Special relativity help: meter stick moving at 0.60c

[positron]

A meter sticks moves with velocity 0.60c relative to an observer. The observer measures the length of the meter stick to be L. The problem states that 0.80<L<1.0 m must always be true.
So far, I have determined that
Gamma = 1/sqrt(1-0.60c^2/c^2) = 0.8.
What I don't understand is why there is a range. If the meter stick is 1 m when it is not moving and 0.8m when it is moving at 0.80c, why is there a range for the length?

 

[robphy]

Is velocity a vector?

 

[quicknote]

The range for the length of the meter stick is due to the effects of special relativity. According to this theory, as an object moves at high speeds, its length in the direction of motion appears to contract for an observer who is stationary relative to the object. This phenomenon is known as length contraction.

In this scenario, the observer is stationary and the meter stick is moving at 0.60c. The observer measures the length of the meter stick to be L. However, if the meter stick was measured by an observer who is moving at the same speed as the meter stick, the length would appear to be shorter due to length contraction.

The range of 0.80<L<1.0m is a result of this length contraction. The actual length of the meter stick is 1m when it is not moving, but as it moves at 0.60c, the observer measures it to be shorter, between 0.80m and 1.0m. This range reflects the varying measurements of the meter stick depending on the relative motion between the observer and the meter stick.

In summary, the range for the length of the meter stick is a consequence of special relativity and the phenomenon of length contraction. It is a fundamental principle of this theory that the length of an object appears to change depending on the relative motion between the observer and the object.