Difference between revisions of "Influence of cosmological expansion on local systems"

If $R=const$, it means there is no expansion. If $r=const$, it means the atom is copmletely "involved" into the cosmological expansion (they say, joins the Hubble flow). Simultaneous alteration of both coordinates corresponds to an intermediate situation. The question whether the intermediate situation can be realized should be considered separately.

Our model includes a fixed nucleus at the origin of the spherical coordinate frame $R,\theta,\varphi$. The position of the electron of mass $m$ in the equatorial plane $\theta= \pi /2$ is given by functions $R(t),\varphi (t)$. As there are only radial forces, its angular momentum $L =m{R^2}\dot \varphi $ is conserved, and for an electron of unit mass the integral of motion is \[L \equiv {R^2}\dot \varphi.\] In the absence of cosmological expansion the equation of motion for $R(t)$ is well-known \[\ddot R - \frac{L^2}{R^3} = - \frac{C}{R^2},\] where $C$ is the constant of electrostatic interaction. Now we have to take into account the effect of expansion. According to $R = a(t)r$, if a point takes part in the cosmological expansion only, i.e has constant comoving coordinates $r,\theta,\varphi$, its radial acceleration is \[\left. {\ddot R} \right|_{expansion} = R\frac{\ddot a}{a}.\] We can treat this terms as a radial force acting on a unit mass and add it to the equation of motion, leading to \[\ddot R = \frac{L^2}{R^3} - \frac{C}{R^2} + R\frac{\ddot a}{a}.\] The solution of this equation $R\big(r(t)\big)$ for some given $a(t)$ allows us to find the radial coordinate of the electron at the given time. Using the integral of motion $L\equiv {R^2}\dot \varphi$, we can find the angular coordinate $\varphi (t)$. Thus we can obtain full description of the orbit both in physical $R,\theta ,\varphi$ and in the comoving $r,\theta ,\varphi$ frames.

In this case \[\ddot R = {L^2}\left( \frac{1}{R^3} - \frac{1}{R^2} \right) + \beta ^2R,\] therefore \[V(R) = \frac{L^2}{2R^2} - \frac{C}{R} - \frac{1}{2}\beta ^2R^2.\]

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  The effective potential for $a = e^{\beta t},\;\beta=0.01$ with different values of $L$.

At small $L$ (or, equivalently, large $\beta$), the potential energy decreases monotonically with $R$. Radial forces push the electron into the region in which the term related to the expansion of the Universe $\beta ^2R^2$ is dominating. As $L$ increases, the effective potential acquires a minimum (see figure) in the neighborhood of $L=1$, in which the electron is captured. Thus our model demonstrates the effect "all or nothing": the electron is either captured in the minimum $(R=const)$, or joins the Hubble flow ($r=const$). It should be stressed, that the chosen exponential form of expansion is unessential. In any case the centrifugal and electrostatic forces decrease with distance, while the cosmological term increases.

Restricting ourselves to circular orbits, we write the equation of motion as \[ \frac{d^2\vec r}{dt^2} - \frac{\ddot a}{a}\vec r = - \frac{GM}{r^2}{\vec e}_r;\] On substitution of $r(t) = r_0 + \delta r(t)$ and $\theta (t) = \omega _0t + \delta \theta (t)$, we arrive to \[ r_0\delta \ddot \theta + 2\delta \dot r\omega _0 = 0;\] \[\delta \ddot r - 3\omega _0^2\delta r - 2\omega _0r_0\delta \dot \theta - \frac{\ddot a}{a}r_0 = 0.\] For a Universe dominated by matter $\left(\ddot a/a = - 2/9t^2 \right)$ so the term $\delta \ddot r$ is negligible and \[r(t) \simeq r_0\left( 1 - \frac{2}{9t^2\omega _0^2} \right).\] Thus the correction due to the cosmological expansion is \[\ddot r = - \frac{4r_0^3}{3t^4}\omega _0^2.\] Using $r_0 =1.5\cdot 10^{11}$m (distance from Earth to Sun), ${\omega _0}= 2\pi /T_0 \simeq 2 \cdot 10^{ - 7}s^{- 1}$ ($T_0$ is the Earth's year) and $t$ the age of the Universe, we obtain $\ddot r \simeq - 3.17 \cdot 10^{ - 47}$m/s$^2$. This quantity should be compared to gravitational acceleration of the Earth caused by the Sun $g = GM_\odot/r_0^2 \simeq 6 \cdot 10^{ - 3}$m/s$^2$, which is 44 orders of magnitude larger.