Clarification of Conclusion of Length Contraction (Without Lorentz Transformations)

Question

In the book, Giancolli $10^{th}$ edition, the author sets up the following scenario to derive the length contraction equation:

Imagine if there was a rocket that traveled a speed of $v$ away from the Earth.

Using this, the author explains the following:

"Ovservers on Earth watch a spacecraft traveling at speed $v$ from Earth to, say, Neptune. THe distance between the planets, as measured by the Earth observers, is $L_0$. The time required for the trip, measured from Earth, is" $$\Delta{t}=\frac{L_0}{v}$$

We know take on the view of an observer in the spaceship:

"The time between departure of Earth and arrival of Neptune (observed from the spacecraft) is the"proper time," since the two events occur at the same point in space. Therefore the time interval is less for the spacecraft observers than for the Earth observers. That is, because of time dilation, the time for the trip as view by the spacecraft is" $$\Delta{t_0}=\Delta{t}\sqrt{1-\frac{v^2}{c^2}}$$

At this point, I have a fair understanding of what they are saying (I suppose). However, they soon make a conclusion that I did not catch on.

"Because the spacecraft observers measure the same speed but less time between these two events, they also measure the distance less. If we let $L$ be the distance between the planets as viewd by the spacecraft observers, then $L = v\Delta{t_0}$..."

From this, I'm taking that since the proper time of the observers on the space ship is less than the time measured from an observer relative to the spaceship by a speed of $v$, the distance of the observer in the spaceship is less than the distance measured by the observers outside of the spaceship.

But they then go on by deriving the following equation and conclusion. $$L = L_0\sqrt{1-\frac{v^2}{c^2}}$$

And

"The length of an object is measured to be shorter when it is moving relative to the observer than when it is at rest."

What it is saying is that the distance measured by an observer relative to the spaceship by a speed $v$ is shorter than the distance measured by an observer in the spaceship Which is the exact opposite of what I just said above unless I'm reading this incorrectly.

Moreover, the equation is saying that the length observed from the outside of the spaceship is greater than the length observed inside. (In order to make this equation jive, $L_0$ should be the length measured from the inside. It's also kind of common sense based of notation, but the author used $L_0$ for the length outside.

Why is this the case, and more importantly how would you derive this equation (with the proper conclusion) without Lorentz Transformations?

Answer 1

I see there's some confusion about which observer is being referred to in the final Giancolli quote.

"The length of an object is measured to be shorter when it is moving relative to the observer than when it is at rest."

To relate back to the Earth-Neptune rocket example, the "object" is the Earth-Neptune distance. This object is moving with respect to the rocket at a speed $v$. Thus, the rocket (the observer) sees this moving object as shorter than an observer standing on Earth. The object is at rest with respect to an observer on Earth, so it appears longer to the Earthling.

The length measured in the object's rest frame is $L_0$ and is termed the proper length. The length $L$ is measured by the rocket and is smaller than $L_0$ (notice that the square root expression is $1/\gamma$ and is smaller than one when $v$ is greater than zero).