4 ms·

This seems to be using a kind of "reactionless drive" I.E. Movement without apparent exterior force. This kind of drive can't produce a net velocity, but can cr

This seems to be using a kind of "reactionless drive" I.E. Movement without apparent exterior force. This kind of drive can't produce a net velocity, but can create repeated small translations. To imagine how one would work in a space ship, imagine a very long ship that acts like a particle accelerator. Take an atom from the front end, accelerate backward to near the speed of light so that it's mass increases say 1000 times. While that atom is flying backward, your ship that pushed on it moves in the opposite direction. Catch it again in the rear, and the ship must return to zero speed because of conservation of momentum, however the ship is now in a slightly different spot. Move that atom back to the front of the ship, but slowly so it's mass stays low. Your ship moves backward, but not quite as far as it moved before. Do this again and again to translate the ship. The net velocity of the ship will always be zero though.

ianai 4y ago
I wonder if there’s a way to do that on an integrated circuit? I’m guessing a little channel carefully designed to create a magnetic field pointing a consistent direction. Electron gun on one end and positive terminal the other. This also starts to resemble how a laser works internally (at least HeNe). Granted, that’s light and massless.
- prox 4y ago
  Wasn’t there a magnetic/levitating robot that did this?
b112 4y ago
Catching the atom, and moving the atom, should result in the ship being returned to its original place. "Being sneaky" by moving things slowly, or catching things, just doesn't work. It will all add up. All movement will be undone.
- Tuna-Fish 4y ago
  You ignored the part where the atom is fired at such high speeds that it sees relativistic increase in mass. I dunno if that works, but that's the part that matters in the explanation above, and that's the part you'll have to address.
  - b112 4y ago
    Momentum is always a thing, regardless of how much velocity mass has. Trying to be sneaky, because it is easy to measure large velocity * mass, instead of slow velocity * mass, doesn't work.
  - TheOtherHobbes 4y ago
    I don't see why that makes a difference. To catch the atom you have to damp its new accelerated momentum, not its original momentum. So the ship will be translated backwards by a correspondingly accelerated amount. All you're doing is converting a slow relativistic acceleration into an equal and opposite fast relativistic deceleration.
  - blamestross 4y ago
    "relativistic mass" is a polite lie for the math. As far as that partial can tell, it can travel at any speed Newtonianly. The "relativistic mass" is just the extra energy needed from our perspective due to relative time dialation. No new mass is created. The force of gravitation caused the energy before you put it into the photon is that same as after you put it into the photon.
- tonyarkles 4y ago
  This is a case where you are definitely correct in theory, but in practice we can cheat and/or get screwed over. While it’s an angular momentum situation and not linear momentum, one method of satellite attitude control is “reaction wheels”. Inside the SV, there’s a mass on a motor; to rotate the satellite about an axis, you spin the reaction wheel on that axis in the opposite direction (equal and opposite momentum). In theory, stopping the reaction wheel should result in the same outcome on the SV, but it doesn’t quite because some of the energy of the reaction wheel was burned off as friction heat. It’s a small effect that adds up over time. Even without this “speed it up to increase its mass” situation the OP is referring to, it seems to me that you could exploit something similar on a spacecraft for linear momentum, by allowing drag on, say, a tennis ball to be asymmetric by shooting it fast in one direction and slowly moving it back in the other. It’s not violating any laws of physics, just exploiting a second-order effect when the first-order effect isn’t going to get you anywhere. You’d eventually need to get rid of the heat, probably just by radiating it out into the void (and, naturally, it wouldn’t be an energy neutral situation, you’d need some kind of power source to make up for the losses)

ianai 4y ago

I wonder if there’s a way to do that on an integrated circuit? I’m guessing a little channel carefully designed to create a magnetic field pointing a consistent direction. Electron gun on one end and positive terminal the other. This also starts to resemble how a laser works internally (at least HeNe). Granted, that’s light and massless.

prox 4y ago
Wasn’t there a magnetic/levitating robot that did this?

b112 4y ago

Catching the atom, and moving the atom, should result in the ship being returned to its original place. "Being sneaky" by moving things slowly, or catching things, just doesn't work. It will all add up. All movement will be undone.

Tuna-Fish 4y ago
You ignored the part where the atom is fired at such high speeds that it sees relativistic increase in mass. I dunno if that works, but that's the part that matters in the explanation above, and that's the part you'll have to address.
- b112 4y ago
  Momentum is always a thing, regardless of how much velocity mass has. Trying to be sneaky, because it is easy to measure large velocity * mass, instead of slow velocity * mass, doesn't work.
- TheOtherHobbes 4y ago
  I don't see why that makes a difference. To catch the atom you have to damp its new accelerated momentum, not its original momentum. So the ship will be translated backwards by a correspondingly accelerated amount. All you're doing is converting a slow relativistic acceleration into an equal and opposite fast relativistic deceleration.
- blamestross 4y ago
  "relativistic mass" is a polite lie for the math. As far as that partial can tell, it can travel at any speed Newtonianly. The "relativistic mass" is just the extra energy needed from our perspective due to relative time dialation. No new mass is created. The force of gravitation caused the energy before you put it into the photon is that same as after you put it into the photon.
tonyarkles 4y ago
This is a case where you are definitely correct in theory, but in practice we can cheat and/or get screwed over. While it’s an angular momentum situation and not linear momentum, one method of satellite attitude control is “reaction wheels”. Inside the SV, there’s a mass on a motor; to rotate the satellite about an axis, you spin the reaction wheel on that axis in the opposite direction (equal and opposite momentum). In theory, stopping the reaction wheel should result in the same outcome on the SV, but it doesn’t quite because some of the energy of the reaction wheel was burned off as friction heat. It’s a small effect that adds up over time. Even without this “speed it up to increase its mass” situation the OP is referring to, it seems to me that you could exploit something similar on a spacecraft for linear momentum, by allowing drag on, say, a tennis ball to be asymmetric by shooting it fast in one direction and slowly moving it back in the other. It’s not violating any laws of physics, just exploiting a second-order effect when the first-order effect isn’t going to get you anywhere. You’d eventually need to get rid of the heat, probably just by radiating it out into the void (and, naturally, it wouldn’t be an energy neutral situation, you’d need some kind of power source to make up for the losses)