if the Earth was electrically charged its orbit would decay due to energy loss by electromagnetic radiation. can you construct an upper limit on the charge asymmetry, i.e. the ratio between electron and proton charge, from the observation that the solar system is very long lived? which planet would provide the best bound?

bonus question: can you show that $(-1)^{n^2}=(-1)^n$?

answer to the bonus question: it's best test the equality for two cases, even and odd $n$: if $n$ is even it can be written as $n=2m$, and the square is even again, $n^2=4m^2$. then, the equality reads $(-1)^{n^2}=(-1)^{4m^2} = 1 = (-1)^{2m}=(-1)^n$. conversely, if $n$ is odd, it can be written as $n=2m+1$ and the square is odd again, $n^2=(2m+1)^2=4m^2+2m+1$. then, the equality reads $(-1)^{n^2}=(-2)^{4m^2+2m+1}=-1=(-1)^{2m+1}$.

ReplyDeleteCQW hopes that there is no mistake... setting all numerical factors to 1 one would write down the power emitted by a Hertzian dipole

ReplyDelete\begin{equation}

P = \frac{p^2\omega^4}{\epsilon c^3}

\end{equation}

using 1. the dipole moment $p$

\begin{equation}

p = e\eta \frac{M}{m_p}r

\end{equation}

which can be constructed from the charge $e$, the charge asymmetry $\eta$, the number of charge carriers $M/m_p$ from the Earth mass $M$ and the proton mass $m_p$ and the orbital radius $r$. then, one gets 2. for the orbit the Kepler law

\begin{equation}

\omega^2 = \frac{GM_\odot}{r^3}

\end{equation}

with the angular frequency $\omega$ and the solar mass $M_\odot$. furthermore, one has 3. the orbital energy

\begin{equation}

E = \frac{GM_\odot M}{r}

\end{equation}

putting everything together leads to an estimate of the time scale $\Delta t=E/P$ of

\begin{equation}

\Delta t = \frac{\epsilon c^3}{e^2\eta^2}\frac{m_p^2}{GM_\odot M}r^3

\end{equation}

the numerical value would be $\Delta t = 10^{-30}/\eta^2$ seconds: therefore, one can constrain $\eta$ to be less than $10^{-23}$ to make $\Delta t$ larger than a tenth of the Hubble time. $c$ is as always the speed of light and $\epsilon$ is the electrical permittivity.

$\eta$ is proportional to $1/\Delta t$, so one needs to look for the largest $M/r^3$ in comparing different planets. there are, perhaps surprisingly, two planets that do well, Venus with this factor being 2.3 in Earth units and Jupiter with 2.26, which makes up with a much larger mass for its larger distance. Earth is the third best planet, followed by Mercury, Saturn, Mars, Uranus and Neptune, in that order.

ReplyDelete