Part 2 of this article
Traveling in an Expanding Universe
Traveling throughout the Milky Way Galaxy, one can assume a flat universe. So we can assume only special relativity to calculate interstellar travel times for all observers 1. This time I want to expand our interstellar travel calculation to the universal scale. On the universal scale, the universe is not flat, however, but is expanding. We can approximate this expanding universe as a so-called flat Friedmann–Lemaître–Robertson–Walker spacetime. Our question is: “How long would it take for a spaceship (with constant thrust) to travel to, say, a nearby galaxy?”
Setting up the Calculation
Since our destination and starting point (Earth) are naturally being pulled apart by spacetime, we need to keep track of their positions. Of course, we have to solve for the trajectory of the spaceship too. Once we have the solutions, we can calculate the elapsed time on the ship.
Due to the complexity of including an expanding spacetime, I suspect only numerical solutions will be possible here. Nevertheless, I can promise some pretty graphs. Let’s start with the action for a massive particle.
$$ S = -mc \int dt \sqrt{-\dot x \cdot \dot x} $$
where we are taking the derivative with respect to $t$, the time parameter of an observer on Earth located at $x_i=0$. Static observers in this frame of reference can be written without any spatial velocity. So we can position the destination at some coordinate distance along an arbitrary axis, say the $x$ axis. The dot product is done with the metric.
$$ -c^2 ds^2 = -c^2 dt^2 + a^2(t) (dx_1^2 + dx_2^2 + dx_3^2) $$
This metric is homogeneous and isotropic. I.e., there is no special point in the universe. The unitless $a$ scale factor captures the expansion of the universe. Using only 1D of spatial direction, we can write the action as
$$ S = -mcL \int d\tau \sqrt{1-a(t)^2 \dot x^2} $$
where we introduced some new unitless variables. $x_1 = L x$ and $v = c \beta$ where $x$ and $\beta$ are unitless. The unitless time is expressed as $t = L/c \tau$ with unitless $\tau$. $L$ is the spatial coordinate distance to our destination.
Here, we set constants appropriately. The variation with respect to $x$ gives us
$$ \ddot x = \frac{\dot a \dot x \left(a^2 \dot x^2-2\right)}{a} + \alpha\left(a\dot x^2-1/a \right)^2 $$
We added the unitless force $\alpha$ which is constant in the rest frame.
Let’s put some numbers to our problem! The nearest galaxy is Andromeda at around $L = 2.50$ Mly 2. I know that Andromeda is actually coming towards us, but for the sake of this article let’s just turn off the attractive force between our two galaxies 😜. $a$ is related to the Hubble Constant, $H$, as $da/dt/a = H = 0.07/\text{Gyr} = 1.75 \times 10^{-4} c/L$. On this scale, $H$ is small. For Centaurus N – the furthest nearby galaxy – $L = 12.3$ Mly, we have $H = 8.616 \times 10^{-4} c/L$. This is still small. Nevertheless, the dimensionless force, $\alpha$, will be huge on account of the large distance in $L$. $\alpha \approx 2.58 \times 10^6$ for Andromeda, and $\alpha \approx 1.27 \times 10^7$ for Centaurus N.
In our equation of motion for the ship, a large $\alpha$ would quickly accelerate the ship to approximately the speed of light in “no time.” So, the nice thing is that for cosmological scales, we can approximate our ship as moving at light speed. If we did not have any acceleration, we would, on the other hand, treat our spaceship as remaining motionless with respect to the comoving coordinate system.
Lightspeed Away!
Calculating time with lightspeed particles is quite easy. Using the metric, we can directly calculate the equation of motion.
$$ -1 + a^2(\tau) \dot x^2 = 0 \implies \dot x = 1/a(\tau) $$
We can integrate this equation to get the coordinate distance.
$$ \int_0^1 dx = 1 = \int_0^{\Delta \tau} \frac {1}{a(\tau)} d\tau $$
Taking $a$ to be linear, we can write $a(\tau) = h\tau + 1$. $h := \left(H \frac Lc\right)$ is the dimensionless version of $H$. Evaluating the time integral, we have
$$ \int_0^{\Delta \tau} \frac {1}{a(\tau)} d\tau = \frac{\log (h \Delta\tau +1)}{h} = 1 $$
This implies that
$$\Delta\tau = \frac{e^h-1}{h}$$
Putting back the units, we have
$$\Delta t = \frac{e^{HL/c}-1}{H}$$
Taking our universe to be a pure de Sitter space, $a = \exp(t H) \equiv \exp(\tau h)$, taking the same integrals, we have
$$\Delta t = \frac{1}{H} \log \left(\frac{c}{c-LH}\right)$$
The first $a$ is a good approximation for our universe at the moment and into the future for some time. The second is more accurate for the long run. An interesting thing to notice is that for a large enough $L$, the time taken goes to infinity. This is the cosmological horizon, $r_h = c/H \approx 14$ Gly. This is a pretty large distance. This length is known as the Hubble length.
Taking the second $a$, let’s calculate the time to Andromeda.
$$ \Delta t = (\text{Gyr}/0.07) \log \left(\frac{1}{1- (2.50,\text{Myr}) * (0.07/\text{Gyr})}\right) \approx 0.52,\text{Gyr} $$
About 500 million years… This might be a few generations of human beings.
If we used the first $a$, it would just be $2.50$ Myr, about the same as just using special relativity. We can see that, as expected, the long-term de Sitter expansion would make intergalactic travel truly impossible.
Nevertheless, the time on the spaceship would be different from the coordinate $t$ time, so perhaps the occupants wouldn’t have to go through that many generations before arriving at their destination.
Further Reading
- I made a Pluto notebook thought it was not used in this notebook.