Special relativity/lorentz transforms

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Homework Statement



Two racers start in between two finish lines of equal distance 'd' , ie a distance 2d sits between the two lines with the racers sitting a distance d from each. Refferees who time the race in the earth frame are able to say that both racers start at time=0s and at the same position of x=0. Both racers run a distance d in opposite direction with constant speed 'u' and in the refs' frame both racers finish at the same time.

Using Lorentz transforms determine the finishing time of each racer in the racer who is traveling left (call him racer one)'s reference frame.


Homework Equations



Lorentz transform equations:

x'=(gamma)(t-(v*x)/c^2)
u'=(u-v)/(1-(u*v)/c^2)
t'=(gamma)(t-(v*x)/c^2)


The Attempt at a Solution



So if we are attempting to find the (t final)' in the primed reference frame, the frame that is moving at a speed of -u I first thought to find the time that it took for each racer to finish in the unprimed frame (ref frame) and then transform it to the primed frame.

so in the unprimed frame it should take each racer t=d/v=d/u time to finish the race.

In order to use the lorentz transform correctly to get to the frame where the speed of the new frame is -u and racer one is travelling at speed=0 I thought to transform racer two's velocity into his velocity in the primed frame.

u(racer 2)'=(u--u)/(1-(-u^2)/c^2)

so then t final for this racer would be the lorentz time transformation as listed above useing this value of u(racer 2)' for gamma.

=(gamma)*(d/u-(-u*d)/c^2))

timefinal(racerone) would have gamma equal to one since he is not moving in this frame so

=(d/u-(-u*d)/c^2))

meaning that the difference is times for finishing the race in the first racers frame is equal to the time racerone thought he finish multiplied by the calculated value of gamma above.

I was told that this is incorrect, but cannot seem to find my mistake.

I don't know how comfortable I am saying that gamma is equal to one for racerone, but since the velocity of this racer is 0 in this new frame why is this not correct?

I'm at a loss,

Thanks for any and all help
 

Answers and Replies

  • #2
vela
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You're mixing up two possible approaches to solving the problem. The Lorentz transformation relates (x,t) to (x',t') where the primed frame is moving with speed v relative to the unprimed frame. In this problem, you have three events. The start of the race is at (0,0). Racer 1 reaches his finish line at (-d, d/u), and racer 2 reaches his finish line at (d, d/u). What you want to do is calculate what the coordinates of these events are in racer 1's frame. You're only moving from one frame to another, so there's only one speed and corresponding gamma to work with.

In the second approach, you look at everything from racer 1's perspective. They both start out at x'=0 and t'=0. You calculate racer 2's new speed and find the distances to the finish lines in this frame. Divide these distances by the corresponding speeds to find when they each finish the race.

You might want to solve the problem both ways to make sure you get the same answer. Although the math used in each method seems different, you should end up with the same results.
 
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Alright so from my understanding of what you said the three events in the unprimed frame are, (x,t)

(0,0)
(-d,d/u)
(d,d/u)

and the gamma that should be used is the one where racerone's velocity is used or

gamma=1/(sqr(1-u^2/c^2))

and then the transforms can be calculated for distance and for time so you get the three events in the racerone frame to be:

(0,0)
(0m, gamma*(d/u-(u*d)/c^2))
(2*d*gamma, gamma* (d/u+(u*d)/c^2))

I can see how racerone's x is zero in his frame since he is not moving in this frame, but it seems odd that the time that the two racers finish is different.

did I make a mistake, or am I just getting confused on some of the rather odd effects of relativity?
 
  • #4
vela
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It's an example of how simultaneity is relative to the observer. To the observer at rest, the two racers crossed the finish lines at the same time, but the racers don't see it that way.

Note that you can simplify the expression for the time racer 1 crosses the finish line:

[tex]\gamma(\frac{d}{u}-\frac{ud}{c^2})=\gamma\frac{d}{u}(1-\frac{u^2}{c^2})=\frac{1}{\gamma}\frac{d}{u}[/tex]

which is just plain old time dilation.
 

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