Question

Appearing in the time-dilation and length-contraction formulas, ${\mathit{\gamma }}_{\mathbf{v}}$ , is a reasonable measure of the size of relativistic effects. Roughly speaking, at what speed would observations deviate from classical expectations by 1 %?

Short answer

The observation deviate from classical expectations at speed of$4.2\times {10}^7\mathrm{m}/\mathrm{s}$

Step-by-step solution

Identification of given information

The given data can be listed below as:

The deviation from classical expectations is $=1\%=0.01.$

Significance of time dilation and length contraction

The time dilation and length contraction are observed due to the object's speed compared to the speed of light. The Lorentz factor is used to find the contraction in length and dilation in time.

Determination of the Lorentz factor for time-dilation and length contraction

The equation to calculate the Lorentz factor is given as:

${\gamma }_v=\frac{1}{\sqrt{1-{\beta }^2}}\mid$

Here, ${\gamma }_v$ is the measure of the size of relativistic effects and its value becomes $\left\{\right(1+1\%)=1.01\}$ due to deviation from classical expectations.

Substitute all the values in the above equation.

role="math" localid="1657542463731" $$\begin{gathered} 1.01=\frac{1}{\sqrt{1-{\beta }^2}} \\ \beta =0.14 \end{gathered}$$

Determination of the speed for deviation in observation from classical expectations

The speed for the deviation in observation from classical expectations is given by formula as:

$\beta =\frac{V}{C}$

Here, is the speed of light and its value is $3\times {10}^8\hspace{0.33em}m/s$.

Substitute all the values in the above equation.

$$\begin{gathered} 0.14=\frac{v}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ v=4.2\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the observation deviates from classical expectations at a speed of$4.2\times {10}^7\hspace{0.33em}m/s.$