Alexander Fufaev
My name is Alexander FufaeV and here I will explain the following topic:

Length Contraction

Here we want to derive a formula for the length contraction using the previously derived time dilation formula.

Imagine you are an observer at rest on Earth. We denote your reference frame by \( \mathrm B \). We denote the time that passes on your clock by \( t \). At time \( t = 0 \) a spacecraft flies past the observer \( \mathrm B \) with a constant velocity \( \class{blue}{v} \). It travels to a planet, let's call it Alpha, and arrives there at some specific time.

Let us denote the reference frame of the spacecraft as \( \mathrm B' \) and the time on the clock in the spacecraft as \( t' \). Now, what does the captain see in the spaceship when he considers himself at rest? From his point of view, the spaceship is resting while the Earth is moving away from him with the velocity \( \class{blue}{v} \) and the planet Alpha is moving towards him with the velocity \( \class{blue}{v} \).

The captain in the spaceship measures on his watch how long it takes him to get from Earth to Planet Alpha. We denote this time span by \( \Delta t' \). Also the observer on earth measures the time span on his clock, in which the spaceship needs from earth to the planet alpha. He denotes his measured time span with \( \Delta t \).

In deriving time dilation using a light clock, we found that the time spans in different reference frames are fundamentally different. Therefore, we cannot simply assume that the time spans \( \Delta t \) and \( \Delta t' \) are equal. We have deduced that two time spans of different reference frames are related by the relative velocity \( \class{blue}{v} \) as follows:

Here \(c\) is the speed of light and the prefactor in the formula is usually abbreviated as \(\gamma\) and called Lorentz factor. Since we know that the magnitude of the relative velocity at which the Earth is moving away from the spaceship is \(\class{blue}{v}\), we can use the time period \( \Delta t' \) to calculate the distance \( \Delta x' \) that the spaceship must travel from the Earth to the planet Alpha:

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fufaev.org Length contraction: Distance between 2 planets as seen by the spacecraft at rest
Abstand zweier Planeten aus Sicht eines ruhenden Raumschiffs \(\mathrm B'\).

And from the point of view of the observer \( \text B \) on Earth the spaceship needs the time span \(\Delta t\) and the spaceship moves away with the velocity \(\class{blue}{v}\). Therefore the travelled distance \(\delta x\) in the reference frame \( \text B \) is to be calculated as follows:

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Abstand zwischen der Erde und dem Planet Alpha aus Sicht eines Ruhebeobachters \(\mathrm B\) auf der Erde.

Now we can substitute the time span from Eq. 1 into Eq. 3.

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Our goal is to find out how the \( \Delta x' \) and \( \Delta x \) distances are related. Therefore we have to express \( \Delta t' \) from Eq. 4 with \( \Delta x' \). To do this, rearrange Eq. 2 for the time period \( \Delta t' \) and insert it into Eq. 4:

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Cancel velocity \( \class{blue}{v} \) and rearrange equation for \( \Delta x' \):

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We are done. If we now insert the Lorentz factor \(\gamma \), we get the formula for the length contraction:

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Since the reciprocal of \(\gamma\) is always less than 1, we can say from Eq. 7 that \( \Delta x'\) is less than \( \Delta x \). In words, the distance traveled \( \Delta x' \) and thus the distance of the Earth and planet Alpha from the perspective of the spaceship must be shorter than from the perspective of the observer on Earth. The distance earth/planet is contracted from the point of view of the spaceship.

Now you should know how to derive time dilation and length contraction. These interesting phenomena can be illustrated with spacetime diagrams.