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.

  • $\begingroup$ I believe that a similar question has been asked here. $\endgroup$ – Jimmy360 Nov 8 '15 at 0:59
  • $\begingroup$ +1 Even though there are mistakes in the idea, this is clever. $\endgroup$ – mmesser314 Nov 8 '15 at 1:35

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.

  • $\begingroup$ thanks, what if you beam the energy to and from Earth with lasers if it won't work self-contained? $\endgroup$ – Minify Nov 8 '15 at 1:53
  • $\begingroup$ If you are adding energy from the outside to accelerate it you don't have a self contained warp drive, $\endgroup$ – Peter R Nov 8 '15 at 3:43
  • $\begingroup$ Isn't that true for all propulsion devices though? That they use some of internal energy to produce thrust? $\endgroup$ – 2012rcampion Nov 8 '15 at 5:25
  • 1
    $\begingroup$ 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. $\endgroup$ – Peter R Nov 8 '15 at 5:40
  • $\begingroup$ 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. $\endgroup$ – 2012rcampion Nov 8 '15 at 7:21

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.

  • $\begingroup$ 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? $\endgroup$ – Minify Nov 8 '15 at 1:55
  • $\begingroup$ @Minify it does increase but not due to mass $\endgroup$ – Jimmy360 Nov 8 '15 at 1:56
  • $\begingroup$ 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. $\endgroup$ – Minify Nov 8 '15 at 2:04
  • $\begingroup$ @minify 1/2 mv^2 is only an approximation valid for the case where v << c. $\endgroup$ – Doug McClean Nov 8 '15 at 3:36

Relativistic mass will not increase the gravitational pull, the gravitational force depends on the rest mass of an object.

  • 4
    $\begingroup$ 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. $\endgroup$ – John Dvorak Nov 8 '15 at 2:32
  • 1
    $\begingroup$ @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. $\endgroup$ – SystematicDisintegration Nov 8 '15 at 7:58
  • 1
    $\begingroup$ @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. $\endgroup$ – SystematicDisintegration Nov 8 '15 at 8:06
  • 1
    $\begingroup$ @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. $\endgroup$ – Oswald Nov 8 '15 at 8:07
  • 1
    $\begingroup$ @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. $\endgroup$ – SystematicDisintegration Nov 8 '15 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.