# Special Theory of Relativity and Lorentz T. question

1. Sep 16, 2012

### estro

1. The problem statement, all variables and given/known data
Particle mu was created in the atmosphere and traveled 2km relative to earth with a speed of 0.8c.
Particle e was created in the same time and traveled parallel to mu with the speed of 0.6c.

How much distance mu traveled relative to e?

2. Relevant equations
Lorentz Transformations.

3. The attempt at a solution
I've got an answer but I'm totally not sure.
https://dl.dropbox.com/u/27412797/classical_mechanics/question_5.jpeg [Broken]

You can't see this in the image but I calculated the relative speed of mu in regards to e and got (5/13)c

Last edited by a moderator: May 6, 2017
2. Sep 16, 2012

### Simon Bridge

mu travels 2km in reference frame of the Earth - how far has it travelled in the reference frame of the e.

That would be length contraction.

3. Sep 16, 2012

### estro

Sure it is, and I calculated this using the Lorentz transformation.
Can you point out what I did wrong?

4. Sep 16, 2012

### Simon Bridge

Can't make head nor tails of your calculations. There are certainly a lot of steps ... but I'd have figured that a length of L in the Earth frame would be L'=L/γ in the e frame

So it remains to find what speed the Earth frame is moving wrt the e frame.

5. Sep 16, 2012

### estro

As far I know you can't just use this formula to calculate the length contraction at this specific instance as both the reference system are actually moving, and you have no choice but to use Lorentz Transformation directly. [BTW the equation you have written is a direct conclusion from Lorentz Transformation.]

I'll try to explain what I did:
In each table I mark two events in the space time coordinates :
1. particle creation.
2. particle destruction.

Table 1 is for particle e in regards to earth.
Table 2 is for particle mu in regards to earth.
Table 3 is for particle mu in regards of e.

6. Sep 16, 2012

### Simon Bridge

Didn't you just tell me that the 2km is measured in the Earth frame?

7. Sep 16, 2012

### estro

Yes, mu traveled 2 km in regards to earth.

What I need to find is the distance mu traveled as seen from e.

But since both particles travel at high speeds I think that I can't simply use the formula you suggested earlier.

Last edited: Sep 16, 2012
8. Sep 16, 2012

### Simon Bridge

If I draw a line on the Earth-frame 2km long, and an observer travelling at 0.6c in the Earth frame measures it, what length do they measure? Does it matter how fast I drew the line?

9. Sep 16, 2012

### estro

He will mesure it like this:

$L=\sqrt{1-(\frac{6}{10})^2}*2000$

10. Sep 16, 2012

### Staff: Mentor

I interpret that to mean: In the rest frame of the e particle, how far has the mu particle traveled in its lifetime.

So: What is its lifetime? What is the relative speed of the particles?

11. Sep 16, 2012

### estro

Yes, exactly!

Actually I put all this data in the table, but I will do it again so it be more clear. [Perhaps I should have done it from the first place...]

Will be back in 15 minutes with more clear attempt.

Last edited: Sep 16, 2012
12. Sep 16, 2012

### estro

Ok I tried to write what I did in detail:
https://dl.dropbox.com/u/27412797/classical_mechanics/relativity.jpeg [Broken]

This answer doesn't make sense, my intuition tells me the final answer should be bellow 500 meters.

Last edited by a moderator: May 6, 2017
13. Sep 16, 2012

### Staff: Mentor

Looking over your work quickly, it looks like you found the lifetime of the mu particle in its own frame (good!), but then for some reason you used its lifetime as measured by earth when finding its lifetime in the e particle frame. That doesn't make sense to me.

Find the proper lifetime. (I think you did this.)
Find the relative speed. (I think you did this.)
Find the observed lifetime in the e frame. (Use the first two results.)
Find the distance traveled in the e frame.

14. Sep 16, 2012

### estro

Oops, I see, will try to fix it now...

15. Sep 16, 2012

### estro

I think that I fixed the observed lifetime in the e frame: https://dl.dropbox.com/u/27412797/classical_mechanics/relativity_take_2.jpeg [Broken]

But now I'm little bit confused as this lifetime is not used in the transformation to calculate the distance traveled in e frame: $x_1^{e}=\gamma(x_1^{\mu}+vt_1^{\mu})$, which means I'll get the same answer as before...:uhh:

Last edited by a moderator: May 6, 2017
16. Sep 16, 2012

### Staff: Mentor

You can do it in several ways. I would transform from the mu frame to the e frame, since the values of t and x in the mu frame are easier to deal with.

(I'll have to look at your work in more detail when I get a moment to see if I misinterpreted what you have already done. :uhh:)

17. Sep 16, 2012

### estro

Done it this way as well, and again 625.

OK, thanks!
BTW, without getting into computations and details, does the the final answer make sense to you, on the intuitive level? Maybe this answer is right and my intuition just wrong?

18. Sep 17, 2012

### estro

I'm still puzzled, by this problem.

19. Sep 17, 2012

### Staff: Mentor

Yes, that answer is correct. And my suggestion of first finding the lifetime in the mu particle frame was silly, since you can just as easily go from the earth frame measurements directly to the e particle frame measurements. Of course you'll get the same answer, but without the need to find the relative speed between the particles.

As far as your intuition goes, why would you question your answer of 625 m? It makes sense that the answer would be less than that seen in the earth frame. After all, in the e particle rest frame, the mu particle is moving slower and has a shorter lifetime.

(Sorry for the delay in getting back to you; was having access problems.)

20. Sep 17, 2012

### estro

Thanks for your help, and last question if you don't mind:

In the earth frame mu traveled 2000 meters, while e traveled 1500 meters, so the difference is 500 meters. However since both particles travel at high speeds I expect length contraction so my intuition tells me the final answer should be less than 500 meters.
How is that makes sense?

Last edited: Sep 17, 2012