Special Relativity

What is special relativity? §

Special relativity is how speed affects mass, space, and time.

Formula Review §

Also see Time Dilation.

Time Dilation §

$$\Delta t=\frac{\Delta t_i}{\sqrt{1-\frac{v^2}{c^2}}}$$

Length Contraction §

$$L=L_i\sqrt{1-\frac{v^2}{c^2}}$$

Practice Questions §

16. You are moving at a speed $\frac{2}{3}c$ toward Randy when he shines a light toward you. At what speed do you see the light approaching you?

Note

Because you are moving toward the light, the light appears to approach you faster than the speed of light. The only option we have that is over c is $\frac{4}{3}c$, so this has to be the answer.

24. Astronaut Rachel leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. Rachel’s destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year). According to Rachel, how long does the trip take?

Note

We can use the time dilation equation to calculate how long the trip takes, as Rachel experiences it.

$$\begin{array}{rl}& \\ \Delta t& =\frac{\Delta t_i}{\sqrt{1-\frac{v^2}{c^2}}}\\ 14.4\left(\frac{1}{0.96}\right)& =\frac{t_i}{\sqrt{1-\frac{(0.96c{)}^2}{c^2}}}\\ 15& =\frac{t_i}{\sqrt{1-0.9216}}\\ 15(\sqrt{1-0.9216})& =t_i\\ t_i& =4.2\end{array}$$

25. At age 21, you set out for a star that is 50 light-years from Earth. How fast would your spaceship have to travel in order to reach that star when you are 61 years old? One light-year is the distance light travels in one year.

Note

Our time experienced ($t$ or $t_i$) is 40 years, or 61-21. The distance we have to travel is 50 light-years, which would mean we want to travel 50 light years in 40 years. We can set up our equation for time in terms of distance and velocity:

$$\begin{array}{rl}v& =\frac{\Delta d}{\Delta t}\\ \Delta t& =\frac{\Delta d}{v}\end{array}$$

Then, we can sub that in as the $\Delta t$ for our actual equation.

$$\begin{array}{rl}\Delta t=\frac{\Delta t_i}{\sqrt{1-\frac{v^2}{c^2}}}& \\ \frac{\Delta d}{v}& =\frac{40}{\sqrt{1-\frac{v^2}{c^2}}}\\ \frac{50}{v}& =\frac{40}{\sqrt{1-\frac{v^2}{c^2}}}\\ \frac{\sqrt{1-\frac{v^2}{c^2}}}{v}& =\frac{40}{50}\\ \frac{1-\frac{v^2}{c^2}}{v^2}& =\frac{1600}{2500}\\ 2500-2500\frac{v^2}{c^2}& =1600v^2\\ \frac{2500\left(\frac{v^2}{c^2}\right)}{1600v^2}& =2500\\ \frac{1.5625}{c^2}& =2500\end{array}$$

What even is this clusterfuck I regret that anyway .58c should be the answer theoretically

19. A spaceship visit a star that is 4.5 light-years from Earth, and the spaceship travels at one-half the speed of light for the entire trip. a. How long did the trip take according to an observer on Earth? b. How long did the trip take according to an observer on the spaceship?

Note

For part A, we need to use the time dilation equation.

$$\begin{array}{rl}t& =\frac{4.5}{\sqrt{1-\frac{(0.5c{)}^2}{c^2}}}\\ & =\frac{4.5}{\sqrt{0.75}}\\ & \approx 5.2\text{ y}\end{array}$$