Question

A meter stick is at an angle of \(45^\circ\) to the \(x\)-axis in its rest frame. The rod moves with a speed of \(\dfrac{c}{\sqrt{2}}\) along the \(+x\)-direction w.r.t. a frame \(S\). The length of the rod in \(S\) is:

• A. \(\dfrac{\sqrt{3}}{2}\)
• B. \(\dfrac{\sqrt{3}}{4}\)
• C. \(\dfrac{1}{2}\)
• D. \(\sqrt{3}\)

Hint:

Only the \(x\)-projection contracts by \(\gamma = \sqrt{2}\); leave the \(y\)-projection alone, then add the two components in quadrature.

Answer 1

The Correct Option is A

Step 1: In the rest frame the stick has length \(L_0 = 1\ \text{m}\) at \(45^\circ\), so its components are
\[L_x^0 = \cos 45^\circ = \frac{1}{\sqrt{2}}, \qquad L_y^0 = \sin 45^\circ = \frac{1}{\sqrt{2}}\]
Step 2: Compute the Lorentz factor for \(v = c/\sqrt{2}\):
\[\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - \tfrac{1}{2}}} = \sqrt{2}\]
Step 3: Only the \(x\)-component contracts; the \(y\)-component is unchanged:
\[L_x = \frac{L_x^0}{\gamma} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2}, \qquad L_y = L_y^0 = \frac{1}{\sqrt{2}}\]
Step 4: Combine the components:
\[L = \sqrt{L_x^2 + L_y^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{2}} = \sqrt{\frac{3}{4}}\]
\[\boxed{L = \frac{\sqrt{3}}{2}\ \text{m}}\]