 So far, we've examined the Carnot cycle in some detail. We've noted the total heat Q-Hot extracted from the hot reservoir, and the total heat Q-Cold delivered to the cold reservoir. We've seen that the reversed Carnot cycle is a precise mirror image in the sense that it completely undoes the forward cycle and restores the system and environment to their previous states. Heat Q-Hot is delivered to the hot reservoir and heat Q-Cold extracted from the cold reservoir. And we've found evidence that this reversibility might be related to the conservation of the quantity transferred heat divided by temperature. To gain more insight into reversibility, let's perform a similar analysis for the Sterling cycle. Like the Carnot cycle, the Sterling cycle employs isothermal expansion at high temperature, converting heat Q-Hot extracted from the hot reservoir into work, and isothermal compression at low temperature, converting work into heat Q-Cold transferred to a cold reservoir. But, whereas the Carnot cycle uses adiabatic expansion and compression to cool in heat, the Sterling cycle cools and heats the gas by direct transfer of heat Q-I equal to 3 halves NK T-Hot minus T-Cold to the cold reservoir in going from state 2 to state 3, and heat Q-I from the hot reservoir in going from state 4 to state 1. As shown at upper right, this results in the Sterling cycle extracting more heat from the hot reservoir, call it Q-Hot Prime, and delivering more heat to the cold reservoir, call it Q-Cold Prime. In each case, these are greater than the Carnot cycle values by Q-I. Now let's look at the reverse Sterling cycle. The isothermal processes are exactly reversed just as in the Carnot cycle. Heat Q-Hot that was extracted from the hot reservoir is now replaced, and heat Q-Cold that was delivered to the cold reservoir is now extracted. But again, the gas is heated by extracting heat Q-I from the hot reservoir and cooled by delivering heat Q-I to the cold reservoir. So, as shown at upper right, the heat delivered to the hot reservoir in the reverse Sterling cycle, labeled Q-Hot double prime, is not the same as the heat extracted in the forward cycle, what we called Q-Hot Prime. The net delivered heat is the isothermal term minus the extracted heat Q-I, minus, because positive heat extracted is negative heat delivered. Likewise, for the net heat extracted from the cold reservoir, labeled Q-Cold double prime, which is the isothermal term minus the delivered heat Q-I. Here positive delivered heat is negative extracted heat, shown here schematically In the forward cycle, heat energy Q-Hot Prime is extracted from the hot reservoir, a portion W is converted into work, and the remainder Q-Cold Prime is delivered to the cold reservoir. Both Q-Hot Prime and Q-Cold Prime are larger than the Carnot cycle values by the term Q-I. In the reverse cycle, work W is done on the system, causing heat Q-Cold double prime to be extracted from the cold reservoir, and heat Q-Hot double prime to be delivered to the hot reservoir. Both Q-Hot double prime and Q-Cold double prime are smaller than the Carnot cycle values by the term Q-I. The net effect of running both cycles is to extract from the hot reservoir heat Q-Hot Prime minus Q-Hot double prime, which equals to Q-I, and deliver to the cold reservoir heat Q-Cold Prime minus Q-Cold double prime, which also equals to Q-I. So the Sterling cycle is not reversible. Let's consider the quantity transferred heat divided by temperature. In the forward cycle, the amount of this delivered to the cold reservoir, Q-Cold Prime over T-Cold, minus the amount extracted from the hot reservoir, Q-Hot Prime over T-Hot is not equal to zero, unlike the Carnot cycle. Instead, it's Q-I over T-Cold minus Q-I over T-Hot, which is greater than zero. So this quantity was not conserved, but instead increased. It seems that the irreversibility of the Sterling cycle might be related to the increase of this somewhat mysterious quantity. It's becoming clear that central to the second law is the concept of reversible and irreversible processes. We need the second law because other laws of physics, such as the first law, cannot distinguish between these two types of processes. The first law tells us that energy is conserved. So for any process in a closed system, the change in energy, delta E, is zero. This means that a plot of energy versus time will be a flat line. If a process runs in the forward direction, delta E is zero. If the process runs in the reverse direction, delta E is zero. The first law is satisfied in both cases. Based only on conservation laws, such as the first law or the conservation of momentum and so on, the reverse of any allowable physical process is also allowable. In the kinetic model we've developed, the fundamental process is elastic collision between spheres. If we run any such collision in reverse, we also have a valid collision. Two particle elastic collisions are always reversible. So it would seem that any physical process resulting from a large number of such collisions should be reversible. Let's do a little thought experiment. Imagine you're at a pub and I ask you to come over to a billiard table to see her trick shot. Six balls have been nicely racked up at one end of the table. And a cue ball lies at the other end. To keep us in the realm of the theoretical ideal, let's assume this pub has somehow obtained ideal, friction-free billiard balls and tables. Your friend strikes the cue ball. It strikes the rack balls and balls scatter all over the table. Understandably, you are unimpressed with this none-too-tricky trick shot. After all, this is exactly what you'd expect billiard balls to do. Now, instead, imagine your friend calls you over to the table where you see seven balls in a more or less random pattern. Then your friend and six of her friends count down together. Three, two, one. They each strike one of the seven balls which scatter around the table. Still not very impressive. I take that back. That did end up being quite impressive. In fact, it looked like the original not-too-tricky shot run in reverse. Billiard balls just don't do that. They can't do that, can they? This leads us to consider reversibility in classical mechanics at a more rigorous level. Consider a system of end particles. At a given time, each particle has a position and a velocity, each specified by three coordinates. For the first particle, let's call these x1, 2, and 3, and v1, 2, and 3. For the second particle, x4, 5, and 6, and v4, 5, and 6, and so on. All the way up to x3n and v3n. For the i-th coordinate, the equation of motion is that vi dot, the rate of change of the i-th velocity, which is the i-th acceleration, ai, equals the corresponding component of force divided by the particle's mass. We assume the force is a function of all the particle positions. The solution of these equations give us each coordinate as a function of time. Here we assume time increases from left to right. Relative to the present position, xi of t, the past is on the left and the future is on the right. And the i-th velocity is the slope of the i-th position curve. For now, to simplify the notation, let's drop the subscript and focus on the evolution of a single coordinate. X of t is the present position. Now consider the position a short time h in the past, x of t minus h, and the position a short time in the future, x of t plus h. The way we solve mechanics problems is to use the equations of motion to relate future position to present and past positions. If we're solving the problem on a computer, we use a small but finite step size h. If we're solving the problem analytically, we employ calculus considering the limit as h grows arbitrarily small. The force law tells us the acceleration and the acceleration tells us how the velocity is changing with time. Let's consider the velocities at the past time t minus h over 2 and at the future time, t plus h over 2. We approximate the past velocity by the slope of the red line which connects the past and present positions. And we approximate the future velocity by the slope of the blue line which connects the present and future positions. This gives us v of t plus h over 2 is approximately x of t plus h minus x of t over h and v of t minus h over 2 is approximately x of t minus x of t minus h over h. Then the present acceleration, a of t, is approximately the change in velocity over the change in time. v of t plus h over 2 minus v of t minus h over 2 all over h. Plugging in our velocity approximations and simplifying, we arrive at a of t is approximately the future position plus the past position minus twice the present position all over h squared. Now we'll replace acceleration by force over mass multiply through by h squared and solve for the future position. This gives us the future position equals twice the present position minus the past position plus h squared over m times the present force. We can use this equation to move into the future as many time steps as desired. Now consider what happens when we reverse a physical process. We use t to denote time increasing from left to right and tau to denote time increasing from right to left. Psi of tau will denote position for the reversed process. Then the present is x of t equals psi of tau. The past position of the forward process x of t minus h equals the future position of the reverse process. Psi of tau plus h and the future position of the forward process x of t plus h equals the past position of the reverse process psi of tau minus h. The equation of motion for the reverse process is simply the forward equation with x and t replaced by psi and tau. Let's rewrite this by substituting x of t minus h for psi of tau plus h x of t for psi of tau x of t plus h for psi of tau minus h and f of x for f of psi. Now compare this to the equation above for the forward process. They are variations of the same equation. The only difference is the future and past position terms have been swapped. This means that for any mechanical process that evolves according to the equations of motion, the reverse process also satisfies the equations of motion. That is, the reverse of every possible mechanical process is a possible mechanical process. So, every possible mechanical process is reversible. Back to our system of n particles and their 3n coordinates. Here we represent four arbitrary coordinates, xi, xj, xk and xl. Suppose we've determined all coordinates for the first two time steps. We can do this if we know the initial positions and velocities of all the particles. Then we use our equations of motion to calculate all coordinates at the next time step and the next and so on for as long as desired. We end up with the complete trajectories of all particles. Now, take the coordinate values at the last two time steps. We can determine these from the final particle positions and velocities if necessary. Then consider time to increase from right to left and apply the equations of motion. The trajectories are precisely retraced in the reverse order. The evolution of the entire system exactly reverses. We say that classical physics is microscopically reversible. Since every object in our experience is composed of simple particles, atoms, the evolution of a macroscopic system is just the net result of the evolution of that system of particles, which is reversible. Therefore, all macroscopic processes should be reversible. But, we've already said that some macroscopic processes, such as the Sterling cycle, are not reversible. How can we make sense of these two contradictory statements? This is known as Lo-Schmidt's paradox and it was a great puzzle for physicists in the late 19th and early 20th centuries.