Natural Science Forum / Physics / Research / March 2008

There is no such thing as a "frame of an observer who accelerates".

What can be seen by an observer can be computed in any system of
coordinates.  Simply compute the light rays which intersect the
worldline of the observer.

Just FYI:

I worked out some results for constant 1g accelerations some years
ago, and published those (and other) results here:

       "Accelerated Observers in Special Relativity",
       PHYSICS ESSAYS, December 1999, p629.

I didn't investigate what those results imply about a light pulse
that is emitted after the traveler departs, so I can't comment
directly on your statements.  But my results might help as a
cross-check on your results.  Two of those results are given in the
abstract of my paper (as items 4 and 5), and I include that
abstract on my webpage:

 http://home.comcast.net/~mlfasf

I'll reproduce those two results here:

 4.  According to an accelerating observer, for a one g
     acceleration occurring when the separation is sufficiently
     great, the object's maximum (in magnitude) rate of ageing
     is greater than the accelerating observer's rate of ageing
     by a factor approximately equal to their separation, as
     measured in the object's frame, in lightyears. If the
     observer is accelerating toward the object, the object will
     be getting older at that rate. If the observer is
     accelerating away from the object, the object will be
     getting younger at that rate.

 5.  If an observer undergoes a constant acceleration forever,
     the object's age will approach a finite limit, according to  
     the observer. If their initial separation has a certain
     critical value, the object's age will never change at all,
     according to the observer.

Item 4 seems to imply that, at least if the traveler starts
sufficiently far away from the emission of the light pulse,
then the pulse can never catch him (because the current time
at the location of the emitter DECREASES monotonically as
the traveler accelerates (according to the traveler).  So in
that case, he would conclude that the pulse is never emitted
at all.

If he starts his acceleration at the location of the emitter,
then the current time at the emitter will INCREASE monotonically
(and will approach a finite limit).  (That result is given
in the paper, but not in the abstract).  So in this case, my
results don't immediately rule out that the pulse can ever catch
him.  I don't know what my results imply about your question in
that case (if anything)...I never investigated that.  But the
equations that I derive for the 1g case might help you check your
conclusions.

    Mike Fontenot

Nop. None of these is correct. We must use the Lorentz transformation
equations. We have in point of fact to differentiate.

(1- v^2/c^2)^-0.5 = v/(1-v^2/c^2)^-1.5

So acceleration will be g(1-v^2/c^2)^1.5

This is in the frame AT WHICH YOU STARTED. In your frame you are, or
course accelerating at g.

(1-v^2/c^2)^-1.5/g between 0 and c is infinite.

The basic special theory tells you that light travels (in your frame)
at the same speed in all directions. The Lorentz equations reflect
that fact.

 - Ian Parker

Thus spake Chalky <chalkyspam@bleachboys.co.uk>

The accelerating observer will see an event horizon at the point of
infinite redshift. I cannot, off the top of my head, remember if the
formula is correct, that this is given by gt = c, where t is time
measured by the accelerating observer, but it sounds right. Red shift
will go to infinity at the event horizon. Inference 3 is therefore
incorrect. The acceleration of the inertial origin is coordinate
dependent, but if we assume that coordinates are used in which the speed
of light is c, then the distance of the origin will go to infinity as
the event horizon is approached.

Regards

Here are a few simple results from my paper, for the
case of a constant acceleration that lasts forever:
___________________________________________________

In my paper, I show that for a constant acceleration,
starting from some initial separation Lo, the CADO
(current age of a distant object) approaches
a finite limit (according to the accelerating twin).

There is a certain critical initial separation, Lc,
for which the CADO never changes at all (no matter
how long the acceleration lasts).  For an acceleration
of 1 ly/y/y (about 0.970g), and for the case where the
acceleration starts from rest, the results are
particularly simple.  In that case, Lc is just
one lightyear, and the total change in the CADO (i.e.,
the difference between the limiting value of the CADO
and the initial CADO) is just

delta_CADO = -(Lo - Lc) years.

(I'm leaving out a factor of c in the above equation,
which is needed for dimensional correctness, but I'm
using units in which c = 1, so the equation is
numerically correct).

For example, if the traveler starts accelerating from
the earth (Lo = 0), then he (the traveler) will
conclude that his home twin's age changes by a total
of Lc = 1 year (i.e., she gets monotonically older
during his trip, but never more than one year older).

If the traveler starts at a separation of 1 lightyear,
then he will conclude that his twin's age never changes
at all.

And if he starts at (say) a separation of 3 lightyears,
then the total age change of his twin will be
-(3 - 1) = -2 years.  So his twin gets monotonically
younger during his trip, but never more than two years
younger.

For a 1g acceleration, the critical distance is only
slightly less than one lightyear (about 0.97 lightyear),
and the equation for delta_CADO remains the same.

    Mike Fontenot

[[...]]

[[This is a broader comment, which doesn't directly address Chalky's
questions.]]

Wikipedia has a nice introduction to physics as seen from the reference
frame of an observer who accelerates at constant (proper) acceleration:
 http://en.wikipedia.org/wiki/Rindler_coordinates

ciao,

On Mar 2, 11:24 pm, Mike Fontenot <mlf...@comcast.net> wrote:

Agreed. In fact, application of the GR time dilation formula to this
reference frame, confirms that the equivalence is exact.

Absolutely . That critical value is c squared /  g (Final formula,
http://physics.nmt.edu/~raymond/classes/ph13xbook/node61.html#vel )

Quite so.

Yes,

(This conclusion was actually stated in MTW)

They certainly have. My final conclusion for d was definitely in
error. The correct formula for this is given by formula 7.22  of the
above hyperlink. Everything else seems to have panned out quite
nicely, despite not initially.being  expressed very well by me

Thanks

The problem is that there is no such thing as "reference frame" for an
accelerated observer.

A reference frame is a rigid set of standard clocks and standard rulers
arranged (in principle) in a 3-d rectangular array; all of the clocks
are mutually synchronized. This cannot be constructed in an accelerated
system, because it is impossible for the clocks to remain synchronized.

One result of this is that if one choses to measure the speed of light
using standard clocks and rulers in an accelerated system, one obtains a
speed of light that is anisotropic, and for some directions ("up" or
"down") it varies in time! Indeed, if one waits long enough then one can
obtain an infinite value, an arbitrarily small value, or even a value
pointed in the obviously wrong direction (the light arrives at a smaller
clock value than when it was emitted).

One can construct an APPROXIMATE reference frame in an accelerated
system, and use it for measurements that are located in both space and
time. But that is of limited use, and cannot be applied to your
examples.

But an accelerated observer can make observations, and SR can predict
what such an observer would see (no "frame"). For instance, for an
observer with constant proper acceleration a, there is a Rindler horizon
located a distance c^2/a behind her. Any light emitted from behind that
horizon will never reach her, and light emitted at that horizon will
appear to be infinitely redshifted. This is your items (1) and (2),
except your equations are incorrect (the horizon is located a distance
behind the observer, not at any particular time).

Your item (3) is wrong -- the uniformly accelerated observer will not be
able to "see" the origin at all after it passes the Rindler horizon.

Note, please, that this is only a gedanken -- constant proper
acceleration lasting forever is not practical.

Tom Roberts

On Mar 2, 11:24 pm, Ilja Schmelzer <ilja.schmel...@googlemail.com>
wrote:

You have my commiserations. It must be a nightmare for you to navigate
around your own office. Even if you don't accept Einstein's
interpretation of the General Principle, you still have to acknowledge
that that is an accelerating reference frame by virtue of the rotation
of the earth