- #1

Rococo

- 67

- 9

## 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

[itex]x'=γ(x-vt)[/itex]

[itex]t'=γ(t-\frac{vx}{c^2})[/itex]

[itex]x=γ(x'+vt')[/itex]

[itex]t=γ(t'+\frac{vx}{c^2})[/itex]

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

## The Attempt at a Solution

Frame S:

Event 1: [itex](x_1, t_1)[/itex]

Event 2: [itex](x_2, t_2)[/itex]

[itex]t_2-t_1=0[/itex]

[itex]x_2-x_1=1000[/itex]

Frame S':

Event 1: [itex](x'_1, t'_1)[/itex]

Event 2: [itex](x'_2, t'_2)[/itex]

[itex]t_2-t_1=?[/itex]

[itex]x_2-x_1=2000[/itex]

Now,

[itex]x'_2 - x'_1 = γ(x_2 - vt_2) -γ(x_1 - vt_1)[/itex]

[itex]2000 = γ[x_2 - x_1 -v(t_2-t_1)][/itex]

[itex]2000 = γ[1000 -0][/itex]

[itex]γ=2[/itex]

Using [itex]γ = \frac{1}{√1-\frac{v^2}{c^2}}[/itex]

I get [itex]v = \frac{√3}{2}c[/itex]

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

[itex]t'_2 - t'_1 = γ(t_2-\frac{vx_2}{c^2}) - γ(t_1-\frac{vx_1}{c^2}) [/itex]

[itex]t'_2 - t'_1 = γ(t_2-t_1-\frac{v}{c^2}(x_2-x_1))[/itex]

[itex]t'_2 - t'_1 = γ(-\frac{v}{c^2}(x_2-x_1))[/itex]

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!

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