ρ is a function of t -|r -r'|/c, so by the chain rule, ∇ρ which is dρ/dr = dρ/d(t -|r -r'|/c) d(t -|r -r'|/c)/dr. Now dρ/d(t -|r -r'|/c) = dρ/dt because as far as the dependence of ρ on (t -|r -r'|/c) is concerned, |r -r'| is a constant. Thus = dρ/dr = dρ/dt d(t_r)/dr = dρ/dt ∇(t_r).
Could you elaborate why at 2:40, the gradient of rho is that expression?
ρ is a function of t -|r -r'|/c, so by the chain rule, ∇ρ which is dρ/dr = dρ/d(t -|r -r'|/c) d(t -|r -r'|/c)/dr. Now dρ/d(t -|r -r'|/c) = dρ/dt because as far as the dependence of ρ on (t -|r -r'|/c) is concerned, |r -r'| is a constant. Thus = dρ/dr = dρ/dt d(t_r)/dr = dρ/dt ∇(t_r).
Is there any chance you'll go over chapters 10,11, and 12? Thanks for the videos.
Very helpful. Thank you!
I'm glad it was useful.
Good explanation but why you mentioned Lorentz Gauge instead of Lorenz Gauge. Remember they are two different scientists right?
It's the fault of the book he's using
its very useful sir.thank you
thanx
super helpful!