Extracting work from magnetic-field-coupled Brownian particles

Thermodynamics of the magnetic-field-coupled Brownian particles is studied. We show that in the presence of the magnetic field, work can be extracted from the reservoir even when the measurement operation and the potential change operation are applied in different spatial directions. In particular, we show that more work can be extracted if the measurements are applied in two different directions simultaneously. In all these cases, we show that the generalized second law involving the measurement information and potential change is satisfied. In addition, we show how the continuous potential change and measurement position affect the work extraction.

Figure 1

$W_1$, $W_2$, $\Delta F$, and $I_m$ with $y_m$. Two different $x$-direction potentials are chosen: (1) $W_1$, $V_x\left(t\right)=\frac{1}{2}[0.5+0.5cos\left(0.5t\right)]x^2$, $t=4\pi$; (2) $W_2$, $V_x\left(t\right)=\frac{1}{2}[1-0.5sin\left(0.5t\right)]x^2$, $t=4\pi$. The potential in the $y$ direction is constant, $V_y=\frac{1}{2}{y}^2$. The thermal relaxation coefficients are $N=1$; the measurement is performed in the $y$ direction; parameters are $u_{y,0}=u_{x,0}=1$, $d_{y,0}=d_{x,0}=0$, and $u_{y,m}=0.01$; the magnetic field takes the value $L(qB/c\gamma )=0.5$.

Figure 2

(a) $W_y$, $W_x$, $W_{x+y}$, $\Delta F$, $I_x$ or $I_y$, and $I_{x+y}$ with measurement accuracy $u_m$ at time $t=4\pi$. (1) $W_y$, the measurement is along the $y$ direction, the position measurement result $y_m=15$. (2) $W_x$, the measurement is along the $x$ direction, the position measurement result $x_m=15$. (3) $W_{x+y}$, the measurements are along the $x$ and $y$ directions, the position measurement results $x_m=y_m=15$. The potential of the $x$ direction is $V_x\left(t\right)=\frac{1}{2}[1-0.5sin\left(0.5t\right)]x^2$, the constant potential on the $y$ direction, $V_y=\frac{1}{2}{y}^2$. (b) $W_y^{{}^{\prime }}$, $W_{xy}$, $\Delta F$, $I_x$ with measurement accuracy $u_m$ at time $t=4\pi$. (1) $W_y^{{}^{\prime }}$, the potential of the $x$ direction is constant, $V_x=\frac{1}{2}{x}^2$. (2) $W_{xy}$, the potential of the $x$ direction is $V_x\left(t\right)=\frac{1}{2}[1-0.5sin\left(0.5t\right)]x^2$. The potential in the $y$ direction is $V_y\left(t\right)=\frac{1}{2}[1-0.5sin\left(0.5t\right)]y^2$. The measurement is along the $x$ direction, and the position measurement result is $x_m=15$. For (a) and (b), the thermal relaxation coefficients are set at $N=1$; the parameters are set as $u_{y,0}=u_{x,0}=1$, $d_{y,0}=d_{x,0}=0$; the magnetic field is $L(qB/c\gamma )=0.5$.

Figure 3

The extracted work $W$ with the change of the magnetic field and the potential change frequency. The potential of the $x$ direction is $V_x\left(t\right)=\frac{1}{2}[1-0.5sin\left(k_{1}t\right)]x^2$; the evolution time is the half-integer period $k_{1}t=\pi$; the constant potential in the $y$ direction is $V_y=\frac{1}{2}{y}^2$. The thermal relaxation coefficients are set, $N=1$; the measurement is done in the $y$ direction; parameters are set as $u_{y,0}=u_{x,0}=1$, $d_{y,0}=d_{x,0}=0$, $u_{y,m}=0.01$, and $y_m=4$.

Figure 4

$W_{\mathrm{ave}}$ and $I_{\mathrm{ave}}$ with $L(qB/c\gamma )$ at time $t=2\pi$. These two quantities $W_{\mathrm{ave}}$ and $I_{\mathrm{ave}}$ average over different $y$ directional measurement outcomes $y_m$. The potential of the $x$ direction is, $V_x\left(t\right)=\frac{1}{2}[1-0.5sin\left(k_{1}t\right)]x^2$, and $W_{\mathrm{ave},1}$, $k_1=1$; $W_{\mathrm{ave},2}$, $k_1=1-0.5y_m$, seen in the inset. The constant potential in the $y$ direction is $V_y=\frac{1}{2}{y}^2$. The thermal relaxation coefficients are set, $N=1$; the measurement is done in the $y$ direction; parameters are set as $u_{y,0}=u_{x,0}=1$, $u_{y,m}=0.1$, $d_{y,0}=d_{x,0}=0$.