7
$\begingroup$

Space is hard, and making a spaceship engine powerful enough to do interesting things requires an absolutely bonkers amount of energy. I know that lots of authors (including me) think of trying to get around the issue by making reactionless drives, but those have the issue of allowing ships to accelerate to arbitrary velocities and destroy planets. I had a slightly different idea.

I've heard that theoretically, we can have a non-physics violating warp drive as long as it can't get above the speed of light. So what if someone invents a warp drive, but one which can only generate "virtual" velocity relative to the ship's "true" reference frame; and only up to a certain relative velocity, like, say, 50 km/s.

  1. If you have a ship in orbit of Earth, and it uses its warp drive to travel at 90% of the speed of light for one minute away from earth, then when the drive shuts off the 0.9c velocity vanishes, and the ship is left with only the velocity it possessed while orbiting earth.

  2. If two ships are stationary relative to each other, and one turns on its drive and attempts to ram the other, the instant they touch the warp bubble collapses and the two ships are left kissing without having damaged or imparted kinetic energy to the other, since they didn't have any relative kinetic energy to begin with.

  3. If a ship wants to get actual velocity without expending fuel, it can do so by using its drive to hover above the surface of a massive object, like a planet, and using its gravity to increase its true velocity, until it shuts off its drive and shoots off like a rock from a slingshot. The caveats are that you can't get a true velocity any higher than your drive's maximum speed (because otherwise you crash into the planet), and you need to be able to slow down at the other end of your trip (because the drive can't cancel your true velocity.)

  4. You can't use the drive to extract infinite energy from a gravity well by lifting an object above the surface and then dropping it through a generator, because raising an object through a gravity well requires additional energy equal to the potential energy gained by the lifted object.

EDIT: It's been pointed out that #3 and #4 contradict each other, since you could hover above a planet to build up 'true' velocity, and then drop into a generator with much greater energy than just the potential energy of the fall. My solution to this is that getting up to a certain 'virtual' velocity requires at least as much energy as the kinetic energy of the mass of the ship traveling at the same 'true' velocity. Though this brings up the issue that the energy would then have to GO somewhere once the drive is shut off, which might be hazardous, as a 20-ton craft going at 50 km/s would have to get rid of an amount of stored energy equal to half of Hiroshima.

My question is: Does this break the universe/physics? How? Does it make sense?

EDIT: Numbered bullet points.

$\endgroup$
11
  • $\begingroup$ What does scare quotes true close scare quotes reference frame even mean? How can one distinguish true reference frames from fake reference frames? $\endgroup$
    – AlexP
    Mar 25 at 1:31
  • $\begingroup$ @AlexP isn't a "true reference frame" the coords where you are in space time and look around as an observer ? I must admit I'm really naive about these topics.. but I've never heard about a fake reference frame. Seems quite uncomfortable to be in a fake place or time. $\endgroup$
    – Goodies
    Mar 25 at 1:48
  • $\begingroup$ @Goodies: The question says that the drive "generate[s] virtual velocity relative to the ship's true reference frame". If there are true reference frames there must also be false reference frames, right? How do we distinguish between them? $\endgroup$
    – AlexP
    Mar 25 at 1:51
  • $\begingroup$ @AlexP maybe a displacement is meant ? some offset, in respect to the reference frame you would exist in ? Suppose you could be able to translate a space ship above a cavity in space time - a place with a large gradient - you'd travel over that place and look "down", it would seem as if you travel faster than light... translate yourself back down, you'd be light years from where you started e.g. 10 days later.. the hypothetical Alcubierre drive does that. $\endgroup$
    – Goodies
    Mar 25 at 1:55
  • $\begingroup$ @AlexP The drive changes the craft's displacement without changing its velocity. You might think of it as a teleporter, which changes your position without changing your direction or speed, except that (from a mathematical perspective) it changes your position an infinitesimally small amount an infinite number of times per second. An outside observer would see that as a change in velocity, but the actual velocity hasn't changed. $\endgroup$ Mar 25 at 2:00

3 Answers 3

6
$\begingroup$

Still breaks planets, but universe is OK

Let's go back one step and look at why a reactionless, sublight drive is useful to an author / universe builder. Alice is on Earth and wants to visit Bob on Ganymede. With real world universe limits, the options are:

  • Take many years to do so using very low thrust methods such as solar sails
  • Take months / years to do so using high-efficiency low thrust methods such as ion drives.
  • Use technomagic drive that expels its reaction mass at near lightspeed - problems are that a) the exhaust from the drive can destroy cities and it lets the spacecraft accelerate to planet-wrecking velocities quite easily; and b) there is still a requirement to commit a large percentage of the ship to fuel storage due to the tyranny of the Tsiolkovsky rocket equation, with the less magical the drive the higher percentage of the ship needed for fuel.

Hence the attraction of a high output reactionless drive - it eliminates the months/years timeframe for interplanetary journeys, it makes the exhaust non-existent (and therefore safe re the Kzinti lesson) and it removes the need for massive fuel tanks of reaction mass.

However, there is a problem with relying on a stutterwarp / warp drive that does not change the velocity (not pseudovelocity) of the ship. Within a solar system, all the bodies are moving at different velocities. Which means that if Alice wants to visit Bob in the example above, when Alice arrives her ship will be moving at somewhere between 16 km/s and 43 km/s with respect to Bob's home city (depending on the relative positions in their orbits of Earth, Jupiter and Ganymede). Without using a reaction drive to make a very significant burn, Alice will either whip past Ganymede without being able to visit at all or will obliterate herself, Bob and everyone else in a large radius in a potentially extinction-event-level impact. Put another way - unless all objects that need to be travelled between are effectively stationary with respect to each other - effectively impossible in any solar system:

  • a warp drive of this type can still be a very dangerous weapon at its destination;
  • a spaceship will rely on a also possessing a reaction drive with associated fuel to match velocity at the end of its journey. (For planets it may be possible to slowly match velocities using the gravity of the planet with repeated passes, but this would not be usable for low mass destinations such as space stations.)

As far as physics goes though, it seems to tick all the boxes - causality is preserved, conservation of momentum, conservation of energy (given that energy is required to "climb" out of a gravity well). So the universe is happy even unfortunate residents at a ship's destination are (very briefly) unhappy.

$\endgroup$
2
  • 2
    $\begingroup$ I was thinking this too, but there are a couple of points to be made. First, 43km/s is a lot better than the hundreds of km/s of delta-v needed for a fusion drive or something similar. An aircraft carrier-sized ship going at 43km/s is only carrying equivalent kinetic energy to about a 20 megaton nuke. Hardly world-shattering, and it can be intercepted easier than a relativistic projectile. Second, you can 'steal' velocity from a planet by using your drive to hang in its shadow long enough for gravity to pull you up to the planet's orbital velocity. $\endgroup$ Mar 25 at 3:23
  • $\begingroup$ Options do exist if you consider nuclear propulsion. We built a nuclear engine with an isp of 710s back in the 1970s and we could likely build an engine with an isp of >1000s today using the same technology. Still, that's nowhere close to the theoretical limit. "Nuclear Lightbulb" drives can get a theoretical "clean" isp of up to 3000s and nuclear pulse propulsion has a theoretical cap somewhere around 100,000s. Granted, it's not sexy in a point-your-ship-and-go-anywhere-way, but it still makes in-system voyages doable in weeks/months instead of months/years. $\endgroup$
    – Dragongeek
    Mar 25 at 12:53
2
$\begingroup$

Points #3 and #4 seem incompatible to me.

Use your drive to hover over the planet, then turn it off and plunge down into a generator. You can get a lot more energy this way than it's going to cost to lift it back up.

I don't believe it's possible to have a no-cost drive that can't be exploited as a perpetual motion machine. And, because it's a generator it can be used for kinetic bombardment. Hover a star fairly close in and you can build up a lot of velocity!

$\endgroup$
5
  • $\begingroup$ Good point. I think I see what you're getting at. As you use the drive to hover above the planet, your 'true' velocity is increasing at a linear rate, but your kinetic energy increases with the SQUARE of your velocity, meaning that your built-up energy increases faster and faster. Keep in mind though that I specified that these drives have a maximum effective warp velocity, and it wouldn't be unreasonable to assume that the amount of energy required to achieve higher warp velocity ALSO increases with the square of the warp velocity. $\endgroup$ Mar 27 at 3:34
  • $\begingroup$ With that in mind, you'll need more and more energy to maintain your position above a planet's surface, until eventually it's impossible to continue doing so. The total energy expended to build up your true velocity might ultimately be no greater than the kinetic energy. Also, it doesn't matter whether you're hovering above a star, a planet, a moon, or a large asteroid. As long as it has appreciable gravity, you'll eventually get the same result. This was a good point to bring up, and I think it's patched, unless I missed something? $\endgroup$ Mar 27 at 3:36
  • $\begingroup$ @FlyingLemmingSoup If you need to expend that much energy on the drive you couldn't use it for takeoff/landing--nothing but a rocket can deliver the sort of energy needed without melting in the process. Once you were in orbit it would let you use nuclear power for low-thrust movement, but that's about it. $\endgroup$ Mar 27 at 19:47
  • $\begingroup$ That's assuming that you're restricted to modern-day power generation technologies. But you're right that even this method of using the drive is extremely power-hungry. Not as power-hungry as an Expanse-esque nuclear thruster, but beyond the ability of modern power generation to supply for the purpose of escaping gravity. $\endgroup$ Mar 28 at 2:45
  • $\begingroup$ @FlyingLemmingSoup It's not just the energy needed, but the energy density needed. Good luck building a nuke-powered craft with a TWR > 1. $\endgroup$ Mar 28 at 2:54
1
$\begingroup$

As presented, you still break physics. It's your first point.

"If you have a ship in orbit of Earth, and it uses its warp drive to travel at 90% of the speed of light for one minute away from earth, then when the drive shuts off the 0.9c velocity vanishes, and the ship is left with only the velocity it possessed while orbiting earth."

How much energy does it take to get up to 0.9 c using the drive? A "normal" drive will need about 9 megatons per kilogram. And then there is an even more embarrassing question - when you turn the drive off and revert to normal velocity, where does all that energy go? For normal drives, the situation doesn't arise: if you arrive at 0.9 c the impact is what dissipates the energy.

If the drive takes essentially no energy, or even just significantly less than a boring old reaction drive, you get a perpetual motion machine (I assume it ignores potential energy just the same as it ignores kinetic energy.) Take a container full of water, and use the drive to transport it to some higher altitude, and release it. Channel the released water into a waterwheel to harvest the energy, then transport it again.

Reactionless drives don't give the advantages you seem to think they do. Just because they don't need reaction mass doesn't mean they move as if by magic. And what you're talking about is not a reactionless drive at all, but rather (for lack of a better word) an inertialess drive.

$\endgroup$
3
  • $\begingroup$ I addressed the 'infinite energy' problem in point four. Using the drive to climb out of a gravity well requires extra energy equal to the potential energy gained. Does that solve the perpetual motion issue, or does it create more physics problems? You've hit the nail on the head when you say that a better term would be 'inertialess drive'; I just used the term 'reactionless drive' because it's a drive that doesn't need reaction mass. I keep calling it a 'warp' drive because it's supposed to be more akin to a hypothetical FTL drive than a conventional thruster. $\endgroup$ Mar 25 at 21:40
  • $\begingroup$ @FlyingLemmingSoup - Your word use is confusing. Inertialess drives go way back, at the very least to the 1930's Lensman novels by E.E, Smith. Reactionless drives get around the Rocket Equation, but they are still subject to conservation of energy and momentum. And while point 2 is an OK first try, you haven't addressed the issue of what happens if a ship simply runs into another object. If the ship has no kinetic energy, it must stop - even if the object is a single hydrogen atom - and space is full of them. $\endgroup$ Mar 26 at 19:37
  • $\begingroup$ Allright, I'll be more careful with my word choice. And the issue with point 2 is exactly the sort of problem I was looking for. In this case, it's probably that the drive creates a bubble of moving space time around the ship, so when a small object, like an atom, enters the bubble, it enters the same reference frame as the ship, and so get carried along with the ship. If the ship hits another object too large to fit into the bubble, like another ship... I don't know, what happens if half of a ship is inside of a moving warp bubble, and half is outside? $\endgroup$ Mar 26 at 20:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .