# Length contraction & relativistic momentum, mass and energy

#### Everything You Need in One Place

Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.

#### Learn and Practice With Ease

Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.

#### Instant and Unlimited Help

Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now!

##### Intros
###### Lessons
1. Length Contraction & Relativistic Momentum, Mass and Energy
2. Introduction to length contraction
3. Relativistic Momentum, Mass, and Energy
##### Examples
###### Lessons
1. A star is 62 light-years away in the Earth frame of reference. How fast would you have to travel so that to you the distance is only 15 light-years?
1. Suppose you decided to travel to a star 60 light-years away at a speed that tells you the distance is only 15 light-years. How many years would it take to make the trip?
1. At what speed will the length of a 1.00m stick look 20.0% shorter?
1. The length of a ruler placed in spaceship in 32cm. The spaceship moves with a speed of 0.080c. what would be the percent decrease in the length of the ruler?
1. What is the relativistic momentum of an antiproton moving at a speed of 0.68c and mass of 1.674 x 10-27 kg?
1. What is the percentage change in the momentum of an electron that accelerates from 0.25c to 0.80c? (mass of electron = 9.12 x 10-31kg)
1. Calculate the rest energy of an electron in joules and in Mev
(1Mev= 1.60 x 10-13J)
1. What would be the momentum and kinetic energy of a proton travelling with a speed of 1.80 x 108m/s. (Proton rest mass = 1.67 x 10-27kg)
1. Calculate the kinetic energy and momentum of a proton travelling 6.24 x 107m/s. What would be the percentage error if the value was calculated using classical formula?
1. A person on a spaceship traveling at 0.60c (with respect to the Erath) observers a comet come from behind and pass her at a speed of 0.60c. how fast is the comet moving with respect to the Earth?
1. A spaceship leaves the Earth traveling at 0.61c. A second spaceship leaves the first at a speed of 0.77c with respect to the first one. Calculate the speed of the second spaceship with respect to the earth if it is fired
1. In the same direction the first spaceship is already moving
2. Directly backwards towards the Earth
###### Topic Notes

In this lesson, we will learn:

• How to find the length contraction?
• How to calculate the relativistic momentum?
• The relation between total energy, kinetic energy and rest energy

Notes:

In addition to the time interval difference in diverse frames of references, lengths and distances also vary from one frame to another.

Let’s consider a spaceship travelling at very high speed from Earth to another planet, and two different frame of references to calculate the length contraction.

1. Observer on earth

2. $L_{0}$: Distance between plants as measured by the observer (proper length)
$\triangle t$: Time required for the trip measured form Earth

$\triangle t = \frac{L_{0}}{v}$

3. Observer in the spaceship

The time interval for the observer in the spacecraft is shorter, since the spacecraft is moving. So the time for the trip according to the time dilation equation would be;

$\triangle t_{0} = \triangle t \sqrt{(1 - v^{2} / c^{2} }$

$L$: the distance between the plants as viewed by the spacecraft observer.

$L = v\triangle t_{0} = v \triangle t\sqrt{(1 - v^{2} / c^{2})} = L_{0}\sqrt {(1 - v^{2} / c^{2})}$

$\gamma =$ $\large\frac{1}{\sqrt{(1 - v^{2} / c^{2}) }}$$\quad \Rightarrow \quad L = \frac{L_{0}} {\gamma}$

therefore; the length of the object is measured to be shorter when it is moving relative to the observer than when it is at rest. (length Contraction)

Note: the length contraction occurs only along the direction of the motion. If the object is moving along $x$-axis, the length is shortened but the height stays the same.

For example, if there is a rectangular painting (1.00m by 1.50m) hung on the wall of a spaceship which is moving at a speed of 0.8c, contraction occurs along the length only.

$L = L_{0}\sqrt {(1 - v^{2} / c^{2}) } =$ (1.50$m$) $\sqrt{(1 - (0.8c)^{2} / c^{2})} =$ 0.9$m$

Relativistic Momentum

In non-relative situation momentum is equal to $p = m_{0}v$ (classical momentum),
In relativistic situation the rest mass ($m_{0}$) increases with the speed and is called relativistic mass ($m_{rel}$)

$m_{rel} =$ $\large \frac{m_{0}} {\sqrt{(1 - v^{2} / c^{2})} }$ $= \gamma \,m_{0}$

therefore, the relativistic momentum can be calculated using the following equation

$p = m_{rel} v = \gamma \, m_{0} v =$ $\large \frac{m_{0} v} {\sqrt{(1 - v^{2} / c{2})} }$

Mass and Energy

The total amount of energy possessed in this case is the total kinetic energy and the rest energy, $E_{total} = E_{k} + E_{0}$

The rest energy by Einstein is defined as $E_{0} = m_{0} c^{2}$

The total energy is also calculated by Einstein’s equation but the mass should be

the relativistic mass $E_{total} =m_{rel} \, c^{2} =$ $\large \frac{m_{0} \,c^{2}} {\sqrt{(1 - v^{2} /c^{2}) }}$

$E_{k} = E_{total} \,- \, E_{0} =$ $\large \frac{m_{0} \, c^{2}} {\sqrt{(1 - v^{2} / c^{2}) }}$ $- \,m_{0} \,c^{2} = m_{0} c^{2}$$\large (\frac{ 1 } { \sqrt{(1 - v^{2} / c^{2})} } - 1)$

$E_{k} = m_{0}c^{2}$ $\large (\frac{1}{\sqrt{(1 - v^{2} / c^{2}})} - 1)$

Relative Velocities do not add simply, as in classical mechanics ($v$ << $c$)
$u = \frac{v \, + \, u^{\prime} }{1 \, + \,v \,u^{\prime} / c^{2}} \qquad$ ($u^{\prime}$ and $v$ are in the same direction)