I was wondering how a realistic fusion-or-something rocket would fly... More specifically, I wonder about the time dilation that would be experienced (say in the classic example of two twins).

So, of course I started by searching the web for some nice calculators... There are two that do basically the same thing, this is one of them:


It assumes some acceleration a for half of the trip and the same a deceleration for the other half. But for large distances your spaceship's velocity climbs really fast to over 95% of c, at which point I'd assume the spaceship will start getting thermalized by blue-shifted cosmic rays and dust... So I'd like to calculate time dilation for a trip like that:

1) launch, accelerate at a

2) reach a given max v and stop the thrust

3) decelerate at a, land

Would that be the same as summing up (1), (3) with the calculator above, with (2) from a calculator that assumes constant velocity? Or if not, what would be a formula for variable relativistic acceleration?

Alternatively, is there a way to calculated a, based on a set max v and distance?

[I'm interested in is the final age difference of twins, one observer, the other on a relativistic rocket with a max-v limit, given the distance and max-v or a.]

Edit: If I wasn't clear before, I'm asking for a given v max, NOT 0.95c or anything near it. For fusion, probably up to 0.2c? But that's not the point - I'm looking to calculate for any reasonably relativistic maximum v.

Edit #2: So I went ahead and tried to do the math... Can someone check if it is conceptually ok? I'm still not sure I can sum time intervals that way... Also, I'm ignoring any jerk from (1)->(2) and (2)->(3).

Indexes refer to which part of the trip I'm talking about. c is speed of light, v, t, d are measured based on stationary observer. T is time on-board. ch and th are hyperbolic function. Acceleration formulas from: http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html Time dilation from memory.

$$T_{tot} = T_1 + T_2+T_3 = 2T_1 + T_2$$

$$v_f = c*th(a*T_1/c) => T_1 = c*th^{-1}(v_f/c)/a$$

$$T_2 = t_2*sqrt(1-(v_f/c)^2) = d_2*sqrt(1-(v_f/c)^2)/v_f $$

$$d_2 = d_{tot} - 2*d_1 = d_{tot} - 2*c^2*(ch(a*T_1/c)-1)/a $$


$$T = 2*c*th^{-1}(v_f/c)/a + sqrt(1-(v_f/c)^2)/v_f*[d_{tot} - 2*c^2*(ch(a*T_1/c)-1)/a]$$

$$T = 2*c*th^{-1}(v_f/c)/a + sqrt(1-(v_f/c)^2)/v_f*[d_{tot} + 2*c^2/a - 2*c^2/(a*sqrt(1-(v_f/c)^2)]$$

$$T = 2*c*ch^{-1}(1/sqrt(1-(v_f/c)^2))/a + d_{tot}*sqrt(1-(v_f/c)^2)/v_f - 2*c^2*(1-sqrt(1-(v_f/c)^2))/(a*v_f)$$

So that's the time on-board T as a function of total distance, acceleration and the top velocity which you can reach without starting to burn... Please tell me if that's correct, and if not, how to correct it?


1 Answer 1


There will never be a relativistic rocket moving at anything like the speed you suggest. I know this is an unpopular thing to say to all you science fiction fans, but that is the truth of the matter. No fuel has anything like the energy to do the job, which is just as well, because at that speed the sparse sprinkling of protons in interstellar space would take the form of a storm of deadly radiation, and if you were unlucky enough to run into an interstellar cloud of gas or dust, your spaceship would swiftly burn up. Interstellar space travel will always be a pipe dream.

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    $\begingroup$ Maybe you misunderstood my question but that's exactly what I'm saying... I'm looking for a calculator like above that would set a Max Velocity... Clearly nothing over 0.8c is even remotely practical... $\endgroup$
    – Rosh
    Commented Jul 24, 2019 at 19:50
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    $\begingroup$ Don't worry, wel'll build space autobahns with the appropriate speed limit: ramprate.com/wp-content/uploads/Speed-of-Light.jpg $\endgroup$
    – safesphere
    Commented Jul 24, 2019 at 21:53
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    $\begingroup$ Maybe instead you could offer an answer to my question? If I knew more than the basics of SR I would think it would be pretty easy to answer my question - as in: "yes, you can add time intervals" or "no, you can't add time intervals. Here's the calculation you need to do instead...". $\endgroup$
    – Rosh
    Commented Jul 25, 2019 at 6:06

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