Help in part of the derivation of Retarded potential.

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    Derivation Potential
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SUMMARY

The discussion focuses on deriving the retarded potential, specifically the equation for the scalar potential \( V_{(\vec r, t)} \) in the context of electromagnetic theory. The participants analyze the transition from the integral form involving charge density \( \rho_{(\vec r', t_r)} \) to its second-order time derivative, leading to the equation \( \nabla^2 V_{(\vec r, t)} \). They also clarify the role of the Dirac delta function in simplifying the volume integral, confirming that it evaluates the charge density at the point where \( \eta = 0 \). The conversation emphasizes the mathematical rigor required in these derivations.

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  • Familiarity with partial differential equations and their applications in physics.
  • Knowledge of the Dirac delta function and its properties in integration.
  • Proficiency in calculus, specifically in evaluating integrals involving variable transformations.
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This discussion is beneficial for physicists, electrical engineers, and students studying electromagnetic theory, particularly those interested in the mathematical foundations of retarded potentials and their applications in wave propagation and field theory.

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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