# Length Contraction Question

1. May 9, 2007

### chris12345

Hello,

I am working on the following problem and cannot figure out if I am doing the problem incorrectly or if my professor gave us the wrong answer. (I think that the former is the issue.)

1. The problem statement, all variables and given/known data[/b]
To a stationary observer two frogs at opposite ends of a 100m long lake jump into the water at exactly the same time. To an observer flying overhead along the length of the lake at a velocity of 0.6c, one frog enters the water sooner than the other.

a) How much sooner?
b) How long is the lake to the flying observer?

2. Relevant equations
&Delta;t' = (&Delta;t - (v/c^2)&Delta;x) / sqrt(1-(v^2/c^2)
&Delta;x' = (&Delta;x - v&Delta;t) / sqrt(1-(v^2/c^2)
(And the inverse of these)
L = Lo sqrt(1-(v^2/c^2))

3. The attempt at a solution
a) Using the Lorentz transformation equations, I got &Delta;t' = 2.5^-7s. This is also what my solution set says.

b) Using the length contraction formula, and saying that the proper length is 100m (because that is the rest length). I get 80m. I think this makes sense because the length is longest in the rest frame.

However (and this is how the solutions say to do it), using the Lorentz transformation and plugging in the known &Delta;x = 100m, and known &Delta;t = 100m, you get 125m. The answer itself does not make sense to me, however working through the problem using the Lorentz transformation seems reasonable. I just don't understand why the two do not agree. And which one is correct?

Any and all help would be much appreciated!

2. May 9, 2007

### Staff: Mentor

First things first: Is it the book that says the answer is 125m? Or does the book just say to solve it using the LT and that's what you get?

Your answer using the length contraction formula is correct; and you should get the same answer using the LT, if you use it correctly.

3. May 9, 2007

### chris12345

Thank you!

The solution set that my professor distributed says that the answer is 125m and he got it using LT. What he wrote exactly is:

l' = x2' - x1' = ((x2 - Vt2) - (x1-vt1)) / sqrt(1- v^2/c^2)
= (x2 - x1) / sqrt(1- v^2/c^2)
= 100 / .8 = 125m

Conceptually this didn't make sense to me, but I am now confused about how to use the LT to solve the problem. When I worked through his method, the process seemed to make sense, but the answer did not.

I even tried using the inverse equation to solve for x' -- but that gave 125m again. Do you have any thoughts as to what I might be doing incorrectly using the LT formula?

4. May 9, 2007

### Staff: Mentor

What the professor has calculated is not the observed length of the lake, but the distance between the two events--frog 1 jumps in the water; frog 2 jumps in the water--as measured by the flying observer. Realize that according to the flying observer, the frogs jumped in at different times! (A meaningful measurement of the length of the lake must measure the end points at the same time according to the frame doing the measurements.)

To see that the LT gives the length contraction formula:
$$\Delta x = \gamma(\Delta x' + v\Delta t')$$

Now set $\Delta t' = 0$.

5. May 9, 2007

### chris12345

Aha! Thank you very much! I see now that the change in x as given by LT is not necessarily the length, but the distance *between the events.* This would have been true whichever LT equation I used.

Very very helpful! Thank you again!