# Intragalactic velocity past the central black hole

In my story people are traveling from somewhere beyond the opposite side of the galaxy to Earth. The distance traveled is approximately 66 million light-years. I would like the travelers to experience 2,000 years of time.

How close to the central black hole must the travelers pass and at what velocity must they travel to achieve this 2,000 period of time in the traveler's time frame?

• Assume the starting point is beyond the opposite edge of the Milky Way galaxy exactly opposite Earth. My 66 million light-year reference is for convenience only. I recognize that the actual parabolic path will change the actual distance traveled.

• Ignore the need to travel around stars, etc., especially near the galactic core. For the purposes of this question, assume the black hole and distance are the only relevant factors (e.g., there are no other gravity wells).

• Assume the time to accelerate and decelerate are effectively instantaneous. How my travelers get up to speed and back down again is not part of the question.

• Bonus points are awarded to the answer that results in a practical set of equations for estimating distance from the black hole and velocity given an arbitrary time experienced by the travelers. In other words, if someone else wants the same kind of results but for the travelers experiencing 8,278 years, those equations would get them to reasonable values for distance from the black hole and velocity. JBH has promised that 250 reputation points will be awarded as a bounty to the best answer that also achieves this goal. (He's about to move to Montana, so if he hasn't posted the bounty by Monday, he might need to be reminded.)

It is assumed that the gravitational effects of the black hole will compound the time dilation. If this is not the case, please explain why.

Image showing start and end points, distance between central black hole and the trajectory of the ship, etc.

• Comments are not for extended discussion; this conversation has been moved to chat. – L.Dutch - Reinstate Monica Jul 4 '19 at 2:58
• Edit explanation: no point 66 million light years away is believed to be any more "on the opposite side of the universe" than any other point 66 million light years away. That doesn't quite make sense. Since it is specified that the ship will travel across the galaxy and passed the central black hole, I made some assumptions and tried to fix. – Loduwijk Jul 16 '19 at 21:23

How close you pass to the black hole is almost completely irrelevant. The galaxy is only 100,000 ly across, and the Earth is only about 25,000 ly from the center. If the aliens are coming from a direction diametrically opposite the Earth across the galactic core, only 75,000 ly of their 66 million ly journey will even be within our galaxy, let alone near the comparatively microscopic supermassive black hole in the center. That's just a little over 0.11% of their journey, to cross the entire Milky Way. Technically, yes, if they choose to make a close pass by the central black hole, that will contribute additional time dilation effects--but at the speeds you would have to travel to make this kind of intergalactic journey, they'll only be in a region where the time dilation effects of the black hole are comparable in magnitude to the pre-existing special relativistic effects for a few minutes of their whole 2,000 year proper time journey.

So, we just need to figure out their speed using the formulas of special relativity.

If the aliens' journey takes time $$\Delta t$$ in our frame, the amount of time that they experience will be given by $$\Delta t\prime = \Delta t\sqrt{1 - \frac{v}{c}^2}$$ Meanwhile, the distance that they observe themselves travelling at relativistic speeds will be given by $$L \prime = L\sqrt{1 - \frac{v}{c}^2}$$

$$\Delta t\prime = 2000\ years$$ and $$L = 66\ million\ ly$$, according to the problem description. Furthemore, $$v = \frac{L\prime}{\Delta t\prime} = \frac{L}{\Delta t}$$; i.e., the velocity of the spacecraft is equal to the distance traveled divided by the time it took to travel it, in either frame, because our speed relative to them is the same as their speed relative to us. Multiplying both sides, we can find that $$L\prime = v\Delta t\prime$$, and we can use that to eliminate a variable and solve for $$v$$:

$$v\Delta t\prime = L\sqrt{1 - \frac{v}{c}^2}$$

$$v^2\Delta t\prime^2 = L^2 - v^2\frac{L}{c}^2$$

$$v^2(\Delta t\prime^2 + \frac{L}{c}^2) = L^2$$

$$v = \sqrt\frac{L^2}{\Delta t\prime^2 + \frac{L}{c}^2}$$

Plugging in the actual numbers, $$v = \sqrt\frac{(66\ million\ ly)^2}{(2000\ years)^2 + \frac{(66\ million \ ly)}{c}^2} = ...$$ something so close to $$c$$ that even Wolfram Alpha doesn't provide enough precision to distinguish it.

With a little extra manipulation we can, however, calculate that the isolated gamma factor ($$\frac{1}{\sqrt{1 - \frac{v}{c}^2}}$$) is 33023, and $$\beta$$ (the velocity as a fraction of light speed) is 0.99999999954.

If you intend them to travel at a more reasonable cruising velocity, no approach to the black holes will help. Trying to rely on the general relativistic time dilation effects of the black hole to compress the proper time of the trip is like travelling at 60 miles and hour for 59 miles, and then wondering how fast you need to go for the last mile to make your average speed over 60 miles equal to 100mph (i.e., it can't be done). The only way the effects of the black hole might matter is if you want them to make a stop there and hang out for some significant amount of Earth-time, while still keeping the subjective trip time under 2000 years.

• Would +2 if I could for leading with "The black hole is irrelevant" – Loduwijk Jul 16 '19 at 21:14

I don't have a solution; what I do have is a possible path to a numerical solution.

For the sake of simplicity and sanity, I will consider the special case of a non-rotating, chargeless, spherically symmetric black hole. This black hole causes space to take a shape described by the Schwarzschild metric. A small test particle - which in this case can be our ship - obeys the following general relativistic equations of motion: $$\ddot{r}=-\frac{G\mathcal{M}}{r^2}+r\dot{\theta}^2-\frac{2G\mathcal{M}}{c^2}\dot{\theta}^2,\quad \ddot{\theta}=-\frac{2}{r}\dot{r}\dot{\theta}$$ where $$r$$ and $$\theta$$ are polar coordinates centered on the black hole and $$\mathcal{M}$$ is its mass. In terms of $$t$$, we're working in the reference frame of the ship, so when we differentiate with respect to time, we're talking about the ship's proper time, rather than the coordinate time measured by an observer infinitely far away from the black hole.

These two equations are all we need to determine the path of the ship, given the initial conditions of the system (namely, the mass of the black hole and the initial velocity vector of the ship). We can vary those initial conditions until we get the result that we want; given that the ship can get arbitrarily close to the speed of light, there's no limit on the time dilation it will experience. It's going to take a lot of trial and error to get it right, but we can get there.

All that said, here are some more details.

## How do we solve the equations of motion?

There's no analytical solution to these equations, so we have to resort to numerical methods. Fortunately, the equations of motion are a pair of coupled, second-order, nonlinear differential equations. Being general relativistic doesn't make them inherently more difficult to solve than the Newtonian versions (though they're slightly different). We can solve them via any method you please. Here are some you might consider:

• Euler's method: This is something you'd only want to do as a sanity check, although it's quite simple to apply in, say, Python, in a few minutes. Say your ship's position, velocity and acceleration at some proper time $$t$$ are $$\mathbf{r}(t)$$, $$\mathbf{v}(t)$$ and $$\mathbf{a}(\mathbf{r}(t))$$. Then the same variables, at a time $$t+\Delta t$$, are $$\mathbf{r}(t+\Delta t)=\mathbf{r}(t)+\mathbf{v}(t)\Delta t$$ $$\mathbf{v}(t+\Delta t)=\mathbf{v}(t)+\mathbf{a}(\mathbf{r}(t))$$ $$\mathbf{a}(t+\Delta t)=\mathbf{a}(\mathbf{r}(t+\Delta t))$$ Euler's method is relatively inaccurate, though, compared to the host of numerical methods at your disposal.
• Runge-Kutta methods: Runge-Kutta methods are a broader set of techniques that have much greater accuracy and are substantially more widely employed than Euler's method (though Euler's method is a special case). The fourth-order Runge Kutta (RK4) is the most common, and also not too hard to apply, but higher-order RK methods involve smaller rounding errors.
• The leapfrog method: This method computes temporary values of position, velocity, etc. in between timesteps and uses them for calculations of the next step, leading to better accuracy than the Euler scheme. Furthermore, it conserves energy.

## A note on time

It's simple to calculate the proper time $$d\tau$$ measured by a ship traveling at a speed $$v$$ in some other reference frame measuring coordinate time $$dt$$, ignoring gravity: $$d\tau=dt\sqrt{1-\frac{v^2}{c^2}}$$ Similarly, the proper time measured by a ship a distance $$r$$ from the black hole is, ignoring motion: $$d\tau=dt\sqrt{1-\frac{2G\mathcal{M}}{rc^2}}$$ To put these together, you have to do a little bit of thinking to find that $$d\tau=dt\sqrt{1-\frac{v^2}{c^2}-\frac{2G\mathcal{M}}{rc^2}}$$ It should be pointed out, though, that we're solving the equations of motion in only one reference frame - that of the ship - and so we likely don't care much about doing this conversion at every point during the integration.

## An idea of the on initial conditions

To get an idea of where we'll need to start looking, consider the case of a ship in traveling 66 million light-years without any sources of gravity nearby. This is an excellent approximation to most of the ship's journey, where it will be far from the black hole and thus essentially not affected by it. For the ship's proper time to be 2,000 years while traveling 66 million light-years, we can see that $$dt$$ should be close to 66 million years, so $$\sqrt{1-\frac{v^2}{c^2}}\approx\frac{2000}{66000000}\approx3.03\times10^{-5}$$ This implies that $$v/c\approx0.9999999995408633$$, a difference from the speed of light of less than one part in a billion.