Suppose an object is in a frame where it is moving at speed $\left|\overrightarrow{\beta_{0}}\right|$ and its length is $L 0$ at that speed. If we double the speed $\left(\overline{\beta_{1}}|=2| \overrightarrow{\beta_{0}}\right)$, then the object's length is compressed by a factor of two $\left(L_{1}=\frac{1}{2} L_{2}\right)$.

Question

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The length contraction formula in special relativity is $L = L_0\sqrt{1-\beta^2}$, where $L_0$ is the proper length (length of the object in its rest frame) and $\beta = v/c$, with $v$ being the velocity of the object and $c$ the speed of light. Doubling the speed does not simply halve the length due to the square root and the squared velocity term in the formula; the relationship between speed and length contraction is not linear.  The formula for calculating $L$ from $L_{R}$ is not linear in $\beta$. For example, if we change $\beta$ from $2 / 5$ to $4 / 5, \sqrt{1-\beta^{2}}$ goes from $\sqrt{21 / 25}=0.92$ to $\sqrt{9 / 25}=0.60$, which is not a factor of 2 .

Suppose an Object Is in a Frame Where It Is ∣β0→∣\left|\overrightarrow{\beta_{0}}\right|​β0​​​

Suppose an Object Is in a Frame Where It Is $\left|\overrightarrow{\beta_{0}}\right|$