We use a rubber balloon model to give a simple explanation of the motion of photons and galaxies in an expanding and collapsing universe. In particular, we study the Friedmann-Lemaitre-Robertson-Walker closed universe. The behavior of the peculiar velocity of galaxies in an expanding universe is also studied with the help of this model.

## REFERENCES

1.

W. M.

Stuckey

, “Can galaxies exist within our particle horizon with Hubble recessional velocities greater than$c$?

” Am. J. Phys.

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), 142

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);W. M.

Stuckey

, “Kinematics between comoving photon exchangers in a closed, matter-dominated universe

,” Am. J. Phys.

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), 554

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G. F. R.

Ellis

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, “Lost horizons

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T. M.

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, C. H.

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, and J. K.

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, “Solutions to the tethered galaxy problem in an expanding universe and the observation of receding blueshifted objects

,” Am. J. Phys.

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), 358

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(2003

).4.

The supernova current data slightly prefer a closed universe, but the data are also consistent with the open possibility. The values of the cosmological constant and mass density favor a universe that expands forever. See Fig. 7 in

A.

Conley

et al., “Measurement of $\Omega M$, $\Omega \Lambda $ from a blind analysis of Type Ia supernovae with CMAGIC: Using color information to verify the acceleration of the universe

,” Astron. J.

644

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), 1

–20

(2006

), $\u27e8$www.arxiv.org/abs/astro-ph/0602411$\u27e9$.5.

Because $vT(t)=(d\u2215dt)(d(t))=cln(et)$, $vT(t)=2\pi $ for $t=tC=tN\u2215e$.

6.

W.

Rindler

, Relativity Special, General and Cosmological

(Oxford U. P.

, New York, 2001

), p. 367

.7.

Besides the particle horizon, there is the related concept of the event horizon (see Ref. 6, pp.

377

–383

), which separates photons that reach us in a finite time (or before the Big Crunch if the universe recollapses), and those that do not. That is, whereas the particle horizon is at a distance beyond which an observer cannot see at the current time, the event horizon is at a distance beyond which the observer will never see. The particle horizon concept is useful in cosmology, and the event horizon is more useful in the study of black holes.8.

Because we cannot see photons emitted before the time at which the matter and radiation were decoupled, for real observations we have to consider another type of horizon, the visual horizon. The distance to the visual horizon is given by Eq. (6) by replacing the Big Bang time 0 by the time at which the cosmic microwave background was emitted. See Ref. 6, p.

377

.9.

See Ref. 6, p.

396

.10.

See Ref. 6, p.

401

.11.

Equation (13) gives the parametric equations of the cycloid. This curve has two remarkable properties: it is a brachistochrone (minimum-time curve) and tautochrone (equal-time curve). This curve also appears in the study of the radial motion of a particle in a central gravitational field when the initial radial velocity is smaller than the escape velocity. In this case, $R$ represents the distance to the gravity center and $t$ the time. It is not a coincidence that the cycloid appears in the context of cosmology. In the Newtonian interpretation, the motion of a galaxy at a distance $R$ from us corresponds to the motion of a particle in the gravitational field of a ball with radius $R$ centered at us and with density equal to the cosmological density (see Ref. 6, pp.

394

–395

).12.

To give a general relativistic proof that $R(t)p(t)=const$, consider a purely radial geodesic $\theta =\varphi =0$, which corresponds to a freely moving particle. This geodesic is a curve along which $\u222bds$ is extremal, where $ds2=c2dt2\u2212R(t)du2=c2d\tau 2$ and $\tau $ is the particle proper time. It follows (see Ref. 6, p.

205

) that the equation of this geodesic is the solution of the Euler-Lagrange equation $(d\u2215d\tau )(\u2202L\u2215\u2202u\u0307)\u2212\u2202L\u2215\u2202u=0$, with $L=c2t\u03072\u2212R(t)u\u03072$ $(\u2009\u0307=d\u2215d\tau )$, which gives $R2u\u0307=const$. If $v(t)=Rdu\u2215dt$ is the velocity of a freely moving particle relative to the local fundamental observer, then $t\u0307=\gamma (v)=(1\u2212v2\u2215c2)\u22121\u22152$ and hence $R2u\u0307=Rvt\u0307=Rv\gamma (v)$, so $Rp=const$.13.

For relativistic velocities the cosmic time $t$ is not equal to the proper time of the particle $\tau $, but $dt\u2215d\tau =\gamma (v)$. Therefore we have to add the factor $\gamma (v)$ to $v$. Thus, $Rp=const$, with $p=mv\gamma (v)$ the relativistic momentum (see also Ref. 12).

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