Length dilation, special relativity

Question attached.

So, from the question, I'm assuming that the time, t, in reference to the detector is the 15ns, as the answer to part a) i) is 1.5m from: L prime = time in reference to detector × speed, so 15ns × 1/3 c =1.5m

Part ii) asks for the length of pulse beam in the frame of reference of the protons which surely must be the length l =l prime × gamma.

What I don't understand is why the markscheme has used tried to find l prime for part ii) I thought L prime was the proper length, I.e the length in reference to a static frame of reference, I.e. The detector?


I've attached the mark scheme

I think you are confused with the proper length. This is the confusing part.

In the frame of reference of the protons, the detector is moving.
In the frame of reference of the detector, the proton is moving.
They are reciprocal of each other.

When the question says that the beam of pulsed proton is moving and is asked to find the length of the pulsed proton beam in the frame of reference of the protons, the measuring ruler that is attached to the observer in the frame of reference of the protons is at rest, while the measuring ruler that is attached to the observer in the frame of reference of the detector is moving.

It would be good that you reread the your study material pertaining this topic at a slower pace. And it is good to reread some of the confusing parts in special relativity a few times.

Of course, that makes perfect sense. Thank you very much. I neglected the detail that you will be static within your frame of reference, hence the length calculated on i) is the 'non proper' length as the detector is the one moving, right?

And to double check, the proper length, Lo, thus doesn't always have to be less than the length? As it depends on the f.o.r you're in?

We covered special relativity in 2 lessons, with limited notes, I shall be doing that now. Cheers