Lorentz transformation for spacetime coordinates at 0.8c

[fredrick08]

Homework Statement

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?

Homework Equations

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

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 I've done just having doubts, could someone please confirm, or help me fix my mistake... thx

 

[fredrick08]

anyone have any ideaS?

 

[fredrick08]

anyone?

 

[Perillux]

please be patient.
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:
https://www.physicsforums.com/showthread.php?t=110015
Your lucky that it didn't affect your answers though because v=0.

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

 

[gabbagabbahey]

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

 

[fredrick08]

ya i know i stuffed that part up... well becasue in my mind v=0 coz the event isn't moving?? and beta=0.8c?

 

[fredrick08]

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

 

[fredrick08]

and t' would equal -2us?

 

[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 doesn't effect the time then the time in both reference fromes should be equal.. thus t"=t'=3.33us??

can anymore ppl confirm this??

 

[Doc Al]

coz if i change it to x'=gamma(x-vt)=(1200-((.8*3x10^8)*2x10^-6))/0.6=1200m?

and t' would equal -2us?

These parts look OK to me.

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

 

[fredrick08]

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

 

[fredrick08]

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

 

[fredrick08]

can i ask doc al, y do u think that v= something when the event isn't moving?? coz its the complete opposite of what me and perillux think..

 

[Doc Al]

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

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

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

I have no idea where you got these values.

 

[Doc Al]

can i ask doc al, y do u think that v= something when the event isn't moving?? coz its the complete opposite of what me and perillux think..

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.

 

[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 doesn't that negative cancel with the square??

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

 

[Doc Al]

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 doesn't that negative cancel with the square??

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

 

[fredrick08]

ok i found another eqn in book under lorentz... I am 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??

 

[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?