From: dr.j.thornburg@gmail-pink.com   
      
   In article , Luigi Fortunati discussed,   
   in Newtonian mechanics and in general relativity (GR), the behavior of   
   an elevator which is (a) initially suspended stationary from an elevator   
   cable or cables, and then (b) free-falling downwards after the cable(s)   
   break.  Luigi then went on to (c) ask some questions about the motion   
   of bodies placed above/below the center of gravity of a free-falling   
   elevator.   
      
   In this article I'll try to analyze the same system Luigi described,   
   and clarify a few of the tricky parts of how this system is modelled   
   in GR.  In this article I'll focus on (a) and (b) above; I'll discuss   
   (c) in a following article.   
      
      
      
   Let's start with the Newtonian perspective, where gravity is a force,   
   and where we measure acceleration with respect to (i.e., relative to)   
   an inertial reference frame (IRF).  To a very good approximation, we can   
   treat the surrounding building housing the elevator shaft as an IRF.   
      
   BEFORE the cable-break, i.e., when the elevator is suspended from   
   the cables, stationary with respect to the surrounding-building IRF,   
   there are 2 (vertical) forces acting on the elevator:   
   * elevator weight (F=mg, pointing down), where m is the elevator's   
     mass and g is the local gravitational acceleration (about 9.8 m/s^2   
     near the Earth's surface)   
   * cable tension (F=T, pointing up, i.e., the cable(s) are pulling up   
     on the elevator)   
   The net (vertical) force acting on the elevator is thus T - mg in the   
   upward direction.   
      
   Since we observe that the elevator is stationary with respect to the   
   surrounding-building IRF, i.e., the elevator's acceleration with respect   
   to that IRF is a=0, we use Newton's 2nd law to infer that the net force   
   F_net acting on the elevator must also be zero:   
     F_net = ma = 0   
   and hence   
     T - mg = 0   
   and hence the cable tension must be   
     T = mg   
      
   Luigi wrote   
   > The cables break and the elevator goes into free fall.   
   > Newton told us that the elevator accelerates and, therefore, there is a   
   > force that makes it accelerate.   
      
   That's correct.   
      
   AFTER the cable-break, when the elevator is in free-fall downwards   
   (let's neglect air resistance for simplicity), the only force acting   
   on the elevator is the elevator's weight (F=mg, pointing down), so the   
   net force acting on the elevator is F_net = mg (pointing down) and the   
   elevator's acceleration with respect to the surrounding-building IRF   
   is a = F_net/m = g (again pointing down).   
      
      
      
   Now let's look at the same system from a GR perspective, i.e., from a   
   perspective that gravity isn't a force, but rather a manifestation of   
   spacetime curvature.  In this perspective it's most natural to measure   
   accelerations relative to *free-fall*, or more precisely with respect   
   to a *freely-falling local inertial reference frame* (FFLIRF).  An   
   FFLIRF is just a Newtonian IRF in which a fixed coordinate position   
   (e.g., x=y=z=0) is freely falling.   
      
   Like Newtonian IRFs, there are infinitely many FFLIRFs at a given   
   position, with differing relative positions, velocities, and   
   orientations, but all these FFLIRFs have zero acceleration and   
   rotational velocity with respect to each other.  If all we care about   
   is acceleration, we often ignore the freedom to choose different   
   relative positions, velocities, and orientations, and refer to "the"   
   FFLIRF.   
      
   Any FFLIRF is (by definition) freely-falling, so it's accelerating   
   *downwards* at an acceleration of g relative to the surrounding-building   
   Newtonian IRF.   
      
   Since the g vector points (approximately) towards the center of the   
   Earth, we see that the FFLIRF *changes* if you go to a different place   
   near the Earth's surface.  For example, my FFLIRF differs from (i.e.,   
   has a nonzero relative acceleration with respect to) the FFLIRF of   
   someone 1000 km away on the surface of the Earth, or even of someone   
   at my latitude/longitude but 1000 km above me.  This why we have the   
   word "local" in the phrase "freely-falling *local* inertial reference   
   frame".  This is related to Luigi's questions (c); I'll elaborate on   
   this in a following article.   
      
   In GR it's easiest to first consider the situation AFTER the cable-break,   
   when the elevator is freely falling (we're neglecting air resistance).   
      
   Luigi wrote:   
   > Then Einstein came along and told us that this is not true and that   
   > there is no force that accelerates the elevator in free fall.   
      
   That's correct.  There are no forces acting on the elevator (remember   
   we're not considering gravity to be a "force"), so the net force acting   
   on the elevator is zero, so Newton's 2nd law   
     a = F_net/m   
   says that a=0, i.e., the elevator has zero acceleration   
   *with respect to (i.e., relative to) a FFLIRF*.   
      
   Since we've already established that a FFLIRF is accelerating *downwards*   
   at an acceleration of g relative to the surrounding-building IRF, we   
   conclude that the elevator is accelerating *downwards* at an acceleration   
   of g relative to the surrounding-building IRF.   
      
   Luigi wrote:   
   > But if there is no force that accelerates the elevator, it means that   
   > the elevator does not accelerate.   
   > And if it does not accelerate, then it moves with uniform speed.   
      
   This two sentences both leave out a key qualification, namely "with   
   respect to a FFLIRF".  That is, a more accurate statement is that if   
   there is no force that accelerates the elevator, it means that the   
   elevator does not accelerate *with respect to a FFLIRF*, and hence it   
   moves with uniform speed *with respect to a FFLIRF*.  The qualification   
   "with respect to a FFLIRF" is essential here -- without it the statement   
   is ambiguous (acceleration with respect to what?).   
      
   Luigi wrote:   
   > But speed is not absolute: it is relative.   
   > And so I ask: is there any reference system with respect to which its   
   > speed is uniform?   
      
   Yes, the elevator's speed is uniform with respect to any FFLIRF.  Since   
   a FFLIRF is accelerating (downwards) with respect to the surrounding   
   buildint's IRF, the elevator's speed is NOT uniform with respect to the   
   surrounding building's IRF.   
      
   Now let's consider  the situation BEFORE the cable-break from a GR   
   perspective.  Now there *is* an external force acting on the elevator,   
   namely the cable tension (F=T pulling up on the elevator).  In the   
   Newtonian perspective we found that T = mg, and this turns out to still   
   be true in GR.   
   	[Aside: What I just wrote is true for weak gravitational   
   	fields like the Earth's.  If we were in a very strong   
   	gravitational field (e.g., close to a neutron star or   
   	black hole) then we might have to be more careful with   
   	many of the statements I'm making.]   
   So, the net force acting on the elevator is F_net = mg, pointing up.   
      
   Newton's 2nd law then says   
     a = F_net/m   
       = mg / m   
       = g   
   i.e., the elevator (which is stationary relative to the surrounding   
   building) must be accelerating *up* at an acceleration of g with respect   
   to (i.e., relative to) any FFLIRF.   
      
   This seems a bit counterintuitive, but in fact it's correct:  Since a   
      
   [continued in next message]   
      
   --- SoupGate-DOS v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)