# Special relativity and magnitude and direction of velocity

1. Feb 13, 2012

### tebes

1. The problem statement, all variables and given/known data
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.

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

3. 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 ?

2. Feb 13, 2012

### 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. :)

3. Feb 13, 2012

### tebes

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 ?

4. Feb 13, 2012

### beth92

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

5. Feb 13, 2012

### Mindscrape

Huh? Your result should be algebraic.
$$t'=\gamma(t-vx/c^2)$$
Did you get?
c^2 t/x=v

6. Feb 13, 2012

### Mindscrape

Perhaps for clarity, the lorentz equations the original poster listed should really be
$$\begin{eqnarray} \Delta x'=\gamma(\Delta x-v \Delta t) \\ \Delta t' = \gamma(\Delta t - v \Delta x/c^2 \end{eqnarray}$$
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.

7. Feb 13, 2012

### 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! :)

8. Feb 14, 2012

### 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