3 ms·
No, “collecting” photons transfers momentum so at best if you have 100% conversion rate you won’t slow down, so you need a separate source for acceleration. Wh
by dogma1138 5y ago
No, “collecting” photons transfers momentum so at best if you have 100% conversion rate you won’t slow down, so you need a separate source for acceleration.
Whilst photons hit your ship from all direction the photons incoming from the direction of travel would be shifted towards higher frequencies so their momentum would be higher than the photons hitting the ship from the opposite direction so unless there is a sufficient difference in the number of photons hitting you from the back to compensate for it you would still not be able to extract energy to move.
This means that even say a solar sail has a limit on acceleration at a certain speed the photons hitting the front of the sail would be blue shifted enough to counteract the acceleration of photons hitting you from the back, even if you convert them to useful energy it wouldn’t matter since you’ll reach equilibrium.
- FeepingCreature 5y ago> No, “collecting” photons transfers momentum so at best if you have 100% conversion rate you won’t slow down, so you need a separate source for acceleration. Given 100% conversion rate you should getting energy via photovoltaic effect, so you use that to accelerate them. > Whilst photons hit your ship from all direction the photons incoming from the direction of travel would be shifted towards higher frequencies so their momentum would be higher than the photons hitting the ship from the opposite direction so unless there is a sufficient difference in the number of photons hitting you from the back to compensate for it you would still not be able to extract energy to move. But that's just "friction". Assuming a photonically uniform environment at rest, just buffer energy at rest, accelerate for a bit, goto 10. Eventually you'll get anywhere. (Assuming a non-uniform environment, you're probably close to a star, in which case gravity outweighs photon pressure anyway.)