I'm having a hard time getting an intuitive understanding of Lorentz Contraction. I understand what it is by definition but I don't 'get it.' I'm not a physicist, just an amateur, so sorry if this question comes across as too naïve.
Okay, I was able to understand Time Dilation with the help of the 'light bouncing off two mirrors' experiment (the one that uses Pythagoras' theorem to derive the equation of time dilation) and I've noticed that this same scenario is also used on a lot of websites to derive Lorentz Contraction. So, I'll use it to present my question:
First, to define the setup:
Let's assume person A is "at rest" and person B goes by in a spaceship traveling at velocity $v$, close to the speed of light $c$. Let's call the time measured by A $t$ and that measured by B $t'$. So, we have:
$$t = t'/\sqrt{1-v^2/c^2}$$
I understand that we can exchange A and B (consider B to be "at rest" instead of A) and come up with the same relation. So, for each observer, the other person's watch seems to move slower.
So, if B were traveling at $v = 0.8c$, $\sqrt{1-v^2/c^2} = 0.6$. This means that in the time interval that A counts off 5 minutes on his watch, B counts off only $5*.6 = 3$ minutes.
Okay, this must all be old hat to you guys but I wrote all this just so that you know how much of this I understand.
Now to get to my question, I can't follow most of the explanations of Lorentz Contraction that use this light-bouncing experiment. Here's why:
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Lorentz Contraction works in the same direction as the motion of the object. So, immediately, the light-bouncing experiment starts making less sense for me because for time dilation it was the transverse (to the movement of the object) motion of light that created the right-angled triangle and allowed us to use Pythagoras in the first place. 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. Moreover, since light always travels at a constant speed of $c$ in vacuum, if they both happened to time the bouncing of the light beam, they'd both measure the same interval between bounces (I'm not saying the bounces would be synchronized on both their watches, just that the interval would be the same).
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I know that Lorentz Contraction means that A will measure the length of B's spaceship (in its direction of motion) as being smaller than what it actually is (or what it is in B's frame of reference). 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. Let's say A times the journey and finds that it takes t seconds on his watch to for the light to cover the distance between the rear and the front. So, for A, B's spaceship is $c*t$ units long. However, since A knows that B's watch is going slower than his own, he can infer that if B sitting in his spaceship had also been timing the beam of light, the time that B measured (say $t'$ seconds) would be $< t$. So, A can deduce that B's spaceship is actually $c*t'$ units long where $$(c*t) > (c*t')$$ or $$\text{A's measure of spaceship length} > \text{B's measure of spaceship length}$$
Now this is precisely the opposite of Lorentz Contraction.
Where did I go wrong?
Best Answer
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.
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.