Re: Gedankenexperiment

PeteK wrote:
I wonder if someone could help me to resolve an apparent contradiction in a "thought experiment" that I was reading ? The experiment involves a train moving at close to light speed relative to a "stationary" observer (I presume that he's standing near the station;-) ). A traveller on the train shines a light directly upwards. The light reflects from a mirror on the ceiling and returns to the light source. The traveller observes that the light travels a distance 2H (where H is the height of the train). To the stationary observer, however the light does not just go up and down; it traces the two sides of a triangle with side D (say) on its round trip. Now, if we assume that the speed of light is constant in all inertial frames, the time for the light to complete its journey as observed by the traveller is 2H/c, but for the stationary observer it is 2D/c, which is greater than the time observed by the traveller (since D must be greater than H if the train is moving). Hence, time goes slower for the moving frame. OK so far.

But note that the "stationary" observer must use _two_ clocks
at different positions to measure the time, while the train observer
can use only one.


Then I modified the experiment as follows:

The stationary observer (let me call him "SO" for the sake brevity) phones the traveller and asks him to shine his light at a point on the ceiling a distance X from the original mirror towards the back of the carriage. The light will again reflect off the ceiling an bounce back to the floor of the carriage. The SO has calculated X so that HE will now observe the light path as straight up and down. The traveller, however will observe that the light will travel along two sides of a triangle, each being longer than H. So we can now conclude that time goes more quickly for the moving frame. That's my apparent contradiction.

Could anyone put me back on the right track; what is wrong with my train of thought ? Sorry, couldn't resist it.

In this case the stationary observer can use only one clock to measure
the time, while the traveller must use _two_ clocks at different positions
to measure the time.

Look at this scenario:

Let's have two synchronized clocks in each of two frames
of reference, let the clocks be a proper distance d from each other
in their respective frames, let the frames move with a relative
speed v.

-d 0 x'
K' frame: B'------A'--> -> v
K frame: A ------B-->
0 d x

There are three events of interest:
E1: A and A' adjacent
E2: A and B' adjacent
E3: B and A' adjacent

Let's calculate what the clocks will show at these events:
E1: A = t1 = 0, A' = t1' = 0 (by fiat, we set the clocks thus)
E2:
In the K' frame, A will be at the position -d at t2' = d/v
LT: t2 = (d/v + (-d)*v/c^2)/sqrt(1 - v^2/c^2)
t2 = (d/v)*sqrt(1 - v^2/c^2)
E3:
In the K frame, A' will be at the position d at t3 = d/v
LT: t3' = (d/v - d*v/c^2)/sqrt(1 - v^2/c^2)
t3' = (d/v)*sqrt(1 - v^2/c^2)

Summing up, the readings of the clocks will be:
E1: A = t1 = 0, A'= t1'= 0
E2: A = t2 =(d/v)*sqrt(1 - v^2/c^2),B'= t2'= d/v
E3: B = t3 = d/v, A'= t3'=(d/v)*sqrt(1 - v^2/c^2)

The symmetry is obvious.
So which clock is running slow or fast relative to which?
The answer depend on how we compare the clocks!
===============================================

In the K-frame, we can measure the rate dt'/dt of
the moving A' clock by comparing the reading of A'
with the _two_ clocks A and B as it passes them:
dt'/dt = (t3' - t1')/(t3 - t1) = sqrt(1 - v^2/c^2)
Conclusion #1:
A' runs slow as measured in the K frame.
=======================================

In the K'-frame, we can measure the rate dt/dt' of
the moving A clock by comparing the reading of A
with the _two_ clocks A' and B' as it passes them:
dt/dt' = (t2 - t1)/(t2' - t1') = sqrt(1 - v^2/c^2)
Conclusion #2;
A runs slow as measured in the K' frame.
=======================================

Conclusion #1 does not contradict conclusion #2.
They state in fact two different things.

But we can draw more conclusions:

We can measure the rate R' at which an observer in K'
will see the co-ordinate time of K run by reading the
clocks A and B as they pass the A' clock:
R' = (t3 - t1)/(t3' - t1') = 1/sqrt(1 - v^2/c^2)
Conclusion #3:
The co-ordinate time of K runs _fast_ as measured in the K'-frame
=================================================================

We can measure the rate R at which an observer in K
will see the co-ordinate time of K' run by reading the
clocks A' and B' as they pass the A clock:
R = (t2' - t1')/(t2 - t1) = 1/sqrt(1 - v^2/c^2)
Conclusion #4:
The co-ordinate time of K' runs _fast_ as measured in the K-frame
=================================================================

There is nothing contradictory between conclusion #3 and #4 either,
as they too state different things.

It is in fact conclusions #1 and #3 and conclusions #2 and #4
respectively that state the same facts.

Paul
.

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