pulse be considered particular in nature? I ask because I have a real problem with wave/particle duality (one of the biggest cop outs in the whole of physics, in my opinion)
A radar pulse is definitely a wave.
It is spatially bounded. (at least in one dimension!)
It has energy.
It is a 'quanta'.
Is this not particle-like? Could a pulse of sufficiently high energy have mass in accordance with E=MC^2 and E=hf?
Apologies for the ignorance of a simple engineer.
| $E=mc^2$ is true only for a particle at rest, the photon is never at rest.
The full equation is
$$E^2=p^2c^2+m^2c^4.$$
A photon has no mass, but two photons can meet to form an electron and a positron, see pair production.
| {
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Radius of centripetal acceleration Suppose you are moving in circle of radius $r$. So there should be centripetal acceleration towards the center. Now you want to decrease the radius of the circle, so someone should apply more centripetal acceleration in order to decrease your radius. But I had read that with the increase in radius, there is increase in centripetal acceleration. What is the correct explanation for without the use of formula? Making it more clear- "My intuition tells me that if I decrease the central force on an object in uniform circular motion, the radius of the circle should increase?"
| Let's use the classic example of a ball revolving around your fingers via a tensed string.
There is a tension in the string because the ball is travelling in a circular motion. The tension force is constantly causing the ball to accelerate toward your fingers. The ball, at all times, wants to travel in a tangentially straight line away from your hand. The tension is constantly preventing this from happening. The tension is related to 3 things; the mass of the ball, the velocity of the ball, and the radius of the string.
If you increase the mass of the ball, the tension will increase to keep it in orbit at its present velocity and radius because more force is require to restrain the greater mass in the existing orbit.
If you increase the radius of orbit and keep the velocity and mass the same, then the orbit is a larger and gentler curve. You could increase the radius so much that the curve would be VERY gentle and theoretically approach the tangential line the ball truly wants to follow. This increased radius and resultant gentler curve causes the restraining tension to DECREASE. If you DECREASE the radius under these same circumstances, the curve is so tight and such a departure from the desired tangential line that the ball wants to follow, then the restraining tension must INCREASE to accomplish this.
| {
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What does "non-normal parity" mean? Nuclear physicists seem to use the term "non-normal parity" a fair bit. Googling for the term yields some 840 results, and the same search on Google Scholar indicates that about 430 of those are research papers.
Unfortunately, those searches seem to only yield research-level papers that just state the term and roll with it, assuming the reader understands it and without stopping to explain it.
So: what does "non-normal parity" mean?
|
A state with parity of $(−1)^L$ is defined as a natural parity state, with $L$ being the total angular momentum of the system, and parity with value of $+1$ or $−1$ are called even or odd, respectively. A state with parity of $(−1)^{L+1}$ is called unnatural parity state.
Source: Ye Ning et al., Natural and Unnatural Parity Resonance States in the Positron-Hydrogen System with Screened Coulomb Interactions. https://doi.org/10.3390/atoms4010003
| {
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Could all the energy in all the photons in the universe account for dark matter? I was hit with fridge logic, and I am curious:
Is it possible that the gravitational influence of photons inside of galaxies (And all throughout the universe) could account for dark matter?
Photons would be most concentrated close to the core and inner galaxy... And have a lessening concentration as they expanded away from galaxies due to the inverse square law..
I am struggling to understand the math involved, because I've not yet reached that level in my education.... But I'm really curious how the total photons stack up vs the measured dark matter in the universe?
| The problem here is the contrasting effects the two forms of matter have.
Photons are massless; they create negligible gravitational fields.
Dark matter, on the other hand, is about 85-90% of all mass in the universe. It's responsible for holding galaxies at constant rotation rates past certain radii (see Galactic Rotational Velocity Curves).
Photons, on the other hand, provide a repulsive force that we call radiation pressure.
In terms of contribution to the energy density of the universe:
Let the critical density $\Omega_c=1$. This is the energy density of flat space, and provides an accurate description of the large-scale universe by current measurements.
$\Omega_m$ is the energy density contribution of barionic matter. It's approximately equal to $0.3$. Of that $0.3$, visible matter contributes about $0.05$.
$\Omega_\Lambda$ is the energy density contribution attributed to the cosmological constant $\Lambda$, and is about $0.7$, or about 70% of the energy in the universe.
You might have noticed that $\Omega_m + \Omega_\Lambda \approx 1 = \Omega_c$. Where's the light/radiation?
$\Omega_r$, the energy density contribution of radiation, is $\approx 0.00001$.
| {
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Are Newton's 1st and 3rd laws just consequences of the 2nd? Can Newton's 1st and 3rd laws be assumed given just $F=ma$. I know that the argument would be, "No, then there would only be 1 law". But I can't think of any situation where 1 and 3 aren't superfluous.
If you just told me $F=ma$:
I would assume nothing else causes an acceleration besides a force. So things not experiencing a force don't change velocity, even when velocity is 0. 1 ✔️
And, when two things that exist interact they use only their mass and acceleration to do so so they both must change in opposite ways. 3 ✔️
| Newton's first law defines the inertial reference frames:
*
*There exist in the universe some very particular reference frames such that, in those very particular reference frames (and only in those and no more) a body not subject to external forces or interactions moves with constant velocity.
Newton's second law states the change in motion in the above defined reference frames (and only in the above)
*
*In the above defined reference frames (see law 1) a body subject to external forces behaves as $$\textbf{F} = \dot{\textbf{p}}$$
I don't see how law 1 is a particular case of law 2, as law 2 is only valid after law 1 defines the inertial reference frames.
If you just told me F=ma...
the above is only valid in the reference frames defined by the first law.
And, when two things that exist interact they use only their mass and acceleration to do so so they both must change in opposite ways.
One does not know in principle how two things interact with each other. In particular that they must both change in opposite ways is a non-trivial statement. There is no a priori reason why it should be so (it could be anything else).
| {
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Interference with a minus sign In Lectures on Physics, by Richard Feynman, pg 3-11, I found the following:
In case of electrons, the interfering amplitudes for exchange interfere with a negative sign.
I was unable to understand this. On reading further, it seems that electrons have spins unlike alpha particles and that causes them them to interfere in this manner.
I am unable to understand the connection between the statement and the reason given. I would be grateful for any clarity on this matter.
I didn't truly understand the + sign either, but it made more sense when I thought about it like either this or that happens to reach the final state, there are two ways to get there, so I need to add the probability amplitudes of both to get the final amplitude. (We always use + to indicate or in probability.)
| This might be something related to particle statistics: Electrons have spin 1/2 and thus belong to the group of fermions. Exchanging fermions causes an additional minus sign for the state. Alpha particles are made up from two protons and two neutrons, all of them spin 1/2 particles. These spins can add up to either zero, one or two and these integer spins belong to the group of bosons. Boson-exchange comes without an additional minus-sign.
| {
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Does time period of a simple pendulum vary if I heat its metallic bob? How does the time period of a simple pendulum with a metallic bob vary if we heat its metallic bob? The pendulum is assumed to be a simple one and air drag is taken to be negligible.
Please provide a proof if possible.
| Let's assume the length of the string remains constant, i.e. only the bob is affected by the heating. This appears to what the question implies. In that case to a first approximation there will be no effect on the period. The period is given by:
$$ \tau = 2\pi\sqrt{\frac{\ell}{g}} $$
and since neither $\ell$ nor $g$ are affected by heating the period does not change.
However I can think of two ways in which the period could change very slightly due to heating, and both are related to the thermal expansion of the bob.
Effect on the bouyant force
Firstly we assume that the downwards force on the bob is given by:
$$ F_d = mg $$
However there is a small bouyant force due to the volume of air displaced by the bob, so the force is actually:
$$ F_d = mg - V\rho g $$
and taking this into account the period is:
$$ \tau = 2\pi\sqrt{\frac{m\ell}{(m-\rho V)g}} $$
where $\rho$ is the density of the air and $V$ is the volume of the bob. If we heat the bob its volume will increase due to thermal expansion so the period will increase.
Effect on the moment of inertia
The other affect arises from the approximation that we treat the bob as a point mass. The equation of motion of the pendulum is:
$$ T = I\ddot{\theta} $$
where $T$ is the torque, $I$ is the moment of inertia and $\theta$ is the angle of rotation. For a point mass, and using the small angle approximation, $T=mg\ell\theta$ and $I=m\ell^2$ so we get:
$$ mg\ell\theta = m\ell^2\ddot{\theta} $$
which rearranges to our usual equation:
$$ \ddot{\theta} = \frac{g}{\ell}\theta $$
However suppose the bob is a sphere or radius $r$. The moment of inertia of a sphere about an axis though the centre is:
$$ I_\text{sphere} = \tfrac{2}{5}mr^2 $$
So using the parallel axis theorem the moment of inertia about the axis a distance $\ell$ away is:
$$ I = \tfrac{2}{5}mr^2 + m\ell^2 $$
In that case the equation of motion is:
$$ mg\ell\theta = -m\left( \tfrac{2}{5}r^2 + \ell^2 \right)\ddot{\theta} $$
and rearranging gives:
$$ \ddot{\theta} = -\frac{g\ell}{\tfrac{2}{5}r^2 + \ell^2}\theta $$
so our period becomes:
$$ \tau = \sqrt{\frac{\tfrac{2}{5}r^2 + \ell^2}{g\ell}} $$
If we heat the bob its radius $r$ will increase due to the thermal expansion so the period will increase with heating.
Summary
So both the effects discussed above mean the period will increase if the bob expands due to heating. However while this sort of exercise is fun to do I must emphasise that these effects will be tiny and in most cases dwarfed by other approximations such as the small angle approximation and ignoring air resistance.
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Wet metal disks stuck together - Casimir Effect or surface tension? I work in a processing plant where round steel cutting blades are used, e.g. 1-2 mm thick and 15-30 cm in diameter. During normal plant operations, these blades are mounted on machinery and rotated at circa 1000 RPM -- thus they are perfectly flat whose outer edges are beveled inward.
I've noticed that often multiple blades tend to stick together, especially when they are wet (being washed and rinsed with water). Sometimes a generous force is needed to "pry" apart two (or more) blades stuck together.
My question is: Is this from the Casimir effect, or just surface tension?
| Surface tension is part of it, but probably it's straight adhesion. Glue. A film of water on (for instance) glass can be wiped off, but usually that just thins the film, the last few microns of water film sticks hard to the surface.
If two flats have a thick layer of water between them, it's a lubricant (you get streamline flow in the water), but a thin layer has significant shear strength due to viscosity. A thin layer is an adhesive.
Pulling the plates apart requires both making new surface area of the film (and that's the surface tension part), and making a thicker spacing of the plates (which happens if the water film transports water to the 'thick' region, by flow in the thin layer). That flow in the thin layer is very slow because of the water's viscosity.
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Difference between analog signal and digital signal There is no perfect generator of a true digital signal in the real world natural phenomenon. Most of the physical signals produced by humans and animals like roaring, speaking, playing, clapping, electrostatic charging, drumming are all analog in nature. Even physical heavenly bodies like sun, earth, planets, moon, rocks, mountains generate signals that are analog in nature like seashore sound waves, light pattern, radioactive decay, gravitational pull, centrifugal force, natural growth rate are all analog in nature.
Correct me if I'm wrong, what i see is that all naturally generated signals are analog in nature. What we see as digital signals are those analog signals sampled in space or time. Is there exists a true generator source of digital signal in nature? Is it correct to say that analog signals are natural and digital signals are man-made?
| You are right. Everything in nature is analog (including man-made "digital" signals). The digitization of signals is a man-made abstraction. Nature just evolves as it does and humans then categorize observed phenomena. The assigning of one signal as a "1" and the other signal as a "0" is a categorization.
It is possible in nature to have a system that has bimodal output (that is outputs are clustered around two qualitatively different behaviors) but there is always some noise and everything is "analog" in the end.
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How is potential energy lost when a water droplet is dropping down slowly on a wall? When a water droplet is on a vertical wall, it usually drops slowly, which is different from free falling. As the dropping speed is slower than free drawing, so I guess some energy must be lost.
I guess it is lost as internal energy, but if it is true, how a water droplet gains internal energy when dropping slowly in microscope view?
| You are not dealing with dry friction here. What is involved is viscous flow within the water droplet. There is a viscous circulation pattern set up within the water droplet, and the flow results in viscous dissipation of the (mechanical) potential energy to internal energy. So, if the water droplet where somehow insulated from its surroundings thermally, its temperature would rise (a tiny amount). In reality, the thermal energy is transferred to the surroundings.
| {
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Why the deflection angle is important for verifying GR? Why the deflection angle is important for verifying GR? What about redshift?
But in particular deflection angle, because we have it in Newtonian Gravity as well.
Please guide me with your answer.
| Newtonion theory did predict the gravitational deflection, but gave an estimation of half of that which GR provided. Experimental evidence showed that GR was a more accurate model of gravitation than the previously universally accepted Newtonion theory.
The details below are a quick summary of this article Gravitational Deflection, but there is far more on that page than my short summary here.
The first calculation of the deflection of light by mass was published by the German astronomer Johann Georg von Soldner in 1801. Soldner showed that rays from a distant star skimming the Sun's surface would be deflected through an angle of about 0.9 seconds of arc, or one quarter of a thousandth of a degree. This angle corresponds to the apparent diameter of a compact disc (CD) viewed from a distance of about 30 kilometers (nearly 20 miles). Soldner's calculations were based on Newton's laws of motion and gravitation, and the assumption that light behaves like very fast moving particles.
In the early 1900's, Einstein developed his theory of general relativity. Einstein calculated that the deflection predicted by his theory would be twice the Newtonian value. When the results of his prediction were confirmed by experiment, Einstein is reported to have suffered heart palpations caused by both excitement and relief that his 10 years work was a success.
The following image shows the deflection of light rays that pass close to a spherical mass. To make the effect visible, this mass was chosen to have the same value as the Sun's but to have a diameter five thousand times smaller (i.e., a density 125 billion times larger) than the Sun's.
According to general relativity, a light ray arriving from the left would be bent inwards such that its apparent direction of origin, when viewed from the right, would differ by an angle (α, the deflection angle; see diagram below) whose size is inversely proportional to the distance (d) of the closest approach of the ray path to the center of mass.
| {
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What is the shape of a deuterium nucleus? What is the shape of a deuterium nucleus?
I can think of two obvious extremes.
A positive proton end intersecting with a neutral neutron end.
Or a cylinder with spherical caps on the ends that is positive on one end and neutral on the other.
| The deuteron has an r.m.s. charge radius $r=2.14$ fm and a static electric quadrupole moment $Q_0=0.2859\ e\cdot\text{fm}^2$. This is related to the nuclear deformation parameter $\beta$, describing prolate approximately ellipsoidal deformations, by
$$\beta \approx 0.080\frac{Q_0}{e\cdot r^2} \approx 0.050.$$
The nuclear shape, in this parametrization, is defined by the surface
$$\frac{R}{R_0} = 1+\beta Y_{20}(\theta),$$
which gives for the ratio of the long axis to the short axes
$$\frac{\text{long}}{\text{short}} \approx 1+\frac{3}{4}\sqrt{\frac{5}{\pi}}\beta \approx 1.047.$$
So the deuteron is very nearly spherical, with its long axis being about 5% longer than its short axes.
This is because the dominant component of the nuclear wavefunction is one in which the neutron and proton each have orbital angular momentum 0 (and total intrinsic spin 1), while there is a small admixture of a state with orbital angular momentum 1 (and intrinsic spin 0).
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Spring force on both sides of spring I am a little confused about springs. I just wanted to know that if I pull an ideal spring of spring constant $k$ such that the spring has been symmetrically pulled and its elongation (total) comes out to be $x$ then would the force on one side by $$F=kx$$ or $$F=kx/2$$ I am a little bit confused and hence I resorted to ask it here.
|
An ideal spring - Three loading cases : free,compressed,stretched.
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Heat pump intuition What is an intuitive explanation for the concept of heat pumps? I know that it is basically a reversed Carnot process. We can for example take an amount of heat $Q_1$ out of a warmer system and transform part of it into work W. The rest goes to the colder system. If we now reverse that process we need to take heat out of the colder system. For this to be done we need the same amount of energy W we got out of the process previously. But here my problem starts: How do you force the energy to come out of the colder reservoir? How can you explain that without just saying that it is an inversed Carnot process?
| You describe the process. For an example, you might think of how a gas refrigerator works. You take a gas and expose it to the cold area, which cools it to the cold temperature. Then you isolate it from the cold area, which costs no energy. You compress it, raising the temperature above that of the hot reservoir, at the cost of physical work. You then connect it to the hot reservoir and let some heat escape. Disconnect it again, then expand it to a temperature below the cold reservoir, recovering some energy in the process. Now connect it to the cold reservoir and you have a cycle with work in and heat moving from cold to hot. This is really just adding details to saying an inverse of the Carnot process.
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Angular Momentum of a Body Constantly Losing Mass My friend has a question: assume that a body such as a star loses mass as it orbits the galaxy. If there is conservation of angular momentum, how is this explained? What happens to its orbital velocity and orbital radius? Thanks.
| Well, the angular momentum is conserved when net torque on a body(or a system) is zero.
Now in your case, the torque produced by the gravitational force between the galaxy and the star is zero as the line of force passes through the axis of rotation.
For simplicity, i am neglecting any effect produced by other celestial objects of the universe.
Now, if the mass separates out tangentially from the star with the same speed, then it will apply no force on the star and as a result the speed of the star will remain unaltered. and if the speed is same, the star and the mass will revolve in the same orbit with same speed.(because the radius of orbit is given by $r=GM/V^2$ where M is the mass of galaxy) However if due to internal forces, the mass separates out with some different speed, then the internal force will change the linear momentum of the star(note that the linear momentum of the star+mass will be conserved for that time when it separates.).
As the Momentum changes, the speed will change and as a result the orbit too will change.
However, in each case the angular momentum of the star+mass will be conserved. Also, if your question is concerned about the angular momentum of the star only(not the mass), then it will not be conserved until the mass separates out with no speed.
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What is the difference between these two ways to calculate average velocity? Average velocity:
$$v_{\rm avg,1}=\frac{v_{\rm final}+v_{\rm initial}}{2}$$
and average velocity:
$$v_{\rm avg,2} =\frac{\rm total\;displacement}{\rm time \;taken}=\frac{\Delta x}{\Delta t} $$
What is the difference between them and when do we use them?
| Your first way of calculating an average velocity is inaccurate and really should be avoided.
Only the second method is accurate. This is a consequence of the underlying calculus of kinematics.
When a object travels (e.g. but not necessarily on a straight line) its velocity is not required to be constant. In fact for the general case we assume $v$ is a function of time, mathematically noted as:
$$\Large{v(t)}$$
Physically the velocity is the first derivative of position ($x$) to time ($t$):
$$\Large{v(t)=\frac{\text{d}x}{\text{d}t}}$$
To find the displacement $\Delta x$ during an interval of time $\Delta t=t_2-t_1$ then $\Delta x$ is calculated by integration:
$$\Large{\Delta x=\int_{t_1}^{t_2}v(t)\text{d}t}$$
This also means that the average velocity $\bar{v}$ can be calculated from:
$$\Large{\bar{v}=\frac{\Delta x}{\Delta t}}$$
This is true regardless of how $v(t)$ evolves over the time interval $\Delta t$.
Taking the "average" by averaging two velocity readings however is meaningless.
**In response to OP's comment about constant acceleration:**
If acceleration is constant the velocity is given by:
$$v=v_0+at$$
Where $v_0$ is the velocity at $t=0$.
After a time interval $\Delta t$ the velocity has become:
$$v_1=v_0+a\Delta t$$
The displacement would be:
$$\Delta x=\int_0^{\Delta t}(v_0+at)dt=v_0\Delta t+\frac12 a(\Delta t)^2$$
The average velocity $\bar{v}$ is:
$$\bar{v}=\frac{\Delta x}{\Delta t}=v_0+\frac12a\Delta t$$
Using the first method:
$$\bar{v}=\frac{v_0+v_0+a\Delta t}{2}=v_0+\frac12a\Delta t$$
So that in the case of constant acceleration we obtain the same result. Note that this is the only case where both give the same result.
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Boltzmann formulation of entropy I could be doing something quite stupid here; this is a tad over my head, so I apologize if this is a stupid question, and if requested to delete it, I will do so.
I know that Boltzmann's formulation of entropy is $S = k \log W$ where $k$ is Boltzmann's constant, $S$ is the entropy of the given system, and $W$ is the number of microstates within the given macrostate - or, to put it another, simpler way, the number of ways you can arrange the pieces of a system such that the overall picture of they system is the same (in, say, a steam cloud, one could say the cloud-ish look is the macrostate and the position of each specific molecule is a particular microstate). Yet another way, this is configurations in phase space.
I read the log relation
$$\log a^b = b \log a$$
Which made me think that Boltzmann's formulation of entropy could be written as $$S = \log W^k$$
I have two questions about this:
*
*Is this right? If not, where did I go wrong?
*If this is right, what are the advantages are writing it the first way versus the other?
Edit:
In light of CuriousOne's comment:
Gibb's formula for entropy is
$$S=-k_{{\text{B}}}\,\sum _{i}p_{i}\ln \,p_{i}$$
Which, by the same log rule gives
$$S = \log p_i^{-k_{{\text{B}}}\,\sum _{i}p_{i}}$$
At which point the same two questions apply. Also in terms of CuriousOne's comment, how is the second equation (in the two cases) physically meaningless?
| The expression $W^k$, where $k$ is Boltzmann's constant, does not make sense mathematically because $k$ is not a number but has dimension of Energy/Temperature. So its numerical value depends on the units you choose. For example if you measure it in Joule/Kelvin then its value is about $1.38 \times 10^{-23}$ but if you measure it in eV/Kelvin it is about $8.62 \times 10^{-5}$. So it is not clear to what power you want to raise the number $W$ when you write $W^k$.
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Kinetic energy of photoelectrons I came across the following question:
A photon of energy $h\nu$ is absorbed by a free electron of a metal having work function $W<h\nu$. Then:
*
*The electron is sure to come out
*The electron is sure to come out with kinetic energy $h\nu-W$
*Either the electron doesn't come out or it comes out with a kinetic energy $h\nu-W$
*It may come out with kinetic energy less than $h\nu-W$
Both options 1. and 2. seem to be correct to me. However, the correct answer provided is 4.! How is this so? Isn't $W$ the amount of energy required for ejection of the most tightly bound electron in the metal? Then why is there a possibility that it won't come out with kinetic energy $h\nu-W$?
| The work function $W$ is the minimum energy of photon that is required to eject the electron from the metal in photoelectric interaction i.e. the energy of least tightly bound electron in the lattice. However if the energy of the photon is higher than W then it can eject the electrons that are more tightly bound in the metal hence photo electrons with energy less than $h\nu -W$ can also be ejected.
| {
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Why is 3D stress tensor acting only on three surfaces? I'm trying to learn about the stress tensor (in 3D)
The tensors are said to have directions (the first subindex $i$ in $\sigma_{ij}$) and specify the surface upon which they act (the second subindex $j$ in $\sigma_{ij}$).
What confuses me is why is it defined only to act on three surfaces, even when the cube has six surfaces?
| That diagram shows only that set of three arrows for clarity: it is the smallest set that displays all the information necessary, but adding arrows to all six faces would make the diagram too crowded.
In general, if you have a material with a given stress tensor $\sigma_{ij}$, and you look at one small internal surface patch of area $A$ and normal vector $\vec n$, then the force $\vec F$ transmitted by the material's stress across that surface patch is given by
$$F_i=\sum_j \sigma_{ij}n_j A.$$
In particular, for the three faces shown with $\vec n=\hat e_k, \ k=x,y,z$, you get a force with $i$th component $\sigma_{ik}$ acting on the face with normal $\hat e_k$. For the back faces, $\hat n=-\hat e_k$ changes direction, so the forces change direction, but they're obviously still there.
(It's a good exercise at this point to draw those arrows for one face and its opposite. You should get normal arrows pointing away from each other, either pushing in (pressure) or pulling out (tension). The tangential arrows ($\sigma_{yx}$ and $\sigma_{zx}$, say) should point away from each other in a shearing motion. The net result should be deformations of the unit cube, but no net force.)
| {
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How to predict length of conductor? A long round conductor of cross sectional area $S$ is made of material whose resistivity depends only on a distance $r$ from axis of the conductor $\rho=\frac{\alpha}{r^2}$, where $\alpha $ is constant .Find the resistance per unit length of such conductor .
My work
$$dR=\frac{\rho l}{2\pi r dr} \space \text{or}\space dR=\frac{\rho dr}{\pi r^2} $$
$$dR=\frac{\alpha l}{2\pi r^3 dr} \space \text{or}\space dR=\frac{\alpha dr}{\pi r^4} $$
$$\int\frac{1}{dR}=\int_0^{a}\frac{2\pi r^3dr}{\alpha l} \space \text{or} \space \frac{1}{dR}=\frac{\pi r^4}{\alpha dr}$$
$$\frac{1}{R}=\frac{\pi a^2 . a^2}{2\alpha l} \space \text{or} \space \frac{1}{R}=\frac{4\pi a^3}{\alpha}$$
Now, question arises that how should I take length of conductor ? Should I take length of cylinder as length of conductor or $dr$ as length of conductor ? This all depends on how current move but question didn't specify anything about how current move .
| I think you have set out the problem incorrectly from the start: Instead of working with the resistance of the shell between $r$ and $r+dr$ and length of material $l$, first work with the conductance, $G(r)$, which is the reciprocal of the resistance. Because the shells are "in parallel" with each other, conductance's add: The incremental conductance due to the shell of radius $r$ and $r+dr$ is $dG(r) = \frac{2 \pi r dr}{\rho l}$, where $\rho$ is the resistivity and given as $\rho = \frac{\alpha}{r^2}$. This gives $dG(r) = \frac{2 \pi r dr}{l \alpha/r^2}$. You can then determine the total conductance as $G = \int_0^R dr \frac{2 \pi r}{l \alpha/r^2}$, where $R$ is the radius of the wire. From this the resistance, $R_{eff} = 1/G$ follows. Then write $R_{eff} = \frac{\rho_{eff} l}{\pi R^2}$.
| {
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Does a higher water volume increase pressure? I am constructing a gravity flow water system. I have 100ft point where I can put my tank. My question is does the size of my tank matter? I am using a 1" pipe. Will I get more pressure if I use a bigger tank? For example what is the difference in pressure if I use a 10 gallon tank or a 50 gallon tank?
| If you have a tank with an outlet at the bottom then the pressure in the outlet is the same as the pressure in the water at the bottom of the tank.
The pressure $P$ at the bottom of the tank is $P=\rho g h$ where $\rho$ is the constant density of water, $g$ is the acceleration due to gravity, and $h$ is the height of your tank.
This shows that for your tank the pressure at the outlet depends only on the height of your tank.
That sounds like an odd result but think about this:
If you swim to the bottom of a 2m deep pool you can feel some pressure. If you're in a small back-yard pool it feels like the same pressure as an olympic sized swimming pool. The volume of the pool isn't affecting the pressure 2 metres down.
| {
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When should you jump off a falling ladder? If you stand on the top of a falling ladder you will hit the ground at a higher speed (and therefore presumedly sustain more injury) if you hold on to the ladder than if you jump off it.
This was solved
here.
Where is the "break even" height on the ladder, from where you will hit the ground with the same speed if you jump off it or if you follow it down? This question just makes an assumption that you would hit the ground more softly if you stay on to the ladder (compared to jumping off it), if you are located at the lower part of it.
I don't think the midpoint is the break even point. I quickly calculated and I think you should stay on the ladder if you are on its midpoint. (The following is just a quick computation, there could be errors in it.)
$$ v_{midpoint}^2 = \frac{1}{2}gl \frac{m + m_L}{\frac{1}{4}m + \frac{1}{3}m_L} $$
$m$ and $m_L$ are the respective masses of man and ladder, $l$ is the length of the ladder.
| If one has the presence of mind, I vote for sliding down the ladder at any cost - gloves and shoes acting as a control brake. Shortening the diameter (height of your feet from ground) will surely reduce any impact. The risk is loss of control in the vertical decent. Gloves might be a recommended safety feature of any ladder climbing to keep confidence while executing a rapid slide. I would think you could get 6 feet lower before getting into the lateral arc. I admit that my own experience is limited to sliding off a wet cedar shingle roof and kicking the ladder away as I slid into it. It was only 10 feet onto damp lawn so I was saved by that and walked away. Yeah, I've had the training but I landed on my butt. No roll was possible.
| {
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Do uncertainty relations for charge or mass exist? Is there a uncertainty relation for charge $q$ of the form
$\Delta q \Delta? \geq \hbar$ in quantum mechanics? From checking the units ($[q] = A\cdot s$) I guess that $?$ would have to be the magnetic flux $[\Phi] = V \cdot s$, so we would have
$$
\Delta q \Delta \Phi \geq \hbar
$$
If I play the same game with mass $m$ the units would suggest a uncertainty relation
$$
\Delta m \Delta \Phi_g \geq \hbar
$$
where $[\Phi_g] = \frac{m^2}{s}$ is the gravitomagnetic flux (which happens to have the same units as the kinematic viscosity, the specific angular momentum, the mass diffusivity and the thermal diffusivity).
Do these two uncertainty relations exist in quantum mechanics?
| Regarding electrical charge the answer is definitely negative: In Quantum Mechanics there exists a so-called superselection rule of the charge which requires that the charge is always definite in every quantum state of any quantum physical system carrying electrical charge. So $\Delta q_\psi =0$ in every state $\psi$ and no Heisenberg relations are possible for whatever choice of a conjugated variable of $q$.
Regarding mass, in non relativistic QM the situation is identical: There is an analogous superselection rule (Bargmann's superselection rule).
In the quantum relativistic realm, for elementary systems the mass observable is actually a Casimir operator and thus, again it is always defined.
For composed systems or non-elementary relativistic systems, the situation is more delicate, but I think the only possibility to define some notion of mass is just the Hamiltonian observable defined with respect to a preferred reference frame at rest with the system. This way one immediately faces the problem of time-energy uncertainty relations which do not have a well-defined status in QM, since there is no self-adjoint operator representing the time observable (this is the so called Pauli's theorem).
| {
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Deceptively simple mass-spring problem? This question is inspired by two other, similar, so far unanswered questions (posed by different OPs).
Mass $m_2$ sits on a incline with angle $\theta$ that provides just enough friction for it not to start sliding down. It is connected by a massless string $S$ and perfect spring (with Hookean spring constant $k$) to mass $m_1$. Pulley $P$ is frictionless and massless. At $t=0$ the spring is not extended at all. Then $m_1$ is released.
Question:
What is the minimum $m_1$ to cause movement of $m_2$ up the incline?
Attempt:
Ignore the spring.
Determine static coefficient $\mu$ first.
\begin{align}m_2g\sin \theta &=\mu m_2g\cos \theta\\ \implies \mu &=\tan \theta\end{align}
To overcome the $m_2g$ component parallel to the inclined and the friction:
\begin{align}m_1g &\gt\mu m_2g\cos \theta+m_2g\sin \theta\\ \implies m_1 &\gt 2m_2\sin \theta\end{align}
But apparently this overestimates $m_1$. It has to be taken into account that $m_1$ starts accelerating before $m_2$ starts moving, because of the spring.
But how? Like several other members I can't see how the work done on the spring affects the minimum $m_1$. Conservation of energy?
| If $m_1$ is too small. The mass fall down and bounces back up. The force on the masses is highest at the lowest point of the bounce. A mass that is just large enough will also bounce back but at the lowest point of the bounce the force will just be large enough to move $m_2$. So we have to find an expression for the force at the lowest point.
Using conservation of energy you can find how far the mass falls by equating the loss in gravitational energy with the gain in energy stored in the spring. From there I guess you should be able to figure it out
| {
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The meaning of covariant but not manifestly covariant What is the most general meaning of the expression covariant, but not manifestly covariant? Suppose I have a general (local) change of coordinates, $x^{\prime} = f(x)$, on an $(n+1)$-dimensional smooth manifold on which some classical fields are defined, say $A_{\alpha}(x_0,x_1,...,x_n)$, which transform into $A_{\alpha}^{\prime}(x^{\prime})$. Suppose the fields $A_{\alpha}(x)$ satisfy some equations of motion, where $x_0 = t$. How should these EOMs look like to be covariant with respect to the given change of coordinates, but not manifestly covariant? Could you explain in plain words the difference between the 2 forms of the EOMs?
If possible, can you write down a practical example of such a situation encountered in physics?
Thx.
| In my experience, we usually call an expression manifestly covariant under some transformation if all the objects which appear in the transformation transform as tensors (or tensor fields) under the transformation.
For example, Maxwell's equations are not manifestly covariant under Lorentz transformations when written in terms of $\vec E$ and $\vec B$ fields, because these do not transform at tensors. On the other hand, Maxwell's equations written in terms of the electromagnetic field strength tensor $F_{\mu\nu}$, 4-vector current $j^\mu$, and '4-derivative' $\partial_\mu$ are manifestly covariant (under Lorentz transformations), because these objects transform as 2-tensors, vectors, and 1-forms, respectively.
Note that it's important to specify which transformations we're talking about. For example, Maxwell's equations in terms of $\vec E$ and $\vec B$ fields are manifestly covariant under rotations.
I don't think the story changes much for local transformations.
| {
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Is an EMF more/same/less in an insulator than in a conductor? Is an EMF (electromotive force) more/same/less in an insulator than in a conductor?
For example: A loop of copper and a loop of plastic in a changing magnetic field.
In which will the emf be the greatest?
| EMF induction depends on change of flux or condition of motional emf if the conditions are same then emf induction is same for all material as faraday's law is being followed irrespective of material medium.
| {
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Does gravitational time dilation happen due to height or difference in the strength of the field? The reason why you have to tune differently the atomic clocks in GPS is because the GPS is higher or because there is less gravity there, or both? In other words in a constant gravitational field which doesn't differ with height, will time dilation still occur?
They say that the reason why people on the first floor age slower than people on higher floors is because as you get further from the earth gravity weakens. Is that true? Does the difference in the field cause that or just the distance?
If pure distance doesn't matter then why do we say that for a spaceship accelerating forward clocks at the front tick faster than clocks at the back since both are accelerating at the same rate?
| The time dilation is due to a difference in the gravitational potential energy, so it is due to the difference in height. It doesn't matter whether the strength of the gravitational field varies, or how much it varies, all that matters is that the two observers comparing their clocks have a different gravitational potential energy.
To be more precise about this, when the gravitational fields are relatively weak (which basically means everywhere well away from a black hole) we can use an approximation to general relativity called the weak field limit. In this case the relative time dilation of two observers $A$ and $B$ is given by:
$$ \frac{dt_A}{dt_B} \approx \sqrt{ 1 + \frac{2\Delta\phi_{AB}}{c^2}} \approx 1 + \frac{\Delta\phi_{AB}}{c^2} \tag{1}$$
where $\Delta\phi_{AB}$ is the difference in the gravitational potential energy per unit mass between $A$ and $B$.
Suppose the distance in height between the two observers is $h$, then in a constant gravitatioinal field with acceleration $g$ we'd have:
$$ \Delta\phi_{AB} = gh $$
If this was on the Earth then taking into account the change in the gravitational potential energy with height we'd have:
$$ \Delta\phi_{AB} = \frac{GM}{r_A} - \frac{GM}{r_B} $$
where $r_A$ and $r_B$ are the distances of $A$ and $B$ from the centre of the earth and $M$ is the mass of the Earth. Either way when we substitute our value of $\Delta\phi_{AB}$ into equation (1) we're going to get a time dilation.
As for the accelerating rocket: the shortcut is to appeal to the equivalence principle. If acceleration is equivalent to a gravitational field then it must also cause a time dilation in the same way that a gravitational field does.
Alternatively we can do the calculation rigorously. The spacetime geometry of an accelerating frame is described by the Rindler metric, and we can use this to calculate the time dilation. The Rindler metric for an acceleration $g$ in the $x$ direction is:
$$ c^2d\tau^2 = \left(1 + \frac{gx}{c^2}\right)^2c^2dt^2 - dx^2 - dy^2 - dz^2 $$
We get the time dilation by setting $dx=dy=dz=0$ to give:
$$ c^2d\tau^2 = \left(1 + \frac{gx}{c^2}\right)^2c^2dt^2 $$
and on rearranging this gives:
$$ \frac{d\tau}{dt} = 1 + \frac{gx}{c^2} $$
which is just the equation (1) that we started with.
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Why my touchscreen doesn't respond to a pencil? I noticed my phone responds to my finger, but not to a plain graphite pencil.
Why is that, what is its physical reason?
How are styli able to approximate the human touch? What kind of conductive materials are being used there?
| Capacitive touch screens work by measuring capacitance at a number of locations on the screen, then interpolating between them to get a precise position. The simplest form of capacitor is two parallel conductive plates on opposite sides of an insulator, with the capacitance being proportional to the area of intersection of the two plates and the dielectric constant of the insulator, and inversely proportional to the distance between the plates. In the case of a touch screen, one plate is a transparent conductor on the back side of the glass, and the other is the user's fingertip. A pencil does not work well for this, even if made of metal, because its tip has a very small area. Styli made for this purpose usually use a tip that is at least 6 mm in diameter, which deforms when you press it against the screen, enlarging the contact area. Depending on the exact measurement technique being used (some touch screens measure mutual capacitance between two conductors on the screen, and others effectively measure capacitance to ground), your stylus may or may not need to be in contact with a human body to work.
| {
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Coupling constants in QFT equations of motion My question is about the coupling constants that appear in the quantum analogue to the equations of motion in a classical field theory (the Schwinger-Dyson equations). As a concrete example say we have the Lagrangian
$$ \mathcal{L} ~=~ \frac{1}{2} \left ( \partial_{\mu} \phi \partial^{\mu} \phi - m^2 \phi^2\right ) - \frac{\lambda}{4!} \phi^4.$$
Then by a short argument (replace $\phi$ by $\phi+\epsilon$ in the path integral) we should get the vacuum expectation value of the equations of motion vanishing
$$ \left \langle (\Box + m^2) \phi + \frac{\lambda}{3!} \phi^3 \right \rangle ~=~ 0.$$
We can also include other operators inside the expectation value if we include the proper contact terms. So in the following (so that the equation is not trivial) imagine that the equation of motion is multiplying some operator.
Now my question is for which $m$ and $\lambda$ is this true? The bare coupling constants don't make sense without a corresponding regularization scheme and scale $\Lambda$, which does not appear in this equation in an obvious way. Likewise the renormalized coupling constants are associated with a scale.
It seems to me these equations must be ill defined and there must be some regularization at some point. And I see that the product of fields at the same point in $\langle\phi^3(x)\rangle$ is problematic. I was wondering if someone could clarify how this equation is interpreted and how the renormalization of the coupling constants comes about?
| Yes, the field equations are ill-defined and must be renormalized. This has been done for simple theories like $\phi^4$ and QED in old work by Brandt and Zimmermann around 1970.
Hovever, their developments have been subsumed into the concept of operator product expansions (OPE), which are the modern analogue and extension of quantum field equations. They essentially correspond to the renormalized version of the equations that one would obtain by multiplying the classical field equations by arbitrary functions of the fields, and then renormalizing the results.
For the renormalized field equations of QED in causal perturbation theory, see Section 4.10 of the 2nd edition of Scharf's book on QED.
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Are Black Holes Neutrino Factories? Are Black Holes Neutrino Factories? That is.. are they an intense source of neutrinos? If so, then the center of almost every galaxy would be a neutrino factory.
| In isolation, black holes probably won't generate reactions powerful enough to produce neutrinos. However, in cases where there is a ring of gas around the black hole (called an accretion disk, and likewise the system is now referred to as an accreting black hole), then they probably will produce neutrinos. Likewise, active galactic nuclei are also strong candidates for neutrino production. My cautionary language is purposeful here, as to date there haven't been any detections of neutrino sources outside of our solar system. Its not that we haven't seen neutrinos from this distance, but rather that we haven't seen enough to zero in on any specific origin, so we are left with largely theoretical calculations.
In regards to dark matter, the discussion begins to get rather involved. Neutrinos are one candidate for something called "hot dark matter". Here, "hot" just means that the particles are moving very quickly. For various reasons, hot dark matter falls a bit short when explaining certain aspects of the early universe. The theory of a mixed hot and cold dark matter was popular up until about the late 1990's, but has since fallen out of favor. Currently, scientists are looking for a combination of cold dark matter and dark energy. I'd recommend checking out the links below for more detailed explanations:
Hot Dark Matter
Mixed Dark Matter
Similar Question
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Kinetic energy of electron in metals Will it be correct to relate temperature of metal with kinetic energy of electron in metal just like as we do to find kinetic energy of gas molecules if we know the temperature by using the following relation:
$$E=\frac32\ k\ T$$
where $E$ = kinetic energy,
$k$ =Botzmann constant, and
$T$ = absolute temperature?
I'm asking this because there is a problem in my textbook as follows:
Compute the typical de Broglie wavelength of an electron in a metal at $27$ °C and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{–10}$ m.
The only way I see is as I said above, is there any problem in that procedure? (for a general case)
| If you use the expression you mention you are implying a classical treatment. Use the expression you mention. You should be able to the wavelength larger than the separation implying a quantum mechanical treatment for the electrons in metals.
| {
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Does cosmic activity outside our observable universe affect us? I am not a scientist or physicist, but i was wondering if the cosmic activities or events which might be on a super massive scale happening outside our observable universe, also affect us?
Like the gravitational waves is due to black holes, which affects us. Can/are there any such events happening outside our observable universe which do affect us?
| The observable universe is named that way, because this is the universe where light has reached us, this is what we can see. Since everything we know of travels at a speed less or equal to the speed of light (including gravitational waves), it should follow that nothing can affect us that is not inside the observable universe or will be inside the observable universe at some time in the future (see also How can we know there is more than just the observable universe for a slightly more technical answer).
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Neutron-Antineutron Annihilation Is the process $n + \bar{n} \rightarrow \pi^{+} + \pi^{-} + \pi^{0}$ possible?
| Nice question. As anna pointed out if neutron and anti-neutron are throwing together with some MeV, no question they will splash into different particles and EM radiation.
But what means annihilation? Usually it means the attraction of charged particles from different sign, for example an electron and a positron. During the approach they emit EM radiation and this process - like it happens between electron and proton or between positron and anti-proton - doesn't stops at some distance.
The neutron and the anti-neutron will not approach by themself and due to noexisting electric fields no annihilation takes place. There is an elaboration of mine about how the approach of charged particles can be modelled and due to this a neutron - antineutron annihilation is impossible (from the above definition of what an annihilation is). The existence of electric fields allows particles to get bonded or to be repealed.
| {
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Why do clouds generally look flat at the bottom? Clouds on a still summers day generally look flat at the bottom and fluffy on top. Why? I was asked the question a couple of days ago and he ventured that it had something to do with the density gradient of air. Is he right? I suspect he is, but wanted a more informed opinion.
| As hot moist air rises, there is a specific height where the gaseous water vapour begins to condense into a mist of tiny suspended liquid droplets.
There is not a specific limit to how far this misty air can be carried upward by air convection (producing billowy cloud tops), but if it falls below that specific height the droplets will sharply start evaporating away into invisibility (since only the non-gaseous, droplet form scatters white light).
The boundary is termed the lifted condensation level or dew point. At greater heights there is less air pressure (because there is less air column weighing down from above). This weakening pressure lets ascending parcels of air push-out or expand, which results in an expenditure of temperature (eventually reaching the point where the water molecules on average no longer have enough kinetic energy left to overcome the intermolecular attraction force). The pressure gradient is also the reason low-density parcels are buoyed upwards. The cloud-forming parcels have low-density because they are warmer than surrounding air (having a comparative surplus of kinetic energy to push out with); these parcels are warmer because of the greenhouse effect (which directly heats the surface while cooling the upper atmosphere) and moister because of evaporation of water (which had eventually separated from the air and fallen to concentrate at the surface).
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Why does image noise happen? Why are digital images shot with a high iso (which means the camera sensor is much more sensitive to light) so much grainer than pictures taken with a lower iso? Here is an example of the type of grain I'm talking about.
At first I thought this was a limit on the cameras precision, but the same thing happens to your eyes when it's dark out— everything gets a grainy look to it. Why is this happening? Is this a property of light itself?
| On the other hand, there are some forms of noise that are a property of the light.
For instance, if you have a detector that is just counting photons per pixel, and during the detection time you average $N$ photons, counting statistics says that number will vary by $\sqrt N$. (For Poisson distributed photons--there are other distributions). If $N$ is too small, you have to integrate longer to shrink $\sqrt N/N$. The noise is a property of the light because the light comes on random bunches (photons).
Another case is coherent light (sound), such as an imaging radar (sonogram). The signal is a sum over many random scattering centers with random phases. In the approximation that they are all equal, the final signal is effectively a random walk of complex phasors. The total amplitude is Rayleigh distributed, so that the intensity is exponentially distributed. That means, where the intensity "should" be distributed as
$$ I = \delta(I-I_0) $$
with mean $I_0$ and standard deviation $0$,
you measure:
$$ I = exp{(-I/I_0 )}$$
with mean $I_0$ and standard deviation $\sqrt{I_0}$. Note that this noise (called speckle noise) is multiplicative--it's a property of the signal such that you cannot reduce it by increasing the signal power. This noise is a result of the light (or sound) being coherent, and as such is a property of the light.
| {
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Volume charge density of an electric dipole
My attempt:
$$\rho(\overrightarrow r) = Q\delta^3 (\overrightarrow r)$$
$\overrightarrow r =$ separation vector from the origin to $\overrightarrow a$
$Q$ = the total charge of the electric dipole
Here's what I don't understand though. The total charge of the dipole is $-q + q=0$, isn't it? Therefore, my volume charge density should be zero, no?
Also, what volume are we talking about here? The question implies that we are either talking about some arbitrary volume defined when attempting to solve the problem or the volumes of the individual point charges, which are zero.
| Note that whatever your answer is, when you integrate over a region of volume $V$ containing the charge $-q$, but excluding the positive charge, your total charge should be $-q$, since there is exactly that much charge in that volume. If there is a total charge $-q$ in a volume $V$, the charge density in that region is $-q/V$. In other words,
$$
\int\limits_{\text{region including $-q$, excluding $+q$}}d^3r\,\rho(\vec r) = -q
$$
Similarly, when you integrate over a region of volume $V$ containing the charge $+q$ but excluding the negative charge, your total charge should be $+q$. Last but not least, when you integrate over a region that contains both charges, your net charge should be zero.
Can you formulate a density function $\rho(\vec r)$ that satisfies these three properties? Remember the sifting property of the three-dimensional delta function:
$$
\int_\mathbf{R^3}d^3r\,\delta^3(\vec r - \vec a) = 1.
$$
Answer: (try to figure it out before revealing the solution)
$\rho(\vec r) = q[\delta^3(\vec r - \vec a) - \delta^3(\vec r)]$
| {
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Why does increasing resistance decrease the heat produced in an electric circuit? If $H=\frac{V^2}{R}{t}$ ,then increasing resistance means decreasing the heat produced.
But, isnt it that the heat in a circuit is produced due to the presence of resistors? Moreover metals with high resistances are used as heating elements ,like Nichrome?
Why does the equation state that the heat produced is inversely proportional to Resistance
| You are just messing up with the equations. In order to state a proportionality relationship between a physical quantity and some other quantities, you need to be make sure that the quantities are all independent.
In your equation, the voltage and resistance are not independent quantities, but voltage is a function of resistance and the current flowing through it. Hence to state a relationship between resistance and heat generated, you need to crack down the voltage into independent quantities- resistance and current. Then the equation reads $\displaystyle{H=I^2Rt}$, which tells us that heat is directly proportional to resistance.
The source of your confusion was that you used voltage, which depends on current and resistance (defining relation between heat and resistance with a quantity involving resistance) to define the proportionality relation between heat and resistance.
| {
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How does the law of conservation of energy explain magnetism? The law of the conservation of energy describes that energy cannot be created nor destroyed however in fact only changes form.
How does this law explain the energy transferred by magnetic fields?
| Magnetism is not explained by conservation of energy. But magnetism is consistent with conservation of energy.
It is commonly stated that the magnetic field does no work. It is true that the magnetic field does no work on charged particles through the Lorentz force law. The force delivered on a charge due to a magnetic field, $\vec{F}=q \vec{v}\times \vec{B}$, is always perpendicular to $\vec{B}$
However, the magnetic field can do work on the electric field, through one of Maxwell's equations: $\frac{d \vec{E}}{dt}=\frac{1}{\mu_0 \varepsilon_0}\nabla \times \vec{B}-\frac{1}{\varepsilon_0} \vec{J}$
So this is one way in which the magnetic field does work. For example, if you have an inductor in a circuit, a constant current produces a constant magnetic field inside the inductor. Once you switch the current off, you actually get some energy back out of the inductor due to the collapsing magnetic field. This is an instance where the magnetic field does work on the electric field, and the electric field in turn does work by creating a current in the wire.
| {
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Formula for lens periscope I want to build a periscope.
To my great surprise even after intense googling I could find little relevant information about periscopes anywhere on the internet.
Let's say I want to build a periscope with height h, apparent height h1, diameter d and viewing angle v. What other variables should I define for a periscope?
Source: Wikipedia.
I am looking for a formula/manual/recipe for calculating the lenses and distances as well as any practical advice on building a DIY periscope.
Thanks.
| What you are trying to do is build the combination is a terrestrial refracting telescope and a periscope.
As the design of the periscope is straight forward so you need to look up terrestrial refracting telescope to get some idea of the linear dimensions between lenses and the focal length of the lenses.
There is some simple theory given at this website.
A search of the internet will give you construction details and suggestions for dimensions.
| {
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Origin of the force of gravity Where does exactly gravity come from when bodies exert this force? Does it come from the center of every object or it arises from their surface?
| It is unknown what physical mechanism "causes" gravity, and indeed the answer to that will depend a lot on the model of gravity that you choose. If you use Newtonian gravity, you assert "action at a distance" that is in some sense caused by the mass itself, but it is known that this is not a complete description, now that we have detected gravitational waves. So one can instead use General Relativity, in which mass/energy creates a curvature in spacetime (that can propagate at the speed of light, as in gravitational waves), but still one does not have a mechanism by which that curvature is "caused." Finally, one can look ahead to new models of gravity, perhaps ones that unify gravity with quantum field theory with the introduction of a gravity-carrying particle called a "graviton." No such theory yet exists, but it seems that when and if it does, the graviton will be able to mediate the force of gravity via the same kinds of "virtual particles" that are now thought to mediate electromagnetic forces (i.e., virtual photons). That would bring us closer to a "cause", though some disparage the concept of a "virtual" particle, given its self-contradictory nature, and also we must admit that no such theory has as yet been successfully created for gravity. But at present, it does seem like gravity is emanated from everywhere that there is mass or energy, so from every piece of an object.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that pseudorapidity equals $\ln \left(\frac{\sqrt{p_z^2 + p_t^2}+p_z}{p_t}\right)$ I found one method to calculate pseudorapidity, $\eta$, using the formula
\begin{equation}
\eta = - \ln \tan \frac{\theta}{2}
\end{equation}
And I know how we can get to this point from Lecture 7 - Rapidity and Pseudorapidity in these notes.
But how can we get from there to this formula?
\begin{equation}
\eta = \ln \left(\frac{\sqrt{p_z^2 + p_t^2}+p_z}{p_t}\right)
\end{equation}
| There is a trigonometric identity for the tangent of a half angle
$$\tan\frac{\theta}{2} = \frac{\sin\theta}{1 + \cos\theta}$$
which you can use to expand the first formula.
Then it's just geometry.
| {
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Why use 2 nonlinear crystals in conjunction for SPDC? I've seen a few articles (ex: 1, 2) that use either two BBO crystals consecutively or a BBO and a KDP crystal to create entangled photon pairs from Spontaneous Parametric Down-Conversion (SPDC). And from this answer, I understand that it's used as a trick because type I conversion doesn't produce entangled photons.
I just don't get what they're achieving by using 2 crystals. Can someone spell this out for me? From my understanding, by using two crystals, you have more chances to create PDC, but I'm not sure how by using two crystals in orthogonal directions that you're producing entangled pairs, contrary to just one crystal.
Thanks for the help all!
| There are some good answers to related questions. So I probably don't need to go into too much detail. Spontaneous parametric down-conversion will usually produce a pair of entangled photons. The questions is, in terms of which degree of freedom are they entangled? Type I phase matching produces the two photons with the same polarization. As a result they are not entangled in terms of polarization. However, they are entangled in terms of there spatial degrees of freedom. That means that if one photon propagates in a particular direction, its correlated photon would propagate in a direction that would ensure momentum conservation.
To get polarization entanglement, as used in many EPR experiments, one needs type II phase matching. Alternatively, one can use some scheme using two BBO's, which is what you are refering to. The two BBO crystals are placed one behind the other and they are oriented differently so that the down-converted photons produced by one have an different polarization from those produce by the other. The combination now gives you a state that is entangled in polarization.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do we prove or disprove that a particle has no internal structure? In many pop physics books I have read that an electron has no internal structure.
How do we know that and how can we rigorously prove that it has no such structure at all?
| We study it using deep inelastic scattering. It was this type of experiment that first revealed the proton had an internal structure, and if the electron has an internal structure it will be this type of experiment that reveals it.
No experiment has yet discovered evidence for an internal structure in the electron, however that doesn't prove the electron has no internal structure. It proves only that any structure is too small scale for us to probe at the energies available. So contrary to what you say in your question, we haven't rigorously proved that the electron has no structure.
There is a long standing idea that the apparently fundamental particles like electrons and quarks might be bound states of fundamental particles called preons, though I must emphasise that there is no experimental evidence for this idea and there is some theoretical evidence against the idea.
We know the electron size must be less than about $10^{-18}$m because we've already probed that length scale in collider experiments. That means any preons have to be confined within $10^{-18}$m and the uncertainty principle implies a correspondingly large uncertainty in momentum. This would suggest the preon bound state has a high energy and therefore a high mass - certainly higher than the electron mass. There are ways around this so it isn't proof that the electron isn't a bound state. However the ways to avoid an excessively high mass are a bit contrived and will get more so as we probe to smaller and smaller length scales.
| {
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Why doesn't electric charge immediately leak off charged objects? I will focus my question with a particular example: a metal sphere, surrounded by vacuum, is given a negative charge. I know that when this charge is great enough, electrons will be emitted from the sphere, but why is the threshold for this so high? As I understand it, the reason an electron stays on the negative sphere despite the electric repulsion is because of the metal's work function. But the work function of metals are typically ~4 eV. Wouldn't this suggest that -4 Volts would be the threshold for electron emission from the sphere in a vacuum? (Or a voltage even closer to zero, because of the thermal distribution of electron energies in the metal.) This seems way too small and I would think the threshold would concern a minimum field strength rather than minimum voltage.
| The threshold for electron emission from a metal is so high because at room temperature the work function of the metal of several 1eV (which acts as an emission barrier) is far above the thermal energy kT=0.026eV of the electrons in the metal. Therefore, at room temperature, only a minuscule fraction of electron can overcome this barrier according to the Fermi distribution. Significant emission can be achieved by heating of the metal which leads to thermionic emission over the barrier, which is used in thermionic cathodes of electron tubes. Another mechanism of electron emission from the metal is the field emission (Fowler-Nordheim emission) caused by high applied electric surface fields which reduce the surface barrier and enable quantum mechanical tunneling through the triangular potential barrier formed by the work function and the surface field, which becomes transparent to the (cold) electrons in the metal at high applied fields in the range of several MV/cm. Therefore it can be expected that with high enough negative charging of the metal sphere the surface electric field strength will reach a critical strength to cause significant field emission of the electrons. For the same applied electrical potential difference, this will occur for spheres with smaller radius. Field emission of electrons is used for field emission cathodes with sharp metal points, e.g., in electron microscopes.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can the work done in bringing a charged particle from infinity to a grounded conductor be zero? Suppose we have a grounded conducting sphere (potential of sphere = 0) and we bring a charged particle to its surface.
Since work done by us would be equal to the change in potential energy of the charge, it will be equal to zero,
As potential at infinity = potential on sphere = 0 therefore change in potential energy = 0.
But this is not possible as when we move the charge, it will induce a negative charge on the sphere, thus for charge to move without acceleration, a negative force (with respect to our displacement) would be required.
Thus we would do negative work all over the path and work done can never be zero.
Am I wrong in assuming that the work done = change in potential in this case? Because only that would make sense.
Any help would be appreciated.
| The potential at a point is the work done in order to bring a unit positive charge from infinity to that point, without any acceleration. Electric potential is a location-dependent quantity that expresses the amount of potential energy per unit of charge at a specified location.
For a conservative field, we have by definition
$$\vec{E}=-\nabla V$$
which means the electric field at a point is equal to the negative gradient of the absolute potential at that point. This assumes the fact that the work done by the external agency in moving a charged particle through a uniform static electric field is the same for all paths, otherwise dependent only on the initial and final positions.
So, the electric field is due to some charge, say $q$ and the potential is observed on some other charge $Q$, a unit test charge, which by definition, should not alter or distort the electric field of the charge $q$.
This means, simply you cannot take into account the negative charge formed on the sphere. To solve such problems, you need to use the method of images
| {
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How can Noether's Theorem be used to prove that the probability density satisfies a continuity equation? How can I use Noether's Theorem to show that the probability density $\rho (x)=|\psi(x)|^2$ for a wave function $\psi(x)$ satisfies the continuity equation $\frac{\partial \rho}{\partial t}+\nabla \cdot\vec{j}=0$, where $\vec{j}$ is the probability current defined in quantum mechanics?
I have solved this problem before by other means but I don't think I understand Noether's Theorem well enough to apply it in this case. Any help would be greatly appreciated.
| First note that Schrödinger's equation can be understood to come from an action. The Lagrangian is
$$L = \int~\mathrm d^3x \,\,\psi^†(x) \left(i \frac{\partial}{\partial t} - \frac{\nabla^2}{2m}\right)\psi(x) - \psi^†(x)\psi(x)V(x)$$
The Euler-Lagrange equation for $\psi^†(x)$ is exactly the Schrödinger equation. Since the dynamics of $\psi(x)$ are determined by Lagrangian mechanics in this way, Noether's theorem applies without any caveats.^^
In particular, this Schrödinger Lagrangian has a $U(1)$ symmetry corresponding to $\psi(x) \mapsto e^{i\alpha}\psi(x)$. The corresponding conserved charge current density is
$$\rho = j^0 = \frac{\partial L}{\partial \dot{\psi}}\delta \psi = \psi^†\psi(x)$$
$$\vec{j}^i = \frac{\partial L}{\partial_i\psi}\delta \psi+\frac{\partial L}{\partial_i\psi^†}\delta \psi^†=\frac{i}{2m}\left((\partial^i\psi^†)\psi-\psi^†\partial^i\psi\right),$$
which is the well-known probability current density.
^^ In non-relativistic quantum mechanics the wavefunction $\psi(x)$ is a "classical" variable in that it is simply a function from space and time to $\mathbb{C}$. Noether's theorem works exactly the same for it as in classical mechanics. In quantum field theory the relevant objects $\psi(x)$ become quantum operators and the usual arguments have to be modified somewhat.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/279635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If you are vacuuming your carpet and you wrap the cord around your body do you become a magnet? If you wrap an active electric cord around your body, do you become an electromagnet?
| Okay, accuse me of having too much time on my hands, but here's what I did:
If you can't tell from the pic. I wrapped the vacuum cord around a steel bar. I turned on the vacuum and tried to pick up the screw. Absolutely nothing not even a hint of attraction so maybe BowlofRed has a point.
In case the comment gets deleted later, here is BowlofRed's comment:
The power cord has two conductors in it. The current is moving in opposite directions in each, so the net current flow through the cord is zero. No net current, no bulk magnetic field. – BowlOfRed
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/279783",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "85",
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