"Daryl McCullough" <stevendaryl3016@yahoo.com> wrote in message
news:d43fcl01n2c@drn.newsguy.com...
>
> Aleksandar,
>
> This is very similar to a standard exercise in Special Relativity
> called the "Pole vaulter paradox".
>

Indeed.

> The resolution is that you can't just use length
> contraction, you have to also take into account
> the timing of events.
>

It is not resolution, but to do you justice, I'll explain you in detail
below.


> In frame K, we have knives at x=0 and at x=1.
> In frame K', we have a rod that extends from x'=0 to
> x'=1. Here's a history of events (not necessarily
> in chronological order)
>

I'll just skip through your diagrams with events, they're ok. Here's where
you make mistake:

>
> Now look at things from the point of view of frame K'
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>
> using the Lorentz tranformations. For our case they are
> (in units where c=1)
>
>     x' = 2(x - .866 t)
>     t' = 2(t - .866 x)
>

That is not the point of view from the K', it's still from K. Your equations

>     x' = 2(x - .866 t)
>     t' = 2(t - .866 x)

are used in K to calculate what are corresponding coordinates of K'. But
this is viewpoint from K. In K' you know what primed coordinates are, and
you use them in *symmetrical* Lorentz transformation to calculate
corresponding coordinates of K, as they are viewed from K'. You didn't use
symmetrical transformation at all, that's why everything's fine in your
analysis. What you actually did is draw diagrams for events for non-primed
and primed coordinates as they are both viewed from K only.

To repeat one more time (although this is not a subtle point that I am
making) you know proper length and position of the rod when you are in K'.
You don't calculate that. But you have no idea what are corresponding
non-primed coordinates in K, and must use the following formulas to
calculate them

x = 2(x' + .866t')
t = 2(t' + .866x')

All that matters from your diagrams is on the fig.1 left on page 4, and you
didn't present the viewpoint from K' at all, which is on the same fig.1
right in  http://www.masstheory.org/lorentz.pdf

The rod will travel unaffected as observed from K, but the poor rod will be
cut in half when observed from K'. C'est la vie... as Scotsmen in New York
say (those who speak French :).

Aleksandar