No, it isn't. The function ρ is zero for x'<-L/2 and x'>L/2 , so it does not give the extent of the retarded distribution. The latter is not only...
You thought wrongly then. You integrated from -L/2/(1-v/c) to +L/2/(1-v/c) which is the width of the retarded distribution. The unretarded...
I am puzzled as to how you get to this conclusion, considering that the 'retarded' charge density (=q*(1-v/c)/L in this case) is exactly what...
There is no integral over time in the definition for the potential. The integral is only over spatial positions. Writing the time argument for ρ only...
It is no surprise that you are getting a different result from a numerical evaluation of your integral because you integral is incorrect. I have...
I asked you to derive it without using any charged density function ρ at all but only the the charge q of each particle. So you are claiming that a...
Then prove it without using any integral over some hypothetical mathematical functions that you claim represent the ensemble of discrete charges....
deleted post (pressed 'Reply' instead of 'Edit' by mistake)
Only that your integral does not give the same result as the discrete sum anymore if the charges are moving. Your argument is a contradiction in...
As far as the static Coulomb potential is concerned (which we are discussing here), it is (at least in some respects) approximately the same if you...
I really don't know on what basis you are saying this. If you have discrete particles, then the only exact representation is given by the discrete...
You should call it just 'integral'. Calling it 'integral equation' hurts anybody who has actually worked with integral equations (as I happen to have...
First of all, you should not be calling the expression for the potential an 'integral equation' (I noticed this already a couple of times before), as...
I am back from my break and ready to resume the discussion. Only that you are not integrating over \rho_{0}(x') but over \rho(x', t_{\mathrm{R}})....
Just another quick one before my break: If you distribute the same amount of charges over a larger region then the density ρ has to decrease...
No, you didn't give any integration bounds at all. That's why you actually haven't given a valid derivation. Obviously, the integration bounds (or...
I wasn't referring to the Lorentz contraction but the apparent contraction by the time-layering effect caused by the retardation here: Take again...
Well, at times when there is no signal, the field is zero anyway.
But you can make the approximation arbitrarily accurate by making r'/r small enough. The point is that your solution still differs by the same factor...
What would be the conceptual mistake with this? If you assume the speed of light does not depend on the speed of the emitter, then the field of the...
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