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enter image description here Why wouldn't this "warp-drive" work? Assuming we can build a flywheel that can survive near light speed and have an onboard power storage or beamed energy that can spin it up to that speed and back down again

1) Telescoping link between two capsules, they are free to move towards each other

2) flywheel in first capsule starts spinning. As it nears light speed from what I've heard about relativity it will start getting more massive instead of actually getting any faster

3) This extra mass will give it a stronger gravitational field right?

4) Both capsules now fall towards each other in accord with Newton's law of gravitation, however because the front capsule is now so much more massive it doesn't move noticeably as opposed to the rear capsule which basically blasts off towards the front capsule

5) After time, t the flywheel slows down, the masses of the two capsules becomes equal again, however the rear capsule stays in constant motion toward the front capsule because of Newton's first law

6) Just before the rear capsule runs into the front one the linkage is made to snap rigid which causes the rear capsule to transfer momentum to the front capsule

7) Because they are the same mass again their new velocity is v/2 where v is the velocity gained by the rear capsule in it's initial gravitational acceleration

Thoughts? I say it's a warp drive because using the visualization of spacetime where it is a flat elastic sheet, the flywheel dents the sheet which sucks in the rear capsule but just before it gets there the dent disappears but then reappears later further down the line to repeat the process.

Minify
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3 Answers3

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You need to accelerate and slow down the flywheel. In order for the system to be self contained, the energy has to come from within your device, The total energy of the device does not change with the accelerating flywheel since the energy has to come from something else in the device.

Peter R
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  • thanks, what if you beam the energy to and from Earth with lasers if it won't work self-contained? – Minify Nov 08 '15 at 01:53
  • If you are adding energy from the outside to accelerate it you don't have a self contained warp drive, – Peter R Nov 08 '15 at 03:43
  • Isn't that true for all propulsion devices though? That they use some of internal energy to produce thrust? – 2012rcampion Nov 08 '15 at 05:25
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    In this case internal energy is being used to increase the kinetic energy of the flywheel. As the kinetic energy increases, the source for the energy, a battery for example is losing the same amount of energy assuming 100% efficiency. The net energy of the system is unchanged. There is not thrust in the system described. He is trying to increase the mass of the system by increasing the kinetic energy of the flywheel, thereby increasing the gravitational force. – Peter R Nov 08 '15 at 05:40
  • But again, isn't the net energy of the system always unchanged? (E.g. for a chemical rocket, the chemical energy of the fuel beforehand is equal to the kinetic energy of the rocket and exhaust afterward.) I think it would be more helpful for you to explain in your answer why conservation of mass/energy means that the gravitational force doesn't change. – 2012rcampion Nov 08 '15 at 07:21
  • @2012rcampion, In a chemical rocket, the net energy is the same, if you include the reaction mass, but in practice, we just care about the rocket's change in energy. For a gravity device, $E=MC^2$, meaning the gravity of a system is the sum of rest mass and "energy mass". So by spinning up the device with (literally a few billion tons of) energy to increase its mass, you're taking energy (and therefore mass) from another part of the system, decreasing the gravity there. – MichaelS Nov 08 '15 at 07:38
  • Right, I'm saying that that (the fact that the gravity includes both mass and energy) is the thing that Minify is neglecting here, and the thing you should explain in your answer. – 2012rcampion Nov 08 '15 at 07:41
  • @2012rcampion ok but remember they are 2 capsules that can move independently of each other in one axis so the system isn't completely contained. If the energy comes from the rear one and decreases it's mass as you say then wouldn't that make this thing work even better because it will accelerate towards the more massive object even faster because a=f/m? – Minify Nov 08 '15 at 11:55
  • The gravitational; attraction between the two capsules will not change. The energy reduced from the rear module will equal the energy added to the front module (again assuming 100% efficiency) so that the gravitation attraction between the modules will be the same as before the transfer of energy. – Peter R Nov 12 '15 at 15:36
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A higher speed does not equal a higher mass. This is something commonly taught to beginning students of relativity as an explanation of relativistic momentum because it is easy to understand (but does lead to misconceptions).

Also, momentum must be transferred to slow down the flywheel.

Jimmy360
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  • Thanks, I didn't know this. But what happens to the kinetic energy as it approaches light speed then if it's not turned to mass? – Minify Nov 08 '15 at 01:55
  • @Minify it does increase but not due to mass – Jimmy360 Nov 08 '15 at 01:56
  • Could you rephrase that please? Are you saying the K.E does increase? But if v is choked off at c and m doesn't increase according to you then how is 1/2 mv^2 still increasing? Sorry for the ignorance, I only know Newtonian mechanics. – Minify Nov 08 '15 at 02:04
  • @minify 1/2 mv^2 is only an approximation valid for the case where v << c. – Doug McClean Nov 08 '15 at 03:36
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Relativistic mass will not increase the gravitational pull, the gravitational force depends on the rest mass of an object.

Courage
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    If so, then photons shouldn't have any gravity. Since they are affected by gravity, this would defy Newton's third law. Also, this would mean that annihilation and pair production cause gravity waves since the total rest mass is changing during the process. So... false. – John Dvorak Nov 08 '15 at 02:32
  • @JanDvorak Spacetime geometry is invariant regardless of the frame of observation, so a frame dependent concept like relativistic cannot possibly alter the EFEs and give a different answer. Sure, some components of the stress energy tensor will change depending on your frame, but the overall prediction by the EFEs will be the same. – SystematicDisintegration Nov 08 '15 at 07:26
  • The gravitational force shouldn't depend on the observer's frame is reference. I'm not an expert in this but I think the photons just follow the path of how the space time is curved, they are not really have gravity. – Courage Nov 08 '15 at 07:31
  • @SystematicDisintegration now, that's the whole point of a relativistic mass, isn't it? Everytime I've heard a motivation for the relativistic mass, it was talking about gravity, not inertia. But it gets worse. Protons have significant rest mass (938 MeV/c2), but they're mostly composed of gluons (which have none). The three valence quarks contribute measly ~10 MeV/c2 of rest mass. Under your theory most of the proton's mass should be non-gravitating. https://en.wikipedia.org/wiki/Proton#Quarks_and_the_mass_of_the_proton – John Dvorak Nov 08 '15 at 07:50
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    @JanDvorak We don't have a working quantum gravity theory, so I'm not going to comment on the proton. As for the motivation for relativistic mass, it has nothing to do with gravity. It was simply introduced to restore many special relativity formulae to their non-relativistic versions, namely the four-momentum and the four-force. Physics can be done perfectly in a framework that considers rest masses only (as is the case nowadays; relativistic mass is an orthodox concept). My first comment stands. – SystematicDisintegration Nov 08 '15 at 07:58
  • Photons have momentum (hint: solar sails, laser thrusters, this inexpensive desktop toy). They are also affected by gravity. This means that gravity can change their momentum. By the conservation of momentum, this also means that the object bending the light's path must have its momentum changed as well. – John Dvorak Nov 08 '15 at 07:58
  • @SystematicDisintegration that protons cause gravity relative to their rest mass can be observed quite easily, since 50% of the Earth mass is made out of protons (and the other half is out of neutrons, which have the same problem). – John Dvorak Nov 08 '15 at 08:01
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    @JanDvorak Photons are affected by gravity because all objects follow geodesics in spacetime, regardless of their rest mass/energy. As I've previously stated, we don't have a working quantum gravity theory, so we can't apply the photon case straight into the EFEs (I've not seen a stress energy tensor constructed for a photon). Even if you apply the EFEs to an object with non-zero rest mass, the momentum and energy of the object will change according to the frame of reference, but the final answer yielded will be unaffected. The EFE is a tensor equation that is invariant under diffeomorphisms. – SystematicDisintegration Nov 08 '15 at 08:06
  • Let's assume for a moment that photons are affected by gravity but don't cause gravitational attraction themselves. Therefore I propose the following (impractical) thruster mechanism that defies the conservation of momentum. Let's grab a black hole (ok, that's hard, so grab the heaviest thing you can grab). Let's grab a pair of mirrors, and place them slightly tilted behind the black hole. Let's grab a powerful laser and aim it so that the photons bounce multiple times between the mirrors. Photons get accelerated forwards due to gravity, but they won't accelerate the black hole. ... – John Dvorak Nov 08 '15 at 08:07
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    @Jan Dvorak, if a particle is travelling close to speed of light with respect to me, then it would obviously see me travelling at close to speed of light (in the opposite direction), so would I appear to the particle as being extremely massive and having a huge gravitational pull? I don't think so. – Courage Nov 08 '15 at 08:07
  • @JanDvorak You are repeatedly trying to combine the standard model with general relativity. GR is a classical theory, and you can't just take some QFT result/prediction and apply it there. – SystematicDisintegration Nov 08 '15 at 08:10
  • ... They will exchange their momentum with the thruster when bouncing off the mirrors. If you aim just right, you'll get a photon that has closed a closed trajectory, meaning its momentum is restored, meaning all of its acceleration went to the mirrors. Since the photons gained more forward momentum than the black hole gained backwards momentum (=none), the net effect is a forward thrust without the thruster emiting any radiation. Violation of the law of conservation of momentum. – John Dvorak Nov 08 '15 at 08:11
  • @SystematicDisintegration are you telling me that the conservation of momentum is a result of quantum mechanics? At which point does my photon gravity thruster assume anything from quantum mechanics? Gravitational lensing is a result of GR and Newton's laws predate both. Or are you denying that the photons have momentum that they can exchange when bouncing off things? – John Dvorak Nov 08 '15 at 08:15
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    @JanDvorak My response addressed your repeated usage of the proton as an example. I'm well aware that conservation of momentum follows from Noether's theorem. You're deviating from the original post. I suggest you post a new question if you have any concerns about the validity of my original statement. – SystematicDisintegration Nov 08 '15 at 08:19
  • @SystematicDisintegration I am not going to dispute your first comment. I just don't know how it relates to photons having their own gravity or not. The answerer wrongly claims that they don't. Conservation of momentum says they do. – John Dvorak Nov 08 '15 at 08:26
  • @JanDvorak No, the post says nothing about photons. It is a well established fact that the source of gravity in GR is the stress-energy tensor, which means both energy and/or mass can produce intrinsic spacetime curvature. The statement "relativistic mass does not determine spacetime curvature" is true under all circumstances (except for the QM ones; it is a consequence of the non-existence of a working quantum gravity theory). An obvious example is the fact that an object moving at relativistic speeds in a frame does not become a black hole even if the relativistic mass would say it should. – SystematicDisintegration Nov 08 '15 at 08:33
  • @SystematicDisintegration so, since gravity is caused by the spacetime curvature, are we both claiming that the answerer is wrong? Good :-) – John Dvorak Nov 08 '15 at 08:35
  • @JanDvorak No. Vishwaas is correct. I just explained it. Vishwaas didn't say that rest mass is the ONLY source of gravity. But he did say that relativistic mass is NOT a source of gravity. I think the distinction should be considered. – SystematicDisintegration Nov 08 '15 at 08:38
  • @SystematicDisintegration how? Is gravity not caused by the spacetime curvature, or does energy only ever change the spacetime curvature in a way that cancels the associated non-rest mass' contributions to spacetime curvature? How does "particles with zero rest mass don't have any effect on space curvature" follow from "both energy and mass can produce intrinsic spacetime curvature"? – John Dvorak Nov 08 '15 at 08:42
  • Speaking of the distinction, are you suggesting the non-rest component of mass causes gravity if carried by massless particles but doesn't cause gravity if carried by massive particles? That sounds unlikely. – John Dvorak Nov 08 '15 at 08:45
  • So, let's replace the photons in my photon gravity thruster with protons or any lightweight balls of mass. If they are moving at relativistic speeds, they will get momentum according to their relativistic speed, so their gravity should give the same momentum to the thruster black hole. – John Dvorak Nov 08 '15 at 08:48
  • @JanDvorak I repeat - "gravity depends on the rest mass of an object" does not mean "gravity ONLY depends on rest mass", i.e. the answer does not say that zero rest mass particles like photons have no effect on spacetime. What the answer does say is that gravity does NOT depend on relativistic mass, and that is correct. – SystematicDisintegration Nov 08 '15 at 08:49
  • @SystematicDisintegration another argument: Let's have a pair of balls moving at relativistic speeds, tight to each other by a really strong rope. The whole ensemble is moving at non-relativistic speeds. The whole ensemble is orbting a star. The whole ensemble has much higher mass than the sum of rest masses of its components (otherwise its mass would be lower than the sum of its component masses). This means that it gains much more momentum from the star's gravity than if the balls inside were at rest. That means that the star should feel a much stronger gravitational pull, too. – John Dvorak Nov 08 '15 at 08:56
  • This is essentially the same argument as for the proton rest mass consisting mostly of the binding energy between the quarks rather than the rest mass of quarks themselves. Note that the source of rest mass of protons is independent on GR... – John Dvorak Nov 08 '15 at 09:00
  • @JanDvorak I do not have enough comment space to answer that. Post it as a new question if you wish to have this problem resolved. – SystematicDisintegration Nov 08 '15 at 09:01
  • @SystematicDisintegration What would be the form of the question? "I have these arguments in favour of relativistic mass causing gravity, but this answer claims otherwise. Which one is right?", then wait for it to get answered, then link it here to document the answer is incorrect? – John Dvorak Nov 08 '15 at 09:05
  • @SystematicDisintegration as for not enough comment space... would you like to move to the chat? – John Dvorak Nov 08 '15 at 09:06
  • In fact, that question has already been asked for photons - yes, they do have gravity. – John Dvorak Nov 08 '15 at 09:12
  • Let us continue this discussion in chat. – SystematicDisintegration Nov 08 '15 at 09:14
  • Conclusion from the discussion so far: the "gravitational pull" mentioned in the answer is distinct from the actual force in the thought experiment's reference frame. It is possible to transform from an inertial reference frame to a different one, dependent on the rotating object's angular velocity, in which the rotating object causes only the space time curvature it would given its rest mass. How that relates to the thought experiment is pending discussion. – John Dvorak Nov 08 '15 at 10:20
  • Okay.. I think I shouldn't have used the word 'force' though – Courage Nov 08 '15 at 10:24
  • @Vishwaas you can edit the answer. An explanation of what exactly doesn't change would be nice. I also don't know how it relates to the question, so you might want to elaborate on that, too. (do take a look at the linked chat) – John Dvorak Nov 08 '15 at 17:54
  • Comments are not for extended discussion; this conversation has been moved to chat. – Manishearth Nov 08 '15 at 20:00