Forums before death by AOL, social media and spammers... "We can't have nice things"
|    sci.physics.research    |    Current physics research. (Moderated)    |    17,516 messages    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
|    Message 17,331 of 17,516    |
|    Luigi Fortunati to All    |
|    Gravity and curvature    |
|    17 Dec 23 16:29:50    |
      From: fortunati.luigi@gmail.com              Gravity is a force and, therefore, is a vector.              Is spacetime curvature a vector? Does it have a direction and a verse?              Luigi Fortunati              [[Mod. note --       In order to answer questions like this, we need to be precise in       our terms. You write that "Gravity is a force". But what do you mean       by "gravity"? Starting with Newtonian mechanics for simplicity,       "gravity" could plausibly mean any of several things:       * gravitational potential energy (which is a scalar in Newtonian mechanics).       * the Newtonian "little g" (which is a 3-vector at any given position        and time       * the *difference* in the Newtonian "little g" between nearby objects        at a given time; this difference is what you can measure about the        gravitational field if you're in a freely falling elevator. This        difference is a 3-vector which depends on the separation between the        nearby objects,        difference = M * separation        where M is a 3x3 matrix and "*" denotes matrix multiplication. This        3x3 matrix M (which is really a rank 2 tensor) provides a complete        description of the local gravitational field at a given position and        time.              In general relativity (GR) things are (not surprisingly) more complicated.       To fully describe spacetime curvature at an event (a point in space, at       a particular time) requires generalizing the 3x3 matrix (rank 2 tensor) M       to the Riemann curvature tensor, which is a 4x4x4x4 4-dimensional matrix       (really a rank 4 tensor), i.e., it's a set of 4x4x4x4 = 256 numbers.       The Riemann curvature tensor has a bunch of symmetries, so it actually       has only 20 independent components.              So in GR, the best answer to your question is that spacetime curvature "is"       the Riemann curvature tensor. This doesn't have a single "direction" any       more than the 3x3 matrix M has a "direction" in Newtonian mechanics.       -- jt]]              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
(c) 1994, bbs@darkrealms.ca