Time difference between 2 events in moving frame

[Rococo]

Homework Statement

Two events occur at the same time in inertial frame S and are separated by a distance of 1km along the x-axis. What is the time difference between these two events as measured in frame S' moving with constant velocity along x and in which their spatial separation is measured to be 2km?

Homework Equations

$x'=γ(x-vt)$
$t'=γ(t-\frac{vx}{c^2})$

$x=γ(x'+vt')$
$t=γ(t'+\frac{vx}{c^2})$

The question should be able to be solved using just Lorentz transformation equations

The Attempt at a Solution

Frame S:
Event 1: $(x_1, t_1)$
Event 2: $(x_2, t_2)$
$t_2-t_1=0$
$x_2-x_1=1000$

Frame S':
Event 1: $(x'_1, t'_1)$
Event 2: $(x'_2, t'_2)$
$t_2-t_1=?$
$x_2-x_1=2000$Now,

$x'_2 - x'_1 = γ(x_2 - vt_2) -γ(x_1 - vt_1)$
$2000 = γ[x_2 - x_1 -v(t_2-t_1)]$
$2000 = γ[1000 -0]$
$γ=2$

Using $γ = \frac{1}{√1-\frac{v^2}{c^2}}$

I get $v = \frac{√3}{2}c$Now to find the time difference in the S' frame:

$t'_2 - t'_1 = γ(t_2-\frac{vx_2}{c^2}) - γ(t_1-\frac{vx_1}{c^2})$
$t'_2 - t'_1 = γ(t_2-t_1-\frac{v}{c^2}(x_2-x_1))$
$t'_2 - t'_1 = γ(-\frac{v}{c^2}(x_2-x_1))$

Putting in the values I get a time difference of -5.77x10^-6
It's negative so this can't be correct, so I like help on where I went wrong!

 

[Andrew Mason]

Now to find the time difference in the S' frame:

$t'_2 - t'_1 = γ(t_2-\frac{vx_2}{c^2}) - γ(t_1-\frac{vx_1}{c^2})$
$t'_2 - t'_1 = γ(t_2-t_1-\frac{v}{c^2}(x_2-x_1))$
$t'_2 - t'_1 = γ(-\frac{v}{c^2}(x_2-x_1))$
Putting in the values I get a time difference of -5.77x10^-6
It's negative so this can't be correct, so I like help on where I went wrong!

Your answer is correct. The sign is immaterial since we only know the absolute value of (x2 - x1)

AM

 

[Rococo]

Your answer is correct. The sign is immaterial since we only know the absolute value of (x2 - x1)

AM

I see, thanks!