Time reversal symmetry in the presence of friction 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?
 A: 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.
A: 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.
A: 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.
