# Homework Help: Length contraction?

1. Oct 5, 2014

### Lucille

1. The problem statement, all variables and given/known data
Sally and Sheldon have identical meter sticks. Sally is on earth, and sheldon is in a spacecraft that moves at 0.5c relative to sally. Sheldon leaves the spaceship at 0.1c relative to the spacecraft in a launch pad, moving away from sally. according to sally, what is the length of sheldon's meter stick?

2. Relevant equations
x=gamma*(x-vt)

3. The attempt at a solution
if in sheldon's frame:
Lo = 1m = length of meter stick
In sally's frame:
L=gamma*(x'-vt'-x-vt)
but t1=t2

so L=Lo/gamma
L= 0.8 m

So Sally sees sheldon's meter stick as 0.8 m long.

2. Oct 5, 2014

### Simon Bridge

You have not defined t1 and t2.
What did you use for v?

3. Oct 5, 2014

### Lucille

I used v=0.6c

But as I was reviewing my notes, I found that
L=Lo/gamma
where Lo is in Sheldon's' frame and L is in Sally's frame - both of Sheldon's meter stick

So would it be:
L=Lo/(1/(sqrt(1-v^2/c^2)
=Lo*sqrt(1-v^2/c^2)
=(1m)*(sqrt 1-(0.6^2))
=0.8 m

4. Oct 5, 2014

### Orodruin

Staff Emeritus
So for the velocity you have used v = 0.5c+0.1c = 0.6c. This is not correct in relativity and you need to apply relativistic velocity addition. Otherwise, it would be correct to use the length contraction formula as you have done, you just need to use relativistic velocity addition to find the velocity to put in.

5. Oct 5, 2014

### Lucille

Would the velocity of the launch craft in the (x,ct) frame be:

u = (u' + v)/(1+u'v/c^2)

where u = velocity of launch craft in x,ct
v = velocity of the spacecraft in x,ct
and u' is the velocity of the launch craft in x,ct

so u = 0.4c?

Then L = gamma*Lo = (1m)*(sqrt(1-0.4^2)) = 0.917m?

6. Oct 5, 2014

### Orodruin

Staff Emeritus
What numbers did you put to get 0.4c? The launch pad is moving faster than the spacecraft and should have a velociy of at least 0.5c.

7. Oct 5, 2014

### Lucille

I used u'= 0.1c v=0.5c

But if u' should be at least 0.5c, would it be u'=0.5c+0.1c = 0.6 c?

So:

u=(u' + v)/(1+u'v/c^2)
where u' = 0.6c, v=0.5c

u= 0.85c?

L=0.53m??

8. Oct 5, 2014

### Lucille

Wait,

I may have substituted in the wrong numbers in the first place

u = (u' + v)/(1-u'v/c^2)
= (0.1c+0.5c)/(1-(0.1c)(0.5c)/c^2)
=0.57c

L=0.82m?

9. Oct 5, 2014

### Orodruin

Staff Emeritus
Sorry, I was a bit sloppy, the launchpad should have a velocity of at least 0.5c in the earth frame.

You should be using what you said first (u'=0.1c and v=0.5c), but it should not evaluate to u=0.4c.

10. Oct 5, 2014

### Orodruin

Staff Emeritus
This looks more in line with what I would expect apart from that there should be a + in the denominator (although your final numbers for u and L are correct so you probably used it like as the - was a +).

11. Oct 5, 2014

### Lucille

Oops, yes, that was supposed to be a plus.

Thank you so much!! :)

12. Oct 6, 2014

### Simon Bridge

Well done and thanks Orodruin.