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tifr-gs-2018/ #2

Open utterances-bot opened 3 years ago

utterances-bot commented 3 years ago

TIFR GS 2018 Solved Paper | Unraveling Physics

TIFR 2018 Graduate admissions entrace test - containing solutions for both integrated PhD and PhD.

http://unravelingphysics.com/tifr-gs-2018/

prateek39 commented 3 years ago

the solution to question 32 is wrong - the time in Earth's frame would be sqrt(3)*l/c. The error in your calculation is when you calculate invariant interval in Earth's frame, \delta(x) will be 0. As the aligning of nose tips and aligning of tails take place at the same place in Earth's frame.

physicophilic commented 3 years ago

Nice job. I confirm that you are right. I should have been more careful. Had the speeds been different in Earth's frame, $\Delta x$ in Earth could have been non-zero. I will correct it and add an acknowledgement. 👍🏼

prateek39 commented 3 years ago

Also, the simplest way to solve the problem is to work in Earth's frame entirely.

physicophilic commented 3 years ago

You can share your simplest calculation if you think so.

physicophilic commented 3 years ago

Your claim with a proof could be a useful addition, but without it - as you can understand - it is inappropriate.

prateek39 commented 3 years ago

In Earth's frame the length of both the spaceships is l' = l/\gamma. Now, the distance travelled by any one of the ships between when their noses are aligned and when their tails are aligned is l'. And the speed in Earth's frame is c/2. So, time in Earth's frame is l'/(c/2) or 2l/(c*gamma).

physicophilic commented 3 years ago

Good job Prateek. Your outline is correct. After defining the events (1) and (2) as I defined above, and drawing them in diagrams your approach indeed becomes way easier than mine. I will certainly add this line of thought to the question when I get time. Thanks for your feedback. 👍🏼