Potential vector of a oscilating dipole:

AI Thread Summary
The discussion focuses on deriving the electric potential of an oscillating dipole as presented in Griffiths' Electrodynamics. Participants explore the transition from one term to another in the context of the time variable t_o, defined as t - r/c. The use of the chain rule is emphasized, particularly in calculating the derivatives of the dipole moment components with respect to spatial coordinates. The conversation clarifies that the gradient operator can be expressed in terms of the time derivative, leading to the conclusion that the x-component of the gradient involves the second derivative of the dipole moment. Overall, the thread delves into the mathematical intricacies of the oscillating dipole's electric potential.
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Homework Statement
I.
Relevant Equations
.
I am passing through some difficulties to understand the reasoning to derive the electric potential of an oscilating dipole used by Griffths at his Electrodynamics book:
Knowing that ##t_o = t - r/c##,
Captura de tela de 2022-05-13 07-55-47.png

What exactly he has used here to go from the first term after "and hence" to the second term? (The one involving ##\ddot p##). Maybe ##\nabla = \nabla t_o \frac{d}{dt_o}## ? But this does not make sense.
 
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A straightforward, but somewhat tedious, method is to work in cartesian coordinates.

Consider the ##x##-component of ##\nabla \left[ \frac {\mathbf{\hat r} \cdot \mathbf{ \dot p}(t_0)}{r} \right]##:

$$\left(\nabla \left[ \frac {\mathbf{\hat r} \cdot \mathbf{ \dot p}(t_0)}{r} \right]\right)_x = \frac{\partial}{\partial x} \left[ \frac{x}{r^2} \dot p_x(t_0) + \frac{y}{r^2} \dot p_y(t_0) +\frac{z}{r^2} \dot p_z(t_0) \right]$$
Expand this out. You can drop terms that fall off faster than ##1/r##.

When calculating ##\large \frac{\partial \dot p_x(t_0)}{\partial x}##, recall that ##t_0 = t-r/c =t-\sqrt{x^2+y^2+z^2}/c##. So, you can use the chain rule to write $$ \frac{\partial \dot p_x(t_0)}{\partial x} = \frac{d \dot p_x}{dt_0} \frac{\partial t_0}{\partial x}$$
Likewise for dealing with ##\large \frac{\partial \dot p_y(t_0)}{\partial x}## and ##\large \frac{\partial \dot p_z(t_0)}{\partial x}##.

You should find that
$$\left(\nabla \left[ \frac {\mathbf{\hat r} \cdot \mathbf{ \dot p}(t_0)}{r} \right]\right)_x = \left( \frac {\mathbf{\hat{r}}}{r} \cdot \mathbf{\ddot p}(t_0)\right) \frac{\partial t_0}{\partial x}$$
Similar results are obtained for the ##y## and ##z## components of ##\nabla \left[ \frac {\mathbf{\hat r} \cdot \mathbf{ \dot p}(t_0)}{r} \right]##.
 
Herculi said:
Homework Statement:: I.
Relevant Equations:: .

Maybe ∇=∇toddto ? But this does not make sense.
Yes it does. That is just the chain rule:
$$
[\nabla \dot{\vec p}(t_0)]_{ij} = \frac{\partial \dot p^j(t_0)}{\partial x^i} = \frac{d \dot p^j}{dt_0}\frac{\partial t_0}{\partial x^i} = \ddot{p}^j(t_0) \partial_i t_0.
$$
 
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