Is the universe like the earth?

Question

Is the universe much like the earth, in that if I were to travel in one and only one direction, I'd eventually end up where I started from?

Answer 1

I'm not an astrophysicist but I'll give it a shot. I'm taking this mostly from Hawking's "A brief history of time" and my own personal studies.

First, the universe is not like the earth. As far as I know this analogy only works when one considers the concept of imaginary time. According to Hawking, when you consider the theory of relativity with a imaginary time (which mathematically is pretty much only considering time as a imaginary number, rather than a real one) the equations that describe space are mathematically similar to those that describe a sphere. This is a trick used to try to make a quantum mechanics concept (Feynman's sum over histories) work with relativity.

This, however, does not mean that we can experience space (or the universe) like a sphere, because we are not capable of experiencing imaginary time. We experience only real time. If we were to experience time as it is defined by imaginary time, we would be able to to travel to the future and past as easily as we would travel from latitude -25º to +25º, and we all know that is not the case.

And in this answer, I'm not even taking into account the expansion of the universe, which would complicate things even further for someone trying to go around the universe. Just for the sake of the argument, imagine going around a sphere that is growing more and more. Even if you take the shortest path possible, depending on the rate of expansion the task becomes either impossible of very hard to say the least.

Hope this helps.

Answer 2

Actually, it does make sense to say that a univere can be like the (surface of the) Earth. A "uniform" (i.e. homogeneous and isotropic) universe can have three distinct geometries. See this post about the shape of a universe. The bottom line of this is that a 2D analogy of a universe can be either flat (like an infinite table), round (like the Earth), or hyperbolic (like a saddle).

Which of the three our Universe is, depends on its average density, and observations suggest that it's flat. That means that you can fly forever in a straight line, forever increasing you're distance to Earth.

However, if the density were just a little higher, it would be "round". This does not mean exactly like a ball, but it does mean that traveling in a straight line away from Earth would eventually in theory bring you back to Earth, just as traveling in a straight line on the surface of the Earth eventually brings you back home. Or to Rome, I've heard.

I say in theory, because for a universe with normal and dark matter, it would actually stop expanding and collapse before you came back. Only if traveling at the speed of light and starting at the time of Big Bang is it possible to return exactly at the Big Crunch.

If the universe contained also dark energy, there is a range of ratios between matter and dark energy that would make such a journey possible, but if there were too much dark energy, that universe would expand faster than it would be possible to traverse the universe.

Answer 3

A Universe with (hyper)spherical spatial geometry is realistic, though we don't see any evidence for it but it can't be ruled out either. As a point of interest, hyperspherical geometry doesn't necessarily mean hyperspherical topology, though it does mean the Universe must be compact i.e. of finite volume.

Now to address more thoroughly though whether it is possible to circumnavigate a spherical (in both geometric and topological senses) Universe:

Nothing can travel faster than light, so in order to find out if it is possible to circumnavigate the Universe we need to know if light travelling radially outward from some point in space will arrive at the same point in space at some later time. If $R(t_1)$ is the radius of curvature of the Universe at time $t_1$, then to circumnavigate the Universe by that time, the light must have traveled at least a proper distance of $2\pi R(t_1)$ from its starting point. Or in other words the radius of the observable Universe must be at least $2\pi R(t_1)$ by time $t_1$ in order for anything to circumnavigate the Universe.

The radius of the observable Universe is given by:

$r_{obs}(t_1) = R(t_1) {\LARGE{\int}}^{t_1}_{ t_ {int}} \frac{cdt}{R(t)}$

Where $t_{int}$ is when the Universe begin (e.g. the big bang)

Therefore the condition for it to be possible to circumnavigate a spherical Universe in a rocket ship is:

${\LARGE{\int}}^{t_{fin}}_{ t_ {int}} \frac{cdt}{R(t)} > 2\pi$

Where $t_{fin}$ is the time when the Universe ends (note in a Universe without beginning $t_{int} = -\infty$ and in a Universe without end $t_{fin} = \infty$).

For simple models with spherical spatial geometry the value of this integral is easy to calculate:

$\begin{array} {|l|l|} \hline \mathbf{MODEL}&\mathbf{VALUE}\\ \hline \mbox{Matter-dominated closed R-W}&2\pi \\ \hline \mbox{Radiation-dominated closed R-W}&\pi \\ \hline \mbox{De Sitter closed-slicing}&\pi\\ \hline \mbox{Einstein static}&\infty\\ \hline \mbox{Eddington-Lemaitre}&\infty\\ \hline \end{array}$

This means that in the case of a Robertson-Walker closed Universe light can just circumnavigate the Universe once from the big bang to the big crunch as long as the pressure is zero. In the de Sitter closed slicing it can only make it half-way around the circumference, even though there is no big bang or big crunch. In the Einstein static Universe you can go round as many times as you like as you have an infinite amount of time to travel the circumference of fixed length. In the Eddington-Lemaitre model the expansion in early times is so slow you can circumnavigate the Universe as many times as you like.