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The electric field around high voltage transmission lines (or "high tension" lines) is extremely high, and can be close to the breakdown threshold of air. That's why the highest voltage lines use multiple (often three) parallel conductors, to increase the effective conductor radius and reduce the peak electric field. Now, introduce a human into that field, ...

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Polarization and gauge symmetry In QFT, the dynamical varible is the four-potential $A_\mu$. The electromagnetic field is defined by $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$, an antysymmetric tensor wich six independent components: 3 for the electric field and 3 for the magnetic field $E^i = - F^{0i}$, $B^i = ... 1 The answer: $$2\partial^\sigma A^\rho.$$ It's not the chain rule but the product rule you use. It is as if the tensor $$\partial_\mu A_\nu$$ were the variable you are differentiating with respect to. To understand the indices, consider a simpler example: $$\frac{\partial}{\partial x^i}(x^j x_j) = \frac{\partial}{\partial x^i}(x^j x^k g_{jk})= \delta^j_i x_j ... 1 Well the derivation of the Snell's law in optics can be done by using Fermat's principle of least time, have a look at Feynman's lecture series volume 1. You can derive an analogous equation for charged particles using the principle of least time by using the potentials in two consecutive areas ( which take the place of the refractive index) this is will ... 0 As Meng Cheng said, you need to solve a wave equation with sources (see, e.g., http://physics.gmu.edu/~joe/PHYS685/Topic6.pdf , eq.(3)). The wave equation's solution can be expressed as some integral (https://en.wikipedia.org/wiki/Wave_equation#Solution_of_a_general_initial-value_problem ), so you do need some integrability condition. 0 You assert that total electromagnetic momentum is small in a small region. Even when this is so, it is not right either morally or mathematically to then throw away a term like \frac{d}{dt} \vec{P}_{em} because something can have a very large time derivative even if it is quite small. A stiff spring oscillating with a small amplitude can have large ... 1 Let u^{\mu} be the 4-velocity of some observer. Then F_{\alpha\beta} = 2E_{[\alpha}u_{\beta]} + \epsilon_{\alpha\beta\gamma\delta}u^{\gamma}B^{\delta} where E^{\alpha}, B^{\alpha} are the electric and magnetic fields relative to this observer and brackets denote antisymmetrization. By definition, u^{\alpha}E_{\alpha} = u^{\alpha}B_{\alpha} = 0. ... 1 You do have a problem, in the sense that even if you had no wire, a loop with self inductance could have lots of different currents running through it. But let's see what we get. We need names to be clear, so let I_w be the current in the wire and let I_l be the current in the loop. Let the flux due to I_w be KI_w and let the flux due to I_l be ... 3 Don't use indices. Use a sophisticated and efficient notation for tensor manipulations: through clifford algebra. Let \gamma^0, \gamma^1, \gamma^2, \gamma^3 be basis vectors. Under the clifford product operation, they obey the following:$$\gamma^\mu \gamma^\nu = \begin{cases}\eta^{\mu \nu} & \mu = \nu \\ -\gamma^\mu \gamma^\nu & \mu \neq ... 0 Here's a systematic way to do this: first recall that the electromagnetic tensor is given by $$F^{\alpha\beta}=\partial^\alpha A^\beta-\partial^\beta A^\alpha$$ where $$F_{\alpha \beta}=\eta^{\mu \nu} F^{\alpha \beta}\eta^{\eta \nu}$$ and $$\partial^\mu=-\nabla,\frac{1}{c}\frac{\partial}{\partial t}$$$A$is the magnetic vector potential and$A^0\$ is ...

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This view would not be accepted by physicists today. Charged particles have mechanical mass, momentum, and energy (rest and kinetic) and the fields have energy and momentum. Total energy is conserved. Total momentum is conserved. Are there cases where it can be sensible to imagine field momentum as an additional mechanical momentum? Sure, consider the ...

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