Re: SRian 'events' versus SRian dilation: ROFFLMFAO!

eleaticus wrote:
Time dilation can be summarized by the common T' = T/gamma, where T is the
'stationary' system elapsed time and T' is the purported 'moving' system
time elapsed time between two events.

Because gamma (=g) is always greater than or equal to one, that equation
says it takes fewer moving system time units to measure the elapsed time
that it does stationary system time units.

In "Tree Paradox" I applied that relationship directly to illlustrate the
poverty of SR's (Special Relativity's) time dilation concept.

As expected, because there almost alway is a contradiction between the BEER
(Basic Equations of Einstein's Relativity) and SR dogma, a polite post was
made by an spr reader that presented without explicitly saying so that one
should examine time dilation in terms of 'events'.

Well, let's examine time dilation (and spatial contraction) in terms of
those events.

The respondent presented one of the BEER and discussed a "time-like"
situation, one where the spatial distances involved are small compared to
the time 'distances'.

In particular, he let x1 and t1 describe the 'event' "axe begins cutting a
tree at one end (down the middle)" and x2 and t2 as the event "axe has cut
all the way down the tree length".

Using the BEER's t'=g(t-vx/cc), he gave:

t1' = g(t1 - v x1 /cc)
t2' = g(t2 - v x2 /cc)

Which, of course, are the wrong equations for that setup, what with the x
values both being the stationary system x-value for the cutting edge of the
axe. (You'd undertand better if you knew the axe just 'sat' there as the
tree ran into it.)

So your "set up" is that the unprimed frame is the axe frame,
while the primed frame is the tree frame.
OK.

Hence, the two events are at a single location at different times, whish is
how he described the time-like situation:

... in the axe frame.

In the tree-frame, the events are at different positions.

t1' = g(t1 - v x1 /cc)
t2' = g(t2 - v x1 /cc).

Thus, the time-like moving system elapsed time is:

T' = t2' - t1' = g(t2- v x1 /cc) - g(t1 - v x1 /cc),

and that reduces to:

T' = g(t2 - t1) = gT,

Which is quite correct, and which _is_ "the SR-dogmatic dilation."

But you don't understand why, do you?

Suppose there are clocks at each end of the tree.
These clocks are synchronized, and show the coordinate time
in the tree-frame. Let us call the clock on the entry side of the tree
A,
and the clock at the other end B.
Suppose also that there is a clock on the axe. Let us call it C.
At event #1, these clocks will show:
Clock A = t1'
Clock C = t1
At event #2, these clocks will show:
Clock B = t2'
Clock C = t2

It does obviously not matter if you consider one or the other or both
of the frames for "moving". As long as the relative speed between
the frames are v, SR must obviously predict exactly the same for
what the clocks will show when they are adjacent.
If it didn't, SR would be inconsistent.

This can be illustrated by letting the tree be "stationary",
and the axe be moving:
(assuming the axe is moving in the postive x' direction)
t1 = g(t1' - v x1' /cc)
t2 = g(t2' - v x2' /cc)

t2 - t1 = g((t2' - t1' ) - v (x2' - x1' )/cc)
Now we have:
(x2' - x1' ) = v (t2' - t1' ) (length = speed * time)
t2 - t1 = g( (t2' - t1' ) (1 - v v/cc)) = (t2' - t1' )/g

T = T'/g or T' = gT

It is the very same scenario regardless of which of
the frames you consider "stationary", so if SR didn't
predict the same result, it would be self contradictory.

just the opposite of the SR-dogmatic dilation: T'=T/g..

The SR-dogmatic dilation in your scenario is T = T'/g.

Your confusion is that you don't understand what is meant
by the phrase "a moving clock is running slow".

Look at what your scenario is.
You are comparing one clock C, to two _different_ clocks A and B.
T is the difference between two readings _of the same clock_,
while T' is the difference between two readings on _different_ clocks.
T is a proper time, T' is not.

In the phrase above, the "moving clock" is the single clock,
while the "stationary frame" is the frame where the _two_
co-ordinate clocks are stationonary.


For this time-like-events situations there is, thus, by SR's very own BEER,
time contraction. It takes more moving system time units than it does
stationary system units to measure the elapsed time between events.

(Need I say "ROFFLMFAO!"?)

But what about "space-like" events intervals?

If the interval between two events is space-like, then it is
impossible for an object to be present at both events.
That's what space-like means.

Here, his equations are appropriate (as much as any application of the
ridiculous BEER):

t1' = g(t1 - v x1 /cc)
t2' = g(t2 - v x2 /cc)

Perhaps the time it takes for an object to travel from one end of a road to
the other end.

Did you mention ROFFLMFAO! :-)


T' = t2' - t1' = g(t2- v x2 /cc) - g(t1 - v x1 /cc),

which gives us

T' = g(t2 - t1 - v x2/cc + v x1 /cc) = gT - g(v x2 /cc - v x1 /cc)

T' + g(v x2 /cc - v x1 /cc) = gT.

Note that in the standard setup x2 > x1 when v>0 and the left side of the
equation actually is greater than T'

When v<0 x2 is still greater than x1 and the v x2 - v x1 difference is still
positive.

Hence, the space-like setup results in a computation of even more time
contraction than does the time-like situation.

ROFFLMFAO!

Indeed.
You have obviously no clue whatsoever. :-)

We have two events, E1 and E2 separated by a space-like interval.
The co-ordinates of these events are:
E1: (x1, t1) in the unprimed frame and
(x1', t1') in the primed frame.
E2: (x2, t2) in the unprimed frame and
(x2', t2' ) in the primed frame.

Of course the temporal intervals between the events
T = (t2-t1) and T' = (t2'-t1') in the respective frames
do not depend on which of the frames you consider
stationary or moving. So how can you imagine that
you in the general case can say anything about
which of them should be shorter than the other?

You can insert these coordinates in either of these
sets of equations:

t1' = g(t1 - v x1 /cc)
x1' =g(x1 - v t1)
t2' = g(t2 - v x2 /cc)
x2' =g(x2 - v t2)

t1 = g(t1' + v x1' /cc)
x1 =g(x1' + v t1')
t2 = g(t2' + v x2' /cc)
x2 =g(x2' + v t2')

Both sets will obviously give the same result.

Please refer to "Einstein's dilation derivation: ROFFLMFAO!" showing how
Einstein's own (1905) derivation of the dilation effect from t'=g(t-vx/cc)
also can be used to derive the obvious time contraction effect from
t'=g(t-vx/cc).

eleaticus
ee-lee-AT-i-cus

Paul

.

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