Help Me Gain an Intuitive Understanding of Lorentz Contraction

If the light is bouncing (off the mirrors) in the same direction as B's spaceship is moving, A would see exactly what B does: a single beam of light bouncing off each mirror alternately, retracing its own path over and over again.

Remember that the mirrors are moving. So when the light beam travels from the rear mirror to the forward mirror, observer A would actually see a light beam having to catch up to a receding mirror. Similarly, when the light beam travels from the forward mirror to the rear mirror, observer A would see the mirror catching up to the light. This means that according to observer A, the light beam travels further each time it goes forward than when it goes backward, as this image shows:

Even the two halves of the light beam's trip are not the same length according to A, so clearly A and B have to measure different intervals for at least one of those halves (and in fact both).

Quantitatively, suppose the relative speed of A and B is $v$ and the distance between the mirrors (as seen by A) is $\Delta x_A$. On the forward trip of the light beam, as observed by A, the position of the light beam is described by $x_\text{light} = ct$ and the position of the forward mirror is described by $x_\text{mirror} = \Delta x_A + vt$. The time it takes for the light to reach the mirror is obtained by setting these equal to each other:

$$\Delta t_\text{forward} = \frac{\Delta x_A}{c - v}$$

On the backward trip of the light beam, observer A sees $x_\text{light} = -ct$ and $x_\text{mirror} = -\Delta x_A + vt$, so

$$\Delta t_\text{backward} = \frac{\Delta x_A}{c + v}$$

Adding it up, you get a total round-trip time of

$$\Delta t_{A,\text{total}} = \frac{2c\Delta x_A}{c^2 - v^2}$$

Now, suppose you want to find the relationship between $\Delta x_A$ and $\Delta x_B$, the proper distance (i.e. as seen by B) between the mirrors. Hopefully it should be clear that if you look at this from B's perspective, you get

$$\Delta t_{B,\text{total}} = \frac{2\Delta x_B}{c}$$

If you believe the time dilation formula (and it sounds like you do), you can write

$$\Delta t_A = \frac{\Delta t_B}{\sqrt{1 - \frac{v^2}{c^2}}}$$

and now combining the last three equations leads you to

$$\Delta x_A = \Delta x_B\sqrt{1-\frac{v^2}{c^2}}$$

Basically, time dilation is able to account for part of the factor of $\bigl(1 - \frac{v^2}{c^2}\bigr)$ difference between $\Delta t_{A,\text{total}}$ and $\Delta t_{B,\text{total}}$, but not all of it. We have to attribute the rest to length contraction.

Since the only constant in all of this is the speed of light, the only way acceptable to all observers is to measure a distance using c as a yardstick. So... imagine that A sees a beam of light start at the 'rear' of B's spaceship and make its way forward to the 'front' of the spaceship.

Actually, that's not the best way to go about measuring distances, for exactly the reason I described above. As you saw, if you time a light beam traveling from the back of the spaceship to the front (or from a rear mirror to a forward mirror), the time you will actually measure is $\Delta t = \frac{\Delta x}{c - v}$, not $\Delta t = \frac{\Delta x}{c}$ as you thought. (Of course, in a reference frame where the distance being measured is at rest, this works fine since $v = 0$.)

The easiest and recommended way to measure the distance of a moving object is by sitting still at a point and recording the times when the front and back of the object pass you. Once you have the time difference, you can determine the length of the object in your reference frame by $\Delta x = v\Delta t$, where $v$ is the object's speed relative to you. Since boosts between different reference frames "mix" time and space, it's easiest to keep your spatial coordinate fixed when you're measuring time, and vice-versa.

Please review this answer first: Einstein's postulates $\leftrightarrow$ Minkowski space for a Layman , and understand space-time pictures. Then you get "time dilation" and "Lorentz contraction" from two simple pictures.

Time Dilation

Time dilation refers a line which is in the direction of the time axis, but tilted a little to the right, and which represents the trajectory of the moving observer. This line has equally spaced dots along it's length, which are the ticks of the moving observer's clock. What is the vertical spacing of these equally spaced dots?

In geometry, they would occur with vertical spacing reduced by a factor of $1\over\sqrt{1+v^2}$. In relativity, they occur with vertical spacing increased by a factor of $1\over\sqrt{1-v^2}$ (in units where c=1). It's the same argument, but for the minus sign in the pythagorean theorem.

Length contraction

Length contraction refers to two parallel lines which are both exactly vertical. These are the two ends of a stationary ruler, and their length is measured perpendicularly between them to be L.

Now if you are a moving observer, your t-axis is tilted by a slope of v relative to the original t axis, and your x-axis is also tilted by a slope of v relative to the original x axis. So the actual length of the segment of your x-axis between the two ends of the rulers is the hypetenuse of a right triangle with sides L and Lv. In geometry, it would be longer by a factor of $\sqrt{1+v^2}$, but in relativity, it's length is $L\sqrt{1-v^2}$.

Time contraction

If you turn the Length contraction picture on its side, so that the x-axis becomes the t axis and the t-axis becomes the (negative of the) x axis, then you get a strange picture of horizontal lines. These represent a line of simultaneous flag-wavers. They are everywhere in space, and they lift a flag up once a second.

If you move through these flag wavers in a rocket, and you look out the window to see how often the flag-wavers seem to wave to you, you see a different flag-waver wave a flag each time you see a flag waved. How often do the waves come?

The answer is that they come more frequently by a factor of $\sqrt{1-v^2}$. This is "time contraction", it is the length contraction picture tipped over in time.

Length dilation

If you tip the time dilation picture by 90 degrees into space, it gives you a simultaneity line for a moving observer. This line is marked up by flag-wavers at 1m who wave their flag exactly once.

If you have a moving system of flag wavers, and they all measure the distance between them to be 1m, and they wave their flags once at exactly the same time as measured by them, how far apart are the flag-waving events as measured by you?

Because it is just a tipped over time-dilation picture, the answer is it is longer by $1\over\sqrt{1-v^2}$, the same time-dilation factor for a spatial interval. You can't get any of this straight without a picture, and it is just as obvious as Euclid's geometry, except you have to get used to minus signs in the pythagorean theorem.