Time Dilation for Moving Rectangle with Clocks at its Corner

[GL_Black_Hole]

Homework Statement

A rectangular structure carries clocks at its four corners. The clocks are synchronized in the structure’s rest frame, in which it has length L =4ft and width W = 3ft. In our laboratory frame the rectangle is moving in the positive x direction at speed v = 0.8c. As the clock at the lower left corner of the rectangle flies past a laboratory laser, the laser fires once, freezing the clock display at zero nanoseconds. The same laser pulse then strikes the rectangle’s upper right clock, disabling that clock too. Determine the reading on the upper right clock after it is disabled.

Homework Equations

The Attempt at a Solution

In the rectangle's rest frame (taken to be coincident with the lab frame at t=0) the laser beam travels along its diagonal with a length of 5 ft. But from the lab frame the length of the rectangle is Lorentz contracted from 4 ft to 2.4 ft. In the time it takes the light to reach the upper right corner, call it $t$ ,the rectangle will have moved $0.8ct$ so the overall distance the light has to travel is the hypoteneuse of a triangle with length $2.4 +0.8ct$ (this is the part I am unsure of) and height 3 ft. The speed of light is the same in all reference frames so $c = 5/\tau = d'/t$ from which $t$ can be found.

 

[TSny]

Hello. Maybe I'm missing something, but isn't the question just asking for the time that you have labeled $\tau$?

 

[GL_Black_Hole]

Possibly. It doesn't directly mention a reference frame. But even so am I correct in how I'm finding the time in the lab frame? Or is there a way to use the Lorentz transformations more directly?

 

[TSny]

Possibly. It doesn't directly mention a reference frame.

In the frame that is moving with the four corner clocks, the clocks are synchronized and run at "normal" rates. So, using this frame, it is easy to deduce the time reading of the upper right clock when it gets zapped by the laser.

But even so am I correct in how I'm finding the time in the lab frame?

Yes

Or is there a way to use the Lorentz transformations more directly?

Yes. But either method gets the answer for the lab time $t$ pretty quickly.

EDIT: By “time in the lab frame” I assume you mean the time it takes for the light to travel between the two corner clocks according to the lab frame. This would not be the reading of the upper right clock when it is hit by the light pulse.

 

[GL_Black_Hole]

Yes that is what I meant by time in the lab frame. Thank you for reinforcing the distinction between the time the clock shows and the time in the lab frame for me.