I'm reading Griffiths Ch. 10. In the 10.3.1 section, there's a proof of this integral
$$ \int \rho(r^\prime, t_r) \mathrm{d} \tau^\prime $$ which is not equal to the charge of the particle, but really I don't understand what is the reason, because if the system is isolated, all density charge will be the same in the time retarded, no? so, it will be equal to the total charge.
I have another question,
Griffiths shows that
$$ \int \rho(r^\prime, t_r) \mathrm{d} \tau^\prime = \frac{q}{1 -\frac{ |\vec{r}-\vec{r'}| \cdot \vec{v}}{c}} $$
And deriving it, he used this example:
he writes $$\frac{L'}{c} =\frac{L'-L}{v}$$
But I think, there are many assumptions that aren't necessarily always right. For example:
- Is the final time the same to the light as to the train? if the Light travels faster, how it can be possible?
I don't know, I'm very confused about it, I will appreciate any help.
