7
$\begingroup$

I was reading a paper on time reversal symmetry, and came across an example of a pendulum swinging in the presence of friction:

When we consider the more realistic physical situation of a swinging pendulum in the presence of friction, we can tell the difference between a forward and a reverse film of this pendulum. Namely, the original (forward) film will show the swinging pendulum losing amplitude until it comes to a standstill. However, the film in reverse direction shows a swinging pendulum whose amplitude is increasing in time. The latter film is clearly unphysical as it does not satisfy the natural laws of motion anymore (assuming there is no hidden source of energy feeding the pendulum). The presence of friction breaks the time-reversal symmetry of the ideal pendulum.

I don't understand this. If I am not wrong, we are only reversing the direction of time, not that of the frictional force (the nature and cause of this force is not relevant either). In the movie played backwards, the frictional force will be along the direction in which the bob swings. This force would be proportional to $v^2$, and would give the bob more momentum consistent with the observation. So what am I missing? How does the presence of friction break time reversal symmetry?

Improve this question
$\endgroup$
2
  • $\begingroup$ If you look at the molecular level, what eventually happens in the forward scenario is that the pendulum gives off its energy to the random molecular motion and when you reverse it, random molecular motion energy changes to energy of the pendulum. by energy I mean kinetic energy. and you can kind of see disorderness increasing in forward flow of time. So everything is consistent in forward and backward direction with the mechanical laws,i.e. with conservation of energy,momentum and angular momentum. The problem with considering friction as a whole force instead of breaking it down to molecular $\endgroup$
    – user23503
    Commented May 20, 2014 at 6:59
  • $\begingroup$ (contd.) level is that the law you associate with friction are approximations, and don't give you the correct picture in all scenarios as to what is happening. It is a correct approximation to say friction opposes relative motion and obeys the law $\mu N$ to predict acceleration but not the correct picture to observe time reversal symmetry as however good an approximation may be, it is still incorrect. $\endgroup$
    – user23503
    Commented May 20, 2014 at 7:03

3 Answers 3

Reset to default
11
$\begingroup$

You have all the elements in your question, your difficulty is about what is meant by "time reversal symmetry". Time reversal symmetry holds if, when "playing backwards", the motion observed obeys the same law. With friction it is not the case : friction opposes movement, when playing backwards it (seemingly) promotes it.

You can also go to equations for this. Let's have a damped oscillator of mass $m$: $$ m \frac{\mathrm d^2x}{\mathrm dt^2} = - c \frac{\mathrm dx}{\mathrm dt} - k x $$ Now play backwards with $\tau=-t$, thus $\mathrm d\tau/\mathrm dt = -1$: $$ m \frac{\mathrm d^2x}{\mathrm d\tau^2} = c \frac{\mathrm dx}{\mathrm d\tau} - k x $$

So, the equation is not the same—unless $c= 0$ : the physics of the backward time phenomenon is not the same if there's friction—while it is the same when there is no friction.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Sachin Kumar
    Commented May 20, 2014 at 9:41
  • 1
    $\begingroup$ @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. $\endgroup$
    – Joce
    Commented May 20, 2014 at 11:50
  • $\begingroup$ $\mathrm d^2\tau/\mathrm dt^2 = 0$ and not 1. and that is also not what we need here. $\mathrm d\tau^2/\mathrm dt^2 = 1$ should be edited in. $\endgroup$
    – Bibek_G
    Commented Aug 7, 2019 at 4:47
2
$\begingroup$

The frictional force $always$ opposes the direction of motion. Try to pull a rock on a sandy beach, or swim in water. It does not matter what direction you move you experience resistance to the motion. So friction is $not$ reversible in the sense that you could 'reverse" it by playing it backwards in time, played forward or backward played, you have resistance.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Suppose along with the bob, I attach a hypothetical device which shows the direction of forces acting on it. Would the directions change in the reversed film? Yes, I understand what you're saying about friction, but my point is that the reverse film would still be consistent with known mechanical laws. $\endgroup$
    – Sachin Kumar
    Commented May 19, 2014 at 16:32
  • $\begingroup$ the reverse film would still be consistent with known mechanical laws - the reverse film would be consistent with mechanical laws, but not with thermodynamical laws. $\endgroup$
    – Johannes
    Commented May 19, 2014 at 16:38
  • $\begingroup$ Exactly! As @Johannes has said it consistency with mechanical laws are not the same as consistency with thermodynamic laws. Newton's laws are approximation to macroscopic reality without taking into account thermal effects. $\endgroup$
    – hyportnex
    Commented May 19, 2014 at 16:53
0
$\begingroup$

It might help if you analyse the situation using thermodynamics. In the forward picture, the pendulum, left to itself, heats its hinge, increasing the entropy of the system. In the reverse picture, we see the same isolated system losing entropy by transforming the disorderly heat into the orderly oscillation of the pendulum. That appears unphysical.

Further, frictional force is usually proportional to $\vec{v}$. Irrespective of its exact dependence on velocity, it is a non-conservative force. Therefore, one cannot extract "work" from it.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.