Understanding a step in the derivation of the length contraction effect in special relativity

Question

I've been trying to figure out how the equation for length contraction is derived in my textbook (Krane, Modern Physics 3e) since a few of the final steps are omitted. The equation in question is:

$$ L = L_0/\gamma = L_0 \sqrt{1-u^2/c^2} $$

Where $u$ is the velocity of the object. I managed to derive myself the previous equations

$$ \Delta t = \Delta t_0\sqrt{1-u^2/c^2} $$

and

$$ \Delta t = \frac{2L_0}{c} \frac{1}{1-u^2/c^2} $$

after which my book states:

Setting the two equations above equation to each other and solving, we obtain: $$ \Delta t = \Delta t_0\sqrt{1-u^2/c^2} = \frac{2L_0}{c} \frac{1}{1-u^2/c^2} \rightarrow L = L_0 \sqrt{1-u^2/c^2} $$

without any appearance of $L$ in the above equations. I have a hunch that it might be implicitly defined as $u\Delta t$, but I'm not certain. What is the missing step here in getting that last equation from the two above?

Answer 1

It isn't clear which bit is foxing you, so let's go through the argument:

This is viewed from the Earth frame i.e. the frame in which the light clock is moving. In this frame the length iof the clock is $L$. In the outward trip the light moves a distance $L+ut_1$ in a time $t_1$, and since light moves at the speed of light we get:

$$ L+ut_1 = ct_1 $$

Likewise for the return trip the light moves a distance $L-ut_2$ in a time $t_2$ so:

$$ L-ut_1 = ct_2 $$

The total time is therefore:

$$ t = t_1 + t_2 = \frac{2L}{c}\frac{1}{1 - u^2/c^2} \tag{1} $$

Now switch to the rest frame of the clock. In this frame the length of the clock is $L_0$. In this frame the light simply moves a distance $2L_0$ in a time $t_0$, and again light moves at the speed of light so:

$$ t_0 = \frac{2L}{c} $$

Now the book uses the previously derived result that:

$$ t = \frac{t_0}{\sqrt{1-u^2/c^2}} = \frac{2L_0/c}{\sqrt{1-u^2/c^2}} \tag{2} $$

This is just the usual equation for time dilation. The time $t$ is the same time in equations (1) and (2), so we just set them equal to get:

$$ \frac{2L}{c}\frac{1}{1 - u^2/c^2} = \frac{2L_0/c}{\sqrt{1-u^2/c^2}} $$

And this rearranges to the final result:

$$ L = L_0\sqrt{1-u^2/c^2} $$

But I have to say that this is a dreadful derivation of the Lorentz contraction because it give you no insight into what actually happens in special relativity. The contraction is not really a contraction it is a rotation in spacetime. Have a look at "Reality" of length contraction in SR to get an idea of what is actually going on.

You might also be interested to look at How do I derive the Lorentz contraction from the invariant interval? to see how the Lorentz contraction is related to the symmetry that underlies special relativity.