Special Theory of Relativity and Lorentz T. question

In summary, the Homework Equations states that mu traveled 2km in reference frame of the Earth with a speed of 0.8c. Meanwhile, particle e was created at the same time and traveled parallel to mu with the speed of 0.6c. Together, the two particles created a wave traveling 3.2cm in the direction of e.
  • #1
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Homework Statement


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?


Homework Equations


Lorentz Transformations.


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

That would be length contraction.
 
  • #3
Simon Bridge said:
mu travels 2km in reference frame of the Earth - how far has it traveled in the reference frame of the e.

That would be length contraction.

Sure it is, and I calculated this using the Lorentz transformation.
Can you point out what I did wrong?
 
  • #4
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
Simon Bridge said:
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.

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
Didn't you just tell me that the 2km is measured in the Earth frame?
 
  • #7
Simon Bridge said:
Didn't you just tell me that the 2km is measured in the Earth frame?

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.
 
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  • #8
If I draw a line on the Earth-frame 2km long, and an observer traveling at 0.6c in the Earth frame measures it, what length do they measure? Does it matter how fast I drew the line?
 
  • #9
He will mesure it like this:

[itex]L=\sqrt{1-(\frac{6}{10})^2}*2000[/itex]
 
  • #10
estro said:
How much distance mu traveled relative to e?
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
Doc Al said:
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?

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.
 
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  • #12
Doc Al said:
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?

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.
 
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  • #13
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
Doc Al said:
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.
...

Oops, I see, will try to fix it now...
 
  • #15
Doc Al said:
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.

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: [itex] x_1^{e}=\gamma(x_1^{\mu}+vt_1^{\mu})[/itex], which means I'll get the same answer as before...:uhh:
 
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  • #16
estro said:
But now I'm little bit confused as this lifetime is not used in the transformation to calculate the distance traveled in e frame: [itex] x_1^{e}=\gamma(x_1^{\mu}+vt_1^{\mu})[/itex], which means I'll get the same answer as before...:uhh:
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
Doc Al said:
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.
...

Done it this way as well, and again 625.

Doc Al said:
...
(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:)

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
I'm still puzzled, by this problem.
Will appreciate any comments.
 
  • #19
estro said:
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?
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. :redface: 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
Doc Al said:
...
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.
...

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?
 
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  • #21
estro said:
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?
While it's true that in the Earth frame the mu is at x = 2000 m at the same time that the e is at x = 1500 m, the other frames do not agree that that they were at those positions at the same time. Recall the relativity of simultaneity.
 
  • #22
Doc Al said:
While it's true that in the Earth frame the mu is at x = 2000 m at the same time that the e is at x = 1500 m, the other frames do not agree that that they were at those positions at the same time. Recall the relativity of simultaneity.

This is a sound argument.
I guess I should solve these problems algebraically without "consulting intuition"...

Thanks again for all the help!
 

What is the Special Theory of Relativity?

The Special Theory of Relativity, also known as the Theory of Relativity or simply Relativity, is a scientific theory developed by Albert Einstein in 1905. It explains how objects and energy behave in the absence of gravity and at very high speeds.

What are the key concepts of the Special Theory of Relativity?

The key concepts of the Special Theory of Relativity include the principle of relativity, which states that the laws of physics are the same for all observers in uniform motion, and the constancy of the speed of light, which states that the speed of light in a vacuum is the same for all observers regardless of their relative motion.

What is the role of the Lorentz Transformation in the Special Theory of Relativity?

The Lorentz Transformation is a mathematical equation that describes how measurements of time and space change for observers in different frames of reference. It is a crucial component of the Special Theory of Relativity, as it allows for the prediction and understanding of how the laws of physics behave in different frames of reference.

How does the Special Theory of Relativity differ from Newton's Laws of Motion?

The Special Theory of Relativity differs from Newton's Laws of Motion in several key ways. Firstly, it takes into account the effects of high speeds and the constancy of the speed of light. Secondly, it replaces the concept of absolute space and time with the idea of relative motion and the relativity of simultaneity. Lastly, it introduces the concept of spacetime, which unifies space and time into a single entity.

In what ways has the Special Theory of Relativity been confirmed by experiments and observations?

The Special Theory of Relativity has been confirmed by numerous experiments and observations, including the famous Michelson-Morley experiment, which showed the constancy of the speed of light, and the Hafele-Keating experiment, which demonstrated the effects of time dilation. Additionally, various astronomical observations, such as the bending of starlight by gravity, have also provided evidence for the validity of the theory.

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