If space is bent into higher than 3 spatial dimension, then inverse square no longer holds - it should be inverse cube at least (if only 1 big dimension "on top" of known 3 exists)?

So, where is 1/r3?

• Two answers - yanniru - November 28, 2001 - 15:12 UTC
  • Richard, I peronally contributed in dismissal of - alexander - November 28, 2001 - 20:42 UTC
    • Reference please to your publication - yanniru - November 29, 2001 - 15:20 UTC
      • For Richard (about gravity on mm scale). - alexander - December 1, 2001 - 02:27 UTC
        • Casmuir effect - yanniru - December 1, 2001 - 19:10 UTC
          • Way far. Casimir is 10^-6 or so. nt - alexander - December 3, 2001 - 22:39 UTC
• Alex = Priest! - luis_hamburgh - November 30, 2001 - 01:00 UTC
  • Luis, stupidity continues? - alexander - November 30, 2001 - 18:47 UTC
    • Alex = 1st non-religious person on my ignore list. :( - luis_hamburgh - December 2, 2001 - 21:15 UTC
      • - luis_hamburgh - December 3, 2001 - 18:26 UTC
      • Luis, if we have inverse square, space is 3-D (for gravity). - alexander - December 3, 2001 - 22:42 UTC
• The explanation of Inverse Square Law - pmartin - November 30, 2001 - 18:43 UTC
  • Paul, don't you see the obvious? - alexander - November 30, 2001 - 19:01 UTC
    • No. At least it is not obvious to me. - pmartin - December 1, 2001 - 01:28 UTC
      • If our space were "imbedded" inverse square would fail. - alexander - December 1, 2001 - 02:07 UTC
        • Not necessarily - pmartin - December 1, 2001 - 05:01 UTC
          • I think, mathematicians did that 100 years ago or so. - alexander - December 3, 2001 - 22:48 UTC