Textbooks often propose a thought experiment for checking whether
there is a transverse length contraction (and prove there is not) but
claim that longitudinal length contraction cannot be checked. In fact,
the verification is just as easy. Example:

Two rods, A and B, travel towards one another with uniform relative
velocity:

A2_______A1-...............
................-B1______B2

The proper length of B is L, that of A is L+X, 0XL(gamma-1).

As A1 reaches B2, A1 bumps into B2 and remains attached to B2 so that
A and B continue their existence as one body in a single frame. The
question is: will A2 reach B1? It can be shown that, if there is
length contraction, the answer of B is YES (before the collision, B
sees the length of A shorter than its own), and that of A is NO (A
always sees the length of B shorter than its own).

Indeed, if there is length contraction (B sees A short and A sees B
short), B will measure A's length as (L+X)/gamma, which, given the
value of X, is smaller
than L. This evaluation is performed BEFORE THE COLLISION between A1
and B2, i.e. B has seen, essentially, the following pictu

.......A2_________A1-.......
B1________________________B2

Clearly, this picture implies that B has also seen A2 passing B1.

In contrast, A sees, both before and after the collision, essentially
the following pictu

A2________________________A1
.................-B1________B2

This means that, according to A, B is always too short (its length is
L/gamma before and L after the collision) for B1 to be able to reach
A2.

Pentcho Valev

"Pentcho Valev" wrote in message m...
Textbooks often propose a thought experiment for checking whether
there is a transverse length contraction (and prove there is not) but
claim that longitudinal length contraction cannot be checked. In fact,
the verification is just as easy. Example:

Two rods, A and B, travel towards one another with uniform relative
velocity:

A2_______A1-...............
...............-B1______B2

The proper length of B is L, that of A is L+X, 0XL(gamma-1).

As A1 reaches B2, A1 bumps into B2 and remains attached to B2 so that
A and B continue their existence as one body in a single frame. The
question is: will A2 reach B1? It can be shown that, if there is
length contraction, the answer of B is YES (before the collision, B
sees the length of A shorter than its own),

yes before the collision, B measures A's lenght as (L+X)/gamma

and that of A is NO (A
always sees the length of B shorter than its own).

I have no idea what you mean with the phrase
"that of A is NO"
but before the collision, A measures B's length to be L/gamma


Indeed, if there is length contraction (B sees A short and A sees B
short), B will measure A's length as (L+X)/gamma, which, given the
value of X, is smaller
than L. This evaluation is performed BEFORE THE COLLISION between A1
and B2, i.e. B has seen, essentially, the following pictu

......A2_________A1-.......
B1________________________B2

Clearly, this picture implies that B has also seen A2 passing B1.

In contrast, A sees, both before and after the collision, essentially
the following pictu

A2________________________A1
................-B1________B2

This means that, according to A, B is always too short (its length is
L/gamma before and L after the collision) for B1 to be able to reach
A2.

Ha, this is exactly the same as the famous U-T problem.
Have a close look at:
lenet-ops.be
and
http://users.pandora.be/vdmoortel/di...Detonation.gif
Everything you need to know and understand is right there on a
silver plate.
It is one of the exercises in the book that I recommended in our last
exchange just before you left last year:
http://groups.google.com/groups?&as_...t01.boi.hp.com
Remember?

Good luck!
Dirk Vdm

On Sat, 29 May 2004 03:58:32 -0700, Pentcho Valev wrote:

Textbooks often propose a thought experiment for checking whether there is
a transverse length contraction (and prove there is not) but claim that
longitudinal length contraction cannot be checked. In fact, the
verification is just as easy. Example:

Two rods, A and B, travel towards one another with uniform relative
velocity:

A2_______A1-...............
...............-B1______B2

The proper length of B is L, that of A is L+X, 0XL(gamma-1).

As A1 reaches B2, A1 bumps into B2 and remains attached to B2 so that A
and B continue their existence as one body in a single frame. The question
is: will A2 reach B1? It can be shown that, if there is length
contraction, the answer of B is YES (before the collision, B sees the
length of A shorter than its own), and that of A is NO (A always sees the
length of B shorter than its own).

"It can be shown" -- right, so why don't you show it?

Continuing in the same vein, IT CAN BE SHOWN that if, in B's FoR, A2
passes B1 before A1 strikes B2, then the information needed by A2 in order
to stop before getting to B1 in A's frame of reference CANNOT GET TO A2
IN TIME to let it perform that stop unless that information travels FASTER
than C in A's frame of reference.

But in keeping with your approach of asserting facts without proving them,
I won't prove it. Believe it, though, it's true, and if you try hard
maybe you can even find the proof for yourself.

In other words, A gets squished because it can't stop fast enough.

(Or you can just deny it, if you prefer, and go on asserting that you
have found a contradiction, but you're missing an opportunity to learn
something if you take that course.)


--
To email me directly, take out nospam and put back foobox.

You forgot to account for RS. Go study.

Paul Cardinale