# Confusion with the Lorentz contraction as explained in "Introduction to Elementary Particles" by D. Griffiths

I am trying to understand how to derive the Lorentz contraction equation using the Lorentz transformations and while I know of a way to do this, I am still confused by a number of things, including the explanation from this book.

As far as I understand, this is how the story goes: we have two inertial reference frames $$S$$ and $$S'$$, with $$S'$$ moving with velocity $$v$$ with respect to $$S$$ along the $$x$$ axis. Additionally, we set the clocks at the origin in each system so they both read 0 when they coincide, so we have $$t = t' = 0$$ when $$x = x' = 0$$. In $$S'$$, we have a stick along the $$x'$$ axis, at rest. One end of the stick is at the origin $$x' = 0$$, and the other end is at $$L'$$, so its length is $$L'$$. Now, let's try to calculate its length in $$S$$, which we will call $$L$$.

To measure the stick in $$S$$, we want one end at the origin, so at $$t = 0$$. We apply the Lorentz transformation alongside the $$x$$ axis to transform the coordinate of the other end of the stick in $$S'$$ to coordinates in $$S$$:

$$x = \gamma(x' + vt')\tag{1}$$ so $$L = \gamma(L'+vt')\tag{2}$$

From the Lorentz transformation on the time axis, we have:

$$t = \gamma(t'+\frac{v}{c^2}x')$$ But we want $$t = 0$$, so: $$\gamma(t'+\frac{v}{c^2}x')=0$$ $$t'+\frac{v}{c^2}x'=0$$ $$t'=-\frac{v}{c^2}x'$$ Plugging this into equation 2, and knowing that in our case $$x'=L'$$, we get:

$$L=\gamma(L'-\frac{v^2}{c^2}L')$$ $$L=\gamma{}L'(1-\frac{v^2}{c^2})$$ $$L=\gamma{}L'\gamma^{-2}$$ $$L=\frac{L'}{\gamma}$$

We get the expected equation for the Lorentz equation. Now I am confused by a number of things here. We defined the frames such that the origins coincide at $$t = t' = 0$$, and in this calculation we want $$t = 0$$ so we should have $$t'=0$$. But if we try to substitute $$t'$$ by $$0$$ in equation 2, we get the incorrect result: $$L=\gamma{}L'$$

Why doesn't this work?

My second question relates to the explanation provided in the book referenced in the title of my question. To arrive at the Lorentz contraction equation, the book says to apply the formula that transforms coordinates in $$S$$ to $$S'$$ which is $$x' = \gamma(x - vt)$$. Shouldn't it use the equation that transforms coordinates on the x axis from $$S'$$ to $$S$$ instead, like I did above? Also, the book doesn't even mention using the equation for the time transformation. Is the book just wrong, or am I missing something here?

• I have understood my confusion in the first question. Specifically, we have $(x',t') = (0,0)$ when $(x,t)=(0,0)$, but that doesn't mean that we always have $t' = 0$ when $t = 0$. Commented Apr 23 at 18:02
My favorite derivation of the length contraction is to rewrite $$x' = \gamma(x-tv)$$ as $$x = \gamma^{-1}x' + tv.$$ This directly shows that the $$x'$$-axis moves with speed $$v$$ but is contracted by the factor $$\gamma^{-1}.$$
• This makes a lot of sense, and I think this answers my follow-up questions as well. Doing things this way, we can get at the result much more directly, and only use the equation to translate $x'$ coordinates in $S'$ to x coordinates in $S$, without having to use the equation for the transformation on the time axis. It seemed more logical for me to use the coordinate transformations from $S'$ to $S$ since we are trying to convert a length in $S'$ to a length in $S$, but this is way easier. Thanks! Commented Apr 20 at 17:18