# Special Relativity Questions

1. Nov 23, 2014

I was not too sure if this was the correct forum, so feel free to move if needed.

1. The problem statement, all variables and given/known data

A spaceship is measured to be exactly 1/3 of its proper length.

(a) What is the speed parameter β of the spaceship relative to the observer's frame?

(b) By what integer factor do the spaceship's clocks run slow, compared to clocks in
the observer's frame?

2. Relevant equations
$L=\frac{L_0}{\gamma} \\ t=\frac{t_0}{\gamma} \\ \gamma = \frac{1}{\sqrt{1-\beta^2}} \\ \beta = \frac{v}{c}$

3. The attempt at a solution
For A i did:
$$L=L_0 \sqrt{1-\beta^2} \\ \frac{L_0}{3}=L_0 \sqrt{1-\beta^2} \\ \frac{1}{3}= \sqrt{1-\beta^2} \\ \frac{1}{9}=1-\beta^2 \\ -\frac{8}{9}=- \beta^2 \\ \frac{8}{9}=\beta^2 \\ \beta = \sqrt{\frac{8}{9}}$$

I am not to sure that is correct. But for part B I was stuck but during typing this up managed to get an integer answer so hopefully it is correct.
$$t=\frac{t_0}{\sqrt{1-\beta^2}} \\ \frac{t}{t_0}=\frac{1}{\sqrt{1-\beta^2}} \\ \frac{t}{t_0}=\frac{1}{\sqrt{1-\frac{8}{9}}} \\ \frac{t}{t_0}=\frac{1}{\frac{1}{3}} =3 \\$$

Would appreciate any help/advice/feedback, thanks :)

2. Nov 23, 2014

### Simon Bridge

Your answers are fine, but it sounds like you don't understand the equations.

$L_0=\gamma L\\ T=\gamma T_0$
Where the 0's index the proper time, for this case.

You were told: $L=\frac{1}{3}L_0$. This means that $\gamma = 3$.
... so you have automatically answered the second question without any further working out: $T=3T_0$

(Pretty much the first thing you want to know in any SR problem is $\gamma$.)

... for the first question, you want $\beta$: $$\gamma = \frac{1}{\sqrt{1-\beta^2}} \implies \beta = \sqrt{1-\frac{1}{\gamma^2}}$$ ... it is best practice to do the algebra with the symbols before putting numbers in.
$$\beta = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} \implies v= 0.9428c$$ ... can you see how much easier that was that what you did?
No worries though everyone does it the hard way at first ;)

3. Nov 24, 2014