Quick question about lorentz transforms

[spacelike]

So there's something that has bothered me for quite some time.

I know the normal equations of time dilation and length contraction are the following:
$$\Delta t'=\gamma \Delta t$$ and $$L'=\frac{L}{\gamma}$$

where the primed variables are in reference frame S' and the unprimed variables are in reference frame S. With frame S' moving with velocity v relative to frame S.


Ok, so far so good. Now when we try to look at the same thing with Lorentz transformations (in other words: derive the above from Lorentz transformations)
we have:
$$t'=\gamma(t-\frac{v}{c^{2}}x)$$
$$x'=\gamma(x-vt)$$

When deriving time dilation we simply take it that the event which lasts duration t in reference frame S, is at x=0. Therefore we immediately get:
$$t'=\gamma t$$
$$\Delta t'=t'_{2}-t'_{1}=\gamma t_{2}-\gamma t_{1}$$
$$\Delta t'=\gamma \Delta t$$

Great, that gives us time dilation as expected.. Now the problem I'm having is with length contraction. So starting again from the lorentz equation:
$$x'=\gamma(x-vt)$$

This time we consider that we measure the length at t=0, leaving us with:
$$x'=\gamma x$$
Now to get length:

$$L'=x'_{2}-x'_{1}=\gamma x_{2}-\gamma x_{1}=\gamma L$$

That means that according to Lorentz transformations:
$$L'=\gamma L$$

However, according to the formula for length contraction which I wrote at the top:
$$L'=\frac{L}{\gamma}$$

These are totally opposite!
what am I doing wrong?

Thanks

 

[jtbell]

Suppose $x_1$ and $x_2$ are the positions of the ends of a rod, in the unprimed frame. In the primed frame, the rod is moving, so to get the length of the rod in that frame, you have to measure the positions of the two ends at the same time in that frame.

However, the $x_1^\prime$ and $x_2^\prime$ that you've gotten from the Lorentz transformation, correspond to different times in the primed frame, even though you measured $x_1$ and $x_2$ at the same time (t = 0) in the unprimed frame. So $x_2^\prime - x_1^\prime \ne L^\prime$.

 

[spacelike]

Thank you!
That cleared it up perfectly
So now I see that I would have to use $x=\gamma(x'+vt)$
and that gives the correct equation for length contraction