special relativity - length contraction Watch

Henny987
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The Stanford linear accelerator has a length of 3.0 km.
Calculate the length, in m, of the accelerator in the reference frame of the electron.

(1– v^2/c^2)^0.5

is 2.1 × 10^-5

The mark scheme says that 3km is proper length (lo). but is it not just the length of someone moving relative to the event because in the reference frame of the electron, the length is the proper length - stationary to event length?
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Eimmanuel
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(Original post by Henny987)
The Stanford linear accelerator has a length of 3.0 km.
Calculate the length, in m, of the accelerator in the reference frame of the electron.

(1– v^2/c^2)^0.5

is 2.1 × 10^-5

The mark scheme says that 3km is proper length (lo). but is it not just the length of someone moving relative to the event because in the reference frame of the electron, the length is the proper length - stationary to event length?
I can understand your confusion because I also have this issue when I first learn special relativity.

I would like to quote a paragraph from a text (underlined are from me):

It is important to emphasize that the proper length and proper time interval are defined differently. The proper length is measured by an observer at rest with respect to the end points of the length. The proper time interval between two events is measured by someone for whom the events occur at the same position. Often, the proper time interval and the proper length are not measured by the same observer. …..
In deriving time dilation, the proper time interval is “defined” by the moving observer and as a result in deriving the length contraction, the proper length is “defined” by the stationary observer on the ground.

I know it is confusing but it is not contradicting!
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