Is My Lorentz Transformation Calculation Correct?

[bckcookie]

Problem in Lorentz transformation(Urgent!)

1. Homework Statement

Please see the attached file and advise me my solution is right or not! Thanks!

2. Homework Equations

~ d/t = v
~ Lorentz transformation : x'=(x-Vt)/(√(1-(V^2)/c^2), t' = [t-(V^2)/(c^2)]/√1-(V^2)/c^2),

3. The Attempt at a Solution

a) According to Cheryl, d/t =v, 15/(c/2) = 10^-7s

From Tom on ground,
i) when the light from the spark reaches mirror A
t1 = L/(c+V) = 15m/(3x10^8 + c/2 ) = 3.33 x 10^-8s

ii) when the light from spark reaches mirror B
t2 = L/(c-V) = 15m/(3x10^8 - c/2 ) = 10^-7s

iii) when the light from spark returns to Cheryl from mirror A
t2 = L/(c-V) = 15m/(3x10^8 - c/2 ) = 10^-7s

iv) when the light from spark returns to Cheryl from mirror B
t1 = L/(c+V) = 15m/(3x10^8 + c/2 ) = 3.33 x 10^-8s

b) How to apply the Lorentz transformation equation in the same four events take place?? please advise!

My approach:

t=10^-7s, d = 15m, v=c/2

x'=(x-Vt)/(√(1-(V^2)/c^2),
= 15-(c/2 x 10^-7)/(√(1-(V^2)/c^2) = 0

t' = [t-(V^2)/(c^2)]/√1-(V^2)/c^2),

= 8.66 x 10^-8s


c) They disagree that ligt from the spark reachs mirror A at the same time as mirror B, because when the light from the spark reaches mirror A of clock slow down and length contract. They also disagree the light returning from mirror A reaches Chery l at the same time as the light returning from mirror B, because the light returning from mirror B of clock slow down and length contract.

d) They agree it, because the clock is not slow down and length contract.

e) Using t' = t√1-v^2/c^2)

=10^-7s√1-c/v^2/c^2)

=8.66 x 10^-8s

"Moving clocks run slow"
ΔT = ΔT0/√1-v^2/c^2

 

[Simon Bridge]

There is something wrong there all right:

eg. for (a)

a) for Cheryl: if the mirror is distance L away, then the time for light to reach the mirror is L/c not the L/(c/2) that you have. c/2 is the speed that Tom is traveling in Cheryl's reference frame, the speed of light is always c.

i) to Tom, on the ground, the mirror is moving away from the place the light comes from, making the light travel a longer distance to get there. This means it takes a longer time but you have it taking a much shorter time.

ii) mirror B approaches the source, so the distance to travel is smaller, so the time is smaller, but you have it come out the same.

You will always do better if you construct your equations from the physics than if you memorize them ... and that's what your answers look like.

 

[bckcookie]

There is something wrong there all right:

eg. for (a)

a) for Cheryl: if the mirror is distance L away, then the time for light to reach the mirror is L/c not the L/(c/2) that you have. c/2 is the speed that Tom is traveling in Cheryl's reference frame, the speed of light is always c.

i) to Tom, on the ground, the mirror is moving away from the place the light comes from, making the light travel a longer distance to get there. This means it takes a longer time but you have it taking a much shorter time.

ii) mirror B approaches the source, so the distance to travel is smaller, so the time is smaller, but you have it come out the same.

You will always do better if you construct your equations from the physics than if you memorize them ... and that's what your answers look like.

Please advise me the updated solution as follows;

a) You are right!
i) d/v =t, where d=15, v=c, so, = 15/c= 5x10^-8s
ii) d/v =t, where d=15, v=c, so, = 15/c= 5x10^-8s
iii) d/v =t, where d=30, v=c, so, = 30/c= 10^-7s
iv) d/v =t, where d=30, v=c, so, = 30/c= 10^-7s

b) The times according to Tom for the above four events, Cheryl's frame as the unprimed frame, Tom's frame as the primed frame. Using Lorentz transfromation equation :
x'=(x-Vt)/(√(1-(V^2)/c^2), t' = [t-(V^2)/(c^2)]/√1-(V^2)/c^2),
v of x-axis: -c/2

i) Since the Δx = 0, then
t' = [t-/√1-(V^2)/c^2),
t' = (15/c) x 0.866
= 5.77 x 10^-8s

ii) I don't know whether it is same as i)? or other approah?, please advise!

iii) Since the Δx = 0, then
t' = [t-/√1-(V^2)/c^2),
t' = (30/c) x 0.866
= 1.155 x 10^-7s

 

[Simon Bridge]

for (ii) - if you used the same method as (i) would t' end up bigger or smaller than t? Does that agree with the geometry?

 

[bckcookie]

I can confirm that the time of (i) is faster than (ii), but i have no idea how to calculate it, please advise!

 

[Simon Bridge]

There is a limit to the assistance I can give you when it's homework like this, at some stage you have to do your own thinking. Consider - the light pulse travels from the source to the mirror in time d/c ... where d is the distance the pulse has to go. This distance is different from the case where the mirror is stationary ...

Hint: work out the geometry of the situation in Tom's reference frame and use the usual physics you are used to. Notice that the mirror is moving towards the pulse, so it's motion makes the distance the pulse has to travel shorter ... which reduces the time.

 

[bckcookie]

Thanks so much! You are right! Done!

 

[Simon Bridge]

Well done.
Caution: - if the difference from the source to the mirror was measured in C's frame, then these distances will be different in T's frame.

 

[bckcookie]

My following approach! Thanks a lot!

a) i) d/v =t, where d=15, v=c, so, = 15/c= 5x10^-8s
ii) d/v =t, where d=15, v=c, so, = -15/c= -5x10^-8s
iii) d/v =t, where d=30, v=c, so, = 30/c= 10^-7s
iv) d/v =t, where d=30, v=c, so, = -30/c= -10^-7s

b) The times according to Tom for the above four events, Cheryl's frame as the unprimed frame, Tom's frame as the primed frame. Using Lorentz transfromation equation :
x'=(x-Vt)/(√(1-(V^2)/c^2), t' = [t-(Vx/(c^2)]/√1-(V^2)/c^2),
v of x-axis: -c/2

In fact i m not sure whether the calc. of x is right:

(ct')^2 + (vt)^2 = (ct)^2
therefore, x= ct' = √3/2(ct)


i) t=15/c, then
t' = [t-(√3(ct)/2)(-c/2) /√1-(V^2)/c^2),
t' = 3.27 x 10^-8s

ii) t=-15/c, then
t' = [t+(√3(ct)/2)(-c/2) /√1-(V^2)/c^2),
t' = 8.273 x 10^-8s


iii) t=30/c, then
t' = [t-(√3(ct)/2)(-c/2) /√1-(V^2)/c^2),
t' = ...


iii) t=-30/c, then
t'= [t+(√3(ct)/2)(-c/2) /√1-(V^2)/c^2),
t'=...

 

[Simon Bridge]

Oh? I'd have length-contracted the mirror-source distance before subtracting vT.
You can check your working by drawing pictures - the normal rules for physics have to apply in each reference frame.

 

[bckcookie]

You are right! I misunderstood my calculation before. I corrected it and submit it to my tutor! Thanks a lot!

 

[Simon Bridge]

No worries.

Now wasn't that fun?!
Hopefully you'll be better at checking your own work now :)