# Tag Info

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Review of the idea we're talking about So let's review the mechanism by which this idea works. (In my view it makes little sense to critique the idea if we don't have a detailed understanding of it.) The idea is that, you've got currents on wires, and those wires (I'll call them wire-1 and wire-2) consist of nucleons and electrons. All of the nuclei are at ...

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First make a parallel plate capacitor with plates of area A and spacing d. Fill the space between the plates with the dielectric whose complex permittivity $\epsilon(\omega)$ you wish to measure. The formula for this capacitance is a complex function of frequency because the permittivity is a complex function of frequency. $$... 3 If you ask around about magnetic fields, you will read seemingly-authoritative articles which say magnetism is a consequence of length contraction. They should say that magnetic fields and magnetic forces are a unique and required element to add to electric fields and forces in order to make it relativistically invariant. Which is a slightly different ... 2 To summarise: it depends on the ratio of refractive indices, the angle of incidence and the polarisation state of the incident wave. Details: The polarisation of the electric field will be perpendicular to the wavevector. Why? Because where there are no free charges then \nabla \cdot \vec{E} = 0. If we represent the wave as \vec{E} = \vec{E_0} f(\omega t ... 2 Hopefully the following gives an idea of a proper derivation of the properties you are interested in. Suppose that the polarisation is a linear function of the applied electric field \begin{equation*} \mathbf{P}(\mathbf{x},t)=\int_{\mathbb{R}^{3}}d\mathbf{y}\int_{-\infty }^{+\infty }ds\chi (\mathbf{x,y},t,s)\mathbf{E}(\mathbf{y},s) \end{equation*} In the ... 2 A general answer is that the field vector is independent from the decay direction, and hence the field vector can be pointing to any direction. However, on the decay direction (that is the -y direction in your formula), all field components decrease. The only constrain for the field vector is that all field components must satisfy the boundary condition at ... 2 The problem for photons is that you cannot observe them time to time. If you do measure their state, photons will collapse to some Eigen state and become classical. In fact, a photon can have a lot of eigenstates in theory. That is special compared to EM for photons. 2 You must include the t because the field configuration can change with time. Perhaps this seems counterintuitive because you are used to thinking about stationary charges and constant currents. However, imagine the following situation: you start with a charge at the origin. Then E will have a peak in magnitude at the origin. Now at some later time you ... 2 Virtual photons and generally virtual particles are mathematical constructs coming from the iconal representation of scattering amplitudes by Feynman diagrams. To make the point clear here is a virtual particle with an enormous mass with respect to the incoming and outgoing real particles. A Feynman diagram is a prescription for writing the mathematical ... 1 A superconducting coil acts mostly as a perfect inductor, so it resists current variations. If it is plugged to an AC voltage source, it will pass an AC current corresponding to its inductance. When shorted, it will continue running the current that was running through it just before shorting (as V=0=L di/dt). So you're going to end up with a constant ... 1 Yes, the field inside a uniformly charged thin spherical shell is zero. Gauss' Law can be used to demonstrate this fact. If there is no field, a charge inside will feel no force. Right? 1 The imaginary part of the permittivity is a measure of the loss in the system. In some literature you will see reference to the "loss tangent" - this is in essence the ratio of the real and imaginary parts of the dielectric constant. The simplest way to measure this is to put the sample of interest in a cavity (for example, make it the dielectric of a ... 1 Use the wave equation (\frac{1}{\epsilon \epsilon_0 \mu \mu_0} \vec{\nabla}^2 - \frac{\partial^2}{\partial t^2}) A = 0 and its solution, i. e. plane waves A = A_0 \exp(\imath ( \omega t - \vec{k} \vec{x})) with ||\vec{k}|| = \frac{\omega}{\sqrt{\epsilon \epsilon_0 \mu \mu_0}}. In words: The bit you already knew was that the (real part of the) wavector, ... 1 I just did it! We have that$$\tan(2\phi)=\frac{2}{\text{cotan}{\phi}-\tan\phi}\Leftrightarrow\Leftrightarrow\arctan\left[\tan(2\phi)\right]=\arctan\left\{\frac{2}{\text{cotan}\left[\arctan\left(\frac{ki}{kr}\right)\right]-\tan\left[\arctan\left(\frac{ki}{kr}\right)\right]}\right\}\Leftrightarrow\Leftrightarrow ...

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If you had ask a slightly different question "Why can't we interact directly with virtual photons" I could say that, in the formulation of Feynman series as scattering off-shell segments of virtual photons and virtual electrically charged matter-antimatter pairs, any virtual particle is by definition not part of the asymptotic states What is an asymptotic ...

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Hawking radiation starts with the creation of a pair of virtual photons, I believe. One is absorbed by the Black Hole and the other could be observed if it were not so faint. My understanding is that in the general case a virtual pair re-combines (annihilates) within a time on the order of a Planck time and cannot be observed.

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No, it will rotate at a speed determined by the load. Witness that the current in, and thus the magnetic field produced by the stator coils is either in-phase with or anti-phase with the rotor current, with the $\pi$-phase change triggered by the split ring commutator. So the torque in each half of the rotor's rotation will throb at twice the AC line ...

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$$\newcommand{\k}[1]{\left( #1 \right)}$$ \begin{align} \nabla\cdot{\vec{J}\over r}={1\over r}\k{\nabla\cdot\vec{J}}+\vec{J}\cdot\k{\nabla\k{{1\over r}}}\\ \nabla'\cdot{\vec{J}\over r}={1\over r}\k{\nabla'\cdot\vec{J}}+\vec{J}\cdot\k{\nabla'\k{{1\over r}}}\\ \nabla'\k{{1\over r}}=-\nabla\k{{1\over r}} \end{align}

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If you use Lumerical or MEEP to do the FDTD calculation, you can simply add a Gaussian source polarized along $x$-direction but propagating along $z$-direction with the given pulse width and $\tau$ and frequency. Alternatively, you can make both $y$ and $z$ components equal to zero for the source. Is this what you want?

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There have been a lot of derivations on the magnetic field of a cylindrical magnet as a homework problem. See for example, this one and the plot below. The gradient line can be obtained by doing a gradient on the scalar potential equation given in the reference.

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This is determined by the type of your generator. If it generates a 3-phase AC current, then reversing the rotating direction doesn't matter since all ABC wired symmetrically only with a phase difference. That is most likely you are using a ac generator. If your generator yields DC current, reversing the rotation direction will reverse the current direction ...

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As pointed out in zeldredge's answer, what is missing is the connection to the photon. However, I do think you can fix this problem. If you couple a system that exhibits natural oscillations at certain frequencies to an external force that oscillates at some frequency $\omega$, then you get a resonance when $\omega$ matches one of the natural frequencies. ...

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