Special relativity and magnitude and direction of velocity

[tebes]

Homework Statement

In the inertial system S, an event is observed to take place at point A on the x-axis and 10^-6 seconds later another event takes place at point B , 900 m further down. Find the magnitude and direction of velocity of S' with respect to S in which these two events appear simultaneous.

Homework Equations

Lorentz's transformations
x = \gamma ( x' + vt')
t = \gamma ( t' + vx'/c^2)

The Attempt at a Solution

Then, I let x/t to solve for velocity and t' = 0 because the two events appear simultaneous. Is attempt correct ?

 

[Mindscrape]

Yeah, that sounds good for the most part. In my mind, you've got your lorentz transformation backwards for the problem at hand, but they will have the same information so however you want to do it. :)

 

[tebes]

Yeah, that sounds good for the most part. In my mind, you've got your lorentz transformation backwards for the problem at hand, but they will have the same information so however you want to do it. :)

Thank you. If I use 3 x 10 ^ 8 m/s as c, I would obtain zero. But If I use c = 299 792 458 m / s, my answer wouldn't be trivial. Does it mean anything physically ?

 

[beth92]

I'm doing a similar problem at the moment, and I don't quite understand why you've set t' to 0?

 

[Mindscrape]

Thank you. If I use 3 x 10 ^ 8 m/s as c, I would obtain zero. But If I use c = 299 792 458 m / s, my answer wouldn't be trivial. Does it mean anything physically ?

Huh? Your result should be algebraic.
<br /> t&#039;=\gamma(t-vx/c^2)<br />
Did you get?
c^2 t/x=v

 

[Mindscrape]

I'm doing a similar problem at the moment, and I don't quite understand why you've set t' to 0?

Perhaps for clarity, the lorentz equations the original poster listed should really be
\begin{eqnarray}<br /> \Delta x&#039;=\gamma(\Delta x-v \Delta t) \\<br /> \Delta t&#039; = \gamma(\Delta t - v \Delta x/c^2<br /> \end{eqnarray}<br />
In the inertial frame, the primed frame, it looks like the two events occur at the same time; there's no difference between the two events.

 

[beth92]

Okay, thanks! I think I've figured it out..

From Lorentz we have t=γ(t'+v/c²x')

So if we have two times in the S frame t₁ and t₂ then the interval

t₂-t₁=Δt=γ(t'₂-t'₁)+γ(v/c²(x'₂-x'₁))

We know that S' observes the events at the same time so t'₁=t'₂

Also from Lorentz: x=γ(x'+vt')

So distance between events A and B in S frame is

x₂-x₁=γ(x'₂-x'₁)+γv(t'₂-t₁)

We know the second term is zero so we can say that x'₂-x'₁=Δx'=Δx/γ

Subbing this into the expression for Δt:

Δt=γ(v/c²(Δx/γ))

γ cancels out and we can rearrange for v=Δtc²/Δx

We know Δx is the distance between A and B w.r.t S frame and is 900m.
Δt is the time between the events in the S frame and is 10^-6s. We know c, so we can just use plug in these numbers and we find

v=10⁸ m/s

If I've done something wrong please let me know! :)

 

[Mindscrape]

Yep, that's the same thing I got. Look at how easy the problem would have been in the primed frame. From my equations:
dt'=0=(dt-vdx/c^2)
So what's in the parenthesis must be zero
v=dt/dx*c^2