12.3.7 Length Contraction

Length Contraction

When objects travel close to the speed of light, to an observer moving relative to that object, it looks as if the object has become shorter
- This is best demonstrated using rulers

Observer H, in their rocket moving close to the speed of light, measures the length of their pencil to be 14 cm
Observer G, at rest on Earth, would measure (with remarkable eyesight) the length of the pencil to be shorter
- They will see lengths contracted in the spaceship from their reference frame, e.g. the length may appear to be 10 cm instead of 14 cm

However, the same occurs the other way around
For observer H on the rocket, it is observer G that is moving relative to them
- Therefore, observer H would measure the length of observer G's pencil as shorter i.e observer H, on the rocket, sees lengths contracted on Earth from their reference frame

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A stationary observer in their own reference frame views lengths as shorter in the moving reference frame

Length Contraction Equation

The length of an object is the difference in the position of its ends
Consider the observers G and H measuring the length of a pencil, which is stationary in reference frame S' (for observer H)

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Observer G measures the length of the pencil to be different to observer H

As observer H is in the moving frame S', they measure the length of the ruler as:

$∆ x' = x_{2}' - x_{1}' = L_{0}$

This is the proper length, L₀as the pencil is not moving relative to observer H
- Both the pencil and observer H are, however, moving relative to observer G
Observer G needs to measure the length of the pencil by measuring the position of its ends at the same time (just like observer H did)
They measure the length of the ruler to be:

$∆ x = x_{2} - x_{1} = L$

This is the observed length, L as the pencil is moving relative to observer G
- Moving between different reference frames tell us how the x and x' are related

We want to find $∆ x$ , the length measured in the reference frame of the stationary observer on Earth (G), who is moving relative to the observer on the rocket (H)
Transforming these distances gives:

$x_{1}' = γ (x_{1} - v t)$

$x_{2}' = γ (x_{2} - v t)$

These are then substituted into the equation for the proper length, L₀:

$x_{2}' - x_{1}' = γ (x_{2} - v t) - γ (x_{1} - v t) = γ (x_{2} - x_{1})$

$L_{0} = γ L$

Therefore:

$L = \frac{L_{0}}{γ}$

Where:
- $L$ = the length measured by an observer moving relative to the length being measured (m)
- $L_{0}$ = the proper length (m)

As $γ$ > 1, this means that the $L < L_{0}$
- In other words, lengths measured from a reference frame moving relative to the object will be measured as shorter than the lengths measured at rest from within their frame of reference

Similar to time dilation, length contraction is also due to Einstein's second postulate
- Both observers G and H must measure the speed of light to be c
- Since the time for observer H will run slower, according to observer G (i.e. t increases), then for c to stay the same, the length of the object, L must decrease

It is important to note that the length has been measured at the same time
- This length is the difference between the ends of the pencil, with both ends measured at the same time

The ruler used in both reference frames is stationary in their own reference frame
- Otherwise, observer G would see the ruler on observer H's rocket contracting as well and wouldn't measure any difference in length

Worked example

A spacecraft leaves Earth and moves towards a planet.

The spacecraft moves at a speed of 0.75c relative to the Earth. The planet is a distance of 15 ly away according to the observer on Earth.

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The spacecraft passes a space station that is at rest relative to the Earth. The proper length of the space station is 482 m.

Calculate the length of the space station according to the observer in the spacecraft.

Answer:

Step 1: List the known quantities

Speed of the spacecraft, v = 0.75c
Proper length of the space station, L₀ = 482 m

Step 2: Analyse the situation

We are trying to find the length of the space station in the reference frame of the observer in the spacecraft
In this observer's reference frame, it is the space station that is moving away from them at 0.75c
Therefore, we a measuring a length in the moving reference frame (relative to the spacecraft) - this is the length, L

Step 3: Substitute values into the length contraction equation

$L = \frac{L_{0}}{γ} = \frac{L_{0}}{\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}}$

$L = \frac{482}{\frac{1}{\sqrt{1 - \frac{{(0.75 c)}^{2}}{c^{2}}}}} = \frac{482}{\frac{1}{\sqrt{1 - {(0.75)}^{2}}}} = 319 m$

Step 4: Check whether your answer makes sense

As the observer in the spacecraft is stationary, the length of the space station they measure should be shorter than the proper length
As the length recorded from the spacecraft is 319 years, and the proper length is 482 m, this length makes sense

Exam Tip

You will not be expected to remember this derivation, but it's helpful to know where all the factors have come from. The time dilation equation is given on your data sheet.

The notion of 'proper length' is incredibly important here, as it depends on the reference frame the length is being measured from.

You will find in some exam questions you can use time dilation or length contraction, you will receive marks for either way.

AQA A Level Physics

Revision Notes

Length Contraction

Length Contraction Equation

Worked example

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Author: Dan MG