Reletivity question

1. Sep 25, 2008

fredrick08

1. The problem statement, all variables and given/known data
An event has spacetime coordinates (x, t) = (1200 m, 2.0 μs) in reference frame S. What are
the event’s spacetime coordinates
(i) in reference frame S′ that moves in the positive x-direction at 0.8c and
(ii) in reference frame S′′ that moves in the negative x-direction at 0.8c?

2. Relevant equations
x'=$$\gamma$$x-vt
t'=$$\gamma$$t-vx/c^2
gamma=1/root(1-beta^2)
beta=v/c

3. The attempt at a solution
ok this is wat i did since v=0 i did.
x'=$$\gamma$$x=(1200/(1-.8^2))=2000m
t'=$$\gamma$$t=(2x10^-6/(1-.8^2)=3.33$$\mu$$s

x"=x-$$\gamma$$x=1200-x'=-800m
t"=t'=3.33$$\mu$$s

im not sure about wat ive done just having doubts, could someone plz confirm, or help me fix my mistake... thx

Last edited: Sep 25, 2008
2. Sep 25, 2008

fredrick08

anyone have any ideaS?

3. Sep 25, 2008

fredrick08

anyone?

4. Sep 25, 2008

Perillux

most of it looks right, but I think you may have made a mistake when you were solving for x".

You also have your basic equations wrong. Check here:

for example it should be
x'=(gamma) (x-vt)
NOT
x'=(gamma)x - vt

5. Sep 25, 2008

gabbagabbahey

For part (i), v=+0.8c while for part (ii) v=-0.8c ...why did you have v=0 before?

6. Sep 25, 2008

fredrick08

ya i know i stuffed that part up... well becasue in my mind v=0 coz the event isnt moving?? and beta=0.8c???

7. Sep 25, 2008

fredrick08

coz if i change it to x'=gamma(x-vt)=(1200-((.8*3x10^8)*2x10^-6))/0.6=1200m????
if thats wat ur ur suggesting gabba.. beat=v/c=>v=beta*c=2.4x10^8m/s?

8. Sep 25, 2008

fredrick08

and t' would equal -2us?

9. Sep 25, 2008

fredrick08

na v has to equal 0... so x'=1200/0.6=2000m
t'=2x10^-6/.6=3.33us
im not 100% sure with x" and t"
but it makes sense to me if x"=x-x'=-800m
and because the direction doesnt effect the time then the time in both reference fromes should be equal.. thus t"=t'=3.33us??

can anymore ppl confirm this??

10. Sep 25, 2008

Staff: Mentor

These parts look OK to me.

Take the same approach for S''. What's the only thing that changes in the Lorentz transformations?

11. Sep 25, 2008

fredrick08

oh ok then ummm the only thing that changes is v??

12. Sep 25, 2008

fredrick08

so x"=x-x'=1200-1200=0m
but does t" still equal t'=2us?

13. Sep 25, 2008

fredrick08

can i ask doc al, y do u think that v= something when the event isnt moving?? coz its the complete opposite of what me and perillux think..

14. Sep 25, 2008

Staff: Mentor

Yes. If for S', v = +.8c, what will v equal in the formulas for S''?

I have no idea where you got these values.

15. Sep 25, 2008

Staff: Mentor

Events don't move. v is the speed of the frame, not the event.

You need to read up on the Lorentz transformations and what they mean.

16. Sep 25, 2008

fredrick08

sorry i kinda just thought wat would make sense to me... so for x" and t" instead of beta=1/root(1-.8^2) its -.8? but then doesnt that negative cancel with the square??

so x"=(1/root(1--.8^2))(1200-(0.8*3x10^8)(2x10^-6))=562m?? i dotn think thats rite..

17. Sep 25, 2008

Staff: Mentor

The value of gamma (not beta!) doesn't change, since the v is squared. But that's not the only place that v appears.

18. Sep 25, 2008

fredrick08

ok i found another eqn in book under lorentz... im sorry my book is really bad, and just formulas and not words... ct'=x'=gamma*t(c-v)=2x10^-6(3x10^8-(0.8*3x10^8))/0.6=200m??? y am i getting different answers for the same things??

19. Sep 25, 2008

fredrick08

ok yes so if x' and t' are correct then all x" and t" is....
x"=(1200-((-.8*3x10^8)*2x10^-6))/0.6=2800m
t"=(2x10^-6-((-.8*3x10^8)*1200)/(3x10^8)^2))/0.6)=8.66us?