# Relativity Problem

1. Jun 3, 2016

### i_hate_math

1. The problem statement, all variables and given/known data
This is an extract of a very long question. The electron is moving relative to the lab at 0.98c.

In the reference frame of the electrons, calculate the time required for an electron to travel the length of a beam pipe which has a proper length of 100 m measured in the laboratory reference frame.

2. Relevant equations
t'=y(t-vx/c^2) where y=1/(1-v^2/c^2)=6 in this case

3. The attempt at a solution
The correct answer is 0.34microsecond, strangely it is, the time interval observed by a person in the lab's reference frame is also ∂t=L0/0.98c=0.34microsecond. Can someone explain to me why this is?

2. Jun 4, 2016

### axmls

How did you get that the correct answer is 0.34 microseconds for the time in the electron's frame? I do not get that.

3. Jun 4, 2016

### i_hate_math

That isnt my answer. It says 0.34microsec on the solutionsheet, but i got different value. It could be that the solution sheet contain errors. Did u use L=L0*(1-v^2/c^2) and then divide L by 0.98c?

4. Jun 4, 2016

### axmls

That is one way of doing it. The other is to use the transformation from $t$ to $t'$ directly. They should produce the same answer. Note, by the way, that you're writing your Lorentz factor incorrectly. The correct equation is $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ but you left out the square root. Aside from that, what answer did you get?

5. Jun 4, 2016

### i_hate_math

I got 1.96 microsecond from the transformation equation, and 0.067microsec using L/0.98c

6. Jun 4, 2016

### i_hate_math

Also, its -1.96 a negative value that i got.

7. Jun 4, 2016

### axmls

Could you show the steps so I can follow your calculations?

8. Jun 4, 2016

### axmls

For reference, your length contraction approach is correct, so you're likely just making a small mistake using the transformation equation.

9. Jun 4, 2016

### i_hate_math

t'=y(t-vx/c^2)
and t=100/0.98c
this gives t'=6(100/0.98c-0.98*100/c)=0.067686 microsec
(looks like i made a mistake while doing the calculation and got -1.96)

the other way to do it
L=L0*lorentz facter=502.5189......
t=L/0.98c=0.067686 microsec

Now I'm fairly confident that his is the correct solution.

10. Jun 4, 2016

### i_hate_math

Thanks heaps for the help!

11. Jun 5, 2016

### Physgeek64

I don't know if this is right, but it gets the quoted solution....

So length, as measured by electron:

$l'=\frac{l}{\gamma}=100 \sqrt{1-0.98^2}= 19.899... m$
So the time to traverse the length of the rod $t= \frac{l'}{v}=\frac{19.899...}{0.98c}= 6.7686... 10^{-8} s$

But then you were asked of the time as measured in the lab frame. Since all moving clocks run slow, this time must be time dilated to convert back to the lab frame:

$t_{lab}= \gamma t= \gamma (6.7686... 10^{-8})= 3.401... 10^{-7} s$ which was the answer required.

However, this comes with no guarantee- I too am learning special rel. this year :)

12. Jun 5, 2016

### i_hate_math

Yeah your calculations would be correct if the question were asking for the time interval in the lab's reference frame.

The question is a bit confusing i must say. Perhaps the guy wrote the solution didn't realise that it is the electron's reference frame that the question concerns.

13. Jun 5, 2016

### Physgeek64

I feel like that's what the question wanted- For you to show that you can work in either fame and the time should be invariant for any given frame (i.e. that the transforms are consistent). That was how I interpreted the question. I feel like this is where punctuation is important had he put a comma such that it read

Then you would have to interpret it in the way I have. However, without this the question is sort of left to interpretation. Write down both, and then it's unambiguous that you knew what you were doing ;)

14. Jun 5, 2016

### axmls

The transit time for the electron (i.e. in the electron's frame) will differ from that of the lab frame. This is due to two equivalent interpretations: 1. the distance the electron must travel is length contracted in its frame, so it has less distance to travel in its frame, or 2. time itself is slowed down in the electron's frame relative to the lab frame. Either way, the electron will experience some passage of time for the duration of the transit, and the lab will measure some time for the duration of the electron's transit. These two times will differ. So no, the time is not invariant between frames.

15. Jun 5, 2016

### Physgeek64

No, as in the calculations should be invariant. If you work out the time in the lab frame this must be identical to the time that is calculated if you worked in the electron frame and then transformed back to the lab frame. This may not have been clear from my wording- but when accompanied with my working it is very clear that this is what I meant..

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