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About DimaMazin

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  • Birthday 01/16/70

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  1. Then redshift of light communication exists between rockets, simultaneously acceleratings in one direction, . And it is bigger on longer distance. Scientists can check it.
  2. We should consider beam scales with very long levers and with massive pans on asteroid. Can there a torque be less than distinction of gravitational attractions as a condition of non-return?
  3. I have tried some experiments, but I can't get non-return like that. Are the beam scales in vacuum on surface of white dwarf ? Is that due to thick rod?Part of mass of top lever is inclined to other side.
  4. Levers and pans have many different g. We should take only g of support point.
  5. You could be more successful with math.
  6. Excuse me. The question was from delusion.
  7. Are force of gravitational acceleration and force of gravitational attraction different forces?
  8. Thanks.
  9. Do you think point of the support is accelerating relative to the beam, and then force of inertia of the bowls is force of the balance?
  10. You have changed ideal balance by your hand.. One bowl has lowered downwards, the second has risen upwards.After the disbalance they came back in balance again.Why? If you solve it with Newton then please.I can't understand why lowered bowl has less force than risen bowl.
  11. Why' after change of balance on scales , balance happens again? How space/time works here?
  12. USA were Great, because they were developing science. It was good thing when people with science were winners. I don't think Trump understands it, but you should.
  13. Then two accelerating rockets should have enough gravitational link for length contraction between them, because cosmological force grows between rockets,accelerating in one direction.
  14. Simultaneity is just choice of math with clocks.
  15. Minkowski cannot be mate when we try to solve Bell's spaceship paradox. Clocks go at the same rate in S and S'. Only they have different readout of simultaneity on distance with speed. Mathematically : simultaneously dt=dt' in both frames but t doesn't equal t' at relative speed due to different simultaneity after acceleration on distance. dt=t-t0 dt'=t'-t0' then t0=t-t'=(r/v)*((gamma-1)/gamma) or t0=t0'+(gamma-1)r/(gamma*v) then r/dt=gamma*v