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Re: Symmetry in Lorentz transformation equations (long)



Stefan Jeglinski wrote:

I think that Lorentz transformation is not valid for accelerated frames.

I thought the issue was not acceleration but whether the local
spacetime was flat (to what ever approximation required). Can a
'relativist' (David?) expand on this for me?

Although I cannot claim to be a relativist, I'll attempt to answer
anyway. (Paul Camp can claim to be a relativist, but I don't think he
contributes to or subscribes to this list any more.)

What the issue is depends on the topic of discussion. The topic we were
discussing was whether or not time is somehow symmetrically equivalent to
space. To show that this is not the case one does not need any curvature
of spacetime, nor any gravitational effects at all. Neither does one
need to consider any transformations to or from any accelerating frame.

It is true that in a flat spacetime, (i.e. a spacetime devoid of
gravitational sources and of gravitational radiation) a transformation to
or from an accelerated frame is not a Lorentz transformation. Such a
transformation can be found, however starting with and generalizing a
Lorentz transformation in any of a number of ways.

For instance, consider the following single example. We make a
transformation from the inertial Cartesian Minkowski coordinates
(ct,x,y,z) (in ordinary flat spacetime) to the accelerated coordinate
system (ct',x',y',z') defined by the transformation equations:

c*t = (z' + (c^2)/g)*sinh(g*t'/c)
x = x'
y = y'
z = (z' + (c^2)/g)*cosh(g*t'/c) - c^2/g

where g is a constant with dimensions of acceleration. The accelerated
primed system is accelerating toward the +z direction in the unprimed
inertial system in such a way that if *at any time* an observer moving
with the primed system 'dropped' an object from rest at the spatial
origin of this accelerated system, then the object would (in the primed
frame), in free fall, spontaneously accelerate toward the -z' direction
with an initial proper acceleration of g. If the observer dropped the
freely falling object from rest at some other location (x',y',z') then
the object would have an initial proper acceleration of
-g/(1 + g*z'/c^2)) along the -z' direction. If the object's initial
velocity in the primed frame was nonzero then its initial proper
acceleration would be different. If the moving observer had a lab (or
elevator?) which was very small in linear extent compared to c^2/g, and
that lab remained centered on the spatial origin of the primed system,
then that observer could conclude that there was a gravitational field
strength g along the -z' direction making the freely falling objects
accelerate downward. Also, if the observer had a weighing scale attached
to the primed spatial origin he would observe that objects weighed on
that scale always had a weight of m*g. If the observer carefully
calculated and looked at the Riemann curvature tensor of spacetime in his
primed coordinate system he would discover that all its components vanish
(showing that his spacetime was actually flat), and he would then know
that the reason for the gravitational field he observes is that his frame
is ever accelerating w.r.t. an inertial frame (such as the unprimed one).

The transformation equations above are definitely *not* a Lorentz
transformation. In fact, the time t' kept by the watch of the observer
above is continuously being ever more dilated w.r.t. the inertial time t
according to the equation: t = (c/g)*sinh(g*t'/c). Also an observer at
rest in the inertial unprimed frame observes the accelerated observer's
lab/elevator moving to the +z direction according to the formula:

z = ((c^2)/g)*(sqrt(1 + (g*t/c)^2) - 1)

having a velocity v = dz/dt given as:

v = g*t/sqrt(1 + (g*t/c)^2) = c/sqrt((c/(g*t))^2 + 1)

If the observer in the inertial unprimed frame remains at rest at the
spatial origin of that frame then the accelerated observer sees the
inertial (freely falling) observer accelerate to the -z' direction
according to the formula:

z' = -((c^2)/g)*(1 - sech(g*t'/c))

having a *coordinate* velocity v' = dz'/dt' given as:

v' = -c*sech(g*t'/c)*tanh(g*t'/c).


If the velocity v of the primed frame lab as observed by the inertial
unprimed observer is written in terms of the primed lab's internal time
t' we get:

v = c*tanh(g*t'/c)

Also, the negation of this expression, -v, is the proper velocity of the
inertial observer as seen in the accelerated frame by an observer who is
locally at rest in the accelerated frame as the inertial observer passes
by that observer's location.

Thus the inertial observer sees the accelerated lab/elevator moving
toward the +z direction with a *rapidity* = g*t'/c which is uniformly
increasing at a *constant* rate in terms of the primed time of the
accelerating lab.

If the metric interval element is written in both (primed and unprimed)
coordinate systems we get:

(c*d[tau])^2 = ((c + g*z'/c)*dt')^2 - (dx')^2 - (dy')^2- (dz')^2

= (c*dt)^2 - (dx)^2 - (dy)^2 - (dz)^2

The Lagrangian L' of a freely falling particle in the primed system is:

L' = -m*(c^2)/[gamma'] where

[gamma'] = 1/sqrt((1 + g*z'/c^2)^2 - (|v'|/c)^2) where

|v'|^2 = (dx'/dt')^2 + (dy'/dt')^2 + (dz'/dt')^2 .

The canonical momentum 3-vector p_vec' is given in terms of the velocity
3-vector v_vec' as:

p_vec' = m*v_vec'*[gamma']

The Hamiltonian (energy) H' of the particle in the primed system is:

H' = [gamma']*m*(c + g*z'/c)^2 =

= sqrt((m*c^2)^2 + (|p'|*c)^2) * (1 + g*z'/c^2)

Compare the twin paradox, which is not resolved by claiming "one twin
has to accelerate at some point, hence..." In fact, there is a nice
problem in Jackson which works the twin paradox while the "moving"
twin is always accelerated - same general result (difference in
aging).

Actually, the accelerated lab/elevator at rest in the primed frame for
the system above is precisely the same case as the problem in Jackson
you refer to (but for only 1 leg of the 4 part trip though).

It's been such a long time since I worked that one out, IIRC
the validity of the approach (which I think still is a Lorentz
transformation at its core) had to do with whether the spacetime was
flat.

The example here definitely has a flat spacetime. Notice though that all
positions z' in the primed system are not equivalent as different z'
values have different amounts of time dilation (even for objects that
remain at rest in the primed frame). This primed coordinate system is
uniformly accelerating *in time*, so that the metric in the primed
coordinate system is independent of the time parameter, but it is *not*
independent of the spatial coordinate z'. This means that the spatial
coordinate system is not uniformly (i.e. homogeneously) accelerating
throughout all space. In fact, the primed coordinate system has a
coordinate singularity at z' = -c^2/g (where t' diverges to infinity and
in the unprimed coordinates z = -c^2/g +/- c*t) beyond which the
accelerated frame cannot be continuously extended even though nothing
anomalous actually happens there in the inertial unprimed coordinate
system.

This coordinate singularity is similar to the one that occurs when one
uses a uniformly rotating coordinate system. In that case the rotating
coordinate system doesn't work for regions of space that are farther from
the rotation axis than r = c/[omega] where [omega] is the angular
rotation rate of the coordinate system.

If one had a spacetime that had the property that all freely falling
objects uniformly and homogeneously throughout spacetime have a globally
constant initial acceleration (from rest), that system would only exist
on an intrinsically curved manifold, and would thus represent a truly
gravitating system whose gravitational field could not be globally
cancelled for all time by a judicious coordinate transformation. An
example of such a system has the metric:

(c*d[tau])^2 = exp(2*g*z/c^2)*((c*dt')^2 - (dz)^2) - (dx)^2 - (dy)^2

This system has everywhere throughout all of time and space the
initial acceleration equal to the constant g in -z direction for any
freely falling (from rest) object. But the spacetime manifold of this
system is intrinsically curved, so there is no coordinate transformation
that will *globally* turn the metric into a pure Minkowski (SR) form.

David Bowman
dbowman@georgetowncollege.edu