# Tag Info

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Yes and no. On the one hand of you have a magnetic field that circulates just like it does around a steady current and the current is anything other than that steady current then the electric field will change: $$\vec \nabla \times \vec B = \mu_0\vec J +\mu_0\epsilon_0\frac{\partial \vec E}{\partial t}$$ is the same as \frac{\partial \vec E}{\partial ...

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I am tempted to give an experimentalist's point of view, though of course I agree with Lubos and Savanah's answers. Consider a disk of radius R, then circumference is 2πR. Now, make this disk rotate at velocity of the order of c(speed of light). Here it is evident that you are defining a center of mass system for a disk and a person/machinery that will ...

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To someone moving on the edge of the disk, the disk is length contracted in the direction of travel. So to them, it is not a circle, no problems with geometry. Euclidean geometry only holds for an inertial plane of simultaneity. By the end of this answer you should know when and how to use Euclidean geometry and what it means (or doesn't). And a collection ...

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This is just relativity of simultaneity again. A similar thing happens if you have a bunch of spaceships in a line that fire their thrusters at a fixed time. Different observers will disagree about whether they fired at the same time and will disagree about the spacing. Always in a consistent way. So I'd like to address the concept of geometry by not having ...

6

No, it doesn't violate the rules of geometry, it violates the rules of Euclidean geometry. Simple conclusion: for an observer fixed to a disk rotating uniformly relative to an inertial frame, the spatial geometry is non-Euclidean; in particular, the ratio of a circle's circumference to its diameter depends both on the circle's diameter and center position. ...

6

What is wrong is the idea that one can actually make the disk rotate; and it will remain perfectly rigid. In reality, what this correct argument shows is that relativity doesn't admit the existence of any perfectly rigid bodies. This is a perfectly basic, settled, and indisputable textbook material that every mature physicist knows. The first sentence of ...

2

The classical electron radius has no basis in fact. And potential energy is associated with a system not with any specific particle, is frame dependent, sometime gauge dependent, and non local son relativistically problematic. It is not a store of energy. In the case of electromagnetism, energy is stored in volumes containing electromagnetic fields (and in ...

1

Putting aside the electron case, which is covered by Vladimir's answer, the energy of a charged sphere at rest is indeed what you are saying it to be. Now, is the uncharged sphere shell mass incomplete? No, it is just not included because is not necessary. In Physics, Energy needs to be accounted completely only if is involved actively in the processes ...

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The oscillating electromagnetic wave, why is the wave oscillating between electric and magnetic waves? I'll focus my answer on electromagnetic radiation. Light - a small part of electromagnetic spectrum - could be produced by accelerating electrons. The easiest way is to take a thin current carrying wire. The electrons inside the wire will bounce on ...

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Quantum mechanics is the reality of things,not classical mechanics(including EM waves). Having this in mind, the photon is not a different thing from the EM wave. They are the same entities. The craziness of quantum mechanics(well one of them) is the wave-particle duality. It's not that the photon travels with the electromagnetic wave, it is the EM wave(as ...

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"Assembling" is a hand-wave explanation, it is not serious. In CED the electron is point-like, but in some calculations (scattering or something) a dimensional quantity $r_0$ arises. It does not mean the electron size, though.

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The voltage V that you're computing would be induced between the upper and lower sides of your orange block, and $L$ in your computation is the distance between these two sides. This voltage is induced between any two opposite points on the upper and lower sides provided they are in the area that is covered by the magnetic field. If you connect a wire with ...

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I'm not sure discontinuity has anything to do with it. If you have two parallel waveguides running in the x direction separated by a distance d with the electricomagnetic field having momentum inside the guides pointing entirely in the x direction then for every little piece of matter inside the waveguide the force density for components other than the x ...

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