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yungman
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This is exact copy from Griffiths Introduction to Electrodynamics 3rd edition page 421. This is regarding to information travel in space. In time varying situation, E depend not only on V, but on A also.
There is a peculiar thing about the scalar potential in the Coulomb gauge: it is determined by the distribution of charge right now. If I move an electron in my laboratory, the potential V on the moon immediately records this change. That sounds particularly odd in the light of special relativity, which allows no message to travel faster than the speed of light. The point is that V by itself is not a physically measurable quantity-all the man in the moon can measure is E, and that involves A as well. Somehow it is built into the vector potential, in the Coulomb gauge, that whereas V instantaneously reflects all changes in [itex]\rho[/itex], the combination:
[tex]\vec E = -\nabra V - \frac{\partial \vec A}{\partial t}[/tex]
does not; E will change only after sufficient time has elapsed for the "news" to arrive.
[ End Quote]
These are my questions:
1) If the information of moving the electron in the lab can only transmit as E to the moon, and it takes time for E to travel to the moon, why then the man on the moon can immediately records the change?
2) Why is potential not a measurable quantity?
3) In the next section about retarded potential:
[tex]V_{(\vec r , t)} =\frac 1 {4\pi \epsilon_0} \int_{v} \frac { \rho(\vec r’\hbox{,} t_r)}{|\vec r - \vec r'|} d\tau [/tex]
This obvious show that V is time dependent. Is this because the information travel through space as E and this potential is obtained at the field point ( the moon in the current case) by calculating
[tex] V_{(\vec r , t)} = -\int \vec E \cdot d\vec l[/tex]
Thanks
Alan
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