View Full Version : Another exasperating question

SergioPL

01-16-2008, 10:36 AM

Recently my doubts relativity are growing faster every day. This is about a problem (apparently) simple. If somebody could help me I will be grateful, I am sorry if someone grows tired from my questions :).

According to my calculations, if you have a rod in X axis, whose length is L, moving respect an observer S with speed V x its length will be 1/gamma*L.

Ok, if this rod suddenly makes a “super acceleration” so that its speed respect S becomes 0, and all the points of the rod change its speed at the same time, in S’s frame the end of the rod with the less X value will stop V*gamma*(L /c) seconds before the other end.

This effect will “lengthen” the rod V^2*gamma*L. Then the length of the rod that will see S will be:

1/gamma*(1+v^2*gamma^2)*L = 1/gamma*(1+(v^2/(1-v^2)))*L = gamma*L

Since the rod will be stopped respect the observer S, its length must be L, not gamma*L. Somebody knows what’s wrong in my approach?

cincirob

01-16-2008, 10:58 AM

Recently my doubts relativity are growing faster every day. This is about a problem (apparently) simple. If somebody could help me I will be grateful, I am sorry if someone grows tired from my questions :).

According to my calculations, if you have a rod in X axis, whose length is L, moving respect an observer S with speed V x its length will be 1/gamma*L.

Ok, if this rod suddenly makes a “super acceleration” so that its speed respect S becomes 0, and all the points of the rod change its speed at the same time, in S’s frame the end of the rod with the less X value will stop V*gamma*(L /c) seconds before the other end.

This effect will “lengthen” the rod V^2*gamma*L. Then the length of the rod that will see S will be:

1/gamma*(1+v^2*gamma^2)*L = 1/gamma*(1+(v^2/(1-v^2)))*L = gamma*L

Since the rod will be stopped respect the observer S, its length must be L, not gamma*L. Somebody knows what’s wrong in my approach?

cinci: This is the reverse of JS Bell's famous rocket and string problem where your rod is the string. If you assume the center of the string in the rocket problem accelerates just like the rockets, then the rockets and the center of the string all follow the same shape of path in the stationary system; that is, they maintain their separation distance throughout the acceleration in the stationary frame. The length of the string reduces according to (1-(v/c)^2)^.5 and in Bell's problem, the string breaks.

Now if you decelerate the rockets and center of the string at the same rate, you will simply reverse the process. As the velcoity approaches zero, the string will approach its rest length and again reach from ship to ship.

I have a little problem following your logic so if I missed the point, please try again. If your problem involves instant deceleration, that is impossible and may be causing the confusion.

SergioPL

01-16-2008, 11:23 AM

Thank you cinci, yes, my problem involves instant deceleration, it's only a model to simplify the problem, it's just an idealization, I guess it's physicaly impossible but I though the problem should be able to to be solved anyway.

My approach is simple, the event "stop" occurs at the same time in both edges of the rod, so in S system, as result of the Lorentz transformation, it will be V*gamma*(L /c) seconds before in the "start" than in the "end" of the rod, so when the "start" of the rod stops, the "end" continues moving during V*gamma*(L /c) and that makes the rod to look longer.

I will read JS Bell problem to understand its logic.

cincirob

01-16-2008, 04:24 PM

Thank you cinci, yes, my problem involves instant deceleration, it's only a model to simplify the problem, it's just an idealization, I guess it's physicaly impossible but I though the problem should be able to to be solved anyway.

My approach is simple, the event "stop" occurs at the same time in both edges of the rod, so in S system, as result of the Lorentz transformation, it will be V*gamma*(L /c) seconds before in the "start" than in the "end" of the rod, so when the "start" of the rod stops, the "end" continues moving during V*gamma*(L /c) and that makes the rod to look longer.

cinci: I have a little trouble thinking about the stop occuring "at the same time". Time for the rod is different than time for the stationary observer.

**************************

I will read JS Bell problem to understand its logic.

cinci: Interesting story if you haven't heard it. He presented it to a room full of physicists and got different answers and a lot of argument when he presented the right one. He was making a point that a lot of physicists may have forgotten the basics. The problem goes like this.

Start with two spaceships separated by a distance L.

Tie a string between them. (make the ships point sized so contraction of theships doesn't confuse the issue.)

According to a stationary observer both ships begin to accelerate at exactly the same instant and at exactly the same acceleration.

Does the string break and, if it does, why?

Good luck

SergioPL

01-17-2008, 09:57 AM

I have read the problem in http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html , the distance between both rockets in the initial system remains constant but the distance of rockets grows because while measured in another system the simultaneity disappears. So the string will break.

I haven't studied it very accurately but I have found it's very smart the way the world-line is described. I would like to explain how I have understood it.

The world-line is x = (1+t^2)^.5, this line is also the result to boost a vector 0t+1x to other system that moves with a generic speed V x (-1<V<1).

This fact means that any point (t,x) in this line meet that t=gamma(V)*V and x=gamma(V). Where V is the speed with that the frame that measures (t,x) sees moving the lab frame.

The world-line starts at 0t+1x, speed is 0 and acceleration is equal to 1. If we derivate the world line we will get that dx/dt = t*(1+t^2)^(-0.5) and d^2x/dt^2 = (1+t^2)^(-1.5)

Since t= gamma(V)*V then (1+t^2) = gamma(V)^2 where V is the speed that boosts vector 0t+1x (0,1) into (t,x). Then it’s easy to see that dx/dt = V, and the acceleration is d^2x/dt^2= 1/gamma(V)^3.

Acceleration is correct, compared with the result of the velocity addition formula. If a body changes its “proper speed” dv x, another body moving with speed V x will see this change is dv’ = (V+dv)/1+V*dv - V = dv/ gamma(V)^2.

Since proper time grows gamma this acceleration spreads another gamma(V) so the final result is that acceleration is gamma(V)^3 times minor in the frame moving with speed V than in the proper frame.

If the acceleration obtained with this model is the same that the result of the velocity addition formula and besides the initial speed is 0, then the evolution of speed and position will be correct so it has been proved (to my eyes) that the world-line is x = (1+t^2)^.5.

So with the selection of an initial position (0,1) and a bit of imagination we can predict the position of the rocket every time. I am impressed! (At first I didn’t trust in the definition of he world-line, as usual in me :) ).

Besides, I have found the solution to my own problem, which it’s easier. The previous result was wrong because I suppose the difference of time between the stop of both edges was V*gamma*(L /c) , that’s the result of a boost with speed V to vector 0t + Lx.

But since the rod changes from the speed V to 0 the delay between the edges is not the result to make a boost with speed V but it’s the “average” of the delay suffered at all the speeds between V and 0 and that is:

Delay = (1/V) *Integral[ v*gamma(v)*L dv] between (V, 0) = (1/V)*(1-1/gamma(V))*L

Then the “start” of the rod continues moving (1- 1/gamma(V)) and thus the final length is its proper length: L :D!!

PS: This result is quite odd but it looks like it works.

Powered by vBulletin® Version 4.2.0 Copyright © 2021 vBulletin Solutions, Inc. All rights reserved.