# Special Relativity frames homework

1. Oct 13, 2008

### tanzl

1. The problem statement, all variables and given/known data
Ted and Mary are playing a game of catch in frame S', which is moving at 0.600c with respect to frame S, while Jim, at rest in frame S, watches the action. Ted throws the ball to Mary at 0.800c (according to Ted) and their separation (measured in S') is 1.8*1012m.

According to Jim, how far apart are Ted and Mary, and how fast is the ball moving?
According to Jim, how long does it take the ball to reach Mary?

3. The attempt at a solution
In this case, I cant use length contraction to calculate the relative length by Jim.
But, it is wrong for time dilation. I got a wrong answer when I calculate the relative time taken by Jim by using t'=$$\gamma$$t.

Another approach is to use Lorentz velocity transformation to calculate the relative velocity.
But I don understand why I can apply time dilation in this case because it works well for other questions.

2. Oct 14, 2008

### tiny-tim

Hi tanzl!

You're not given t, so why would time dilation be relevant?

Hint: what is the formula for converting separation from one frame to another?

3. Oct 14, 2008

### tanzl

Actually there are another part of the question before this
(b) According to Mary, how long does it take the ball to reach her?

I calculated the time in the reference frame of Mary.
Then, I convert it into Ted Frame using time dilation.
But, somehow it is incorrect.

There is a typo error in my first post. It is correct when I use length contraction but not time dilation.

4. Oct 14, 2008

### tiny-tim

ah … it all makes sense now!

ok … show us what you tried, and where you're stuck, and then we'll be able to see where the problem is.

5. Oct 14, 2008

### tanzl

6. Oct 14, 2008

### tiny-tim

Hi tanzl!

I don't follow what time you're applying the time dilation to.

Time dilation applies to Ted's clock. Time dilation also applies to Mary's clock.

But to compare calculations of speed, you need to subtract the time on Mary's clock from the time on Ted's clock, which will only work if their two clocks started out at the same time … which, in Jim's frame, they didn't!

7. Oct 14, 2008

### tanzl

I dont really get what you mean. Maybe, I will try to explain my idea 1st.

Lets say both Mary and Jim start their clock immediately after Ted throws the ball.
When the ball reaches Mary (or Mary and the ball moving towards each other in Jim's frame), both of them stops their clock.
Lets say Mary measures time t and Jim measures time t'.

When they compare their time, they should find out that
t' = $$\gamma$$t. Right?

8. Oct 14, 2008

### tiny-tim

Yes … but length contraction only applies as between equal observed-times …

that is, for Jim, the length contraction can only be applied between two events which are at the same time as time is measured by Jim

so it applies to the length between Ted's position when Ted throws, and Mary's position at the same time as measured by Jim, which is not Mary's position when she starts her clock.

To put it another way, Jim says "Mary started her clock too late!"

9. Oct 14, 2008

### tanzl

I think I get your idea of equal observed time. Length contraction works because Jim sees and takes two point in the space at the same time and measure their distance. So, no matter when Jim is doing the measurement, Jim will always get the same length.

I dont really get this part. One way I can understand this is that light travel at finite speed. So, lets say a photon from both Mary and Jim needs time t to travel to Jim. So, Jim starts the clock after time t. But, it also takes time for a photon to travel from Ted to Mary, t'. So, Mary should starts her clock after time t'. From the reference of Jim, Mary starts her clock after time t+t' which is not actually immediately after Ted throws the ball.

But when the ball reached Mary, Mary stops the clock immediately but Jim stops his clock after time t. So, from the reference of Jim, Mary stops at the right time. I am not sure whether this is correct o not but that is what I can understand . I think time dilation doesnt work because they are not actually measuring the same event.

Suppose that there is a ball that starts and stops clock automatically. I think time dilation works in this case.

Hmmm... I thought of a pitfall in my explanation. Says that three of them are travel at the same speed ( so same reference frame). Light still takes time to travel between Mary and Ted , so I can still draw the same conclusion. But, it is obviously not true.

10. Oct 15, 2008

### tiny-tim

Yes, in those parapgraphs, you've got it exactly right!

I think the problem with this next paragraph is that we're misunderstanding each other on how Jim and Mary synchronised their clocks in this particular example:
Yes, that is what I mean … "Mary's position at the same time as measured by Jim" means Mary's position at Jim's time t, which as you say is not Mary's position when she starts her clock!

Loosely speaking, the time-axis is tilted along the train, so that events at the same time on the train are progressively earlier (I think) as you go towards the front of the train.
Sorry, I'm not following this … same frame means no dilation or contraction.

11. Oct 15, 2008

### tanzl

If I follow the same reasoning from my last post (photon takes time to travel from Ted to Mary), then time dilation will be observed. But if three of them are in the same frame, a photon still need to take some time to travel from Ted to Mary. So, time dilation should also be observed in this case but it is obviously wrong. So, that is why I think that my explanation is incorrect.

12. Oct 15, 2008

### tiny-tim

oh I see now!

ok, you originally said "Lets say both Mary and Jim start their clock immediately after Ted throws the ball", which does give the correct analysis.

But instead ("Mary starts her clock after time t+t'") you calculated as if they start their clock immediately after they see Ted throw the ball, which is different.

13. Oct 15, 2008

### tanzl

I am trying to use relative velocity transformation to solve this problem.
By velocity transformation, from the reference frame S',
vx' = $$\frac{v_x-v}{1-\frac{v_xv}{c^2}}$$

where
vx'=velocity of the ball as observed by Ted = 0.8c
vx= velocity of the ball as observed by Jim
v = relative velocity between Ted and Jim in reference frame S = 0.6c

But it is incorrect...

14. Oct 15, 2008

### tiny-tim

Well, it should be right …

what did you do, and what is the correct result?

15. Oct 15, 2008

### tanzl

I just plug in the value to get vx but my answer is different.

vx' is -0.8c in the answer booklet.
vx= -0.385c

But I thought that vx should have positive sign since the ball should move in the same direction from both frame S and S'.

Last edited: Oct 15, 2008