A Post-Newtonian Expansion Including Radiation Damping for a Collisionless Plasma

Abstract

We study the dynamics of many charges interacting with the Maxwell field. The particles are modeled by means of nonnegative distribution functions $f^+$ and $f^-$ representing two species of charged matter with positive and negative charge, respectively. If their initial velocities are small compared to the speed of light, $\mathrm{c}$, then in lowest order, the Newtonian or classical limit, their motion is governed by the Vlasov–Poisson system. We investigate higher-order corrections with an explicit control on the error terms. The Darwin order correction, order $|\bar{\mathrm{v}}/\mathrm{c}|^2$, has been proved previously. In this contribution, we obtain the dissipative corrections due to radiation damping, which are of order $|\bar{\mathrm{v}}/\mathrm{c}|^3$ relative to the Newtonian limit. If all particles have the same charge-to-mass ratio, the dissipation would vanish at that order.

Appendix: Some Explicit Derivatives and Integrals

We point out some formulas that have been used in the previous sections. For $p\in \mathbb {R}^3$ and $z\in \mathbb {R}^3$, we have

$$\begin{aligned} \nabla _{z}\left( \left| z\right| ^{-1}\right)&= -\left| z\right| ^{-2}\bar{z} \end{aligned}$$

(6.1a)

$$\begin{aligned} \nabla _{z}\left( \left| z\right| ^{-1}p\right)&= -\left| z\right| ^{-2}\bar{z}\otimes p \end{aligned}$$

(6.1b)

$$\begin{aligned} \nabla _{z}\left( \left| z\right| ^{-1}\bar{z}(\bar{z}\cdot p)\right)&= \left| z\right| ^{-2}\left( -3(\bar{z}\cdot p)\bar{z}\otimes \bar{z} +(\bar{z}\cdot p)\,\mathrm{id}+\bar{z}\otimes p\right) \end{aligned}$$

(6.1c)

$$\begin{aligned} \nabla _{z}z&= \mathrm{id}\end{aligned}$$

(6.1d)

$$\begin{aligned} \nabla _{z}\left( \left| z\right| ^{-1}\bar{z}\left( \bar{z}\cdot p\right) ^2\right)&= \left| z\right| ^{-2}\left( -4\bar{z}\otimes \bar{z} \left( \bar{z}\cdot p\right) ^2+\left( \bar{z}\cdot p\right) ^2\,\mathrm{id}+2\left( \bar{z}\cdot p\right) \bar{z}\otimes p\right) \end{aligned}$$

(6.1e)

$$\begin{aligned} \nabla _{z}\left( \left| z\right| ^{-1} \left( \bar{z}\cdot p\right) p\right)&= \left| z\right| ^{-2} \left( -2\bar{z}\otimes p \left( \bar{z}\cdot p\right) +p\otimes p\right) \end{aligned}$$

(6.1f)

$$\begin{aligned} \nabla _{z}\left( \left| z\right| ^{-1}\bar{z}\right)&= \left| z\right| ^{-2}\left( -2\bar{z}\otimes \bar{z}+\mathrm{id}\right) \end{aligned}$$

(6.1g)

$$\begin{aligned} \nabla _{z}\bar{z}&=\left| z\right| ^{-1}\left( -\bar{z}\otimes \bar{z}+\mathrm{id}\right) \end{aligned}$$

(6.1h)

$$\begin{aligned} \nabla _{z}\left( \bar{z}(\bar{z}\cdot p)\right)&= \left| z\right| ^{-1} \left( -2\bar{z}\otimes \bar{z}(\bar{z}\cdot p) + \left( \bar{z}\cdot p\right) \,\mathrm{id}+\bar{z}\oplus p\right) \end{aligned}$$

(6.1i)

For $z\in \mathbb {R}^3$ and $r>0$, an elementary calculation yields

$$\begin{aligned} \int _{|\omega |=1}|z-r\omega |^{-1}\,d\omega =\left\{ \begin{array}{ccc} 4\pi r^{-1} &{}\ \ : &{}\ \ r\ge |z| \\ 4\pi |z|^{-1} &{}\ \ : &{} \ \ r\le |z| \end{array}\right. . \end{aligned}$$

(6.2a)

Differentiation with respect to z gives

$$\begin{aligned} \int _{|\omega |=1}|z-r\omega |^{-3}(z-r\omega )\,d\omega =\left\{ \begin{array}{ccc} 0 &{}\ \ : &{} \ \ r>|z| \\ 4\pi |z|^{-2}\bar{z} &{}\ \ : &{}\ \ r<|z| \end{array}\right. . \end{aligned}$$

Similarly,

$$\begin{aligned} \int _{|\omega |=1}|z-r\omega |\,d\omega =\left\{ \begin{array}{ccc} 4\pi r+\frac{4\pi }{3}z^2 r^{-1} &{}\ \ :&{}\ \ r\ge |z| \\ 4\pi |z|+\frac{4\pi }{3}r^2 |z|^{-1} &{}\ \ :&{}\ \ r\le |z| \end{array}\right. , \end{aligned}$$

and thus by differentiation,

$$\begin{aligned} \int _{|\omega |=1}|z-r\omega |^{-1}(z-r\omega )\,d\omega =\left\{ \begin{array}{ccc} \frac{8\pi }{3r}\,z &{}\ \ :&{}\ \ r>|z| \\ 4\pi \bar{z} -\frac{4\pi }{3}r^2|z|^{-2}\bar{z} &{}\ \ :&{}\ \ r<|z| \end{array}\right. . \end{aligned}$$

(6.2b)

Finally, for $z\in \mathbb {R}^3\setminus \{0\}$, also

$$\begin{aligned} \int |z-v|^{-1}|v|^{-3}v\,dv=2\pi \bar{z} \end{aligned}$$

(6.2c)

can be computed.

At last, we shall prove formulae (3.13)–(3.15). Using charge conservation $\partial _t \rho _0+\nabla \cdot j_0=0$ and integration by parts, we find

$$\begin{aligned} \dot{D}_0(t)=\int j_0 (t, x)\, dx=D^{[1]}_0(t)\,. \end{aligned}$$

Exploiting the Vlasov equation (VP) and integration by parts again leads to

$$\begin{aligned} \ddot{D}_0(t)=\dot{D}^{[1]}_0(t)=\int E_0(t, x) (f_0^+(t, x) +f^-_0(t, x))\,dx=D^{[2]}_0(t)\,. \end{aligned}$$

With regard to the third time derivative, we use the notation $E_0^{\pm }=-\int \left| z\right| ^{-2}\bar{z}\rho _0^{\pm }\,dz$ and compute using the transformations $y=x+z$ and $w=x-y$

$$\begin{aligned} \dot{D}^{[2]}_0(t)= & {} \partial _t \int (E_0^+-E_0^-)(\rho _0^++\rho _0^-)(t, x)\,dx\\= & {} \iint \left( \nabla _z \left| z\right| ^{-1}\right) (\partial _t \rho _0^+ -\partial _t \rho _0^-)(t, x+z) (\rho _0^++\rho _0^-)(t, x)\,dz\,dx\\&+\int (E_0^+-E_0^-)(\partial _t\rho _0^++\partial _t\rho _0^-)(t, x)\,dx\\= & {} \iint \left( \nabla _y\left| y-x\right| ^{-1}\right) (\rho _0^++\rho _0^-)(t, x) (\partial _t \rho _0^+-\partial _t \rho _0^-)(t, y)\,dx\,dy\\&+\int (E_0^+-E_0^-)(\partial _t\rho _0^++\partial _t\rho _0^-)(t, x)\,dx\\= & {} \iint \left( -\nabla _w\left| w\right| ^{-1}\right) (\rho _0^++\rho _0^-)(t, w+y) (\partial _t \rho _0^+-\partial _t \rho _0^-)(t, y)\,dw\,dy\\&+\int (E_0^+-E_0^-)(\partial _t\rho _0^++\partial _t\rho _0^-)(t, x)\,dx\\= & {} -\int (E_0^++E_0^-)(\partial _t\rho _0^+-\partial _t\rho _0^-)(t, y)\,dy\\&+\int (E_0^+-E_0^-)(\partial _t\rho _0^++\partial _t\rho _0^-)(t, x)\,dx\\= & {} 2\int (E_0^+\partial _t \rho _0^{-}-E_0^-\partial _t \rho _0^+)(t, x)\,dx\,. \end{aligned}$$

Using charge conservation $\partial _t\rho _0^{\pm }+\nabla \cdot j_0^{\pm } = 0$ and partial integration, we, e.g., compute

$$\begin{aligned} \int E_0^+\,\partial _t\rho _0^-\,dx&= -\int E^+ \nabla \cdot j_0^-\,dx =\iint \frac{z}{\left| z\right| ^3}\rho _0^+(x+z)\nabla _x\cdot j_0^-(x)\,dz\,dx\\&=-\iint \frac{z}{\left| z\right| ^3}\left( \nabla _x\rho _0^+ (x+z) \cdot j_0^-(x)\right) \,dz\,dx\\&=\int \left( -\int \frac{z}{\left| z\right| ^3}\otimes \nabla _z\rho _0^+ (x+z)\,dz \right) j_0^-(x)\,dx\,. \end{aligned}$$

Now, we have a closer look onto the inner integral: Using integration by parts and principal values, we find

$$\begin{aligned} H^+(t, x)&:= -\int \frac{z}{\left| z\right| ^3}\otimes \nabla _z\rho _0^+(x+z)\,dz\\&=-\lim _{\eta \rightarrow 0}\int _{\left| z\right|>\eta }\frac{z}{\left| z\right| ^3} \otimes \nabla _z\rho _0^+(x+z)\,dz\\&=\lim _{\eta \rightarrow 0}\left( \int _{\left| z\right| >\eta } \left( -3\left| z\right| ^{-5}z\otimes z+\left| z\right| ^{-3}\mathrm{id}\right) \rho _0^+(t, x+z)\,dz\right. \\&\quad \qquad \left. +\eta ^{-4}\int _{\left| z\right| =\eta }z \otimes z \rho _0^+(t, x+z)\,ds(z)\right) \\&= \oint \left| z\right| ^{-3}\left( -3\bar{z}\otimes \bar{z}+\mathrm{id}\right) \rho _0^+(t, x+z)\,dz +\frac{4\pi }{3}\rho _0^+(t, x)\,. \end{aligned}$$

Note that the kernel $H(z)=-3\bar{z}\otimes \bar{z}+\mathrm{id}$ is bounded on $\mathbb {R}^3{\setminus }\{0\}$, is homogeneous of degree zero and satisfies $\int _{|z|=1}K(z)\,d\sigma (z)=0$. Thus, using the Calderón–Zygmund inequality, $H^+$ is well defined for smooth functions $\rho _0^+$ with compact support and for $1<p<\infty$ can be extended to a bounded linear operator mapping $L^p(\mathbb {R}^3)$ to $L^p(\mathbb {R}^3)$, see Stein (1970). With regard to the boundary integral, note that z / |z| is the inner normal. Furthermore,

$$\begin{aligned} \int _{|{z}|=\eta }z_iz_j\, ds(z)=0 \end{aligned}$$

if $i\ne j$ and

$$\begin{aligned} \int _{|{z}|=\eta }z_i^2\, ds(z)=\frac{1}{3}\int _{|{z}|=\eta }| {z}|^2\, ds(z)=\frac{4\pi }{3}\eta ^4\,. \end{aligned}$$