Tree Paradox
eleaticus
grows more rings at a regular, relatively frequent rate. It and its rings
are perfectly symmetrical around its long axis, and it is headed in the
direction of its long axis straight onto the cutting end of a large-enough
ax head that will perfectly halve the tree along the long axis, killing the
tree and stopping the ring growth.
Let there be a 'station' at rest with respect to the ax-head observers to
which the tree communicates its number of rings as it narrowly passes the
station.
The tree, having previously been in the unchanging neighborhood and with
perfect knowledge of itself knows that it will grow exactly 120 more rings
before it gets chopped and dies.
Relative to the station the tree is moving at v~.94281c, so gamma=3 in the
station's view.
Therefore the station Relativists know darn well that time has slowed for
the tree to just a third of the station time.
The number of new rings at chop-time, thus, must be only 40.
Relative to the tree another observer is approaching from the opposite
direction at v~.96824c relative to the tree, so gamma=4 with respect to the
tree in its view.
The number of new rings at chop-time, thus, must be only 30.
Relative to the tree another observer is receding from the tree along the
tree's vector so that the net is v~.866c, so gamma=2 with respect to the
tree in its view.
The number of new rings at chop-time, thus, must be only 60.
As the tree commits hari-kari on the ax the ax begins the emission of such
strong lights in relevant directions that the two moving observers can
accurately see the tree rings and count them, and, of course, the station is
busy transmitting the number of rings the tree had before it began the
growth of the 120 new rings.
So, question, these figures being straight Special Relativity dogma, how
many new rings are counted/observed by each of:
A. the ax's observers
B. the approaching observer
C. the receding observer
ROFFLMFAO!
Surely you don't think that Special Relativity's time dogma isn't complete
nonsense? Only one answer is possible because there was only one tree doing
only one time-thing and the tree knew that answer all along.
The paradox isn't the contrast between the nonsensical SR time dilation
effects, it is that otherwise(?) intelligent people could maintain their
cult beliefs once they are released from thrall to their professors.
If there is a time effect of motion it can't be Special Relativity's.
See also "Einstein's dilation derivation: ROFFLMFAO!"
That article will show you, using Einstein's own method, how silly it is to
reach any conclusion about time effects based on Einstein's
t'=gamma(t-vx/cc).
eleaticus
ee-lee-AT-i-cus
RP
eleaticus wrote:
> Let there be a space-going tree, the gedanken-virtue of which is that it
> grows more rings at a regular, relatively frequent rate. It and its rings
> are perfectly symmetrical around its long axis, and it is headed in the
> direction of its long axis straight onto the cutting end of a large-enough
> ax head that will perfectly halve the tree along the long axis, killing the
> tree and stopping the ring growth.
The tree is redundant, just use a clock that stops as it passes some
marker.
>
> Let there be a 'station' at rest with respect to the ax-head observers to
> which the tree communicates its number of rings as it narrowly passes the
> station.
Clock reading passed to bystanders.
>
> The tree, having previously been in the unchanging neighborhood and with
> perfect knowledge of itself knows that it will grow exactly 120 more rings
> before it gets chopped and dies.
Redundant, all observers will predict the same number of rings.
>
> Relative to the station the tree is moving at v~.94281c, so gamma=3 in the
> station's view.
>
> Therefore the station Relativists know darn well that time has slowed for
> the tree to just a third of the station time.
>
> The number of new rings at chop-time, thus, must be only 40
No it mustn't. Wrt the station observer the distance traversed is 3
times greater, thus offseting the slowed growth rate exactly and
allowing time for 120 rings to grow during the tree's travel at v.
> Relative to the tree another observer is approaching from the opposite
> direction at v~.96824c relative to the tree, so gamma=4 with respect to the
> tree in its view.
>
> The number of new rings at chop-time, thus, must be only 30.
Again, you must account for differences in space-like displacement
measured by different observers. They won't agree on either the time
required, nor on the distance travelled, but in each case there will be
precisely time for 120 rings to grow.
> Relative to the tree another observer is receding from the tree along the
> tree's vector so that the net is v~.866c, so gamma=2 with respect to the
> tree in its view.
>
> The number of new rings at chop-time, thus, must be only 60.
Same as above.
>
> As the tree commits hari-kari on the ax the ax begins the emission of such
> strong lights in relevant directions that the two moving observers can
> accurately see the tree rings and count them, and, of course, the station is
> busy transmitting the number of rings the tree had before it began the
> growth of the 120 new rings.
>
> So, question, these figures being straight Special Relativity dogma, how
> many new rings are counted/observed by each of:
>
> A. the ax's observers
120
> B. the approaching observer
120
> C. the receding observer
120
Richard Perry
eleaticus
"RP" <no_mail...@yahoo.com> wrote in message
news:1145223124.2...@v46g2000cwv.googlegroups.com...
>
> eleaticus wrote:
> > The tree, having previously been in the unchanging neighborhood and with
perfect knowledge of itself knows that it will grow exactly 120 more rings
> Redundant, all observers will predict the same number of rings.
> > Relative to the station the tree is moving at v~.94281c, so gamma=3 in
the station's view.
> > Therefore the station Relativists know darn well that time has slowed
for the tree to just a third of the station time.
> > The number of new rings at chop-time, thus, must be only 40
> No it mustn't. Wrt the station observer the distance traversed is 3
> times greater, thus offseting the slowed growth rate exactly and
> allowing time for 120 rings to grow during the tree's travel at v.
In general, and in particular, I very much appreciate your response.
In the particular, because you bring up the two views of the distance, which
I make much of in a post elsewhere and which will show up here in many
segments..
But, you do realize that you are saying that if the tree synchronizes clocks
with the station which is synchronized with the ax, that the tree's clock
will show the same time as the ax's clock upon the tree's arrival at the ax?
And that an initially synchronized travelling twin will arrive at a site
that has its clock synchronized with his stationary twin's and find that his
clock reads the same as the site's and his twin's, and that when he returns,
completing the round trip, his clock will read the same as his twin's?
And you do realize that you come close to saying the whole gamma bit is
wrong?
That if light travels at c and you and gamma are correct then any distance
it travels should be contracted in its view to at least close to zero so, by
your logic, in the statioanry observer's view the light has to travel for a
distance damn near infinitely greater than the light 'thinks'? Which would
make c damn near infinite?
The least that could be said, if you are right in saying the stationary view
is that the moving object has to travel the stationary object's distance, is
that if you let light travel without its own space-contraction then you have
to discard gamma as universal law.
Only what the ax sees should be counted in what the ax sees. Only what the
tree
sees should be counted in what the tree sees.
eleaticus
ee-lee-AT-i-cus
RP
eleaticus wrote:
> "RP" <no_mail...@yahoo.com> wrote in message
> news:1145223124.2...@v46g2000cwv.googlegroups.com...
> >
> > eleaticus wrote:
> > > The tree, having previously been in the unchanging neighborhood and with
> perfect knowledge of itself knows that it will grow exactly 120 more rings
> before it gets chopped <by the 'stationary' ax> and dies.
>
> > Redundant, all observers will predict the same number of rings.
>
> > > Relative to the station the tree is moving at v~.94281c, so gamma=3 in
> the station's view.
>
> > > Therefore the station Relativists know darn well that time has slowed
> for the tree to just a third of the station time.
>
> > > The number of new rings at chop-time, thus, must be only 40
>
> > No it mustn't. Wrt the station observer the distance traversed is 3
> > times greater, thus offseting the slowed growth rate exactly and
> > allowing time for 120 rings to grow during the tree's travel at v.
>
> In general, and in particular, I very much appreciate your response.
>
> In the particular, because you bring up the two views of the distance, which
> I make much of in a post elsewhere and which will show up here in many
> segments..
>
> But, you do realize that you are saying that if the tree synchronizes clocks
> with the station which is synchronized with the ax, that the tree's clock
> will show the same time as the ax's clock upon the tree's arrival at the ax?
Observers in rectilinear motion wrt each other cannot syncronize their
clocks wrt both frames. An ability to do so would automatically
require t=t', implying Galilean Relativity. Even when syncronized wrt
one of the frames, there would be no contradiction in the fact that the
two clocks will agree upon their passing one another. Wrt the frame in
which they are not syncronized there willl now be a disagreement about
their closing velocity wrt each other, which will offset the difference
in displacement as measured by the two observers. You've only replaced
ticking rate differences with velocity differences.
We can take it one step further and recalibrate the measuring sticks so
that wrt the tree's frame both itself and the ax will measure the same
time-like and space-like displacements. Now there will be no offsets,
and obviously no disagreements. There is also no contradiction raised
by this arbitrary recalibration. It would be a fallacious conclusion
that "the clocks both agree and disagree", since the former occurs in
one context, and the latter in another".
> And that an initially synchronized travelling twin will arrive at a site
> that has its clock synchronized with his stationary twin's and find that his
> clock reads the same as the site's and his twin's, and that when he returns,
> completing the round trip, his clock will read the same as his twin's?
It will indeed, but he will nevertheless be younger than his twin
because he cannot recalibrate his own physiological processes. There is
no contradiction. The traveling twins clock will age less than what
it's reading implies, by definition of *recalibration*.
>
> And you do realize that you come close to saying the whole gamma bit is
> wrong?
No.
>
> That if light travels at c and you and gamma are correct then any distance
> it travels should be contracted in its view to at least close to zero so, by
> your logic, in the statioanry observer's view the light has to travel for a
> distance damn near infinitely greater than the light 'thinks'? Which would
> make c damn near infinite?
c is infinitely greater than zero, which is not the same as saying that
c equals infinity.
> The least that could be said, if you are right in saying the stationary view
> is that the moving object has to travel the stationary object's distance, is
> that if you let light travel without its own space-contraction then you have
> to discard gamma as universal law.
No observer can travel at c because of that law. By assuming a frame
of reference moving at c wrt another frame you are applying a non-real
initial condition. Might as well assume that it was a pixie and that it
was moving at 2000c.
>
> Only what the ax sees should be counted in what the ax sees. Only what the
> tree
> sees should be counted in what the tree sees.
And they will both see the same invariants.
Richard Perry
Spoonfed
You say gamma = 3 in the first case.
So t1' = 3(t1-v x1 /cc);
t2' = 3(t2-v x2 /cc);
x1, t1 represents the position where and time when the axe strikes the
top of the tree
x2, t2 represents the position where and time when the axe gets to the
bottom of the tree.
t1', and t2' represent the time passed for the tree and will be
represented by the 120 rings.
The two events are timelike, meaning they will appear to have the
shortest separation in the reference frame where the two events occur
at the same point in space--the reference frame of the tip of the axe.
A single set of observers would see the tree coming through the scene,
splitting itself on the stationary axe.
A few sets of observers (those with velocity between the velocity of
the tree and the axe) would see the axe and the tree approaching one
another. This group would be greatly outnumbered by the others.
Most possible reference frames are outside those finite limits; either
moving down, faster than the axe, or up, faster than the tree.
Observers moving down, faster than the axe, see the axe receding while
the tree overtakes the axe.
Observers moving up, faster than the tree, see the tree receding as the
axe overtakes it.
Either way, both the time separating the events, and the space
separating the events is increased.
Galileo could have told you that the space separating the events was
much greater if not viewed from the frame of the axe.
eleaticus
"Spoonfed" <jonatha...@spoonfedrelativity.com> wrote in message
news:1145306157....@v46g2000cwv.googlegroups.com...
> t'=gamma(t-vx/cc)
>
> You say gamma = 3 in the first case.
>
> So t1' = 3(t1-v x1 /cc);
> t2' = 3(t2-v x2 /cc);
>
> x1, t1 represents the position where and time when the axe strikes the
> top of the tree
>
> x2, t2 represents the position where and time when the axe gets to the
> bottom of the tree.
>
> t1', and t2' represent the time passed for the tree and will be
> represented by the 120 rings.
>
> The two events are timelike, meaning they will appear to have the
> shortest separation in the reference frame where the two events occur
> at the same point in space--the reference frame of the tip of the axe.
I don't know whether you are presenting new information that is not directly
responsive to my presentation, or are responding.
If the latter, you misunderstood the setup. The time involved was for the
tree to pass from the station to the ax.
(I too spell the word 'axe' but the cursed spell checker kept insisting I
was wrong, so I finally gave up.)
In the equations you gave, you misrepresented the situation you specified in
your verbal text: 'same point in space'. The correct equations for the
relevant ax location, per that setup, are for just one x, the cutting-edge
location.
So t1' = 3(t1-v x /cc);
t2' = 3(t2-v x /cc);
(BTW, that is certainly an excellent way to present the v x /cc.)
And T' = t2' - t1' = 3(t2-t1) = 3T and T' /3 = T, which is NOT dilation, it
is contraction, showing the number of moving system units as three times the
stationary system's.
Which, of course, is what I pointed out in one part of 'Einstein's dilation
derivation: ROLMFAO!"
And if you wish to posit a stationary observer who can NOT measure the whole
moving object at the one time (which certainly violates the gedanken freedom
to posit observers whereever needed, whenever):
x1' = 3(x1-v t1 /cc);
x2' = 3(x2-v t2 /cc);
L' = x1' - x2' <> 3L, and L' /3 <> L, which is to say, hardly the standard
SR dogma/concept of shrinkage with gamma=3.
So, we must give SR gedankenen an instantaneous measurement of moving
lengths or give up SR dogma about the amount of contraction.
But given the obvious intelligent and competent set of observers, one at
each clock perhaps or 'just' receiving useful reports at one location for
use in calculation, so a moving length can be 'measured at one instan't:
x1' = 3(x1-vt /cc);
x2' = 3(x2-vt /cc);
And we have the standard dogma: L' = 3L, and L' /3 = L, showing the number
of moving system units as three times the number of stationary system units.
So, it is by positing a competent observer (or system thereof) that confirms
SR spatial contraction dogma vis a vis x'=g(x-vt).
Back to your equations, which are written to demand an incompetent observer
(or system thereof):
t1' = 3(t1-v x1 /cc);
t2' = 3(t2-v x2 /cc);
T' = t2' - t1' = 3 ( t2- t1 - v x2 /cc + v x1/cc)
= 3T - 3 (v x2 - v x1) /cc
So, insisting that the system of observation is incompetent, we do not get
straight temporal contraction,.
But that setup is not correct for a competent observer (system). There is no
necessity to not posit, for one example, a transmission by the stationary
system location at both t2 and t1 as to the time of the event at that
position, which data can be collected at leisure and computed, with the
obvious anywhere-you-wish x:
So t1' = 3(t1-v x /cc);
t2' = 3(t2-v x /cc);
And T' = t2' - t1' = 3(t2-t1) = 3T and T' /3 = T, which is NOT dilation, it
is contraction, showing the number of moving system units as three times the
stationary system's.
Yes, there is only an insistence on incompetency to support insistence on
the two-locations formulae.
Which, of course, again, is what I pointed out in one part of 'Einstein's
dilation derivation: ROFFLMFAO!"
Thanks for the paragraphs about the relative velocity 'who sees what'.
eleaticus
ee-lee-AT-i-cus
Dirk Van de moortel
"Spoonfed" <jonatha...@spoonfedrelativity.com> wrote in message news:1145306157....@v46g2000cwv.googlegroups.com...
>
> You say gamma = 3 in the first case.
>
> So t1' = 3(t1-v x1 /cc);
> t2' = 3(t2-v x2 /cc);
>
> x1, t1 represents the position where and time when the axe strikes the
> top of the tree
>
> x2, t2 represents the position where and time when the axe gets to the
> bottom of the tree.
>
> t1', and t2' represent the time passed for the tree and will be
> represented by the 120 rings.
You are talking in terms of events. Bad idea.
Eleaticus (Oren Webster) doesn't even understand what the
variables mean in the equations he throws at you or vice versa.
Before you try to explain something to this malicious retard,
you might enjoy finding out how he feels about this:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
and then about these:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/VectorFiasco.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Schzoid.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/ShowFormula.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Periodically.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CorruptIdiocy.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/.html
Otherwise, enjoy ;-)
Dirk Vdm
Hexenmeister
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:_C81g.379284$lV7.11...@phobos.telenet-ops.be...
|
| "Spoonfed" <jonatha...@spoonfedrelativity.com> wrote in message news:1145306157....@v46g2000cwv.googlegroups.com...
| > t'=gamma(t-vx/cc)
| >
| > You say gamma = 3 in the first case.
| >
| > So t1' = 3(t1-v x1 /cc);
| > t2' = 3(t2-v x2 /cc);
| >
| > x1, t1 represents the position where and time when the axe strikes the
| > top of the tree
| >
| > x2, t2 represents the position where and time when the axe gets to the
| > bottom of the tree.
| >
| > t1', and t2' represent the time passed for the tree and will be
| > represented by the 120 rings.
|
| You are talking in terms of events. Bad idea.
Thanks. :-)
http://www.androcles01.pwp.blueyonder.co.uk/Dork/badevents.htm
Androcles.
Spoonfed
run slow) or you move your eye from clock to clock nearest you (you see
the clocks run fast.) Your tree-ring and axe example is a version of
the latter type, except the surfaces of the clock faces merge one to
another in a continuum.
It is exactly the same situtation in reverse. If the tree watched you
and JUST you, it would see you moving in slow motion. If it watched
you and all the hypothetical observers in your frame as they made their
nearest approach, it would see the set seeming to age fast.
This time contraction/dilation symmetry you've been ROFFLYFAO about is
actually pretty essential to the argument. If this symmetry weren't
there, then it wouldn't make any sense.
eleaticus
sauce for the system B goose/
eleaticus
ee-lee-AT-i-cus
Hexenmeister
"Spoonfed" <jonatha...@spoonfedrelativity.com> wrote in message news:1145767484.7...@j33g2000cwa.googlegroups.com...
| It just depends on whether you watch a single clock (you see that clock
| run slow) or you move your eye from clock to clock nearest you (you see
| the clocks run fast.) Your tree-ring and axe example is a version of
| the latter type, except the surfaces of the clock faces merge one to
| another in a continuum.
|
| It is exactly the same situtation in reverse. If the tree watched you
Where are the eyes of trees situtaterated? Do trees have eight eyes like spiders?
You are droolin', doolin.
Androcles.