Timeline for Why doesn't the acceleration of an electron along the line of sight from the observer contribute to the electric field?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2018 at 10:08 | vote | accept | Prem | ||
| S Jan 17, 2018 at 8:58 | history | suggested | user2299067 | CC BY-SA 3.0 |
Fixed grammar
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| Jan 17, 2018 at 8:40 | answer | added | Voulkos | timeline score: 2 | |
| Jan 17, 2018 at 7:09 | review | Suggested edits | |||
| S Jan 17, 2018 at 8:58 | |||||
| Jan 17, 2018 at 6:03 | history | edited | Prem | CC BY-SA 3.0 |
added 290 characters in body
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| Jan 17, 2018 at 6:00 | comment | added | Prem | @Triatticus But the general solution of Maxwell's equation the equation appears to be saying that a moving charge simply radiates its scalar and vector potentials in all directions, and then an observer just takes the space and time derivatives of these vectors to obtain the electric fields. In this picture, even if the electron is accelerating directly towards the observer, $$\frac {\partial \mathbf{A}}{\partial t}=\int\frac{q}{4\pi\epsilon_{0} c^2}\frac{\partial \frac{\mathbf{v}}{r_{12}}}{\partial{t}}\,dV_2$$ Which means even the acceleration along the line of sight will act. | |
| Jan 17, 2018 at 5:52 | comment | added | Prem | @Frobenius But the general solution to Maxwell's equation says that$$\mathbf{A}(1,t)=\int\frac{q\mathbf{v}(2,t-r_{12}/c)}{4\pi\epsilon_{0} c^2r_{12}}\,dV_2$$which means that$$\frac {\partial \mathbf{A}}{\partial t}=\int\frac{q}{4\pi\epsilon_{0} c^2}\frac{\partial \frac{\mathbf{v}}{r_{12}}}{\partial{t}}\,dV_2$$ Here, the acceleration resulting from $\frac{\partial \frac{\mathbf{v}}{r_{12}}}{\partial{t}}\,dV_2$will also have a component along $\mathbf{r'}$, not just perpendicular to it, as Feynman is telling. So, my question is- what happened to the component of $\mathbf{a}$ along $\mathbf{r'}$? | |
| Jan 16, 2018 at 22:51 | comment | added | Voulkos | We have also seen that if the velocity $\:\upsilon\:$ of a charge is always much less than c, and if we consider only points at large distances from the charge, so that only the last term of Eq. (21.1) is important, the fields can also be written as $$ \boldsymbol{E}=\dfrac{q}{4\pi\epsilon_{0}c^{2}r'} \begin{bmatrix} \text{acceleration of the charge at } (t-r'/c) \\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\text{projected at right angles to } r' \end{bmatrix} \tag{21.1$^\boldsymbol{\prime}$} $$ (Feynman Lectures, Volume 2, $\S$21-1) | |
| Jan 16, 2018 at 21:23 | comment | added | Triatticus | A particle traveling away from or towards you looks identical to a particle standing still, the field radiated looks unchanged from a stationary particle which is what the first term yields | |
| Jan 16, 2018 at 18:55 | history | asked | Prem | CC BY-SA 3.0 |