Help in math in retarded potential.

[yungman]

This is copy from Griffiths Introduction to Electrodynamics page 424.


V = \frac 1 {4\pi\epsilon_0} \int \frac {\rho_{(\vec r',t_r)}}{ \eta } d \tau \hbox{'} \Rightarrow \; \nabla V = \frac 1 {4\pi\epsilon_0} \int \nabla \left ( \frac {\rho_{(\vec r',t_r)}}{\eta}\right ) d \tau \hbox{'} = \frac 1 {4\pi\epsilon_0} \int \left [ (\nabla \rho) \frac 1 {\eta} + \rho \nabla \left ( \frac 1 { \eta} \right ) \right ] d \tau \hbox{'}

Where \eta = | \vec r - \vec r’|



Please help me in deriving the following equation:

1) \nabla \rho = \dot{\rho} \nabla t_r = -\frac 1 c \dot {\rho} \nabla (\eta)

Where \dot{\rho} = \frac {\partial \rho}{\partial t}



2) Also why is \nabla (\eta) = \hat {(\eta) }

Thanks

 

[yungman]

Anyone can help?

 

[vanhees71]

This is in the Lorenz gauge (older books name it Lorentz gauge, but historically it is more just to name it after the Danish physicist Ludvig Lorenz instead of the Dutch physicist Hendrik A. Lorentz). Then the scalar and the vector potential are both retarded. In the following, I set the speed of light, c=1, to make the formulae easier to read. I also use Heaviside-Lorentz units (i.e., rationalized Gauss units), leading to the more natural settings \epsilon_0=\mu_0=1.

The complete scalar and vector potential reads

V(t,\vec{x})=\int_{\mathbb{R}^3} \dd^3 \vec{x}' \frac{\rho(t-|\vec{x}-\vec{x}'|,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|},

\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \dd^3 \vec{x}' \frac{\vec{j}(t-|\vec{x}-\vec{x}'|,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.

The fields are given by

\vec{E}(t,\vec{x})=-\vec{\nabla} V(t,\vec{x})-\partial_t \vec{A}(t,\vec{x}),

\vec{B}(t,\vec{x})=\vec{\nabla} \times \vec{A}(t,\vec{x}).

To get the fields, you have to do the derivatives of the retarded potentials. This is not difficult but a bit tedious. The easist way is to use the Ricci calculus in Cartesian coordinates. So we define the differential operators

\partial_j=\frac{\partial}{\partial x_j}, \quad \partial_j'=\frac{\partial}{\partial x_j'},

where the indices j \in \{1,2,3\} label the components of vectors. Also in this calculus, the Einstein summation convention is useful, i.e., if an index appears twice, one has to sum over it.

Now it's easy to answer your questions. The retarded time is

t_r=t-|\vec{x}-\vec{x}'|.

Then

\partial_j t_r=-\partial_j|\vec{x}-\vec{x}'|.

Now we have

|\vec{x}-\vec{x}'|=\sqrt{(x_k-x_k')(x_k-x_k')},

and you get the derivative by the chain rule

\partial_j|\vec{x}-\vec{x'}|=\frac{2(x_j-x_j')}{2 \sqrt{(x_k-x_k')(x_k-x_k')}}=\frac{x_j-x_j'}{|\vec{x}-\vec{x}'|}.

Now the gradient is just the vector with the components given by these partial derivatives. Thus, we finally get

\vec{\nabla} \eta=\vec{\nabla} |\vec{x}-\vec{x}'|=\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|}=\hat{\eta} \quad \text{with} \quad \hat{\eta}=\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|}.

This answers your second question. Now back to the first. What's calculated there is \vec{\nabla}\rho(t_r,\vec{x}').

Again we use the Ricci calculus first, and with the chain rule we find indeed

\partial_j \rho(t_r,\vec{x}')=[\partial_t \rho(t,\vec{x}')]_{t=t_r} \partial_j t_r=-\dot{\rho}(t_r,\vec{x}') \hat{\eta}.

Of course, if you set c not to 1, you get an additional 1/c in this expression.

 

[yungman]

Thanks Vanhees71