Help in part of the derivation of Retarded potential.

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The discussion focuses on deriving the retarded potential and clarifying two specific steps in the proof. For the first part, participants confirm that taking the second partial derivative with respect to time and dividing by c² leads to the desired equation. In the second part, the integration involving the Dirac delta function simplifies to evaluating the density function at the point where the delta function is zero, which resolves the confusion about the absence of η in the final expression. There is also a discussion about the relationship between the variables t and t_r, noting that they differ by a constant. Overall, the participants seek clarity on the derivation steps and the implications of the variables involved.
yungman
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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
 
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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'.
 
Born2bwire said:
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
 
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