10 ms·

You could actually reach almost anywhere in the galaxy within a single human lifetime (the traveler's lifetime that is) using only 1-G constant acceleration: ht

You could actually reach almost anywhere in the galaxy within a single human lifetime (the traveler's lifetime that is) using only 1-G constant acceleration: https://en.wikipedia.org/wiki/Space_travel_using_constant_acceleration https://en.wikipedia.org/wiki/Space_travel_using_constant_ac... No FTL required. We just need to figure out how to use propellant efficiently enough among a few other slightly difficult challenges.

macspoofing 8y ago
The challenge is still communication (it's going to take you 100k years to send a message from one end to another) and synchronization (it took you 60 years to get to your destination, but hundreds of thousands of years have passed for your peers)
- TheSpiceIsLife 8y ago
  This can be solved by simply not communicating.
  - macspoofing 8y ago
    That's true. Or just dealing with the time-lag.
magduf 8y ago
The problem with this is that, by the time you get anywhere, a ridiculous amount of time has passed back on Earth, and also on the destination planet (since the reference frame for the other planet is very close to Earth's). So if you travel around to hundreds of planets, pretty soon all the stars have gone nova and you're nearing the heat-death of the universe, even though you may not have aged much.
- rwcarlsen 8y ago
  Minor concerns.
  - dorgo 8y ago
    Limiting believes
- state_less 8y ago
  If we make it for another hundred or few thousand years more without killing ourselves, I'd imagine we might be able to shoot self replicating ships out at near C, then beam copies of ourselves to each ship, expanding out everywhere at near C. We might be able to be everywhere. I imagine we could 'live inside' a projected VR world that looks like earth or some other massive online world unless we wanted to go out camping at whatever planet we were orbiting.
  - saiya-jin 8y ago
    copy of me ain't me but another human, i wouldn't consider it much closer compared ie to my best friend
    - state_less 8y ago
      Consider an arm that has 'fallen asleep' or placed under local anesthetic. It feels foreign, but you take care of it still as if it were your own. One has to simply extend their idea of themselves outside of what their neurons are directly touching to see the linkage between another human and themselves. Here is a case where two people are brought in more direct contact, which seems to me to hold the door open for future possibilities. It makes it easier to see how two might merge, at least in a rough accidental sort of way. https://kottke.org/17/11/these-conjoined-twins-can-share-each-others-thoughts-vision https://kottke.org/17/11/these-conjoined-twins-can-share-eac... https://www.nytimes.com/2011/05/29/magazine/could-conjoined-twins-share-a-mind.html https://www.nytimes.com/2011/05/29/magazine/could-conjoined-... Have a good one.
- civilitty 8y ago
  The wikipedia article says you can travel the diameter of our galaxy (including coming to a stop relative to your destination) in 24 years ship time and 226,000 years Earth time. The sun will enter the end of its lifecycle in about 7.6 billion years so you'd have to make over 30 thousand such trips from one end of the galaxy to the other, putting you at over 700,000 years old. Traveling between galaxies is another matter, however. The distance from the edge of the Milky Way to Andromeda is about 2.5 million light years. That journey would take 55 years ship time or slightly more than 2.5 million years Earth time and Andromeda really isn't that far on the scale of galaxies. Leaving a solar system might still allow some sort of interplanetary trade (in historic items mostly, since they'd be pretty old by the time they got anywhere) but leaving a galaxy means cutting all ties forever. Don't worry about it. You can travel from the Milky Way to Andromeda and back 1500 times before the sun explodes.
  - mirimir 8y ago
    I wish that Vinge had written more about the Qeng Ho. And re looping the galaxy, I'm reminded of Peter Watts' The Freeze-Frame Revolution. Mission time is hundreds of millions of years.
  - LandR 8y ago
    Space Travel calculator for anyone wanting to work out the time dilation stuff http://nathangeffen.webfactional.com/spacetravel/spacetravel.php http://nathangeffen.webfactional.com/spacetravel/spacetravel...
- MLij 8y ago
  I never really understand this. From the Wikipedia link: "At a constant acceleration of 1 g, a rocket could travel the diameter of our galaxy in about 12 years ship time, and about 113,000 years planetary time." But that's from the frame of reference of planet earth. If you are on the spaceship, earth will be moving away from you so time on earth will appear to be slower than your time. So if you come back to earth you will appear younger than should be to the people that stayed, but to you people that stayed will appear younger than should be. Frames of reference are equivalent, aren't they? Maybe I'm rambling, but I can't figure this out.
  - rwcarlsen 8y ago
    https://en.wikipedia.org/wiki/Twin_paradox#Resolution_of_the_paradox_in_special_relativity https://en.wikipedia.org/wiki/Twin_paradox#Resolution_of_the...
    - MLij 8y ago
      Thank you!
    - rajacombinator 8y ago
      Great article!
  - Sean1708 8y ago
    Minute Physics did a good couple of videos on this[0,1], pretty much the same info as is in the Wiki but might be easier to digest for some people. [0]: https://www.youtube.com/watch?v=Bg9MVRQYmBQ https://www.youtube.com/watch?v=Bg9MVRQYmBQ [1]: https://www.youtube.com/watch?v=0iJZ_QGMLD0 https://www.youtube.com/watch?v=0iJZ_QGMLD0
  - raattgift 8y ago
    It comes down to geometry. In space, the shortest distance between two points is a straight line, but in spacetime things are different. The last paragraphs of explanation below are essentially what the wiki editors have done at https://en.wikipedia.org/wiki/Twin_paradox#Difference_in_elapsed_time_as_a_result_of_differences_in_twins'_spacetime_paths https://en.wikipedia.org/wiki/Twin_paradox#Difference_in_ela... only without acceleration/deceleration phases, since they aren't helpful in understanding the geometry of the problem. Moreover, I'd like to lead you to the same result more gently than wikipedia-browsing would. On a flat Euclidean plane, between two points there is a single shortest path. We call that a "straight line", and we can assemble it from line elements which are tiny little steps from the starting point to the ending point. Using Cartesian coordinates (x,y), we can write the line element dS as dS = sqrt(dx^2 + dy^2) -- we do the squaring and take the square root in order to deal with sign differences. We can also write it as dS^2 = dx^2 + dy^2. Dx and dy are tiny steps along the x or y axes, just as dS is the line element: a tiny step along the line. A plane is a 2-space. Flatness means that the line element above applies everywhere in the plane, even if it goes to infinity in all directions. On a flat Euclidean 3-space, our line element becomes dS^2 = dx^2 + dy^2 + dz^2. Again, we have a straight line assembled from tiny displacements in one or more of the three spatial dimensions. Flatness, again, means that this line element applies on a line between any two points. In the 2-space and 3-space cases, the straight line is the uniquely shortest possible line. If we deviate slightly even at one tiny step when building line out of line elements, the line itself is longer. The form of these line elements are called metrics, and they apply everywhere in each the spaces described above. They have a signature, which can be found by the operators between the coordinates on the right-hand-side of the metric: two + in the 2-space case, three +es in the 3-space case: note that the + in front of the leading term (dx^2) is implicit. Explicitly, we could say dS^2 = +dx^2 + dy^2 + dz^2 Or dS^2 = 0 + dx^2 + dy^2 + dz^2. Because of the squaring and square-rooting, we can alternatively reverse the signs and say dS^2 = -dx^2 - dy^2 - dz^2. We get the same result, but then the signature is three minuses rather than three plusses. That's space. Now let's consider spacetime. A Lorentzian spacetime's timelike dimension has the opposite sign. So the Lorentzian extension of a 3-space into a 4-spacetime takes the signature from (+,+,+) to (+,+,+,-) or from (-,-,-) to (-,-,-,+). So in flat spacetime, in Minkowskian coordinates (they're the flat spacetime analogues of Cartesian coordinates (x,y,z) as (x,y,z,t)) we can write the line element as dS^2 = dx^2 + dy^2 + dz^2 - k^dt^2 or as dS^2 = k^2dt^2 - dx^2 - dy^2 - dz^2. "k" here is a constant; it can be set to the value 1, in which case it vanishes. It can be set to a different numerical value, representing a conversion constant between lengths in the spacelike axes and durations in the timelike axis. In our Lorentzian spacetime, the constant is better known as "c". One thing to observe here is that if "c" is large than the [+/-]c^2dt^2 in the line element dominates. Say we mark off spatial distances in light-nanoseconds and timelike durations in nanoseconds, so we can make "k" vanish (this is effectively setting c to 1). If your start point is t=0,x=0,y=0,z=0 and your end point is t=onebillion,x=0,y=0,z=0 you've moved one light second into the future and zero light seconds left/right, up/down, or forward/backward. Using (+,-,-,-) sign convention, S is a large number, because we are subtracting nothing from it. However, light travels at one light-nanosecond per nanosecond, so light shining to the left would go from x=0,y=0,z=0,t=0 to t=onebillion,x=onebillion,y=0,z=0. The length of the line is then zero, exactly. Finally, if we have two objects at t=0, one at t=0,x=0,y=0,z=0 and the other at t=0,x=onebillion,y=0,z=0, then when we calculate dS^2 we get negative one billion. A Lorentzian (one sign difference) metric lets us categorize based on this. For the (+,-,-,-) signature, a positive value for dS^2 is timelike, a zero value is null or light-like (since c relates to the speed of light), and a negative value is space-like. The absolute value of dS^2 is smallest when dS^2 = 0. Null intervals are thus the "shortest line" through a flat Lorentzian spacetime. Timelike paths are longer. The origin and rendezvous points in the "twin paradox" or "relativistic galaxy cruiser who comes back", in light years, differ either not at all (or only a small small fraction if we leave/return right near Earth, since the planet moves around a bit in a year). Non-travelling twin is always at or near x=0,y=0,z=0, but moves a full year along t. Most of the path non-traveller's t is changing but the x,y and z are non-changing. Travelling twin, however, moves away from x=0 while moving "left" away and then moves back towards x=0 when moving "right" back. t is changing throughout traveller's trip, of course, but so is x. We can look at it like this, with non-traveller t,x and traveller t',x': start: t=0,x=0; t'=0,x'=0. In the very next step, measured in very small units (say, light-attoseconds), and with traveller already accelerated to c exactly, we have t=1,x=0; t'=1,x'=1. In the next step, t=2,x=0; t'=2,x'=2. And so forth. The values for dS for these first two steps are non-traveller 1, 2 while traveller is 0, 0. When we integrate all these, non-traveller's total is about 3 * 10^26 (that's about the number of attoseconds in a year) while traveller's total is about zero (or exactly zero if we assume instant turn-around, and rendezvous spatial coordinates are x=0,y=0,z=0). These totals are path-lengths. Non-traveller's path length is extreme, traveller's is zero. Proper times -- what one's cells, wristwatch, and other clocks one carries at all times reflect -- follows these path lengths. The non-traveller ages a year, the travelling-at-exactly-c traveller does not age at all. Usually you take the traveller as massive and make it travel at a bit less than c, so that the total path length is timelike but short (rather than lightlike and thus null) compared to the extremal timelike path of the non-traveller. Thus the traveller's proper time does pass a little, but not nearly as much as the non-traveller's. Additionally, the traveller is often cast as accelerating and decelerating gently, so there are periods where the x coordinate is changing slowly with respect to t, and periods where the x coordinate is changing close to how quickly t is changing. The massive gently-accelerating (and decelerating) traveller is what is described at the wikipedia link way above. Almost finally, the geometry does not depend on choice of coordinates. One can use arbitrary units to tick off durations the timelike axis, and can choose spatial coordinates that are spherical or cylindrical rather than Cartesian-like. The metric for flat spacetime changes form with a change of coordinates[1], but the resulting path lengths and travel times work out exactly the same. Finally, the metric of curved spacetime is different from that of flat spacetime; the reason you get time dilation around a black hole is because path lengths change because the line elements have a radial dependency such that the paths of objects moving tangentially past a black hole are shorter the further away from the black hole the tangent lies. Gravitation, then, is geometrical. - -- [1] see, for example, below (10.4) at http://ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Spacetime_%20of_%20General_Relativity.htm http://ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Spacet...
- 34381 8y ago
  Once our minds are digital, it will require a rethinking of the concept of self. Send a copy of you to the stars and keep a copy at home enjoying terrestrial life. Meet back up in 20 billion years and have a beer with an old friend.
- girvo 8y ago
  I quite liked how “A Deepness In The Sky” tried to tackle this, by making the galaxy travelling traders permanently outside of the existing societies, as time dilation means they will have to be!
  - taneq 8y ago
    Same with the Revelation Space universe's lighthugger starships. The crews were seen as quite alien by the rest of their culture. That's also the first place I saw the (in hindsight obvious) idea that a ship that can accelerate at over 1G can land on (or hover over) a planetary surface. Everyone else seems to have kept the "starships are for outside atmospheres only" trope.
    - simias 8y ago
      That's assuming that the thrusters can still output 1G within the atmosphere and that the hull of the ship can withstand the outside pressure and the weather (humidity, strong winds etc...). Also a huge ship firing thrusters continuously in the atmosphere might have adverse side effects (for instance radiation, toxic smoke, heat etc... depending on the technology). Might make sense to optimize large ships for space only and use smaller vehicles to taxi back and forth from the surface of planets.
      - clarkmoody 8y ago
        Another Vinge goodie is the effect of the "main torch" of a starship on the permafrost of a planet as it hovers around looking for a landing site.
      - taneq 8y ago
        Yeah, but if we're allowing thrusters that can generate a 1G acceleration for years at a time without using the entire ship as reaction mass, those are much smaller "gimmes". :)
    - magduf 8y ago
      >Everyone else seems to have kept the "starships are for outside atmospheres only" trope. Engineering a ship that can land on a planet with gravity and an atmosphere is much bigger challenge than engineering a ship that never does anything besides go into orbit. It makes it much easier to create a really, really, really huge ship. You're not going to build a Death Star that can land on a planet's surface, unless you can come up with some kind of materials that we have no idea about now. So why bother? It's easy enough to make a giant ship for interplanetary travel with smaller shuttles for landing.
mortenjorck 8y ago
Are you familiar with The Expanse? Based on the way G-forces are presented in the TV adaptation (I presume the book does something similar), I take it the Epstein Drive is supposed to be a constant-acceleration design, which in the story is limited to intra-solar travel (though presumably the generation ship Nauvoo was designed to use it for interstellar). I mention this because your comment made me think: Even staying within the solar system, how short would a trip between Earth and Mars have to be to start to pose relativistic problems? I wonder at what point frequent travelers would start aging noticeably more slowly. (If this is dealt with in the books, now I’ll really have to read them.)
- wyattpeak 8y ago
  At relativistic speeds, Mars and Earth are ludicrously close - between 4 and 20 minutes apart at light speed. So even travelling arbitrarily close to the speed of light, at the greatest possible distance, you could only gain 20 minutes on the folks back home per trip. A round trip weekly at average distance (just going with 12mins/trip because the details get complicated), 48 weeks a year for 40 years would leave you just over a month younger than those who stay put.
TheSpiceIsLife 8y ago
And, if you can work out how to do 1G acceleration, you can probably work out how to 2 or 3 or 4G acceleration. That's going to get uncomfortable, but most people should be able to cope with 2 or 3G acceleration for short periods.
anonytrary 8y ago
> I never thought of c being a real problem to space travel. > We just need to figure out how to use propellant efficiently enough among a few other slightly difficult challenges Both of these sentiments are pretty common misconceptions I've seen on HN. First, if c was infinite, the energy required to accelerate a constant amount would grow quadratically. This alone is a tough problem. Maybe we could "just" solve it. On the other hand, if c was finite, the energy required to accelerate would grow exponentially. This is more than a "slightly difficult" problem.

macspoofing 8y ago

The challenge is still communication (it's going to take you 100k years to send a message from one end to another) and synchronization (it took you 60 years to get to your destination, but hundreds of thousands of years have passed for your peers)

TheSpiceIsLife 8y ago
This can be solved by simply not communicating.
- macspoofing 8y ago
  That's true. Or just dealing with the time-lag.

magduf 8y ago

The problem with this is that, by the time you get anywhere, a ridiculous amount of time has passed back on Earth, and also on the destination planet (since the reference frame for the other planet is very close to Earth's). So if you travel around to hundreds of planets, pretty soon all the stars have gone nova and you're nearing the heat-death of the universe, even though you may not have aged much.

rwcarlsen 8y ago
Minor concerns.
- dorgo 8y ago
  Limiting believes
state_less 8y ago
If we make it for another hundred or few thousand years more without killing ourselves, I'd imagine we might be able to shoot self replicating ships out at near C, then beam copies of ourselves to each ship, expanding out everywhere at near C. We might be able to be everywhere. I imagine we could 'live inside' a projected VR world that looks like earth or some other massive online world unless we wanted to go out camping at whatever planet we were orbiting.
- saiya-jin 8y ago
  copy of me ain't me but another human, i wouldn't consider it much closer compared ie to my best friend
  - state_less 8y ago
    Consider an arm that has 'fallen asleep' or placed under local anesthetic. It feels foreign, but you take care of it still as if it were your own. One has to simply extend their idea of themselves outside of what their neurons are directly touching to see the linkage between another human and themselves. Here is a case where two people are brought in more direct contact, which seems to me to hold the door open for future possibilities. It makes it easier to see how two might merge, at least in a rough accidental sort of way. https://kottke.org/17/11/these-conjoined-twins-can-share-each-others-thoughts-vision https://kottke.org/17/11/these-conjoined-twins-can-share-eac... https://www.nytimes.com/2011/05/29/magazine/could-conjoined-twins-share-a-mind.html https://www.nytimes.com/2011/05/29/magazine/could-conjoined-... Have a good one.
civilitty 8y ago
The wikipedia article says you can travel the diameter of our galaxy (including coming to a stop relative to your destination) in 24 years ship time and 226,000 years Earth time. The sun will enter the end of its lifecycle in about 7.6 billion years so you'd have to make over 30 thousand such trips from one end of the galaxy to the other, putting you at over 700,000 years old. Traveling between galaxies is another matter, however. The distance from the edge of the Milky Way to Andromeda is about 2.5 million light years. That journey would take 55 years ship time or slightly more than 2.5 million years Earth time and Andromeda really isn't that far on the scale of galaxies. Leaving a solar system might still allow some sort of interplanetary trade (in historic items mostly, since they'd be pretty old by the time they got anywhere) but leaving a galaxy means cutting all ties forever. Don't worry about it. You can travel from the Milky Way to Andromeda and back 1500 times before the sun explodes.
- mirimir 8y ago
  I wish that Vinge had written more about the Qeng Ho. And re looping the galaxy, I'm reminded of Peter Watts' The Freeze-Frame Revolution. Mission time is hundreds of millions of years.
- LandR 8y ago
  Space Travel calculator for anyone wanting to work out the time dilation stuff http://nathangeffen.webfactional.com/spacetravel/spacetravel.php http://nathangeffen.webfactional.com/spacetravel/spacetravel...
MLij 8y ago
I never really understand this. From the Wikipedia link: "At a constant acceleration of 1 g, a rocket could travel the diameter of our galaxy in about 12 years ship time, and about 113,000 years planetary time." But that's from the frame of reference of planet earth. If you are on the spaceship, earth will be moving away from you so time on earth will appear to be slower than your time. So if you come back to earth you will appear younger than should be to the people that stayed, but to you people that stayed will appear younger than should be. Frames of reference are equivalent, aren't they? Maybe I'm rambling, but I can't figure this out.
- rwcarlsen 8y ago
  https://en.wikipedia.org/wiki/Twin_paradox#Resolution_of_the_paradox_in_special_relativity https://en.wikipedia.org/wiki/Twin_paradox#Resolution_of_the...
  - MLij 8y ago
    Thank you!
  - rajacombinator 8y ago
    Great article!
- Sean1708 8y ago
  Minute Physics did a good couple of videos on this[0,1], pretty much the same info as is in the Wiki but might be easier to digest for some people. [0]: https://www.youtube.com/watch?v=Bg9MVRQYmBQ https://www.youtube.com/watch?v=Bg9MVRQYmBQ [1]: https://www.youtube.com/watch?v=0iJZ_QGMLD0 https://www.youtube.com/watch?v=0iJZ_QGMLD0
- raattgift 8y ago
  It comes down to geometry. In space, the shortest distance between two points is a straight line, but in spacetime things are different. The last paragraphs of explanation below are essentially what the wiki editors have done at https://en.wikipedia.org/wiki/Twin_paradox#Difference_in_elapsed_time_as_a_result_of_differences_in_twins'_spacetime_paths https://en.wikipedia.org/wiki/Twin_paradox#Difference_in_ela... only without acceleration/deceleration phases, since they aren't helpful in understanding the geometry of the problem. Moreover, I'd like to lead you to the same result more gently than wikipedia-browsing would. On a flat Euclidean plane, between two points there is a single shortest path. We call that a "straight line", and we can assemble it from line elements which are tiny little steps from the starting point to the ending point. Using Cartesian coordinates (x,y), we can write the line element dS as dS = sqrt(dx^2 + dy^2) -- we do the squaring and take the square root in order to deal with sign differences. We can also write it as dS^2 = dx^2 + dy^2. Dx and dy are tiny steps along the x or y axes, just as dS is the line element: a tiny step along the line. A plane is a 2-space. Flatness means that the line element above applies everywhere in the plane, even if it goes to infinity in all directions. On a flat Euclidean 3-space, our line element becomes dS^2 = dx^2 + dy^2 + dz^2. Again, we have a straight line assembled from tiny displacements in one or more of the three spatial dimensions. Flatness, again, means that this line element applies on a line between any two points. In the 2-space and 3-space cases, the straight line is the uniquely shortest possible line. If we deviate slightly even at one tiny step when building line out of line elements, the line itself is longer. The form of these line elements are called metrics, and they apply everywhere in each the spaces described above. They have a signature, which can be found by the operators between the coordinates on the right-hand-side of the metric: two + in the 2-space case, three +es in the 3-space case: note that the + in front of the leading term (dx^2) is implicit. Explicitly, we could say dS^2 = +dx^2 + dy^2 + dz^2 Or dS^2 = 0 + dx^2 + dy^2 + dz^2. Because of the squaring and square-rooting, we can alternatively reverse the signs and say dS^2 = -dx^2 - dy^2 - dz^2. We get the same result, but then the signature is three minuses rather than three plusses. That's space. Now let's consider spacetime. A Lorentzian spacetime's timelike dimension has the opposite sign. So the Lorentzian extension of a 3-space into a 4-spacetime takes the signature from (+,+,+) to (+,+,+,-) or from (-,-,-) to (-,-,-,+). So in flat spacetime, in Minkowskian coordinates (they're the flat spacetime analogues of Cartesian coordinates (x,y,z) as (x,y,z,t)) we can write the line element as dS^2 = dx^2 + dy^2 + dz^2 - k^dt^2 or as dS^2 = k^2dt^2 - dx^2 - dy^2 - dz^2. "k" here is a constant; it can be set to the value 1, in which case it vanishes. It can be set to a different numerical value, representing a conversion constant between lengths in the spacelike axes and durations in the timelike axis. In our Lorentzian spacetime, the constant is better known as "c". One thing to observe here is that if "c" is large than the [+/-]c^2dt^2 in the line element dominates. Say we mark off spatial distances in light-nanoseconds and timelike durations in nanoseconds, so we can make "k" vanish (this is effectively setting c to 1). If your start point is t=0,x=0,y=0,z=0 and your end point is t=onebillion,x=0,y=0,z=0 you've moved one light second into the future and zero light seconds left/right, up/down, or forward/backward. Using (+,-,-,-) sign convention, S is a large number, because we are subtracting nothing from it. However, light travels at one light-nanosecond per nanosecond, so light shining to the left would go from x=0,y=0,z=0,t=0 to t=onebillion,x=onebillion,y=0,z=0. The length of the line is then zero, exactly. Finally, if we have two objects at t=0, one at t=0,x=0,y=0,z=0 and the other at t=0,x=onebillion,y=0,z=0, then when we calculate dS^2 we get negative one billion. A Lorentzian (one sign difference) metric lets us categorize based on this. For the (+,-,-,-) signature, a positive value for dS^2 is timelike, a zero value is null or light-like (since c relates to the speed of light), and a negative value is space-like. The absolute value of dS^2 is smallest when dS^2 = 0. Null intervals are thus the "shortest line" through a flat Lorentzian spacetime. Timelike paths are longer. The origin and rendezvous points in the "twin paradox" or "relativistic galaxy cruiser who comes back", in light years, differ either not at all (or only a small small fraction if we leave/return right near Earth, since the planet moves around a bit in a year). Non-travelling twin is always at or near x=0,y=0,z=0, but moves a full year along t. Most of the path non-traveller's t is changing but the x,y and z are non-changing. Travelling twin, however, moves away from x=0 while moving "left" away and then moves back towards x=0 when moving "right" back. t is changing throughout traveller's trip, of course, but so is x. We can look at it like this, with non-traveller t,x and traveller t',x': start: t=0,x=0; t'=0,x'=0. In the very next step, measured in very small units (say, light-attoseconds), and with traveller already accelerated to c exactly, we have t=1,x=0; t'=1,x'=1. In the next step, t=2,x=0; t'=2,x'=2. And so forth. The values for dS for these first two steps are non-traveller 1, 2 while traveller is 0, 0. When we integrate all these, non-traveller's total is about 3 * 10^26 (that's about the number of attoseconds in a year) while traveller's total is about zero (or exactly zero if we assume instant turn-around, and rendezvous spatial coordinates are x=0,y=0,z=0). These totals are path-lengths. Non-traveller's path length is extreme, traveller's is zero. Proper times -- what one's cells, wristwatch, and other clocks one carries at all times reflect -- follows these path lengths. The non-traveller ages a year, the travelling-at-exactly-c traveller does not age at all. Usually you take the traveller as massive and make it travel at a bit less than c, so that the total path length is timelike but short (rather than lightlike and thus null) compared to the extremal timelike path of the non-traveller. Thus the traveller's proper time does pass a little, but not nearly as much as the non-traveller's. Additionally, the traveller is often cast as accelerating and decelerating gently, so there are periods where the x coordinate is changing slowly with respect to t, and periods where the x coordinate is changing close to how quickly t is changing. The massive gently-accelerating (and decelerating) traveller is what is described at the wikipedia link way above. Almost finally, the geometry does not depend on choice of coordinates. One can use arbitrary units to tick off durations the timelike axis, and can choose spatial coordinates that are spherical or cylindrical rather than Cartesian-like. The metric for flat spacetime changes form with a change of coordinates[1], but the resulting path lengths and travel times work out exactly the same. Finally, the metric of curved spacetime is different from that of flat spacetime; the reason you get time dilation around a black hole is because path lengths change because the line elements have a radial dependency such that the paths of objects moving tangentially past a black hole are shorter the further away from the black hole the tangent lies. Gravitation, then, is geometrical. - -- [1] see, for example, below (10.4) at http://ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Spacetime_%20of_%20General_Relativity.htm http://ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Spacet...
34381 8y ago
Once our minds are digital, it will require a rethinking of the concept of self. Send a copy of you to the stars and keep a copy at home enjoying terrestrial life. Meet back up in 20 billion years and have a beer with an old friend.
girvo 8y ago
I quite liked how “A Deepness In The Sky” tried to tackle this, by making the galaxy travelling traders permanently outside of the existing societies, as time dilation means they will have to be!
- taneq 8y ago
  Same with the Revelation Space universe's lighthugger starships. The crews were seen as quite alien by the rest of their culture. That's also the first place I saw the (in hindsight obvious) idea that a ship that can accelerate at over 1G can land on (or hover over) a planetary surface. Everyone else seems to have kept the "starships are for outside atmospheres only" trope.
  - simias 8y ago
    That's assuming that the thrusters can still output 1G within the atmosphere and that the hull of the ship can withstand the outside pressure and the weather (humidity, strong winds etc...). Also a huge ship firing thrusters continuously in the atmosphere might have adverse side effects (for instance radiation, toxic smoke, heat etc... depending on the technology). Might make sense to optimize large ships for space only and use smaller vehicles to taxi back and forth from the surface of planets.
    - clarkmoody 8y ago
      Another Vinge goodie is the effect of the "main torch" of a starship on the permafrost of a planet as it hovers around looking for a landing site.
    - taneq 8y ago
      Yeah, but if we're allowing thrusters that can generate a 1G acceleration for years at a time without using the entire ship as reaction mass, those are much smaller "gimmes". :)
  - magduf 8y ago
    >Everyone else seems to have kept the "starships are for outside atmospheres only" trope. Engineering a ship that can land on a planet with gravity and an atmosphere is much bigger challenge than engineering a ship that never does anything besides go into orbit. It makes it much easier to create a really, really, really huge ship. You're not going to build a Death Star that can land on a planet's surface, unless you can come up with some kind of materials that we have no idea about now. So why bother? It's easy enough to make a giant ship for interplanetary travel with smaller shuttles for landing.

mortenjorck 8y ago

Are you familiar with The Expanse? Based on the way G-forces are presented in the TV adaptation (I presume the book does something similar), I take it the Epstein Drive is supposed to be a constant-acceleration design, which in the story is limited to intra-solar travel (though presumably the generation ship Nauvoo was designed to use it for interstellar). I mention this because your comment made me think: Even staying within the solar system, how short would a trip between Earth and Mars have to be to start to pose relativistic problems? I wonder at what point frequent travelers would start aging noticeably more slowly. (If this is dealt with in the books, now I’ll really have to read them.)

wyattpeak 8y ago
At relativistic speeds, Mars and Earth are ludicrously close - between 4 and 20 minutes apart at light speed. So even travelling arbitrarily close to the speed of light, at the greatest possible distance, you could only gain 20 minutes on the folks back home per trip. A round trip weekly at average distance (just going with 12mins/trip because the details get complicated), 48 weeks a year for 40 years would leave you just over a month younger than those who stay put.

TheSpiceIsLife 8y ago

And, if you can work out how to do 1G acceleration, you can probably work out how to 2 or 3 or 4G acceleration. That's going to get uncomfortable, but most people should be able to cope with 2 or 3G acceleration for short periods.

anonytrary 8y ago

> I never thought of c being a real problem to space travel. > We just need to figure out how to use propellant efficiently enough among a few other slightly difficult challenges Both of these sentiments are pretty common misconceptions I've seen on HN. First, if c was infinite, the energy required to accelerate a constant amount would grow quadratically. This alone is a tough problem. Maybe we could "just" solve it. On the other hand, if c was finite, the energy required to accelerate would grow exponentially. This is more than a "slightly difficult" problem.