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Q1
: What is the significance of the other solution
to
the differential equation namely,
f (v) = (1+k2v2) ½
A:
The other solution pertains to a situation where
moving
clocks run faster. This is only a
hypothetical
case, as in this situation, even though
the symmetry is maintained between the frames, there is no maximum limit
for velocity. Also in this solution the event coordinates x,t behave as if
they correspond to the x,y coordinates of planar
geometry and uniform motion is a rotation of the
axis.
If moving clocks did run faster, the illusion
that
the event coordinates x,t obey the rules of planar
geometry, would not make the events
situated
on a plane. Such a view would be an
illusion.
The events are instantaneous and they will
not
exist beyond an instant.
The
transformations for this case would be
x'
= (x-vt)/ (1+k2v2) ½
and
t' = (t+k2vx)/
(1+k2v2) ½
The
inverse of this 2x2 transformation will be identical
to itself with the sign of v reversed, that is
x
= (x'+vt')/
(1+k2v2) ½
and
t = (t'-k2vx')/
(1+k2v2) ½
Some
may suggest that this is same as the Lorentz transformation when k is
replaced by i/c where i is
the square root of –1 and c
is the velocity of light. But
the physical significance of this one and the Lorentz are completely
different.
Q2
Are all inertial frames isotropic?
No.
Only the absolute rest frame is isotropic. If the moving frame were
isotropic, then there would be no asynchronicity in that frame for
spatially separated clocks. This would mean that there is no
asynchronicity between that frame and the absolute rest frame. The moving
frame erroneously assumes that it is stationary and isotropic and
concludes again erroneously that his spatially separated clocks are
synchronous.
Q3
From: ES -
I am
reading your paper http://www.easyrelativity.com/contents.pdf and am
totally confused by your meaning in step 12. What am I failing
tounderstand?
"He separates two clocks by adistance x, by moving one of them at a
velocity v." Is the distance x measured inframe K or M? "But since
the frame M isalready moving at V, the absolute velocity of the moving
clock will beV+v." Your formula for addingvelocities seems Galilean. Are
you are intentionally contradicting the rules of special relativity?
Please clarify what you're saying.
"The time taken for theseparation will be
x/v as seen by K, (x/v) f(V) as seen by the clock atx=0 in M and (x/v)
f(V+v) as per the clock that is being separated."
My confusion in the previous two sentences prevents me from understanding
this line clearly !
E S
Answer
Dear ES,
"Is the distance x measured in frame K or M?"
In my & Lorentz's conception, there is only one absolute distance between
two points. When two clocks are getting separated in frame 'M', after the
separation, the absolute distance between them is a unique value say x.
This distance will be correctly measured only by observers in K.
Yes, I am using the normal formulae used for adding velocities before the
advent of relativity. There is no reason to assume they are incorrect. And
for calculting time taken for crossing a distance, I use the formula
"distance divided by velocity (or speed in 1 dimension)"; if a clock runs
a little slow, then that factor is to be accounted for. For the clock
being separated in M, it runs slow by a factor
f(V+v) while clocks which are 'stationary' in M run at a rate f(V); the
actual time taken is x/v (as seen by K which is the standard time) and
clocks which are stationary in M measure this time as (x/v)f(V) and the
clock which is moving in M, measures this time as (x/v)f(V+v). I hope this
clarifies your question. I thank you for your interest and time. -
Chandru
Q4 What is x' in terms of x and t?
Dear E, - The rulers in the moving frame contract by sqrt(1-v^2/c^2).
Because of this any distance measured by these contracted rulers will
appear increased by the factor 1/(sqrt(1-v^2/c^2)).
Further because of the motion, the origin of
M is shifting constantly and at any given time t it will be at vt (as seen
from K). And therefore any spatial point located at x will be separated by
a distance (x-vt) from the origin of M. This distance of (x-vt) will be
measured by the rulers in M as (x-vt)/(sqrt(1-v^2/c^2)) as explained in
the previous paragraph.
Therefore x' = (x-vt)/(sqrt(1-v^2/c^2))
I hope this answers your question.
Thanks and Best Regards
Chandru
The First section is very
interesting.
The reason I say is that, as we all
know, we frequently apply the technique of proof by absurdity, i.e.
assuming something to be true and then try to logically prove if it leads
to contradiction at sometime, and if can we must conclude that our
original assumption to be erroneous.
Applying it to the case of K and K'
frames, for any "normal" human being it is "impossible" that two things
slow down with respect to each other. Here also we are talking about human
observation and since the theory leads to contradiction by observation, we
must conclude that the original assumption must be wrong. If not, we
contradict the method itself.
What do you think of that?
N K
Quick Reply. Hope it makes sense.
Yes. Something is wrong somewhere. If the original assumption was wrong,
then the conclusion is that moving clocks do not run slow. Then why do
they appear as running slow? May be the synchronicity is wrong in the
frame K. But if moving clocks do not run slow it is so easy to achieve a
set of synchronised clocks in frame K by just running one moving clock
across frame K and setting the time in all the clocks of frame K as that
of the moving clock as it passes over eachclock in frame K.
Chandru
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