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*From*: "A. R. Marlow" <marlow@beta.loyno.edu>*Date*: Tue, 2 Apr 1996 18:28:46 +0000 (GMT)

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

...

My apologies for my original misconstruction of Part (b) -- I agree with

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

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

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