Subfield: Modern Physics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Field-drive is a locomotion mechanism that is analogous to general relativistic warp-drive. In this mechanism, an active particle continuously climbs up the field-gradient generated by its own influence on the environment so that the particle can bootstrap itself into a constant non-zero velocity motion. Consider a field-drive in one-dimensional (the $\mathrm{O} x$ axis) environment, where the position of the particle at time $t$ is given by $X(t)$ and its instantaneous velocity follows from:
$$
\frac{\mathrm{d}}{\mathrm{d} t} X(t)=\left.\kappa \frac{\partial}{\partial x} R(x, t)\right|_{x=X(t)}
$$
in which $\kappa$ is called the guiding coefficient and $R(x, t)$ is the field-value in this space. Note that, the operation ... $\left.\right|_{x=X(t)}$ means you have to calculate the part in ... first, then replace $x$ with $X(t)$. For a biological example, the active particle can be a cell, the field can be the nutrient concentration, and the strategy of climbing up the gradient can be chemotaxis. The cell consumes the nutrient and also responses to the local nutrient concentration, biasing its movement toward the direction where the concentration increases the most. If the nutrient is not diffusive and always recovers locally (e.g. a surface secretion) to the value which we defined to be 0 , then its dynamics can usually be approximated by:
$$
\frac{\partial}{\partial t} R(x, t)=-\frac{1}{\tau} R(x, t)-\gamma \exp \left\{-\frac{[x-X(t)]^{2}}{2 \lambda^{2}}\right\}
$$
where $\tau$ is the timescale of recovery, $\gamma$ is the consumption, and $\lambda$ is the characteristic radius of influence. Before we inoculate the cell into the environment, $R=0$ everywhere at any time. What is the smallest guiding coefficient $\kappa$ (in $\mu \mathrm{m}^{2} / \mathrm{s}$ ) for field-drive to emerge, if the parameters are $\tau=50 \mathrm{~s}, \gamma=1 \mathrm{~s}^{-1}$, and $\lambda=10 \mu \mathrm{m}$.



<img_4247>

Solution:
Assume that we inoculate the cell into the environment at position $x=0$ and $t=0$. The field dynamics at $t>0$ can be rewritten as:
$$
\begin{array}{r}
\frac{\partial}{\partial t} R(x, t)+\frac{1}{\tau} R(x, t)=\exp \left(-\frac{t}{\tau}\right) \partial_{t}\left[\exp \left(+\frac{t}{\tau}\right) R(x, t)\right]=-\gamma \exp \left\{-\frac{[x-X(t)]^{2}}{2 \lambda^{2}}\right\} \\
\Longrightarrow \exp \left(+\frac{t}{\tau}\right) R(x, t)=\int_{0}^{t} d t^{\prime} \exp \left(+\frac{t^{\prime}}{\tau}\right)\left(-\gamma \exp \left\{-\frac{\left[x-X\left(t^{\prime}\right)\right]^{2}}{2 \lambda^{2}}\right\}\right) \\
\Longrightarrow R(x, t)=-\gamma \int_{0}^{t} d t^{\prime} \exp \left\{-\frac{t-t^{\prime}}{\tau}-\frac{\left[x-X\left(t^{\prime}\right)\right]^{2}}{2 \lambda^{2}}\right\} .
\end{array}
\tag{39}
$$
If the cell can field-drive at a constant velocity $W>0$, then after a very long time $t \rightarrow+\infty$ we expect the cell will be in a steady-state, moving at this velocity. For consistency, this field-drive velocity $W$ should related to the field gradient evaluated at $x=X(t)$ such that:
$$
W=\left.\kappa \partial_{x} R(x, t)\right|_{x=X(t)}
\tag{40}
$$
From Eq. (39) we obtain:
$$
\begin{aligned}
W & =\left.\kappa \partial_{x}\left(-\gamma \int_{0}^{t} d t^{\prime} \exp \left\{-\frac{t-t^{\prime}}{\tau}-\frac{\left[x-X\left(t^{\prime}\right)\right]^{2}}{2 \lambda^{2}}\right\}\right)\right|_{x=X(t)} \\
& =\left.\frac{\kappa \gamma}{\lambda^{2}} \int_{0}^{t} d t^{\prime}\left[x-X\left(t^{\prime}\right)\right] \exp \left\{-\frac{t-t^{\prime}}{\tau}-\frac{\left[x-X\left(t^{\prime}\right)\right]^{2}}{2 \lambda^{2}}\right\}\right|_{x=X(t)} \\
& =\frac{\kappa \gamma}{\lambda^{2}} \int_{0}^{t} d t^{\prime}\left[X(t)-X\left(t^{\prime}\right)\right] \exp \left\{-\frac{t-t^{\prime}}{\tau}-\frac{\left[X(t)-X\left(t^{\prime}\right)\right]^{2}}{2 \lambda^{2}}\right\} .
\end{aligned}
$$
We then use the steady field-drive condition $X(t)-X\left(t^{\prime}\right)=W\left(t-t^{\prime}\right)$ at $t \rightarrow+\infty$ and define $t^{\prime \prime}=t-t^{\prime}$, so that the temporal integration $\int d t^{\prime \prime}$ will run from 0 to $+\infty$ :
$$
\begin{aligned}
W & =\frac{\kappa \gamma}{\lambda^{2}} \int_{0}^{t} d t^{\prime}\left[W\left(t-t^{\prime}\right)\right] \exp \left\{-\frac{t-t^{\prime}}{\tau}-\frac{\left[W\left(t-t^{\prime}\right)\right]^{2}}{2 \lambda^{2}}\right\} \\
& =\frac{\kappa \gamma}{\lambda^{2}} \int_{0}^{+\infty} d t^{\prime \prime}\left(W t^{\prime \prime}\right) \exp \left[-\frac{t^{\prime \prime}}{\tau}-\frac{\left(W t^{\prime \prime}\right)^{2}}{2 \lambda^{2}}\right] .
\end{aligned}
\tag{41}
$$

For the set of parameter values $(\kappa, \tau, \gamma, \lambda)$ when the field-drive mechanism start to emerge, we can treat the field-drive velocity as infinitesimal small $W=0^{+}$. Thus, divide both sides of Eq.(41) by $W$, we can arrive at:
$$
1=\left.\frac{\kappa \gamma}{\lambda^{2}} \int_{0}^{+\infty} d t^{\prime} t^{\prime \prime} \exp \left[-\frac{t^{\prime \prime}}{\tau}-\frac{\left(W t^{\prime \prime}\right)^{2}}{2 \lambda^{2}}\right]\right|_{W=0^{+}}=\frac{\kappa \gamma}{\lambda^{2}} \int_{0}^{+\infty} d t^{\prime} t^{\prime \prime} \exp \left(-\frac{t^{\prime \prime}}{\tau}\right)=\frac{\kappa \gamma \tau^{2}}{\lambda^{2}}
$$
Hence, the smallest guiding coefficient that give us field-drive, for $\tau=50 \mathrm{~s}, \gamma=1 \mathrm{~s}^{-1}, \lambda=10 \mu \mathrm{m}$ :
$$
\kappa=\frac{\lambda^{2}}{\gamma \tau^{2}}=4 \times 10^{-2} \mu \mathrm{m} / \mathrm{s}
$$

Final answer: $4 \times 10^{-2}$ $\mu \mathrm{m} / \mathrm{s}$