Help in part of the derivation of Retarded potential.

[yungman]

This last part of the steps in proofing

$$V_{(\vec r, t)} = \frac 1 {4\pi \epsilon_0} \int_{v'} \frac {\rho _{(\vec r',t_r)} }{\eta} d\tau' \;\hbox { where } \;\eta=|\vec r -\vec r'|$$.

In the last step:

$$\nabla^2 V_{(\vec r , t) } = \frac 1 {4\pi \epsilon_0} \int_{v'} \left [ \frac 1 {c^2} \frac {\ddot{\rho} _{(\vec r',t_r)} }{\eta} - 4\pi \rho _{(\vec r',t_r)} \delta^3 (\vec {\eta}) \right ] d\tau' \;= \;\frac 1 {c^2} \frac {\partial^2 V_{(\vec r , t)}}{\partial t^2}\; - \;\frac {\rho_{(\vec r',t_r)} }{\epsilon_0}$$



My questions are:

1) How do I go from $$\frac 1 {4\pi \epsilon_0} \int_{v'} \frac 1 {c^2} \frac {\ddot{\rho} _{(\vec r',t_r)} }{\eta} d\tau' \;= \;\frac 1 {c^2} \frac {\partial^2 V_{(\vec r , t)}}{\partial t^2}$$ ?

Only way I can come up with is:

$$V_{(\vec r, t)} = \frac 1 {4\pi \epsilon_0} \int_{v'} \frac {\rho _{(\vec r',t_r)} }{\eta} d\tau' \;\Rightarrow \; \frac {\partial^2 V_{(\vec r , t)}}{\partial t^2} = \frac 1 {4\pi \epsilon_0} \int_{v'} \frac {\ddot{\rho} _{(\vec r',t_r)} }{\eta} d\tau'$$

Does anyone have a better way to derive this?

2) I cannot verify the second part:

$$\frac 1 {4\pi \epsilon_0} \int_{v'} 4\pi \rho _{(\vec r',t_r)} \delta^3 (\vec {\eta}) \right ] d\tau' = \frac {\rho _{(\vec r',t_r)} }{\epsilon_0}$$

Notice $$\delta ^3(\eta)}$$? But the final part $$\;\frac {\rho _{(\vec r',t_r)} }{\epsilon_0} \;$$ has no $$\eta$$ in it?

Can anyone help?

thanks

 

[Born2bwire]

This all seems fairly trivial but I do not readily perceive what your integrating variables are. I assume that they are over space.

For part one, they took the second order partial derivative with respect to time of both sides and then divided both sides by c^2.

As for part two, it's just the volume integral of a function scaled by the dirac delta distribution which equals the value of the function evaluated where the dirac delta's arguments are zero. Since \eta is zero when r = r', then we simply evaluate \rho at r'.

 

[yungman]

This all seems fairly trivial but I do not readily perceive what your integrating variables are. I assume that they are over space.

For part one, they took the second order partial derivative with respect to time of both sides and then divided both sides by c^2.
That's what I came up as shown in my work, that I took the 2nd partial derivative respect to t. I was hoping that I can derive straight from the last step instead of take the answer and back up by taking the derivative on both side. But I would be happy if that is the only way.
As for part two, it's just the volume integral of a function scaled by the dirac delta distribution which equals the value of the function evaluated where the dirac delta's arguments are zero. Since \eta is zero when r = r', then we simply evaluate \rho at r'.

Thanks for the reply

I am still a little unsure about the part, I have to think a little more. Also from the original equation $\rho(\vec r, t_r)$ means it is function of $t_r$. But the answer is function of t only. Do I just look at this that $t_r$ differ from t by a constant so it is a function of t?

Thanks

Alan