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*From*: "Rauber, Joel Phys" <RAUBERJ@mg.sdstate.edu>*Date*: Tue, 02 Apr 96 15:29:00 PST

Here is my solution to part (b) of the posted problem which was as follows:

______________________________

A spaceship has a length of 200m in its own reference

frame. It is traveling at 0.95c relative to Earth.

Suppose that the tail of the spaceship emits a flash

of light. (a)In the reference frame of the spaceship,

how long does the light take to reach the nose?

(b)In the reference frame of the Earth, how long does

this take? Calculate the time directly from the motions of

the spaceship and the flash of light, and explain

why you cannot obtain the answer by applying the

time-dilation factor to the result from Part (a).

(from Ohanian's Principles of Physics)

_______________________________

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

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