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Re: relativity problem

On Tue, 2 Apr 1996, Rauber, Joel Phys wrote:
...

unprimed frame = spaceship frame

primed frame = earth frame,
which moves at speed .95 in the negative x direction relative to the rocket

assume the two frames are coincident at t=t'=0 and the origin of either
frame is the location of the tail of the rocket at this time.

I'll measure time in units of meters (c=1) and convert to SI units at the
end of the calculation.

Event A = emission of light pulse at the tail of the rocket
Event B = detection of light pulse at the head of the rocket

We will use the idea of the invariant interval to solve the problem; i.e.
the interval between A and B is an invariant.

* denotes multiplication
^ denotes eponentiation
_ denotes a subscript

Interval_AB^2 = delta x ^2- delta t ^2 = delta x'^2 - delta t'^2 = 0

delta x = 200 meters , delta t = 200 meters , as discussed in
other postings

The Lorentz transformation says that:

delta x' = gamma * .95*delta t + gamma * delta x = gamma * (1.95)* 200
= 3.203*1.95*200=1249 meters

Note: gamma=1/sqrt(1-.95^2)= 3.203
substitute this into the expression for the interval and solve for delta t'
and you get

delta t' = 1249 meters (or just realize the interval is light-like)

which upon dividing by the speed of light in SI units gives

delta t' = 4.16 micro seconds.

In short I think the analysis in the original posting (second method) was
correct and uses more physical based reasoning then the above, but I wanted
to give the more mathematically based solution; and therefore the stated
answer in the book is incorrect.

I believe that you can not just use a time dilation factor (which the book
apparently did!!) because both dilation and and contraction effects are
occuring in this problem.

Joel Rauber
rauberj@mg.sdstate.edu

My apologies for my original misconstruction of Part (b) -- I agree with
the above post, and the other different methods of arriving at the same
thing. Both observers see a nonzero spatial separation of emitter and
receiver in the direction of motion of the spacecraft, and this has to
enter the calculation of the time in Earth's frame.

To clarify the difference, the book answer to Part (b) solves a different
problem: Suppose a light beam is emitted from floor to ceiling in a
spacecraft (i. e., orthogonal to the direction of motion Earth observes
for the spacecraft). If the separation between floor and ceiling measured
in the spacecraft frame is 200 m (a large spacecraft!), compute
a) the time between emission and absorption as measured on the spacecraft;
b) the time between the same two events as determined in Earth's frame.

In this second problem, the further calculation of c*T21 would then give
640.5 m as the spatial separation of the events of emission and absorption
as observed from Earth.

A. R. Marlow E-MAIL: marlow@beta.loyno.edu
Department of Physics PHONE: (504) 865 3647 (Office)
Loyola University 865 2245 (Home)
New Orleans, LA 70118 FAX: (504) 865 2453