Re: Tree Paradox


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

.

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