Timeline for Time reversal symmetry in the presence of friction

10 events

when toggle format	what		by	license	comment
Sep 3, 2019 at 10:16	history	edited	Joce	CC BY-SA 4.0	deleted 40 characters in body
Aug 7, 2019 at 4:47	comment	added	Bibek_G		$\mathrm d^2\tau/\mathrm dt^2 = 0$ and not 1. and that is also not what we need here. $\mathrm d\tau^2/\mathrm dt^2 = 1$ should be edited in.
Nov 20, 2018 at 15:52	history	edited	Joce	CC BY-SA 4.0	clarify
Jan 27, 2016 at 8:21	history	edited	Joce	CC BY-SA 3.0	edit (thx!) changed one negation, restored meaning.
Jan 25, 2016 at 8:52	history	edited	user36790	CC BY-SA 3.0	added 99 characters in body
May 20, 2014 at 11:50	comment	added	Joce		@Sachin: Really focus on the word "symmetry": if it is symmetric, then after a mirror reflection (t -> -t), you should be able to superimpose the two "images", taht is, the trajectory. Said another way, you shouldn't be able to tell that the film is being played backwards. However, when there's friction, you can: there is no mirror symmetry.
May 20, 2014 at 9:41	comment	added	Sachin Kumar		Yes, this clarifies things. But in the reverse film, the sign of the frictional force doesn't change. My main problem was with the explanation that time reversal is equivalent to a movie played backwards, which it isn't.
May 20, 2014 at 9:39	vote	accept	Sachin Kumar
May 20, 2014 at 6:48	history	edited	Joce	CC BY-SA 3.0	added 9 characters in body
May 19, 2014 at 19:47	history	answered	Joce	CC BY-SA 3.0