# Time dilation difference of two spacecraft at different relativistic speeds, if the faster one circles back

I have the tables explaining time dilation for a single ship at different relativistic speeds, but am unsure how to modify that to work out for two ships at different velocities, and the two different dilation effects:

I have two spacecraft travelling together at relativistic speeds, say 70%c. One, ship A, accelerates to 99.5%c or so, so that time dilation was more dramatic for them. Fuel and shielding are not a problem. These guys are good at gravity manipulation so acceleration not a problem.

If it finds a problem and turns back to rejoin ship B, how much further ahead from ship B (which is moving the same direction but slower) could it have gone if 100 years passed on ship B? How much time (from their point of view) passed on ship A?

The story is all about the travelers, so no outside non-relativistic observers.

I have two story points to cover: a) A mother and son are on the different ships. When the mother returns, the son is very old. b) Ship A finds something ahead which needs to be avoided - so how far ahead that is could affect the story and time left to avoid the problem after ship A returns with the bad news.

Any pointers would be appreciated, thanks!

• This question, in a somewhat different form, was originally posted on Physics. – a CVn Mar 8 '18 at 20:23
• Do you care about time outside the context of the two spaceships, or could you set your reference frame to ship A? – Jakob Lovern Mar 8 '18 at 21:30
• So that @JBH's answer still has relevance to my statement while I have to delete my answer, the point I was making is that time dilation in relativity is a function of relative energy, not relative speed. This is a complicated equation, but JBH's answer does a very good job of filling in the blanks with what theory and maths we have at hand but as he states, it's entirely theoretical at this point and the idea that humans can live longer because they're travelling faster may not be true because we don't know how entropy applies at relativistic speeds. – Tim B II Mar 8 '18 at 22:33
• Your second story point makes a significant difference. If all you've gotta do is get ship A to go to 'x,' then you design the new flight path as a circle and keep velocity at .995c. By contrast, if you need to stop at the planet, then you'll need to get to .995c and then drop down to whatever the planet's velocity is for however long, then rejoin ship B at whatever its velocity is. YAY ACCELERATION! – Jakob Lovern Mar 9 '18 at 1:32
• Hi everyone thanks for the help. I did ask this on physics but they sent me here :-). The story is only about the travelers in the ships so no other observers outside those two timelines (if that's the right word), primarily focussed on ship b, the unexpected return of ship A is catalyst for next stage of the adventure. The problem ahead is undecided at the moment. The flight path may be constant velocity away and then back, not a circle but would involve a tight turn to come back and another I guess to rejoin ship B. – Kris Puzulis Mar 9 '18 at 23:04

You haven't provided enough information to complete the calculations, because distance is an issue. Let's have some fun anyway.

Oh, wait, the info's there after all. Sorry!

I like this site, which gives you several calculators, including one that tells you how much time is compressed given your speed relative to light.

So, we have two ships both travelling at 0.7c. the relative speed between them is zero, so they're experiencing time wasting away at the same rate. But that's not a useful point of reference. If we're standing on a planet thinking about them, on the other hand, they'd be experiencing time at about 71% of the rate experienced by us.

Now, one of the ships needs to deal with a problem ahead, so it accelerates to 0.995c. Let's assume magic occured and it accelerated instantly. Suddenly, it's experiencing time at 10% the rate of the planet.

So, from the perspective of the planet, it would seem that if ship A is experiencing time at 10% and ship B is experiencing time at 71%, so ship B should be aging at 7X the rate of ship A — from the perspective of the planet.

But all this is moot unless enough time distance has passed to make a difference. How far ahead is this issue? How long was the 0.995c round trip?

Duh, you said Ship B spent 100 years. So, 100/7 = 14.29 years at 0.995c.

NOTE: The website I reference makes a point that's important. Light, traveling at 1.0c, experiences 100% time dilation, which means no matter where it started from, it (the photon) thinks it arrived at its destination instantly. Despite the fact that it (the photon) would have experienced a loss of energy, however small, over the distance it travelled. In other words, as Tim B II states, the relationship really is between energy states, not speed, and degeneration may be a whole lot more complicated than the math equating speed with time suggests. In other words, maybe it seemed to you like you only travelled 5 years out of an actual 50, but your body may have lost "50 years" worth of life energy anyway. It would put an interesting spin on the story, and you would join the ranks of authors who have had a lot of fun thinking about how all this manifests over the decades. But, until we can actually push something to relativistic speeds and measure what really happens, we just don't know what will really happen.

Heaven help the monkey that makes that first trip....

• I'm assuming that in your note where you say "maybe it seemed to you like you only travelled 5 years out of an actual 50, but your body may have lost '50 years' worth of life energy anyway", that's meant to be some fictional hand-waving. I don't think it's sufficiently clear that that is meant to be fictional, though. – David Z Mar 9 '18 at 6:06
• Hi. Thanks for your reply! I've seen the site you mentioned but because the calculators all compare a moving object against a stationary reference I had a bit of a block about how to modify that for the two ships. Perhaps it is simpler than I'd thought. I'll have a go – Kris Puzulis Mar 9 '18 at 22:45
• @DavidZ, it isn't any more fictional than any other assumption lacking empirical testing. We know photons lose energy over distance regardless 100% time dilation. That a mere human may also lose energy over distance is a hypothesis waiting for empirical evidence to either substantiate it or disprove it. If you want to call that aspect of the scientific method fiction, you're welcome to do so. – JBH Mar 10 '18 at 5:22
• Photons do not lose energy as they travel, not in special relativity. And the idea you're describing, about humans losing "life energy" over distance, is pseudoscientific nonsense. It's perfectly fine for it to exist in a fictional world which doesn't obey the real-world laws of physics, but it is not real, and I think it's important to be clear about that. – David Z Mar 10 '18 at 5:32
• @DavidZ, no, it's a creative way to describe entropy and the fact that we have no idea how entropy is expressed at relativistic speeds. I don't care if you call it life energy or cellular decay. I merely made a point. We know electrons lose energy over distance. It's an ignorant arrogance to assume we know enough about relativity to boldly declare that organic compounds won't. But, as I said, you can call it what you want. – JBH Mar 10 '18 at 14:39

Going by real-world physics, this situation is actually quite a bit simpler than it may seem because you only care about the perspective of people on the ships, not the people on the planet (or whatever) they are moving at 70% of light speed with respect to.

Imagine yourself, the "narrator", moving along with ship B. In your reference frame, both ships start out just sitting there at rest. Ship A changes its velocity, i.e. it accelerates up to some speed and flies away, then later it comes back. As long as the acceleration happens quickly, this is precisely the classical twin paradox.

The speed at which ship A winds up moving, in your reference frame (where B stays at rest), can be calculated from the velocity addition formula: $$\frac{u - v}{1 + \frac{uv}{c^2}} = \frac{0.995c - 0.7c}{1 - 0.995\times 0.7} = 0.972c$$ This is enough information to figure out the total distance ship A will have traveled, as seen by ship B: $$0.972c\times 100\ \text{years} = 97.2\ \text{light years}$$ Half of this distance is covered on the way out, and half on the way back, so ship A has gone $48.6\ \mathrm{ly}$ ahead, as seen by ship B.

Furthermore, the time dilation factor a.k.a. Lorentz factor corresponding to this speed is $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.972^2}} = 4.26$$ which means that, if the ships meet again when $100\ \text{years}$ have passed for ship B, the amount of time that will have passed on ship A is $\frac{100\ \text{years}}{4.26} = 23.5\ \text{years}$. For your situation, where the mother is on ship A and the son is on ship B, it is entirely realistic that the son will be biologically older than the mother (or honestly, dead, unless you assume a longer-than-average human lifetime) when she gets back.

Of course, you'd also want to know how long ships A and B have left before they run into "the Problem" that ship A found. (Assuming they can't just change course to avoid it... maybe ship B is disabled, and ship A doesn't want to leave it behind for sentimental reasons or some such thing. Doesn't really matter for the calculations.) That really depends on how fast and in which direction the Problem is moving, and by adjusting the velocity of the Problem, you can give them as much or as little time as you want.

But I suspect you meant for the Problem to be at rest in the "planet reference frame", i.e. the same reference frame in which the ships are moving at $0.7c$ to begin with. If that's the case, then consider how this situation looks from ship B:

• There is a Problem out there in space approaching you at $0.7c$.
• Ship A spent 100 years (ship B time) making a round trip to the Problem and back, which means that 50 years (ship B time) have passed since ship A encountered the Problem.
• At the time of that encounter, the Problem was $48.6\ \mathrm{ly}$ (ship B distance) away.

Traveling at $0.7c$, it will take the Problem $\frac{48.6\ \mathrm{ly}}{0.7c} = 69.4\ \text{years}$ after the encounter to reach ship B, of which 50 years have already passed, so they have $19.4\ \text{years}$ left to prepare when ship A returns. And at that time, the Problem will be $0.7c\times 19.4\ \text{years} = 13.6\ \mathrm{ly}$ away.

• Thank you for that great answer! That really helps. The only factor I think to add is that Ship B was still moving, 70 ly, so the trip back for Ship A was shorter. Over the 100 years, I think this means Ship A could have got a bit further ahead before it turned around. As the return trip would be much less than half the 100 years, so the outward trip could be much more than half. Your calculations and formulae are really useful thanks, I'll see if I can work this out! – Kris Puzulis Mar 13 '18 at 13:26
• I considered this from a reference frame where ship B was at rest, though. So ship B wasn't moving. You did say that the 100 years was measured by ship B's clocks, right? If so, then it would be wrong to include ship B's motion in the calculation. – David Z Mar 13 '18 at 20:16