Describe the length of an electron's journey in its own frame of reference

[bq1892]

Homework Statement

How far does an electron travel in its own frame of reference if a laboratory observer measures its speed to be 0.8c and the length of its journey 150m

Relevant Equations

Lv = Lo/γ

Lv = Lo / γ

1/γ =√(1-v^2/c^2) = √(1-0.8^2) = 0.6

Therefore Lv = Lo x 0.6 = 150 x 0.6 = 90m

Therefore electron travels 90m in its own frame of reference (answer key solution)

However, shouldn't the electron be assigned rest length, Lo, as its frame of reference is at rest with itself instead of Lv?

 

[Halc]

First of all, why an electron? That's a quantum particle that may have a measured momentum at one point from which a speed can be inferred, but it doesn't really have a defined position after that, and so no meaningful 'length of travel'. I'm just saying it could have been better expressed as some classic object.

However, shouldn't the electron be assigned rest length, Lo, as its frame of reference is at rest with itself instead of Lv?

Any object 'in its own frame of reference' is stationary by definition and therefore goes nowhere. That goes for a classic object at least. The idea of a stationary electron is pretty, well, oversimplified at least.

The question didn't ask about the length of the electron, in either frame.

 

[bq1892]

First of all, why an electron? That's a quantum particle that may have a measured momentum at one point from which a speed can be derived, but it doesn't really have a defined position after that, and so no meaningful 'length'. I'm just saying it could have been better expressed as some classic object.Any object 'in its own frame of reference' is stationary by definition and therefore goes nowhere. That goes for a classic object at least. The idea of a stationary electron is pretty meaningless.

The question didn't ask about the length of the electron, in either frame.

Mb, should have expressed it better. I meant the length of the electron's journey as measured in its frame of reference.