# Tag Info

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Photons come with chirality, so you should consider angular momentum conservation as well. For $1\gamma \to 2\gamma$ scattering, this will not be possible. (I'm assuming production of collinear photons only; it's obvious when two are not collinear, energy and momentum conservation will be violated)

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It is unfortunate that the physics of magnetism got saddled with several different *-hand rules, and that they use different hands. Let's pull them apart: Fleming's left-hand rule gives you the direction of the force that acts on a current if you know the magnetic field. Image source This rule applies to motors, i.e. devices which use currents in a ...

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In circuits with nontrivial inductance, Kirchoff's law of voltages is precisely an expression of Faraday's law, $$\oint \vec{E} \cdot d\vec{r} +\frac{d\Phi}{dt} =0.$$ The first part are the resistive and capacitive voltages as would be measured in an electrostatic setting, as well as the EMF of any sources in the circuit. The second term, ${d\Phi}/{dt}$, ...

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Recall that the Faraday tensor in this form is a linear mapping that maps a charged particle's contravariant four-velocity to the latter's rate of change, wrt proper time (modulo scaling by invariant rest mass $m$ and invariant charge $q$): $$m\,\frac{\mathrm{d} v^\mu}{\mathrm{d}\tau} = q\, F^\mu{}_\nu\,v^\nu\tag{1}$$ Now let's think of a particle's ...

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Inside an ideal conductor, $\vec E=0$ has to hold. The idealized inductor is usually taken to be an ideal conductor, so there is actually no voltage drop in the sense of $\int\mathrm dr\cdot\vec E$ across an ideal inductor. In that sense you are right that Kirchhoff's loop rule $\oint\mathrm dr\cdot\vec E=0$ is not applicable in the presence of ...

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Actually, I presume you are confused between induced and electrostatic electric fields. Kirchhoff's second law follows from the conservativity of the electrostatic electric field. But the E you are using in the equation is the induced electric field which has properties totally different from the electrostatic electric field. Work done in a closed loop by ...

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The above information is totally false. Notice no authoritative sources are cited. This statement violates the laws of physics: " circuit delivers electrical energy to an antenna where it is converted to electromagnetic radiation which propagates away at the speed of light." The first clue that it is gross error is the obviously false statement that RF ...

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Gravitationally time dilated muscles are weak, and gravitationally time dilated magnets are equally weak. If muscles and magnets were not equally weakened by gravitational time dilation, then a person falling while holding magnets would notice something changing. Now I choose to calculate how much gravitationally time dilated muscles are weakened: Let's ...

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You can try gnuplot, it might have a slight learning curve at the beginning, but overall it's very versatile: http://www.gnuplotting.org/equipotential-lines/ http://gnuplot.sourceforge.net/demo_canvas/vector.html

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Light is an emergent phenomenon from the confluence of innumerable individual photons with the energy h*nu, where nu is the frequency of the emergent macroscopically beam. The question then reduces imo to "does a photon see/interact with a rotating "transparent" disk the same way it does with a gravitational field" . Checking on the special relativity ...

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Indeed, an infinitely long and thin wire with a current $I$ has a magnetic field given by (in SI units): $$B = \frac{\mu_0 I}{ 2\pi r}$$ Now suppose your wire has a finite radius. Ampère's law shows that as long as cylindrical symmetry is mantained, the field depends only on the current and not on the detailed properties such as the wire's radius or the ...

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You have a box of pebbles. There are some pebbles that do not move at constant velocity when placed closed to one another. We repeat the experiment many times with these pebbles and characterize their motion. We make a plot of their acceleration as a function of their position and we notice that the acceleration is always proportional to the inverse distance ...

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It's not actually circular, speaking purely logically. One says "An X is a field produced by things having Y." The other says "Y is a property of matter that causes it to respond to an X." Putting these two things together reveals simply that some matter has Y, which produces an X, which affects other bits of matter which have Y's. In other words we are ...

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It is not acceptable to change a question to invalidate two existing good answers to your question, you should have asked a new question. There are two types of particles in the universe, the fermions and the bosons. The fermions take up space in the sense that you can't have too many in the same region of (phase) space. And the bosons basically mediate ...

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Recall that science is descriptive of nature. These words are defined so that they describe the effects of experiments (generally the descriptions that stick are the simplest ones that cover the range of behaviors actually seen). As a result, though they are circular in structure, they absolutely do not rely on circular logic for their validity: they rely ...

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Definitions in physics are always somewhat circular, because they are not really definitions. Instead, they're descriptions of the world. The way to make sense of "circular definitions" is usually to think about the experiments that led to those definitions. Suppose you're Charles-Augustin de Coulomb. One day you discover that by rubbing objects on each ...

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Some facts - not a definitive answer: A calculation of GPS power is given at this link and a more in depth analysis at this link. With about 1 kW available from the solar panels for most of the orbit, it seems that "available power" is not the limiting factor. The second link does imply that balancing the power from all satellites is important to ensure ...

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Hint: A semi-infinite uniform magnetic field would be one which would be described by something like this: $B=0$ (for $x<0$) and $B=constant$ (for $x\ge0$). The question describes a situation of this sort: (The $X$s indicate that the region has a magnetic field directed into/out of the page): You need to know how the trajectory of a charged particle ...

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The electron will turn to one side and the proton will turn to the other side. Say, the electron will describe a spiral path to the left (this depends from the direction of the magnetic field), then the proton will describe a spiral path to the right. This happens because both particles have a magnetic dipole moment and an intrinsic spin. Aligning the ...

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Yes, these thermally generated currents (Johnson noise) generate magnetic fields. This means that even non-magnetic materials generate a very-small magnetic noise if they are conductive. This actually places a limit on very-sensitive magnetic field measurements in shielded environments because the shields are usually conductive. The following Review of ...

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The electron having mass will exhibit gravitational attraction and begin moving again, however; unless there is a larger object near the electron to it's nucleus it will orbit around it's original host and eventually regain it's original velocity as long as the Lagrange point is not breached.

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Your argument is correct, in fact you can use it to show that the Hall conductivity is set by the density. However, there is an underlying assumption when you apply Galilean transformation, that is translation invariance. In reality there are impurities that can backscatter electrons and cause the current to dissipate. So your argument immediately fails for ...

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Your work is fine (depending on your units) and what you were asked to show is wrong. Though I do object to saying you have a force equal to $ma_c,$ I would just say that a net force orthogonal to the velocity makes it go in a circle of radius $r$ where $F=mv^2/r.$ And the problem is famous. The fact that the frequency doesn't depend on the velocity or the ...

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Yes you are right. You end up having a varying electric field which generates a varying magnetic field which in turn generates an electric field etc... This causes a particular type of radiation called black body radiation.

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What happens if the particle is on the circular path? Hint. There must be something compensating the Lorentz force. EDIT: I think your identity for the angular velocity is wrong anyways. It should be $$\omega = \frac{qB}{m}$$

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No. The charge enclosed by a surface is really just the number of protons inside minus the number of electrons inside all multiplied by the charge of the proton (which is a constant). If the electric field has a large density in one region then it has less in another so that total flux over a closed surface is still the net enclosed by the whole surface is ...

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It wouldn't be a static dipole field. It would have some dynamics associated with it, and there would be net radiation. The no-hair theorem only applies to late-time, evolved spacetimes. And is about the horizon itself, not about apparent observations made by distant observers. Less rambly, more bullet-pointy answer: Distant observers would see a ...

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After the hypothetical split, 2 photons with the same energy would be propagating at an angle ok with momentum conservation. Then there would be a rest frame where the angle is 180 degrees. Now if you stay in this restframe and go back in time before the split, your single photon would be at rest. However, that is not possible: According to relativity, speed ...

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As the value of $b$ increases the resistance between the outer and the inner shells will converge to $1/4 \pi \sigma a$. If we consider the outer shell to be at the "infinity", the resistance between the "infinity" and the inner shell will be $1/4 \pi\sigma a$. We can think of the situation in which there are two shells in the infinite sea of poorly ...

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I am not totally sure of this answer, which is why I asked the question. However, I think the answer is that only relatively short wavelengths can pass through an annular aperture. Specifically, I think that if the outer radius of the annulus is R and the width is W, where W << R, the maximum wavelength that passes through is approximately 2R ...

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A photon is an elementary particle. As much elementary and as much particle as the electron . A single elementary particle has a fixed mass and cannot emit another particle without violating energy conservation, because its mass is fixed. In the center of mass of a massive elementary particle, electron, there is no energy for an emission , for a ...

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It looks like you're trying to find a vector field $\vec{A}_m$ such that $\vec{E} = \nabla \times \vec{A}_m$. This is only possible in regions of space that are charge-free: the divergence of the curl of a vector field is always zero, so we necessarily have $$\frac{\rho}{\epsilon_0} = \nabla \cdot \vec{E} = \nabla \cdot (\nabla \times \vec{A}_m) = 0.$$ ...

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I don't think there is any universal "intuition" to tap into, aside from that which comes from practice. You perhaps need to explore different physics texts in the electromagnetic department. I for example loathed Jackson as a learning text: it is comprehensive and useful as a reference for refreshing knowledge, but not good at conveying it. Volume 2 of the ...

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I've only really done antennas in undergrad, and the most we really looked at was phased arrays of half wavelength antennas and looking at the resulting field distribution far from the source, so take what I say with a grain of salt. One interesting thing to look at might be genetic optimization processes for antenna design. I believe that there has been ...

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{1} Here is a link to a paper by Patrick C. Crane on the transmission line disturbances to RF: http://www.faculty.ece.vt.edu/swe/lwa/memo/lwa0168.pdf. I didn't read the paper, but it seems to be an extensive analysis, and mentions at least 3 other references on the subject. On page 3, you'll find a summary of effects on three types of RF: "The phenomena ...

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Yes it will remain constant. Because of this property magnetic strength remains constant when cut transversally.

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In terms of a dielectric, it means there is a linear constitutive relation between the vectors. $${\boldsymbol D} = \epsilon_0 {\boldsymbol E} + {\boldsymbol P}$$ Or $${\boldsymbol D} = \epsilon_r \epsilon_0 {\boldsymbol E}$$ where $\epsilon_r \epsilon_0$ is a scalar relative and vacuum permitivitty. This way, there is now a linear relation between the ...

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resistor is an electric component which is used for the purpose to restrict the electron flow. if you run with 10Km/hour speed now i will stop you is it possible to stop you suddenly. No it is not possible. so that resistor dissipate some energy in the form of heat.

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It "runs out" when it reaches thermal equilibrium with it's environment. In that case, as @Kevin points out, the dissipation equals the absorption from the environment. If you pass a current through the resistor, it will heat up and that will drive extra current/voltage noise in the resistor. FYI: Resistive cooling of ions in a Penning trap uses a cooled ...

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It works just like every other kind of thermal energy. If a resistor can give out energy to the environment, it can also receive it. For example, if it gives it out by radiating, it can also absorb radiation; if it gives it out by having its fast-moving atoms smash into air molecules, then fast-moving air molecules can also smash into it. When it's in ...

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There are two classes of methods one could practicably use. 1. Method grounded on Ampère's law If your flux is time-varying, then, in air or dielectric (no conductivity), Ampère's law will give you the time derivative of the electric displacement flux through your loop (call it $\Gamma$): \oint_\Gamma ...

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In dV=-E.dr, E is electrostatic electric field and the negative sign implies that the electric potential V decreases with increasing electric field E. This is the significance of this negative sign. In E is not electrostatic, but induced electric field. This field is induced due to time-varying magnetic fields or motional electromotive force causing a ...

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I'm struggling to identify whether a scalar is a Lorentz-scalar. E.g: ∂iAii∈1,2,3. How do I determine if this is a Lorentz-scalar or not? If got the same problem with tensors. How do I differentiate between a tensor and a Lorentz-tensor? By "tensor" do you mean a tensor in three space dimensions? If so, then a tensor is never a Lorentz tensor. A Lorentz ...

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By measuring the shape of the transient output envelope it is possible to infer the phase of a pulse of known frequency and length if the pulse is passed through a bandpass filter. You just have to make sure that the filter is not energized before the pulse hits it.

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Im somehow not able to comment on anna v answer so here goes my cmment How isnt the magnetic field non static. Acc. To The relation the force is dependent on the the magnitudes of current. Lets suppose some constant current i1 is flowing in the below placed wire and we bring the wire 2 also carrying some dc constant current i2 and place it above the i1 ...

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Here is the misunderstanding: Now, if the magnetic force is greater than mg, the wire moves up. Now magnetic force is up and displacement is up too which means that work done by magnetic force should be positive. The statement that a magnetic field does no work is of a static magnetic field. You are positing a changing magnetic field by the word ...

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You can find some information about that on John D. Norton's website. Einstein thought of this at the age of sixteen. Here's another article: "If I pursue a beam of light with the velocity c (velocity of light in a vacuum), I should observe such a beam of light as an electromagnetic field at rest though spatially oscillating. There seems to be no such ...

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Is it true that superconducting electromagnets don't need any power? So... energy to created a magnetic field via a superconductor would be zero? They have to be at very low temperatures, that takes a lot of energy. The ATLAS Barrel Toroid was first cooled down over a six-week period in July-August to reach –269°C . It was then powered up ...

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What you have here is basically the B-field seen in the frame of a charge with velocity $\mathbf{v}$, moving in an electric field that is $\mathbf{E} \perp \mathbf{v}.$ You can derive this expression by considering the relativistic field transformations of $\mathbf{E}$ and $\mathbf{B}$ in a moving frame, I'll only show you the most important steps, for a ...

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I just want to reply to your simpler example, since it's something we can hopefully both agree on without too much confusion. You say An electron is travelling at high speed (say,0.9 C). If is moving near another electron, (proton, positron or a live wire) it can make anything move, acquire KE, it can do work. When v approaches c, the attractive magnetic ...

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