# Derivation Length Contraction

**Level 2**requires school mathematics. Suitable for pupils.

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:

**Traveled distance of the spaceship from the point of view of the spaceship**

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:

**Traveled distance of the spaceship from the point of view of the earth**

Now we can substitute the time span from Eq. 1

into Eq. 3

.

**Distance traveled equals velocity times Lorentz factor times time span**

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

:

**Distance traveled equals velocity times Lorentz factor times distance per velocity**

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

**Distance in the spaceship reference frame is the reciprocal of the Lorentz factor times the distance from earths reference frame.**

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

**Formula for length contraction**

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.