Length Contraction in Relativity

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Homework Help Overview

The problem involves length contraction in the context of special relativity, specifically examining the length of an accelerator tube as measured from different frames of reference: that of an electron traveling at a high speed and that of an observer on Earth.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the assignment of variables L and L0 in the context of the problem, questioning whether the length of 1.20 m should be considered the proper length or the contracted length. There is exploration of the implications of measuring lengths from different frames of reference.

Discussion Status

Some participants have provided clarifications regarding the definitions of proper length and contracted length, indicating a productive exchange of ideas. There is a recognition of confusion surrounding the assignment of variables, but no explicit consensus has been reached.

Contextual Notes

Participants express uncertainty about the nature of the variables and the proper length in relation to the frames of reference involved. The discussion reflects the complexities of applying relativistic concepts to the problem.

bjgawp
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Homework Statement


An electron travaels in an accelerator tube at a speed of 0.997 c relative to Earth. In the frame of reference of the electron, the length of the tube is 1.20 m. What is the length of the tube relative to earth?

Homework Equations


L = Lo / γ
Textbook answer: 15.5 m


The Attempt at a Solution


I know it is a simple plug-in problem but when I plug in the numbers, they don't work out and I'm guessing that I'm assigning the numbers wrong. For this problem, would 1.20m be L or Lo? I'd assume it to be Lo since that's the proper length in terms of the electron's reference frame. However, the numbers do not work out then - indicating that L = 1.20 m. This does not make sense to me since 1.20 m is the length measured from the electron's frame, is it not?
 
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Yes, 1.2m is the length as measured from the electron's frame. In your formula above, L0 is the proper length of the tube as measured in the frame we are working in, and L is the length as observed by the moving observer. So, suppose we are in the frame at rest relative to the earth. Then, L0 is the proper length relative to the earth, L is the length which the electron sees, and v (in the factor gamma) is the relative velocity between the Earth and the electron.
 
Hmm. So if 1.20m is from the electron's frame, wouldn't that mean it is assigned to L0? But, since this is the proper length we're dealing with where we have to measure the tube at the same time, it has to be from the Earth's frame .. doesn't it? Sorry, I guess I don't understand the assignation/ the nature of the variables.

L = L0 / γ

Either:
L = (1.20m)√(1 - 0.997²)
L = 0.093 m

Or:
1.20m = (L0)√(1 - 0.997²)
L0 = 15.5 m (textbook answer)
 
bjgawp said:
Hmm. So if 1.20m is from the electron's frame, wouldn't that mean it is assigned to L0? But, since this is the proper length we're dealing with where we have to measure the tube at the same time, it has to be from the Earth's frame .. doesn't it? Sorry, I guess I don't understand the assignation/ the nature of the variables.

The 0 in L0 stands for the proper value, the value in its own restframe. Clearly from the electron's rest frame the tube is not at rest and thus it cannot be assigned L0.
 
Ah, it just clicked. Thanks guys. Relativity is so fun :rolleyes:
 

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