# Homework Help: Clocks out of Sync (Special Relativity)

1. Oct 8, 2008

### Fusilli_Jerry89

There are 2 identical 1m rulers (same rest length), and one if going past the other at some velocity. Each ruler has a clock every 10 cm. According to the fixed clock, the moving ruler's clocks will appear to be moving slow, and out of sync. (if t=0 when the rulers pass each other, then the front of the moving ruler will have t' equal to some negative time, the clock at the back of the moving ruler will read some positive t', and the clock in the middle will read t' = 0 when the fixed ruler's clock reads t=0.

My question is, how do we apply time dilation (or lorentz transformations) to this problem in order to find out exactly how far ahead in time the back of the ruler is, and how far in time behind the front of the ruler is.

2. Oct 8, 2008

3. Oct 8, 2008

### Fusilli_Jerry89

K, so say we have two 4 meter long rulers. One is going past the other at 3/5 c. From one ruler's frame: both backs of the 2 rulers line up at x=x'=0 and t=t'=0. If we want to find the time that the contracted ruler's front clock will read to the other ruler, is this how we would do this?:

gamma = 5/4

lorentz transormations:

x'=(5/4)(x-vt)
x' is 4 since this is the length of the moving ruler in it's frame.
4 = (5/4)(x) (since t = 0 all along the ruler in it's own frame)

hence x = 16/5 m

I could have also just used length contraction 4/(5/4) = 16/5 m

Next: I used the lorentz transformation for time:

t'=(5/4)(t-vx/c^2)
t' = 0 since this is the time all along the ruler in the moving ruler's frame
v= 3c/5
x = 16/5 m (as calculated above)

0 = (5/4)(t - (3/5c)(16/5))
t = 48/25c sec

Is this right, or am I missing something?

Thx.

4. Oct 8, 2008

### Fusilli_Jerry89

wait a sec, I think I may have gotten my t's backwards.. Did I?

5. Oct 8, 2008

### Fusilli_Jerry89

so I reversed my t's and did the following:

Looking for t'

t' = (5/4)(-vx/c^2) (since t = 0 everywhere on the ruler)

t' = (5/4)(-48/25c)
= -12/5c sec <------------ which makes sense since it should be a negative time anyway... I think I forgot about that above lol.

Just to check, which method is right?? (I think it's this one)

6. Oct 8, 2008

### granpa

uncontracted meter stick=5 m
gamma=5/4
v=3/5

in the stationary meter sticks frame at t=0:

one end of stationary meter stick (0,0)
other end of stationary meter stick (5,0)

one end of moving meter stick (0,0)
other end of moving meter stick (5/gamma,0)=(4,0)

in the moving meter sticks frame:
first end of moving meter stick (gamma*(0-v(0)), gamma*(0-v(0)))=(0,0)
other end of moving meter stick (gamma*(4-v(0)), gamma*(0-v(4)))=(5,-3)

the whole thing boils down to -gvd
(gamma*d)=uncontracted length

Last edited: Oct 8, 2008
7. Oct 8, 2008

### granpa

btw. nice choice of gamma/velocity.

8. Oct 8, 2008

### granpa

the whole thing boils down to -gamma*d*v
(gamma*d)=uncontracted length (that should be easy to remember)