Lorentz transform for two events with v=3c/5

[strangequark]

I have a problem from my modern physics class I'm hoping to get some insight on... I got an answer, but it seems, well, odd...

Homework Statement

Inertial frame S' moves with speed v=$$\frac{3c}{5}$$ in the +xdirextion past inertial frameS. Event A is a synchronizing event. Event B occurs at t=0 in Frame S and at position x'=1 meter in frame S'. Given:

For Event A:
$$x_{A}=0$$
$$x_{A}'=0$$
$$t_{A}=0$$
$$t_{A}'=0$$

For Event B:
$$x_{B}'=1 meter$$
$$t_{B}=0$$

Find,
$$x_{B}$$ and $$t_{B}$$'

Homework Equations

The relevant equations are the two lorentz transforms:

$$x'=\gamma(x-vt)$$

and

$$t=\gamma(t'+\frac{vx'}{c/^{2}}$$

The Attempt at a Solution

For $$x_{B}$$:
Applying equation 1 I get,

$$\frac{x'}{\gamma}+vt=x$$
Then $$x=\frac{4}{5} meters$$

and for $$t_{B}$$' I get (applying equation 2),

$$t'=\frac{t}{\gamma}-\frac{vx'}{c^{2}}=-2.001 x 10^{-9} seconds$$

I'm confused about the second answer. Did I do something wrong? or does this just mean that in frame S' the event happens $$-2.001 x 10^{-9}$$ seconds prior to t=0 in frame S?

Heeeeeeeelp please!

 

[learningphysics]

You didn't do anything wrong. The event happens $$2*10^{-9}$$ seconds before t'=0 (that is in the S' frame). It just means that in the S' frame, this event happens before the synchronization event.

or does this just mean that in frame S' the event happens seconds prior to t=0 in frame S?

I wouldn't put it that way. You can't really say that an event happens in frame S' before that same event happens in frame S... before and after can only be used when comparing events in the same frame...

 

[strangequark]

OK, that makes sense... thanks a lot, i really appreciate it.