Energy consumption for shortcuts to adiabaticity

Shortcuts to adiabaticity let a system reach the results of a slow adiabatic process in a shorter time. We propose to quantify the “energy cost” of the shortcut by the energy consumption of the system enlarged by including the control device. A mechanical model where the dynamics of the system and control device can be explicitly described illustrates that a broad range of possible values for the consumption is possible, including zero (above the adiabatic energy increment) when friction is negligible and the energy given away as negative power is stored and reused by perfect regenerative braking.

Figure 1

Overhead crane composed of a load of mass $m$ and a trolley of mass $M$ connected through a rope of constant length $l$. The red solid arrows represent the active force ${\mathcal{F}}_a$ and the friction force ${\mathcal{F}}_r$ acting on a rightward-moving trolley.

Figure 2

The total power $\mathcal{P}$ to control $x\left(t\right)$ for different $M$ and friction coefficients and load power $P$. Symbols represent $\mathcal{P}$ using the small-oscillation approximation, Eq. (10), and lines represent using the exact Eq. (4). (a) Power $P$ of the load (symbols) and total power $\mathcal{P}$ (lines) in the $M=\gamma =0$ limit for $t_f=7$ s (green line and circles) and $t_f=$ 8 s (black line and diamonds). (b) Total power $\mathcal{P}$ with friction, $\gamma =15$ kg/s, $t_f=7$ s, for different values of the trolley mass: $M=0$ kg (green solid line and circles), $M=10$ kg (red solid line and squares), and $M=20$ kg (blue solid line and triangles). $m=10$ kg, $l=5$ m, $d=10$ m, $q\left(0\right)=0$ m, $\dot{q}\left(0\right)=0$ m/s, and $g=9.8{\mathrm{m}/\mathrm{s}}^2$.

Figure 3

Effect of the trajectory $q\left(t\right)$ on the total power $\mathcal{P}$ for different trolley masses. $q\left(0\right)=\dot{q}\left(0\right)=0$ (dashed lines), $q\left(0\right)=0.2$ m, and $\dot{q}\left(0\right)=0.1$ m/s (circles). (a) $q\left(t\right)$ for different initial conditions. (b) Corresponding power consumed for different $m/M$ ratios: $M=2$ kg (blue long-dashed line and small circles) and $M=100$ kg (green short-dashed line and big circles). $m=1$ kg, $l=5$ m, $d=10$ m, $t_f=7$ s, $\gamma =0$ kg/s, and $g=9.8{\mathrm{m}/\mathrm{s}}^2$.

Figure 4

Contour surface of the energy consumption $\mathcal{E}$ as a function of the CS variables $M$ and $\gamma$ for different values of the $\eta$ parameter: (a) $\eta =1$ and (b) $\eta =-1$. $m=10$ kg, $l=5$ m, $d=10$ m, $t_f=9$ s, $q\left(0\right)=0$ m, $\dot{q}\left(0\right)=0$ m/s, and $g=9.8$ m/${\mathrm{s}}^2$.

Figure 5

(a) Designed function $\xi =\alpha +x$ as a function of time. Polynomial interpolation used in Sec. 2 (red dashed line) and optimal solution to minimize the energy consumption (blue solid line). (b) Trolley displacement $x$ as a function of time. Parameter values: $t_f=8$ s, $l=5$ m, $d=10$ m, $g=9.8$ m/$s^2$, and $k_0=-1$.

Figure 6

Energy excitation of the load versus initial angle (with load initially at rest) in an inversely engineered transport process with $d=10$ m, $t_f=10$ s, $l=5$ m, and $m=10$ kg, using additional free parameters in the ansatz for $\alpha \left(t\right)$. The scaling factor is the kinetic energy for a constant-velocity process, $K_0=m{d}^2/\left(2t_f^2\right)$. Black dashed line: process without additional parameters; red dotted line: one parameter added to minimize excitation in ${\theta }_i={20}^{\circ }$ ($b_8=-3513.3$); solid blue line: minimization for the excitation in ${\theta }_i={20}^{\circ }$ and ${\theta }_i={45}^{\circ }$ using two free parameters ($b_8=-13862$ and $b_9=2941.5$).