Bubble dynamics and load mapping

Source: supplied project-theory book, Stage 3. The complete bilingual chapter follows, with its derivations and qualifications. Textbook source (Markdown) · Original Markdown.

来源：所给项目理论教材阶段 3。下方完整收录双语章节及推导和限定。教材原文（Markdown） · 原始 Markdown。

Keep the source’s evidence states: optical normalization is an accounting check; activation is uncalibrated; R(t) is prescribed; near-wall loading and cohesive release remain conditional.

保留来源证据状态：光学归一化属于核算检查；激活尚未标定；R(t) 为规定输入；近壁加载及黏聚释放仍有条件。

3. Bubble dynamics 3.1 Continuity determines the liquid velocity 3.2 The velocity potential gives the pressure field 3.3 Applying the wall condition gives Rayleigh–Plesset Why this equation does not yet predict laser-induced growth 3.4 Deriving the Rayleigh collapse time 3.5 Deriving the prescribed smooth trajectory 3.6 Mapping bubble pressure onto a load disk 3.7 What the signed pressure does—and does not—mean

3. Bubble dynamics

气泡动力学

From a moving radius to a pressure field. Corresponding to Theory 03, source page 3.

从变化的半径到压力场。对应 Theory 03，原始材料第3页。

This is the central mechanics derivation.

这是整套理论中最核心的力学推导。

The spherical model assumes a bubble of radius $R(t)$ , approximately uniform internal pressure $p_B(t)$ , and an unbounded incompressible surrounding liquid. Let the liquid density and dynamic viscosity be $\rho_l$ and $\mu_l$ , and its distant pressure be $p_\infty(t)$ . We neglect the correction to interface kinematics caused by mass transfer.

球对称模型假设气泡半径为 $R(t)$ ，内部压力 $p_B(t)$ 近似均匀，周围为无界、不可压缩液体。液体密度和动力黏度分别记为 $\rho_l$ 和 $\mu_l$ ，远处液体压力为 $p_\infty(t)$ 。这里忽略质量传递对界面运动学条件造成的修正。

These are the classical assumptions behind the radial flow and Rayleigh–Plesset framework; they are not assumptions of a fully resolved near-wall bubble. [R1]

这些是径向流动与 Rayleigh–Plesset 框架背后的经典假设，并不等同于对近壁气泡进行完全解析时所采用的模型。[R1]

3.1 Continuity determines the liquid velocity

连续性方程确定液体速度

For a purely radial, incompressible flow,

对于纯径向不可压缩流动，

\nabla\cdot\mathbf u=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2u)=0.

3.1 Continuity determines the liquid velocity / 连续性方程确定液体速度; u: liquid or solid particle velocity; r: liquid observation radius / embryo radius

Thus,

因此

r^2u=C(t).

3.1 Continuity determines the liquid velocity / 连续性方程确定液体速度; r: liquid observation radius / embryo radius; u: liquid or solid particle velocity; C: gas-content constant / damping; t: physical time

At the bubble wall, the liquid follows the interface:

在气泡壁面处，液体跟随界面运动：

u(R,t)=\dot R.

3.1 Continuity determines the liquid velocity / 连续性方程确定液体速度; u: liquid or solid particle velocity; R: instantaneous wall radius; t: physical time; Ṙ: radial wall velocity

Therefore,

于是

C(t)=R^2\dot R,

3.1 Continuity determines the liquid velocity / 连续性方程确定液体速度; C: gas-content constant / damping; t: physical time; R: instantaneous wall radius; Ṙ: radial wall velocity

and

从而

\boxed{u(r,t)=\frac{R^2\dot R}{r^2}}.

3.1 Continuity determines the liquid velocity / 连续性方程确定液体速度; u: liquid or solid particle velocity; r: liquid observation radius / embryo radius; t: physical time; R: instantaneous wall radius; Ṙ: radial wall velocity

The interpretation is simple: the same instantaneous volume flux passes through every concentric spherical surface,

其物理含义很简单：通过每一个同心球面的瞬时体积流量都相同，即

4\pi r^2u=4\pi R^2\dot R.

3.1 Continuity determines the liquid velocity / 连续性方程确定液体速度; π: circle geometry constant; r: liquid observation radius / embryo radius; u: liquid or solid particle velocity; R: instantaneous wall radius; Ṙ: radial wall velocity

The entire radial velocity field is therefore determined by one coordinate, $R(t)$ .

因此，整个径向速度场由单个坐标 $R(t)$ 决定。

3.2 The velocity potential gives the pressure field

由速度势求得压力场

Introduce a velocity potential $\Phi$ such that

引入速度势 $\Phi$ ，使其满足

u=\frac{\partial\Phi}{\partial r}.

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; u: liquid or solid particle velocity; Φ: velocity potential; r: liquid observation radius / embryo radius

Choosing $\Phi\to0$ at infinity,

取无穷远处 $\Phi\to0$ ，则

\boxed{\Phi(r,t)=-\frac{R^2\dot R}{r}}.

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; Φ: velocity potential; r: liquid observation radius / embryo radius; t: physical time; R: instantaneous wall radius; Ṙ: radial wall velocity

The unsteady Bernoulli relation is

非定常 Bernoulli 关系为

\frac{\partial\Phi}{\partial t}+\frac12u^2+\frac{p-p_\infty}{\rho_l}=0.

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; Φ: velocity potential; t: physical time; u: liquid or solid particle velocity; p: pressure at the indicated location; p∞: ambient liquid pressure; ρ_l: liquid / layer / droplet density

The time derivative is taken at a fixed spatial location $r$ :

时间偏导数是在固定空间位置 $r$ 处计算的：

\frac{\partial\Phi}{\partial t}=-\frac{R^2\ddot R+2R\dot R^2}{r}.

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; Φ: velocity potential; t: physical time; R: instantaneous wall radius; R̈: radial wall acceleration; Ṙ: radial wall velocity; r: liquid observation radius / embryo radius

Also,

同时，

\frac12u^2=\frac{R^4\dot R^2}{2r^4}.

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; u: liquid or solid particle velocity; R: instantaneous wall radius; Ṙ: radial wall velocity; r: liquid observation radius / embryo radius

Consequently,

因此

\boxed{p'(r,t)\equiv p(r,t)-p_\infty(t)=\rho_l\left[\frac{R^2\ddot R+2R\dot R^2}{r}-\frac{R^4\dot R^2}{2r^4}\right]}.

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; p': pressure at the indicated location; r: liquid observation radius / embryo radius; t: physical time; p: pressure at the indicated location; p∞: ambient liquid pressure; ρ_l: liquid / layer / droplet density; R: instantaneous wall radius; R̈: radial wall acceleration; Ṙ: radial wall velocity

This reconstructs the pressure-mapping equation shown on source page 3. [S1, p. 3]

这就重构了原始材料第3页所示的压力映射方程。[S1，第3页]

Define

定义

A(t)=R^2\ddot R+2R\dot R^2,\qquad B(t)=R^4\dot R^2.

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; A: area / volume-acceleration coefficient; t: physical time; R: instantaneous wall radius; R̈: radial wall acceleration; Ṙ: radial wall velocity; B: kinetic pressure coefficient

Then

则

p'=\rho_l\left(\frac{A}{r}-\frac{B}{2r^4}\right).

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; p': pressure at the indicated location; ρ_l: liquid / layer / droplet density; A: area / volume-acceleration coefficient; r: liquid observation radius / embryo radius; B: kinetic pressure coefficient

The leading term has an especially useful interpretation. Since

主导项具有一个特别有用的物理解释。由于

V=\frac{4\pi}{3}R^3,

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; V: bubble or observation volume; π: circle geometry constant; R: instantaneous wall radius

we have

可得

\ddot V=4\pi\left(R^2\ddot R+2R\dot R^2\right)=4\pi A.

Thus,

因此

\frac{\rho_lA}{r}=\frac{\rho_l}{4\pi r}\ddot V.

3.2 The velocity potential gives the pressure field / 由速度势求得压力场; ρ_l: liquid / layer / droplet density; A: area / volume-acceleration coefficient; r: liquid observation radius / embryo radius; π: circle geometry constant; V̈: bubble or observation volume

Pressure is closely connected to bubble-volume acceleration, not merely to bubble size or whether the bubble is growing.

压力与气泡体积的二阶时间导数密切相关，而不仅仅取决于气泡尺寸，或气泡是否正在增长。

That is why an expanding bubble can produce a negative pressure perturbation while its expansion decelerates.

因此，一个正在膨胀的气泡，在膨胀过程减速时，也可能产生负的压力扰动。

3.3 Applying the wall condition gives Rayleigh–Plesset

由壁面条件得到 Rayleigh–Plesset 方程

Evaluate the pressure relation at $r=R$ :

在 $r=R$ 处计算上述压力关系：

p_l(R,t)-p_\infty=\rho_l\left[R\ddot R+2\dot R^2-\frac12\dot R^2\right].

3.3 Applying the wall condition gives Rayleigh–Plesset / 由壁面条件得到 Rayleigh–Plesset 方程; p_l: liquid-side pressure; R: instantaneous wall radius; t: physical time; p∞: ambient liquid pressure; ρ_l: liquid / layer / droplet density; R̈: radial wall acceleration; Ṙ: radial wall velocity

Therefore,

因此

\boxed{p_l(R,t)-p_\infty=\rho_l\left(R\ddot R+\frac32\dot R^2\right)}.

Here $p_l(R,t)$ is the liquid pressure immediately outside the interface. It differs from the bubble pressure because of surface tension and viscous normal stress.

这里的 $p_l(R,t)$ 是紧邻界面外侧的液体压力。由于表面张力和黏性法向应力的存在，它与气泡内部压力并不相同。

Let $\sigma$ denote surface tension. The normal-stress condition is

以 $\sigma$ 表示表面张力。法向应力条件为

p_B-p_l(R)=\frac{2\sigma}{R}+\frac{4\mu_l\dot R}{R}.

3.3 Applying the wall condition gives Rayleigh–Plesset / 由壁面条件得到 Rayleigh–Plesset 方程; p_B: total interior bubble pressure; p_l: liquid-side pressure; R: instantaneous wall radius; σ: interface tension / solid stress; μ_l: dynamic viscosity / absorption coefficient; Ṙ: radial wall velocity

The $2\sigma/R$ term follows from the force balance on a hemispherical interface. The viscous term follows from

其中， $2\sigma/R$ 项由半球形界面的受力平衡得到。黏性项来自

\left.\frac{\partial u}{\partial r}\right|_{r=R}=-\frac{2\dot R}{R}

3.3 Applying the wall condition gives Rayleigh–Plesset / 由壁面条件得到 Rayleigh–Plesset 方程; u: liquid or solid particle velocity; r: liquid observation radius / embryo radius; R: instantaneous wall radius; Ṙ: radial wall velocity

and the Newtonian normal stress $2\mu_l\partial u/\partial r$ .

以及牛顿流体的黏性法向应力 $2\mu_l\partial u/\partial r$ 。

Substituting yields

代入可得

\boxed{\rho_l\left(R\ddot R+\frac32\dot R^2\right)=p_B-p_\infty-\frac{2\sigma}{R}-\frac{4\mu_l\dot R}{R}}.

3.3 Applying the wall condition gives Rayleigh–Plesset / 由壁面条件得到 Rayleigh–Plesset 方程; ρ_l: liquid / layer / droplet density; R: instantaneous wall radius; R̈: radial wall acceleration; Ṙ: radial wall velocity; p_B: total interior bubble pressure; p∞: ambient liquid pressure; σ: interface tension / solid stress; μ_l: dynamic viscosity / absorption coefficient

This is the Rayleigh–Plesset equation printed in the source. [S1, p. 3]

这就是原始材料给出的 Rayleigh–Plesset 方程。[S1，第3页]

Its structure should look familiar:

它的结构应当十分熟悉：

\underbrace{\rho_l(R\ddot R+\tfrac32\dot R^2)}_{\text{liquid inertia / 液体惯性}} = \underbrace{p_B-p_\infty}_{\text{pressure driving / 压力驱动}} - \underbrace{2\sigma/R}_{\text{surface tension / 表面张力}} - \underbrace{4\mu_l\dot R/R}_{\text{viscous resistance / 黏性阻力}}.

One subtlety is that viscosity can appear in the wall stress even though we used the potential-flow pressure relation: for this incompressible spherical velocity field, the bulk viscous-force term vanishes, while the interfacial viscous stress does not.

一个细节是：虽然采用了势流压力关系，黏性仍然可以出现在壁面应力中。对于这一不可压缩球对称速度场，液体内部的黏性力项为零，但界面处的黏性应力并不为零。

Why this equation does not yet predict laser-induced growth

为什么该方程还不能预测激光诱导增长

Rayleigh–Plesset determines $R(t)$ only after the internal pressure is specified consistently.

只有在以自洽方式给定内部压力之后，Rayleigh–Plesset 方程才能确定 $R(t)$ 。

For a laser-heated, vaporizing droplet, that requires additional thermal and phase-change information. The source does not supply a closed internal-pressure, heat-transfer, and mass-transfer model. Instead, it uses Rayleigh–Plesset inversely:

对于受到激光加热并发生汽化的液滴，这需要额外的热学和相变信息。原始材料没有提供闭合的内部压力、传热与传质模型，而是反向使用 Rayleigh–Plesset 方程：

\boxed{p_{B,\mathrm{required}}(t)=p_\infty+\rho_l\left(R\ddot R+\frac32\dot R^2\right)+\frac{2\sigma}{R}+\frac{4\mu_l\dot R}{R}}.

Why this equation does not yet predict laser-induced growth / 为什么该方程还不能预测激光诱导增长; p_B,required: pressure at the indicated location; t: physical time; p∞: ambient liquid pressure; ρ_l: liquid / layer / droplet density; R: instantaneous wall radius; R̈: radial wall acceleration; Ṙ: radial wall velocity; σ: interface tension / solid stress; μ_l: dynamic viscosity / absorption coefficient

Given a prescribed $R(t)$ , this asks what internal pressure would be required to support that trajectory. It is a consistency check, not a forward prediction of activation and growth. [S1, p. 3]

在给定人为规定的 $R(t)$ 后，这个式子回答的是：维持该轨迹需要怎样的内部压力？它属于一致性检查，而不是对激活与增长的正向预测。[S1，第3页]

3.4 Deriving the Rayleigh collapse time

推导 Rayleigh 塌缩时间

The collapse-time estimate comes from a simpler limiting problem.

塌缩时间估计来自一个更简单的极限问题。

Assume constant bubble pressure $p_v$ , constant external pressure $p_\infty>p_v$ , negligible surface tension and viscosity, and an initially stationary bubble of radius $R_{\max}$ . Define

假设气泡内部压力恒为 $p_v$ ，外界压力恒为 $p_\infty>p_v$ ，表面张力与黏性可忽略，且气泡初始静止、半径为 $R_{\max}$ 。定义

\Delta p=p_\infty-p_v>0.

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Then

则

R\ddot R+\frac32\dot R^2=-\frac{\Delta p}{\rho_l}.

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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Set $v(R)=\dot R^2$ . Along the trajectory,

令 $v(R)=\dot R^2$ 。沿着运动轨迹有

\ddot R=\frac12\frac{dv}{dR},

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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因此

R\frac{dv}{dR}+3v=-\frac{2\Delta p}{\rho_l}.

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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Multiplication by $R^2$ gives

两边乘以 $R^2$ ，得到

\frac{d}{dR}(R^3v)=-\frac{2\Delta p}{\rho_l}R^2.

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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Using $v(R_{\max})=0$ ,

利用 $v(R_{\max})=0$ ，得到

\boxed{\dot R^2=\frac{2\Delta p}{3\rho_l}\left[\left(\frac{R_{\max}}{R}\right)^3-1\right]}.

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

During collapse, take the negative square root. The time to reach zero radius is

在塌缩阶段，应取负平方根。到达零半径所需的时间为

t_R=\int_0^{R_{\max}}\frac{dR}{\sqrt{\dfrac{2\Delta p}{3\rho_l}\left[\left(\dfrac{R_{\max}}{R}\right)^3-1\right]}}.

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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With $x=R/R_{\max}$ ,

令 $x=R/R_{\max}$ ，则

t_R=R_{\max}\sqrt{\frac{3\rho_l}{2\Delta p}}\int_0^1\frac{x^{3/2}}{\sqrt{1-x^3}}\,dx.

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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Evaluation of the dimensionless integral gives

计算该无量纲积分可得

\boxed{t_R\approx0.915R_{\max}\sqrt{\frac{\rho_l}{\Delta p}}}.

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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This is the origin of the coefficient $0.915$ , rather than an empirical fitting factor.

这就是系数 $0.915$ 的来源；它不是经验拟合系数。

The source reports $t_R=0.735\ \mu\mathrm s$ for its reference case and uses it only as a timescale check. Its prescribed trajectory does not actually collapse to zero radius. [S1, p. 3]

原始材料给出的参考情景结果为 $t_R=0.735\ \mu\mathrm s$ ，并仅将其用于时间尺度检查。材料中人为规定的轨迹实际上并不会塌缩到零半径。[S1，第3页]

3.5 Deriving the prescribed smooth trajectory

推导规定的平滑轨迹

The source constructs a smooth radius history instead of solving a forward cavitation problem.

原始材料构造了一条平滑的半径历程，而没有求解正向空化问题。

We want a function $S(\xi)$ , $0\le\xi\le1$ , satisfying

需要构造函数 $S(\xi)$ ，其中 $0\le\xi\le1$ ，并满足

S(0)=0,\qquad S(1)=1,

3.5 Deriving the prescribed smooth trajectory / 推导规定的平滑轨迹; S: smoothstep / capillary scale

and

以及

S'(0)=S'(1)=0,\qquad S''(0)=S''(1)=0.

3.5 Deriving the prescribed smooth trajectory / 推导规定的平滑轨迹; S': smoothstep / capillary scale; S'': smoothstep / capillary scale

These six conditions determine a fifth-degree polynomial:

这六个条件确定了一个五次多项式：

\boxed{S(\xi)=10\xi^3-15\xi^4+6\xi^5}.

3.5 Deriving the prescribed smooth trajectory / 推导规定的平滑轨迹; S: smoothstep / capillary scale; ξ: normalized branch time

Its derivatives are

其导数为

S'(\xi)=30\xi^2(1-\xi)^2,

S''(\xi)=60\xi(1-\xi)(1-2\xi).

Let $\Delta R=R_{\max}-R_0$ , with growth time $t_g$ and shrinkage time $t_c$ . Then

令 $\Delta R=R_{\max}-R_0$ ，增长时间为 $t_g$ ，收缩时间为 $t_c$ 。则

\boxed{R(t)=\begin{cases}R_0+\Delta R\,S(t/t_g),&0\le t\le t_g,\\[4pt]R_{\max}-\Delta R\,S((t-t_g)/t_c),&t_g<t\le t_g+t_c.\end{cases}}

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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The zero endpoint derivatives prevent artificial velocity and acceleration jumps. This is the stated purpose of the prescribed trajectory. [S1, p. 3]

端点处一阶与二阶导数为零，可避免速度和加速度出现人为跳变。这正是原始材料构造该规定轨迹的目的。[S1，第3页]

Because $S'$ has its maximum at $\xi=1/2$ ,

由于 $S'$ 在 $\xi=1/2$ 处达到最大值，

S'_{\max}=\frac{30}{16}=1.875.

3.5 Deriving the prescribed smooth trajectory / 推导规定的平滑轨迹; S'_max: smoothstep / capillary scale

Therefore,

因此

\dot R_{\max,\mathrm{growth}}=1.875\frac{\Delta R}{t_g},\qquad |\dot R|_{\max,\mathrm{collapse}}=1.875\frac{\Delta R}{t_c}.

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Using the supplied values

代入所提供的数值

R_0=1\ \mu\mathrm m,\quad R_{\max}=8\ \mu\mathrm m,\quad t_g=0.7\ \mu\mathrm s,\quad t_c=0.6\ \mu\mathrm s,

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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gives

得到

\dot R_{\max,\mathrm{growth}}=18.75\ \mathrm{m\,s^{-1}},

3.5 Deriving the prescribed smooth trajectory / 推导规定的平滑轨迹; Ṙ_max,growth: bubble wall radius; m: mass or areal mass; s: in-plane radial coordinate

\boxed{|\dot R|_{\max,\mathrm{collapse}}=21.875\ \mathrm{m\,s^{-1}}}.

3.5 Deriving the prescribed smooth trajectory / 推导规定的平滑轨迹; Ṙ: radial wall velocity; m: mass or areal mass; s: in-plane radial coordinate

This exactly reproduces the reported maximum wall speed. It follows from the chosen polynomial and timings—not independently predicted cavitation dynamics. [S1, p. 5]

这精确复现了原始材料报告的最大壁面速度。该数值来自选定的多项式与时间参数，而不是独立预测得到的空化动力学结果。[S1，第5页]

3.6 Mapping bubble pressure onto a load disk

将气泡压力映射至载荷圆盘

Let the bubble centre be a distance $d$ from a plane. Use $s$ for radial distance along that plane. Then the distance from the bubble centre to a point on the plane is

设气泡中心距某平面的距离为 $d$ ，以 $s$ 表示该平面上的径向距离。则气泡中心到平面上某点的距离为

r=\sqrt{d^2+s^2}.

3.6 Mapping bubble pressure onto a load disk / 将气泡压力映射至载荷圆盘; r: liquid observation radius / embryo radius; s: in-plane radial coordinate

Substitution into the spherical pressure field gives

代入球对称压力场，得到

p'(s,t)=\rho_l\left[\frac{A(t)}{\sqrt{d^2+s^2}}-\frac{B(t)}{2(d^2+s^2)^2}\right].

3.6 Mapping bubble pressure onto a load disk / 将气泡压力映射至载荷圆盘; p': pressure at the indicated location; s: in-plane radial coordinate; t: physical time; ρ_l: liquid / layer / droplet density; A: area / volume-acceleration coefficient; B: kinetic pressure coefficient

The average over a disk of radius $L$ is

在半径为 $L$ 的圆盘上的平均压力为

\bar p(t)=\frac{1}{\pi L^2}\int_0^L p'(s,t)\,2\pi s\,ds.

3.6 Mapping bubble pressure onto a load disk / 将气泡压力映射至载荷圆盘; p̄: pressure at the indicated location; t: physical time; π: circle geometry constant; L: length / latent heat / load disk radius; p': pressure at the indicated location; s: in-plane radial coordinate

The required integrals are

所需积分为

\int_0^L\frac{s}{\sqrt{d^2+s^2}}\,ds=\sqrt{d^2+L^2}-d,

3.6 Mapping bubble pressure onto a load disk / 将气泡压力映射至载荷圆盘; L: length / latent heat / load disk radius; s: in-plane radial coordinate

and

以及

\int_0^L\frac{s}{(d^2+s^2)^2}\,ds=\frac12\left(\frac1{d^2}-\frac1{d^2+L^2}\right).

Hence,

因此

\boxed{\begin{aligned}\bar p(t)=\rho_l\bigg[&\frac{2A(t)}{L^2}\left(\sqrt{d^2+L^2}-d\right)\\&-\frac{B(t)}{2L^2}\left(\frac1{d^2}-\frac1{d^2+L^2}\right)\bigg].\end{aligned}}

3.6 Mapping bubble pressure onto a load disk / 将气泡压力映射至载荷圆盘; p̄: pressure at the indicated location; t: physical time; ρ_l: liquid / layer / droplet density; A: area / volume-acceleration coefficient; L: length / latent heat / load disk radius; B: kinetic pressure coefficient

This reconstructs the load-disk average on source page 3. [S1, p. 3]

这就重构了原始材料第3页给出的载荷圆盘平均压力。[S1，第3页]

As a check, letting $L\to0$ recovers the axis value,

作为检查，令 $L\to0$ ，可以恢复轴线上的压力值：

\bar p\to\rho_l\left(\frac{A}{d}-\frac{B}{2d^4}\right).

3.6 Mapping bubble pressure onto a load disk / 将气泡压力映射至载荷圆盘; p̄: pressure at the indicated location; ρ_l: liquid / layer / droplet density; A: area / volume-acceleration coefficient; B: kinetic pressure coefficient

3.7 What the signed pressure does—and does not—mean

带符号压力的含义与局限

The signs follow from the acceleration terms, not simply from the sign of $\dot R$ . Growth acceleration, growth deceleration, collapse acceleration, and collapse deceleration can generate different pressure branches.

压力的正负号来自加速度相关项，而不只是由 $\dot R$ 的符号决定。增长加速、增长减速、塌缩加速与塌缩减速可能产生不同的压力分支。

The model consequently produces a time-dependent signed load. Replacing it by a constant positive pressure would change the structural problem. The source specifically emphasizes the relative timing of these branches. [S1, p. 3]

因此，该模型产生的是一个随时间变化、具有正负号的载荷。如果用恒定正压替代它，就会改变结构问题本身。原始材料特别强调这些分支之间的相对时序。[S1，第3页]

However, there are two important limits.

不过，这里存在两项重要限制。

First, the reported stand-off ratio is

第一，原始材料报告的距壁比为

\gamma=\frac{d}{R_{\max}}=1.0625.

3.7 What the signed pressure does—and does not—mean / 带符号压力的含义与局限; γ: stand-off / heat-capacity ratio; R_max: maximum expansion radius

With $R_{\max}=8\ \mu\mathrm m$ , this implies $d=8.5\ \mu\mathrm m$ . The bubble is close to the boundary, whereas the pressure expression was derived for an unbounded spherical flow. Evaluating it on a plane does not enforce a real wall boundary condition. The source therefore does not claim validated near-wall pressure, jetting, or shock prediction. [S1, p. 3]

当 $R_{\max}=8\ \mu\mathrm m$ 时，可得 $d=8.5\ \mu\mathrm m$ 。此时气泡非常接近边界，但压力表达式是在无界球对称流动下推导的。仅仅在一个平面上计算该表达式，并不能满足真实壁面的边界条件。因此，原始材料并未声称已经预测出经验证的近壁压力、射流或冲击波。[S1，第3页]

Second,

第二，

p_{\mathrm{absolute}}=p_\infty+p'.

3.7 What the signed pressure does—and does not—mean / 带符号压力的含义与局限; p_absolute: pressure at the indicated location; p∞: ambient liquid pressure; p': pressure at the indicated location

The source explicitly flags its negative source branch as failing the reference absolute-pressure check. That branch cannot simply be treated as a validated physical load and passed downstream. [S1, p. 5]

原始材料明确指出，其负压源分支未通过参考情景的绝对压力检查。因此，不能直接将该分支视为经验证的物理载荷，并传递给下游模型。[S1，第5页]

This is a limitation of this calculation, not a universal statement that liquids cannot sustain tension: metastable liquids can sustain negative pressure under appropriate conditions. Such behaviour requires its own nucleation and metastability description, which is absent here. [R2]

这是当前计算本身的限制，并不是说液体普遍不能承受拉伸：在适当条件下，亚稳态液体能够承受负压。但描述这种行为需要相应的成核与亚稳态模型，而当前模型中没有这些内容。[R2]

Integrated calculation and model check

综合计算与模型检查

Use the declared parameter set in each part. The values are teaching scenarios unless specifically identified as inherited project inputs. Each reference panel follows its question.

使用每小题明确给出的参数。除明确标为沿用项目输入外，数值均为教学情景。各参考解答紧随题目。

Use the source R0 = 1 µm, Rmax = 8 µm, tg = 0.7 µs and tc = 0.6 µs. Derive the maximum growth and shrinkage speeds from its quintic smoothstep. Compare with the Rayleigh timescale.
使用来源 R0 = 1 µm、Rmax = 8 µm、tg = 0.7 µs、tc = 0.6 µs，从五阶平滑函数求最大增长和收缩速率，并与 Rayleigh 时间尺度比较。
Reference answer / 参考答案
$S(\xi)=10\xi^3-15\xi^4+6\xi^5,\quad S'(\xi)=30\xi^2(1-\xi)^2,\quad |\dot R|_{\max}=1.875\Delta R/t_{\mathrm{branch}}$
The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.
图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。
Growth maximum = 18.75 m/s, shrinkage maximum = 21.875 m/s at ξ = 1/2. For inherited ρ = 998.2 kg/m³ and Δp = 98,986 Pa, the empty-cavity scale is 0.7348 µs. The chosen 0.6-µs branch is prescribed, not a forward RP solution.
最大增长速率 = 18.75 m/s，最大收缩速率 = 21.875 m/s，均在 ξ = 1/2。沿用 ρ = 998.2 kg/m³、Δp = 98,986 Pa 时，空腔时间尺度为 0.7348 µs。选择的 0.6 µs 分支是规定输入，并非正向 RP 解。
At the growth midpoint R = 4.5 µm, Ṙ = 18.75 m/s and R̈ = 0, take d = 8.5 µm, disk L = 10 µm, ρ = 998.2. Compute A, B and disk-average p′. Does evaluating this on a plane impose the wall condition?
增长中点 R = 4.5 µm、Ṙ = 18.75 m/s、R̈ = 0。取 d = 8.5 µm、圆盘 L = 10 µm、ρ = 998.2，计算 A、B 及圆盘平均 p′。在平面上计算是否就施加了壁面条件？
Reference answer / 参考答案
$A=R^2\ddot R+2R\dot R^2,\quad B=R^4\dot R^2,\quad \bar p=\rho\left[\frac{2A}{L^2}(\sqrt{d^2+L^2}-d)-\frac{B}{2L^2}\left(\frac1{d^2}-\frac1{d^2+L^2}\right)\right]$
The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.
图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。
A = 3.1641×10⁻³ m³/s², B = 1.4416×10⁻¹⁹ m⁶/s², p̄′ ≈ 0.28633 MPa. This samples the unbounded spherical field. It does not enforce wall impermeability, resolve a jet or include the film’s mechanical response.
A = 3.1641×10⁻³ m³/s²，B = 1.4416×10⁻¹⁹ m⁶/s²，p̄′ ≈ 0.28633 MPa。这是对无界球形场的采样，不施加壁面不可穿透条件，不解析射流，也不包含膜机械响应。
Suppose a later source branch gives p′ = −150 kPa at ambient 101.325 kPa. Calculate absolute pressure and explain the inverse pressure consistency check.
假设随后源分支给出 p′ = −150 kPa，环境为 101.325 kPa，求绝对压力，并解释反向压力一致性检查。
Reference answer / 参考答案
$p_{\mathrm{abs}}=p_\infty+p',\quad p_{B,\mathrm{required}}=p_\infty+\rho(R\ddot R+\tfrac32\dot R^2)+2\sigma/R+4\mu\dot R/R$
The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.
图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。
pabs = −48.675 kPa. The prescribed source fails its stated reference absolute-pressure screen. Metastable tension can exist, but no such nucleation closure was supplied here. Inverse RP estimates the internal pressure required by R(t); it does not show the laser generates it.
pabs = −48.675 kPa。规定源未通过其既定参考绝对压力筛查。亚稳拉伸可以存在，但此处没有提供相应成核闭合。反向 RP 估算 R(t) 所需内部压力，并不证明激光产生了它。

Brief conceptual synthesis

简短概念综合

After using the reference panels, give a short explanation of the physical mechanism, the strongest assumption, and one observation that could falsify this chapter’s model. Preparation of later chapters does not require waiting for that reply.

核对参考解答后，简述物理机制、最强假设，以及一个可证伪本章模型的观测。后续章节编写无需等待该回答。