Re: [Phys-L] another DIY relativity experiment

[David Bowman <David_Bowman@georgetowncollege.edu>]

Regarding Bob S's question:

> David,
>
> Using the transformation (from your post):
> x' = x
> y' = y
> z' = (z + (c^2)/g)*cosh(g*t/c) - (c^2)/g
> t' = (z/c + c/g)*sinh(g*t/c)   ==>
>
> The third equation says that an event at z = -(c^2)/g has the
> primed coordinate z' = -(c^2)/g, a constant in time.
> This seems to contradict the second statement of :
>
> "Note (from the equation for t') a clock at rest w.r.t. the
> spaceship at z = -(c^2)/g is infinitely dilated and stopped
> relative to frame K'.
> Such a clock is moving at speed c in the K' frame."
>
> What am I missing?
> Thanks,
>
> Bob Sciamanda

Part of the problem is I accidentally misspoke, er, miswrote, and referred to the wrong equation in the parenthetical phrase you quoted above.  The rest of the problem is one needs to be more careful than you have been in determining the behavior in frame K'.  What I ought to have said in that phrase and rest of the quote was:  "Note (from the equation for the invariant proper time increment dτ using K-frame coordinates) a clock at rest w.r.t. the spaceship at z = -(c^2)/g is infinitely dilated and stopped relative to frame K'.  Such a clock is moving at speed c in the K' frame."

In particular, the actual relevant equation is (c^2)*(dτ)^2 = ((c + g*z/c)^2)*(dt)^2 -((dx)^2 + (dy)^2 + (dz)^2).  So we can see from this equation for an object (i.e. the "clock") at rest w.r.t. the spaceship at z = -(c^2)/g has its RHS vanish.  This is because the spatial increments dx, dy, dz are zero because the "clock" is at rest in frame K.  And the time-like contribution is also zero because any nonzero coordinate time increment dt is multiplied by the vanishing pre-factor coefficient
(c + g*z/c)^2. 
Since the RHS vanishes for such a "clock" that means the LHS must also vanish.  But the LHS is the elapsed proper time on the "clock".  Thus the "clock" doesn't age in frame K.  Since the proper time is invariant the "clock" doesn't age in *any* frame, including frame K'.  Some clock! This means our "clock" is necessarily a massless null object.  (Recall any massless object, like  a photon, doesn't age, either.)  The fact that the "clock" experiences no proper time means its path is necessarily a null path through space-time.  We can see its space-time trajectory in frame K' in at least a couple of ways.  One way it so look at the corresponding equation for the elapsed proper time as seen in frame K', i.e.
(c^2)*(dτ)^2 = (c^2)*(dt')^2 -((dx')^2 + (dy')^2 + (dz')^2).
Because the LHS of the former proper time equation in K vanishes that means the LHS of this latter proper time equation in K' also vanishes because both LHSs are the very same invariant quantity.  Now since the object in frame K is at rest that means dx = dy = 0.  Since our coordinate transformations between frames obey x' = x and y' = y we know dx' = dy' = 0 also.  Putting in all these zero quantities into our latter proper time equation gives: 
0 = (c^2)*(dt')^2 -(dz')^2.  Rearraranging this equation and taking a square root gives |dz'/dt'| = c.  But we already know that objects on null paths must move at speed c, so we really ought not be surprised by this result.  Another way to see the same thing is to look at the explicit equation I gave for the velocity of an object which is at rest in K as seen in frame K'.  That equation is:

v'(t') = c/sqrt(1 + ((z_0 + (c^2)/g)/(c*t'))^2) .

Recall here z_0 is the initial but constant fixed z coordinate of the object in frame K.  When such an object has that fixed z = z_0 = -(c^2)/g then we get v'(t') = c for all t'.

BTW, when you wrote "The third equation says that an event at z = -(c^2)/g has the primed coordinate z' = -(c^2)/g, a constant in time." that is not really correct.  The third equation by itself only is *part* of a multi-dimensional/multi-equation transformation from R^4 to R^4 on space-time.  The RHS of that equation may superficially look like a constant, but it only really describes how to get the z' coordinate for an event in frame K from a (z,t) coordinate pair in frame K for that event.  To  see how z' varies, or not, in time in frame K' we need to write the whole trajectory in *frame K'* and not just look at one equation of the multi-dimensional transformation between two *different* frames.  To do that we need to eliminate coordinate t in frame K from the RHS of the 3rd equation and instead write it in terms of time t' in frame K' *before* we move to the singular limit of z = -(c^2)/g in frame K. To do this we first take the 4th equation which says

t' = (z/c + c/g)*sinh(g*t/c)

and rearrange it to be solved for t, *after* which we then substitute it back into the 3rd equation to eliminate coordinate t from it.  Rearranging the 4th equation to be solved for t gives:

t = (c/g)*arcsinh(t'/(z/c + c/g)) .

Substituting this into the 3rd equation to eliminate t gives:

z' = (z + (c^2)/g)*cosh(arcsinh(t'/(z/c + c/g))) - (c^2)/g

Using the hyperbolic identity cosh(Z) = sqrt(1 + sinh^2(Z)) in the above equation gives:

z' = (z + (c^2)/g)*sqrt(1 + (t'/(z/c + c/g)^2) - (c^2)/g, and finally:

z'  = sqrt((z + (c^2)/g)^2 + (c*t')^2) - (c^2)/g .

Now (not before) we can take this equation to see how z' varies with t' for a fixed value of z.  Now when we use the fact that our fixed z = -(c^2)/g the above equation boils down to simply:

z ' = c*t' - (c^2)/g.

This equation does *not* say z' is constant in time, but rather, that z' increases at velocity c.

David Bowman