Natural Science Forum / Physics / Relativity / July 2009

"Pole and Barn for Dummies"

I was challenged by "Mike" to write this. It is as if he is forcing
Chateau Latour 1982 down my throat.

I don't know why Mike thought this problem deep enough for a
challenge. There are many treatments already published, and all I have
examined (outside of this newsgroup!) are correct. So I have nothing
new really to contribute  --

-- except for a brief discussion on how to tell which frame has the
primes in the Lorentz Transformation. This detail has tripped up many
a student, including our voluble friend with the lion here in s.p.r.

The reader bored with the Pole and Bar "Paradox" may wish to read
instead a short dissertation of mine on the subject "How to tell which
frame gets the primes"  It is published as

"PrimedFrames.txt"

in my shared online directory at

http://tinyurl.com/greenba.
----------------------
A word about writing for "Dummies" for the two or three of you
thinking of taking offense.  (You know who you are.)    Any bookstore
will show you hundreds of volumes titled in this fashion. The
originator of the series had trouble finding a publisher, because all
the stupid publishers thought no one would buy them.

On the contrary, who would buy "Architcture for Experts"?  Not many.
You would be afraid you would not understand. If it were for dummies,
on the other hand, you would be reassured that there are no pre-
requisites, and that you WILL understand. I have learned a lot from
books in that series.
--------------------------

The Pole and Barn question:

This is about the Lorentz Contraction, which says that a straight
wooden pole which is 15 m long when at rest, will be only 9.9 m long
when moving at 0.75 c to the right, the length being measured in a
frame of reference B such as a barn 10 m long, front door (left) to
back door (right), fixed to the earth.

Thus the speeding pole will momentarily fit into the barn while in
flight through it. If one slammed the doors of the barn at exactly the
right moment, the pole would be trapped. But to avoid flying splinters
that would frighten the horses, one quickly opens the doors again in
time for the pole to escape undamaged. Precision work with nanosecond
timing!

The curiosity arises when we regard the pole to be a rest and the barn
is moving leftward at 0.75 c.  Then the barn is much shorter than the
pole when measured in the pole frame of reference P.  How can one
flick the doors of the barn as in this scenario?

The answer is that, while in frame B the two doors are operated
simultaneously, events that are simultaneous in one frame will not be
simultaneous in another frame if there is some distance between the
events and the second frame is moving with respect to the first in the
line of the two events.

In the barn frame of reference B, the first scenario above is correct.
But in the pole frame P, the two events are not simultaneous.  In fact
the moving back door is flicked (closed and opened) just before it
hits the right end of the pole, and the moving front door is flicked
just after it clears the left end of the pole.
---------------------------------------
To show that this is really what the Lorentz Transformation predicts,
we will show the LT equations and plug in reasonable numbers.

For the first scenario, we let the pole at rest have ends at two x
coordinates, values 0 and 15 m  in frame P. Then we set the pole in
motion at 0.75c and ask what is the length now in frame B?  Lorentz
and Einstein teach us that given the P-values 15 and 0, we can relate
them to B-values of time and space by

15 = gamma(x1 - v * t1) for the right end
0 = gamma( x2 - v * t2) for the left end

where x's are the coordinates of the pole in B at B-time t of the
observation, v = 0.75c, and
gamma = 1/sqrt(1 - 0.75 ^2) = 1.51.

So, how long is the pole in B?  The coordinate x is changing with
time.  To measure the length of the moving pole in terms of
coordinates, you have to measure the ends at the same time.  So t1 =
t2.

Subtract the second equation from the first, and you get

15 = gamma (x1-x2),
(x1-x2) = 15/gamma,
or (x1-x2) = 9.9 meters

which once again yields the Lorentz Contraction.

(I have purposely avoided the notation x' and t' at this stage, for
pedagogical reasons. The words control the meanings of x and t.)

-----------------------------------
Now for the pole's point of view  The pole is 15 m long. The barn is
flying to the left at 0.75 c.

The barn's doors are 10 meters apart in B.
Let the doors have coordinates in B of 0 and 10 in meters at rest.
What are there coordinates in P?

10 = gamma(x1 + v  t1) for the rear door of the barn
0 = gamma(x2 + v * t2) for the front door of the barn

where x is the door coordinate in P at P-time t,  v = -0.75c
gamma = 1.51 (again)

We can ask two questions, each of which has to be treated on its own.

(1) How long is the barn, between doors?

Again, to measure the length of a moving body, you have to note the x
coordinates at the same time.  So t1=t2, and

10 = gamma (x1-x2)
(x1-x2) = 10/gamma = 6.62 m

just as we derived the pole contraction. And yes, the pole is more
than twice the length of the barn, in P.

(2) When in P-time do the doors flick? There is more than one way to
get the answer.

Simplest. We know the B spatial coordinates of the barn doors and the
B-time of their flicking.  We want to find the P-time of flicking of
each door.

There are four equations in the LT with one spatial dimension, one for
x and one for t on the left-hand side, and the  corresponding pair for
the inverse transformation.
.
For esthetic reasons, let us choose units in which c=1. Without yet
specifying which frame gets the primes and which has naked
coordinates, the equations are

(1) x' = gamma(x - vt), the only one we have used so far,
(2) t' = gamma(t - vx)
(3) x = gamma(x' + vt' )
(4) t = gamma(t' + vx' )

(In MKS units, the term xv becomes xv/cc.)

v is defined as the velocity of the primed system w.r.t.  the naked
system, and is positive in the direction of increasing x and x' . The
function gamma is still 1/sqrt(1-v^2).
.
We can get the answer to question 2 quickest from (4),  the Inverse LT
time equation of the LT. But we have to decide which frame gets the
primes.

Pay close attention, children, for this is where many people fumble
many relativity problems!

The answer comes from elementary physics.  In slow motion, gamma
approaches 1, and equations (1) and (3) look familiar.

Originally the LT applied to events, not to objects. But if we imagine
a pin stuck into an object, the pin represents a continuous stream of
events in time. The pin has coordinates in both frames, but in only
one frame is it at rest.

With gamma nearly 1, equation (1), x' = x - vt, is classical.  It says
how a pin fixed to an object at primed point x' on an object at rest
moves with time with respect to the naked frame.  The primes go to
coordinates in the REST FRAME of the object for both (1) and (2). If v
is positive, the object moves to the right. Primes go to frame P in
the first scenario, since pole is moving in B and P is its rest frame.

In the inverse transform (3),  x = x' + vt' says that a pin fixed to
an object at point x at rest in the naked frame is moving  w.r.t. the
primed frame. This is the rule that applies in the second scenario.
The primes go with the frame IN WHICH THE OBJECT IS MOVING for both
(3) and (4). If v is positive, the object goes to the left. In the
second scenario, the barn is moving, and the primes go with frame P
again.

I have found this rule very useful in complicated scenarios.

So in question 2, concerning the P frame, the pole is at rest and the
barn is moving.  Therefore the P-frame get the naked coordinates.. The
primed coordinates specify the position and times of the door events
in the B-frame.

The two door flicks have coordinates in both systems. In the P frame,
as the barn moves with v = 0.75c, it encounters the resting pole first
with open doors.  The first flick event occurs at the back door just
as the door approaches the right end of the pole.

Thus In P: x1= 15, x2 = 0,  t1 and t2 unknown

In B: x1' = 10 , x2'  = 0,  t1' = t2' = 0

v = -0.75c

Applying equation 4, we have

t1 = gamma( t1' + vx1'/cc)
   = 1.51( 0 - 0.75*10/c)
   = -1.51* 0.75 * 10 / (3*10^8) sec
   = -37.75 nanoseconds

t2 = gamma( t2' + vx2'/cc)
   = 1.51( 0 - 0.75* 0/c)
   = 0

t1 = -37.75 nanoseconds means:

That the flick at the back door occurred earlier than the flick at the
front door, as expected.

-----------------------------------------------------------------------------

Summary:  The long pole gets through the short moving barn with its
flicking doors because the back door flicks 37.2 nanoseconds earlier
than the front door flicks.

The key to these puzzles is knowing which frame has the primes and
which doesn't.

Having this rule would have saved Androcles the embarrassment of
having predicted the  Einstein Dilation.

Uncle Ben