# Radiated power from a 'source volume' outside which charge & current are zero (i.e. derive radiated power from Jefimenko's equations)

In classical electrodynamics, what is the radiated power from a generalized source (consisting of charge density $\rho$ and current density $\vec{J}$) in vacuum?

Let us define $V_s$ to be the smallest sphere that completely encloses all source points -- that is, it is the smallest sphere outside which $\rho\left(\vec{r},t\right) = \vec{J}\left(\vec{r},t\right) = 0$ for all $t$. Let us say $\mathcal{R}_s$ is the radius of $V_s$.

Now let us define another sphere $V$ concentric with $V_s$. Let us say the radius of this sphere is $\mathcal{R}$, and $\mathcal{R} > \mathcal{R}_s$.

So what is power crossing the surface of $V$ at any instant? In other words, what is the value of the following? $$P\left(\mathcal{R},t\right) = \oint_{\partial V} \vec{S}\left(\vec{r},t\right) \cdot \space d\vec{s}\left(\vec{r}\right) = \frac{1}{\mu_0} \oint_{\partial V} \left( \vec{E}\left(\vec{r},t\right) \times \vec{B}\left(\vec{r},t\right) \right) \cdot \space d\vec{s}\left(\vec{r}\right)$$

Below, I have provided an answer for the 'far field' approximation ($\mathcal{R} \gg \mathcal{R}_s$). Are there 'sanitized' expressions for cases where this approximation does not hold?

Even if there isn't a 'clean' expression for the general case, what I am looking for is the following. Let us say that there are two instants $t_1$ and $t_2$ such that for all $\vec{r} \in V_s$ we have $\rho\left(\vec{r}, t_1\right) = \rho\left(\vec{r}, t_2\right)$ and $\vec{J}\left(\vec{r}, t_1\right) = \vec{J}\left(\vec{r}, t_2\right)$. If we now define: $$\mathcal{P}_{\infty}\left(t\right) = \lim_{\mathcal{R}\to\infty} P\left(\mathcal{R},t + \frac {\mathcal{R}} {c}\right) \\ \mathcal{P}_{\delta}\left(\mathcal{R},t\right) = P\left(\mathcal{R},t + \frac {\mathcal{R}} {c}\right) - \mathcal{P}_{\infty}\left(t\right)$$ ... can the following be proven? $$\int_{t_1}^{t_2} \mathcal{P}_{\delta}\left(\mathcal{R},t\right)\space dt = 0$$ Also, is it true (and can it be proven) that if $\hat{n}\left(\vec{r}\right)$ is the unit normal at any $\vec{r} \in V$, there will always be one axis along which $\vec{S}\left(\vec{r},t\right) \cdot \hat{n}\left(\vec{r}\right) = 0$?

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For $\mathcal{R} \gg \mathcal{R}_s$, if we define: $$\vec{Y}\left(t\right)= \iiint_{V_s} \vec{J}\left(\vec{r},t\right) \space dV\left(\vec{r}\right) ,\qquad \vec{Z}\left(t\right)= \iiint_{V_s} \frac{\partial \vec{J}\left(\vec{r},t\right)} {\partial t} \space dV\left(\vec{r}\right) = \frac {d \vec{Y}\left(t\right)} {dt}$$ and also $$t_{\mathcal{R}} = t - \frac{\mathcal{R}} {c}$$ Assuming that the origin is at the center of $V$, so that $\hat{n}\left(\vec{r}\right) = \hat{r}$, the results are: $$\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right) = - \frac { \left( \vec{Y}\left(t_\mathcal{R}\right) \times \hat{n}\left(\vec{r}\right) \right) \cdot \left(\vec{Z}\left(t_\mathcal{R}\right) \times \hat{n}\left(\vec{r}\right) \right) } {16 \pi^2 \epsilon_0\mathcal{R}^3 c^2} - \frac { \left| \vec{Z}\left(t_\mathcal{R}\right) \times \hat{n}\left(\vec{r}\right) \right|^2 } {16 \pi^2 \epsilon_0\mathcal{R}^2 c^3} \\ P\left(\mathcal{R},t\right) = \frac {1} {6 \pi \epsilon_0 c^2} \left( \frac {\vec{Y}\left(t_{\mathcal{R}}\right)\cdot\vec{Z}\left(t_{\mathcal{R}}\right)} {\mathcal{R}} + \frac {\left|\vec{Z}\left(t_{\mathcal{R}}\right)\right|^2} {c} \right) \\ \mathcal{P}_{\infty}\left(t\right) = \frac {\left|\vec{Z}\left(t\right)\right|^2} {6 \pi \epsilon_0 c^3} ,\qquad \mathcal{P}_{\delta}\left(\mathcal{R},t\right) = \frac {\vec{Y}\left(t\right)\cdot\vec{Z}\left(t\right)} {6 \pi \epsilon_0 \mathcal{R} c^2}$$

## Derivation

Given the following definitions: $$\vec{R} = \vec{r} - \vec{r}_s, \qquad R = \left | \vec{R} \right |, \qquad t_r = t - \frac {R} {c}$$ From Jefimenko's Equations, the value of $\vec{E}$ and $\vec{B}$ for any $\vec{r}$ and any $t$ is as follows: $$\vec{E}(\vec{r},t) = \frac {1} {4 \pi \epsilon_0} \iiint_{V_s} {\left( \frac {\rho (\vec{r}_s, t_r)} {R^3} \vec{R} + \frac {1} {R^2 c} \frac {\partial \rho (\vec{r}_s, t_r) } {\partial t} \vec{R} - \frac {1} {R c^2} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \right)} \space dV\left(\vec{r}_s\right)$$ $$\vec{B}(\vec{r},t) = \frac {\mu_0} {4 \pi} \iiint_{V_s} {\left( \frac {\vec{J} (\vec{r}_s, t_r)} {R^3} \times \vec{R} + \frac {1} {R^2 c} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \times \vec{R} \right)} \space dV\left(\vec{r}_s\right)$$ Using this, we can write down the expression for $\vec{S}\left(\vec{r},t\right)$, which has six terms, each of which is a product of two volume integrals. We are, however, interested in $\vec{S}\left(\vec{r},t\right) \cdot \hat{n}\left(\vec{r}\right)$, where, if we designate the center of $V$ (and $V_s$) as $\vec{r}_0$: $$\hat{n}\left(\vec{r}\right) = \frac {\vec{r}-\vec{r}_0} {\left|\vec{r}-\vec{r}_0\right|}$$ Now if we choose a frame of reference where $\vec{r}_0$ is the origin, we find that $\hat{n}\left(\vec{r}\right) = \hat{r}$. When $\mathcal{R} \gg \mathcal{R}_s$, we can consider $\dfrac{\left|\vec{r}_s\right|}{\left|\vec{r}_s\right|} \approx 0$. Thus, irrespective of $\vec{r}_s$ under this 'far field' approximation, $\vec{R}\approx\vec{r}-\vec{r}_s$, and is effectively independent of $\vec{r}_s$, so it can be taken outside of the volume integrals.

Then, because $\left(\vec{a}\times\left(\vec{b}\times\vec{a}\right)\right)\cdot\vec{a}=0$ four of the six terms in $\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right)$ turn out to be zero.

Next, we substitute $\vec{Y}$ and $\vec{Z}$ into the expression for $\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right)$ and get the expression listed above.

To help us with the surface integral, we define a spherical coordinate system $(\mathscr{r},\theta,\phi),\space 0\le \mathscr{r}\lt\infty,\space 0\le\theta\le\pi,\space 0\le\phi\le 2\pi$ with its center at $\vec{r}_0$, and orient it such that: $\vec{Z}\left(t\right)$ is $( |\vec{Z}\left(t\right)|, 0, 0)$ and $\vec{Y}\left(t\right)$ is $( |\vec{Y}\left(t\right)|, \gamma\left(t), 0\right)$. Note that the orientation of this coordinate system changes with time. In this coordinate system, therefore, $\hat{n}\left(\vec{r}\right)$ at any instant $t$ has to be represented in terms of the retarded time as $\left(1,\vartheta\left(\vec{r},t_r\right), \varphi\left(\vec{r},t_r\right)\right)$

Working out $\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right)$ in this coordinate system, we get: $$\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right) = \frac {\left|\vec{Z}\left(t_{\mathcal{R}}\right)\right|^2} {16 \pi^2 \epsilon_0 {\mathcal{R}}^2 c^3} \sin^2\vartheta\left(\vec{r},t_{\mathcal{R}}\right) + \frac {\vec{Y}\left(t_{\mathcal{R}}\right)\cdot\vec{Z}\left(t_{\mathcal{R}}\right)} {16 \pi^2 \epsilon_0 {\mathcal{R}}^3 c^2} \sin^2\vartheta\left(\vec{r},t_{\mathcal{R}}\right) \\ - \frac {\left|\vec{Y}\left(t_{\mathcal{R}}\right)\times\vec{Z}\left(t_{\mathcal{R}}\right)\right|} {32 \pi^2 \epsilon_0 {\mathcal{R}}^3 c^2} \sin2\vartheta\left(\vec{r},t_{\mathcal{R}}\right) \cos\varphi\left(\vec{r},t_{\mathcal{R}}\right)$$

Finally, we compute the surface integral. For a sphere $V$ of radius $\mathcal{R}$ we get: $$\oint_{\partial V} \sin^2\vartheta\left(\vec{r}\right) \space ds(\vec{r}) = \frac {8 \pi \mathcal{R}^2} {3},\qquad \oint_{\partial V} \sin 2\vartheta\left(\vec{r}\right) \space\cos\varphi\left(\vec{r}\right) \space ds(\vec{r})= 0$$

Substituting, we get our result.

## Observations

For the record,because of the way we've defined $V_s$ and because of conservation of charge, $Q_s$ is the (constant) total charge inside $V_s$, $$\iiint_{V_s} \rho\left(\vec{r},t\right) \space dV\left(\vec{r}\right) = Q_s ,\qquad \iiint_{V_s} \frac{\partial \rho\left(\vec{r},t\right)} {\partial t} \space dV\left(\vec{r}\right) = \frac {d Q_s} {dt} = 0$$ ... but we don't need these terms in the derivation.

Also, $\int_{t_1}^{t_2} \mathcal{P}_{\delta}\left(\mathcal{R},t\right) \space dt = 0$ because: $$\int_{t_1}^{t_2} \vec{Y}(t)\cdot\vec{Z}(t) dt = \int_{t_1}^{t_2} \vec{Y}(t)\cdot\frac{d\vec{Y}}{dt}(t) dt = \frac12 \int_{t_1}^{t_2} \frac{d}{dt}|\vec{Y}(t)|^2 dt = \frac{1}{2} \left[|\vec{Y}(t)|^2\right]_{t_1}^{t_2}$$ and $$\forall\vec{r}\in V: \vec{J}\left(\vec{r},t_1\right) = \vec{J}\left(\vec{r},t_2\right) \implies \vec{Y}\left(t_1\right) = \vec{Y}\left(t_2\right)$$

The value of $\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right)$ is zero whenever $\vec{Z}\left(t_{\mathcal{R}}\right) \times \hat{n}\left(\vec{r}\right) = 0$, that is, wherever the unit normal is parallel to $\vec{Z}\left(t_{\mathcal{R}}\right)$

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My previous answer addresses the case where $\mathcal{R} \gg \mathcal{R}_s$, so $\dfrac{\left|\vec{r}_s\right|}{\left|\vec{r}\right|}$ for all $\vec{r}_s\in V_s$ and $\vec{r}\in \partial V$. What if it isn't so?

Unfortunately, things no longer stay simple. If we define the following new integrals: $$\vec{W}_3\left(\vec{r},t\right)= \iiint_{V_s} \left( \frac {\vec{a}} {\left|\vec{r}-\vec{a}\right|^3} {\rho\left(\vec{a}, t - \dfrac{\left|\vec{r}-\vec{a}\right|} {c} \right)} \right) \space dV\left(\vec{a}\right) ,\\ \vec{X}_2\left(\vec{r},t\right)= \iiint_{V_s} \left( \frac{\vec{a}} {\left|\vec{r}-\vec{a}\right|^2} \frac{\partial} {\partial t} \rho\left(\vec{a}, t - \dfrac{\left|\vec{r}-\vec{a}\right|} {c} \right) \right) \space dV\left(\vec{a}\right) ,\\ \vec{Y}_{3}\left(\vec{r},t\right)= \iiint_{V_s} \left( \vec{J}\left(\vec{a}, t - \dfrac{\left|\vec{r}-\vec{a}\right|} {c} \right) \times \frac {\vec{r} - \vec{a}} {\left|\vec{r}-\vec{a}\right|^3} \right) \space dV\left(\vec{a}\right) ,\\ \vec{Z}_{2}\left(\vec{r},t\right)= \iiint_{V_s} \left( \frac{\partial} {\partial t} \vec{J}\left(\vec{a}, t - \dfrac{\left|\vec{r}-\vec{a}\right|} {c} \right) \times \frac {\vec{r} - \vec{a}} {\left|\vec{r}-\vec{a}\right|^2} \right) \space dV\left(\vec{a}\right) ,\\ \vec{Z}_1\left(\vec{r},t\right)= \iiint_{V_s} \left( \frac {1} {\left|\vec{r}-\vec{a}\right|} \frac{\partial} {\partial t} \vec{J}\left(\vec{a}, t - \dfrac{\left|\vec{r}-\vec{a}\right|} {c} \right) \right) \space dV\left(\vec{a}\right)$$ We can derive that: $$\vec{S}\left(\vec{r},t\right)\cdot \hat{r} = \frac {1} {16 \pi^2 \epsilon_0 \mathcal{R} } \left( \vec{Y}_3 \left(\vec{r},t\right) + \frac {\vec{Z}_2 \left(\vec{r},t\right)} {c} \right) \cdot \left( \vec{W}_3 \left(\vec{r},t\right) \times \vec{r} + \frac {\vec{X}_2 \left(\vec{r},t\right) \times \vec{r}} {c} - \frac {\vec{Z}_1 \left(\vec{r},t\right) \times \vec{r}} {c^2} \right)$$

The problem, now, is that unlike $\vec{Y}$ and $\vec{Z}$, the functions $\vec{W}_3$, $\vec{X}_2$, $\vec{Y}_3$, $\vec{Z}_2$ & $\vec{Z}_1$ all contain $\vec{r}$ as a parameter, and so cannot be brought out of the surface integral over $\partial V$.

If we could take the approximation $\vec{r} - \vec{a} \approx \vec{r}$, we could say: $$t_{\mathcal{R}} = t - \frac{\mathcal{R}} {c} ,\\ \vec{W}\left(t\right)= \iiint_{V_s} \rho\left(\vec{r},t\right) \space\vec{r} \space dV\left(\vec{r}\right) ,\qquad \vec{X}\left(t\right)= \iiint_{V_s} \frac{\partial \rho\left(\vec{r},t\right)} {\partial t} \vec{r} \space dV\left(\vec{r}\right) = \frac {d \vec{W}\left(t\right)} {dt} ,\\ \vec{Y}\left(t\right)= \iiint_{V_s} \vec{J}\left(\vec{r},t\right) \space dV\left(\vec{r}\right) ,\qquad \vec{Z}\left(t\right)= \iiint_{V_s} \frac{\partial \vec{J}\left(\vec{r},t\right)} {\partial t} \space dV\left(\vec{r}\right) = \frac {d \vec{Y}\left(t\right)} {dt}$$ ... and from there ... $$\vec{W}_3\left(\vec{r},t\right)=\frac {\vec{W}\left(t_\mathcal{R}\right)} {\mathcal{R}^3} ,\qquad \vec{X}_2\left(\vec{r},t\right)=\frac {\vec{X}\left(t_\mathcal{R}\right)} {\mathcal{R}^2} ,\qquad \vec{Z}_1\left(\vec{r},t\right)=\frac {\vec{Z}\left(t_\mathcal{R}\right)} {\mathcal{R}} ,\\ \vec{Y}_3\left(\vec{r},t\right)=\frac {\vec{Y}\left(t_\mathcal{R}\right) \times \vec{r}} {\mathcal{R}^3} ,\qquad \vec{Z}_2\left(\vec{r},t\right)=\frac {\vec{Z}\left(t_\mathcal{R}\right) \times \vec{r}} {\mathcal{R}^2}$$ Finally leading us to: $$\vec{S}\left(\vec{r},t\right)\cdot \hat{r} = \frac {1} {16 \pi^2 \epsilon_0 \mathcal{R}^2} \left( \frac {\vec{Y}\left(t_\mathcal{R}\right) \times \hat{r}} {\mathcal{R}} + \frac {\vec{Z}\left(t_\mathcal{R}\right) \times \hat{r}} {c} \right) \cdot \left( \frac {\vec{W}\left(t_\mathcal{R}\right) \times \hat{r}} {\mathcal{R}^2} + \frac {\vec{X}\left(t_\mathcal{R}\right) \times \hat{r}} {\mathcal{R} c} - \frac {\vec{Z}\left(t_\mathcal{R}\right) \times \hat{r}} {c^2} \right)$$ So while $P_{\delta}\left(\mathcal{R},t\right)$ is still very well defined, and physically I still expect that $\int_{t_1}^{t_2} P_{\delta}\left(\mathcal{R},t\right) ~dt = 0$ for the definition of $t_1$ and $t_2$ given in the question, I cannot take the mathematics forward.

As to the axis with a zero power flow, it also appears from the expression above that if we consider the medium distance and near fields, there won't be one in general.

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