<style> .text-center{ text-align: center; //文字置中 } .text-left{ text-align: left; //文字靠左 } .text-right{ text-align: right; //文字靠右 } </style> # 動力海洋學基本教材 ## CH 13 熱帶海洋的波動 > Typo > pp437 line 6: $U_n^-$ > pp440 last line: (13.4.27)應為(13.4.25) > 熱帶海洋具有以下特色 1. 柯氏力弱 2. 密度垂直構造不同(海溫高、混合層厚、層化強烈$\Rightarrow$理論分析上用斜溫層為邊界分成上下兩層) 3. 風場差異(東風帶、上升強烈):對流旺盛、易生颱風改變海洋上層狀態 ### 13.1赤道波導(Equatorial wave guide) 旋轉狀態波動:Inertial gravity waves, Kelvin waves(傳播方向右側需要邊界$_{NH}$), Rossby waves 若背景流靜止,波數向量與固有頻率具有關係(運動學保守方程): $$ \begin{align*} \frac{d\textbf{k}}{dt} &= \frac{\partial\textbf{k}}{\partial t}+\textbf{c}_g\cdot\nabla\textbf{k} = -\nabla\Omega|_{\textbf{k},t}\\ \frac{d\mathbf{\sigma}}{dt} &= \frac{\partial\mathbf{\sigma}}{\partial t}+\textbf{c}_g\cdot\nabla\mathbf{\sigma} = \frac{\partial\Omega}{\partial t}|_{\textbf{k},t} \end{align*} $$ 以下使用正壓討論、結果適用斜壓: #### 正壓慣性重力波頻散:受到$f$影響 $$ \sigma = \sqrt{f^2(y)+c_0^2k^2} = \Omega(\textbf{k},f(y)) $$ <p class="text-right">13.1.1</p> 代入慣性重力波群速$\mathbf{c}_g = c_{gx}\hat{\mathbf{x}}+c_{gy}\hat{\mathbf{y}} = \frac{c_0^2k_1}{\sigma}\hat{\mathbf{x}}+\frac{c_0^2k_2}{\sigma}\hat{\mathbf{y}}$可得: $$ \frac{d\textbf{k}}{dt} = -\nabla\Omega|_{\textbf{k},t} = -\frac{\beta f}{\sigma}\hat{\mathbf{y}},\ \frac{d\sigma}{dt} = 0 $$ 因此傳播時,頻率、$k_1$維持不變,$k_2$則與南北移動有關、離赤道越遠越小(波長增加。) 臨界緯度(13.1.1;$k_2=0$、$f=2\Omega_0\sin\varphi$): $$ \varphi_c = \sin^{-1}\left(\frac{\sqrt{\sigma^2-c_0^2k_1^2}}{2\Omega_0}\right) $$ #### 正壓Rossby wave頻散:受到$f$影響 $$ \sigma = \frac{\beta k_1}{k_1^2+k_2^2+f^2/c_0^2} = \Omega(\textbf{k},f(y)) $$ <p class="text-right">13.1.7</p> 代入運動學保守方程: $$ \frac{d\textbf{k}}{dt} = -\nabla\Omega|_{\textbf{k},t} = -\hat{\mathbf{y}}\beta\frac{\partial\Omega}{\partial f}|_{\textbf{k},t} = -\frac{2f\sigma^2}{c_0^2k_1}\hat{\mathbf{y}},\ \frac{d\sigma}{dt} = 0 $$ 因此傳播時,頻率、$k_1$維持不變(與頻率反向),$k_2$則與南北移動有關、離赤道越遠越小(波長增加。) $$ \mathbf{c}_g = c_{gx}\hat{\mathbf{x}}+c_{gy}\hat{\mathbf{y}} = \beta\frac{k_1^2-(k_2^2+\lambda^{-2})}{k_1^2+k_2^2+\lambda^{-2}}\hat{\mathbf{x}}+\beta\frac{2k_1k_2}{k_1^2+k_2^2+\lambda^{-2}}\hat{\mathbf{y}} $$ #### 正壓Kelvin wave:需要邊界 遠離邊界衰減。 赤道恰可為邊界。 ### 13.2赤道海洋基本動力方程組 #### 海水運動基本控制方程組 \begin{gather*} \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}-fv = -\frac{1}{\rho_T}\frac{\partial p_T}{\partial x}+\frac{\partial }{\partial z}\left(\nu_z\frac{\partial u}{\partial z}\right)\\ \frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}+fu = -\frac{1}{\rho_T}\frac{\partial p_T}{\partial y}+\frac{\partial }{\partial z}\left(\nu_z\frac{\partial v}{\partial z}\right)\\ \frac{\partial p_T}{\partial z}+g\rho_t = 0\\ \nabla\cdot\textbf{u} = \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} = 0\\ \frac{d\rho_T}{dt} = \frac{\partial \rho_T}{\partial t}+u\frac{\partial \rho_T}{\partial x}+v\frac{\partial \rho_T}{\partial y}+w\frac{\partial \rho_T}{\partial z} = 0 \end{gather*} > $\nu_z$:垂直黏性係數 靠近赤道時可以採用赤道$\beta$平面近似、並假設靜止海洋進行簡化: ![image](https://hackmd.io/_uploads/SyaHwXMAa.png =70%x) > $\Gamma$:水平摩擦應力 ##### 布氏近似 $$ \nu_z = \frac{A_z}{N^2},\ N^2 = -\frac{gd\bar{\rho}}{\rho_0dz} $$ > $A$:Const. $N$:Freq. \begin{gather*} \frac{\partial u}{\partial t}-fv = -\frac{1}{\rho_0}\frac{\partial p}{\partial x}+\frac{\partial }{\partial z}\left(\frac{A_z}{N^2}\frac{\partial u}{\partial z}\right)\\ \frac{\partial v}{\partial t}+fu = -\frac{1}{\rho_0}\frac{\partial p}{\partial y}+\frac{\partial }{\partial z}\left(\frac{A_z}{N^2}\frac{\partial v}{\partial z}\right)\\ \frac{\partial p}{\partial z}+g\rho = 0\\ \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} = 0\\ \frac{d\rho}{dt}+w\frac{\partial \bar{\rho}}{\partial z} = 0 \end{gather*} #### 分層模型 ![image](https://hackmd.io/_uploads/rysq57zC6.png) 密度代替垂直座標($z$) ##### 假設密度固定 水層壓力、水層介面遞迴: $$ \begin{gather} p_i(z)=p_{i-1}(z)+\Delta\rho g(z_{i-1}-z),\ p_1(z) = \rho_0 g(\eta-z)+p_a\\ \Rightarrow P=p+\rho gz\ \mathsf{(Montgomery\ potential)}\\ z_{i-1} = z_i+h_i,\ z_n=-H,\ z_0 = \eta=\sum^n_{i=1}h_u+z_n\\ \rho_i=\rho_0+(i-1)\Delta\rho \end{gather} $$ ###### 最簡化斜壓模型(減重力模型;1 1⁄2層模型) - 共兩層用斜溫層分界(100 m),假設下層無限深 - 下層水體不動(動能有限) ##### 假設密度遞減、摩擦不可忽略 \begin{gather*} \frac{\partial u}{\partial t}-fv = -\frac{1}{\rho}\frac{\partial p}{\partial x}+\frac{1}{\rho}\frac{\partial \Gamma_x}{\partial z}\\ \frac{\partial v}{\partial t}+fu = -\frac{1}{\rho}\frac{\partial p}{\partial y}+\frac{1}{\rho}\frac{\partial \Gamma_y}{\partial z} \end{gather*} > $\Gamma$: 海面風應力 垂直變化尺度遠小於水平變化尺度(垂直項重要): \begin{align} \frac{\partial \tau_{xx}}{\partial x}+\frac{\partial \tau_{xy}}{\partial y}+\frac{\partial \tau_{xy}}{\partial z}\sim \frac{\partial \tau_{xz}}{\partial z}&=\frac{\partial \Gamma_{x}}{\partial z}\\ \frac{\partial \tau_{yx}}{\partial x}+\frac{\partial \tau_{yy}}{\partial y}+\frac{\partial \tau_{yz}}{\partial z}\sim \frac{\partial \tau_{yz}}{\partial z}&=\frac{\partial \Gamma_{y}}{\partial z} \end{align} 強風混合:混合層頂至混合層底 $$ \left(\frac{\partial \Gamma_{x}}{\partial z}, \frac{\partial \Gamma_{y}}{\partial z}\right)=\left\{ \begin{matrix} \left(\frac{\tau_x}{h_{mix}},\frac{\tau_y}{h_{mix}}\right)&,z\ge-h_{mix}\\ 0&,z<-h_{mix} \end{matrix} \right. $$ #### 控制方程簡化 ##### 假設密度變化小($\rho_0$近似、單層海洋) \begin{align} \frac{\partial u}{\partial t}-fv=-g\frac{\partial \eta}{\partial x}+\frac{\tau_x}{\rho_0H}-\frac{\tau_{Bx}}{\rho_0H}\\ \frac{\partial v}{\partial t}+fu=-g\frac{\partial \eta}{\partial y}+\frac{\tau_y}{\rho_0H}-\frac{\tau_{By}}{\rho_0H}\\ \frac{\partial\eta}{ \partial t }+H\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=\frac{(P-E)}{\rho_0} \end{align} ##### 假設密度變化大(強烈分層eg熱帶、1 1⁄2層層海洋、$|\eta|\ll H$) \begin{align} \frac{\partial u}{\partial t}-fv=-g'\frac{\partial \eta}{\partial x}+\frac{\tau_x}{\rho_0H}\\ \frac{\partial v}{\partial t}+fu=-g'\frac{\partial \eta}{\partial y}+\frac{\tau_y}{\rho_0H}\\ \frac{\partial\eta}{ \partial t }+H\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=\frac{(P-E)}{\rho_0} \end{align} > 用上層的厚度取代底深、水層厚度起伏替換水面起伏、忽略摩擦力、$H$為上層。 > $g'$ represents? ### 13.3方程組降維(分離變數) 分為兩個函數:$F(z)$、$G(x,y,t)$ 不管波動為何,只要(1)海底為水平且(2)流體處於靜水平衡(3)密度穩定分層,分離變數法可將運動分解為垂直模態組合,然後求出各模態之垂直構造及時間-水平變化函數。 > 密度不需均勻、斜壓不適用 \begin{gather*} \frac{\partial U_n}{\partial t}-fV_n = -g\frac{\partial\eta_n}{\partial x}\\ \frac{\partial V_n}{\partial t}+fU_n = -g\frac{\partial\eta_n}{\partial y}\\ \frac{\partial\eta_n}{\partial t}+H_n\left(\frac{\partial U_n}{\partial x}+\frac{\partial V_n}{\partial y}\right)=0 \end{gather*} > <!-- 令 \begin{align} \left(\begin{matrix}u\\v\end{matrix}\right)&=\left[\begin{matrix}U(x,y,t)\\V(x,y,t)\end{matrix}\right]F(z)\\ w&=W(x,y,t)G(z)\\ \frac{p}{\rho_0}&=g\eta(x,y,t)F(z) \end{align} 控制方程: \begin{gather*} \frac{\partial U}{\partial t}-fV = -g\frac{\partial\eta}{\partial x}\\ \frac{\partial V}{\partial t}+fU = -g\frac{\partial\eta}{\partial y}\\ -\frac{1}{W}\left(\frac{\partial U}{\partial x}+\frac{\partial V}{\partial y}\right)=\frac{G_z}{F}\equiv\frac{1}{h}=const.\\ \Rightarrow \frac{\partial U}{\partial x}+\frac{\partial V}{\partial y}+\frac{W}{h}=0 \end{gather*} <p class="text-right">13.3.2~5</p> 同時 \begin{align} &\because \frac{\partial^2}{\partial z\partial t}+\rho_oN^2w=0,\\ &\therefore \frac{\partial\eta}{\partial t}+\frac{N^2G}{gF_z}W=0, F_z=\frac{dF}{dz}\\ &\Rightarrow -\frac{1}{W}\left(\frac{\partial \eta}{\partial t}\right)=\frac{N^2G}{ghG_{zz}}W \end{align} > 13.3.4: $F_z=hG_{zz}$ 2. 淺水近似 --> ### 13.4赤道陷波 赤道$\beta$平面近似、水平底、無摩擦、無驅動外力、不考慮熱擴散效應、海水不可壓縮、靜水平衡: \begin{gather*} \frac{\partial U_n}{\partial t}-\beta yV_n = -g\frac{\partial\eta_n}{\partial x}\\ \frac{\partial V_n}{\partial t}+\beta yU_n = -g\frac{\partial\eta_n}{\partial y}\\ \frac{\partial\eta_n}{\partial t}+H_n\left(\frac{\partial U_n}{\partial x}+\frac{\partial V_n}{\partial y}\right)=0 \end{gather*} > $f=\beta y$ > $\beta = 2.288\times10^{11}$ [ms]$^{-1}$ #### equatorial Kelvin waves 赤道海洋微波導(由西向東傳) 代入$V_n=0$: \begin{align*} \frac{\partial U_n}{\partial t}&=-g\frac{\partial\eta_n}{\partial x}\\ \beta yU_n &= -g\frac{\partial\eta_n}{\partial y}\\ \frac{\partial\eta_n}{\partial t}&+H_n\frac{\partial U_n}{\partial x}=0\\ \Rightarrow \frac{\partial^2 U_n}{\partial t^2}&-gH_n\frac{\partial^2 U_n}{\partial x^2}=0\\ \Rightarrow \frac{\partial^2 U_n}{\partial t^2}&-c_n^2\frac{\partial^2 U_n}{\partial x^2}=0 \end{align*} > solution: $U (x,y,t)=U_n^+(x+c_n t,y)+U_n^−(x−c_n t,y)$ > 傳遞途中波型不變、遠離赤道振幅減小 ![IMG_0833](https://hackmd.io/_uploads/H17b8VM0a.jpg =40%x) 另外, \begin{align*} &\frac{\partial^2 U_n}{\partial t\partial y}-\beta\frac{\partial\eta_n}{\partial x}=0\\ \Rightarrow& \partial_y\partial_xU^+_n-\frac{\beta y}{c_n}\partial_xU^+_n=0\\ \Rightarrow& \partial_y\partial_xU^-_n+\frac{\beta y}{c_n}\partial_xU^-_n=0\\ \Rightarrow& \left\{\begin{matrix} U^+_n=& F^+(x+c_nt\exp\frac{\beta y^2}{2c_n})&,\ DNE\\ U^-_n=& F^-(x+c_nt\exp\frac{-\beta y^2}{2c_n})& \end{matrix}\right. \end{align*} > 赤道變形半徑$\lambda_n=\sqrt{c_n/\beta}$ > 頻散關係$\sigma = c_nk$ > 等速東傳、非離散型赤道陷波、相速群速相等 #### 其他赤道陷波 解$V_n$非零: ![image](https://hackmd.io/_uploads/BJgnykUAa.png =40%x) ![image](https://hackmd.io/_uploads/rJAnyJIR6.png =75%x) 對$t$積分一次: ![image](https://hackmd.io/_uploads/SyEfg180T.png =60%x) WLOG, 可求出$U_n$。由赤道波導預期$x$方向為傳播、$y$方向模態構造。 控制方程:![image](https://hackmd.io/_uploads/ByfjlyUAa.png =40%x) 臨界值:![image](https://hackmd.io/_uploads/Hkkhg1U0T.png =60%x) > $|y| >|y_c|$:指數成長或衰減 > $|y| <|y_c|$:震盪 > ![image](https://hackmd.io/_uploads/SkZClJL0T.png =60%x) **適用equatorial Kelvin waves之外所有赤道陷波:** 東西向流速:$U_{nj}=\frac{iA_{nj}\sqrt{c_n\beta}}{2}\left\{\frac{\sqrt{2(j+1)}D_{j+1}(\xi)}{(\sigma-kc_n)}+\frac{\sqrt{2j}D_{j-1}(\xi)}{(\sigma+kc_n)}\right\}\exp(i\chi)$ 水位:$\eta_{nj}=\frac{iA_{nj}c_n\sqrt{c_n\beta}}{2g}\left\{\frac{\sqrt{2(j+1)}D_{j+1}(\xi)}{(\sigma-kc_n)}-\frac{\sqrt{2j}D_{j-1}(\xi)}{(\sigma+kc_n)}\right\}\exp(i\chi)$ ### 13.5赤道的動力屬性與傳播行為 頻散關係: \begin{gather} \sigma^2=c^2_n[k^2+\frac{2j+1}{\lambda_e^2}]+\frac{\beta k}{\sigma}c^2_n\\ \sigma^{*2}=c^2_n[k^{*2}+(2j+1)]+\frac{k^*}{\sigma^*}\\ k^*=-\frac{1}{2\sigma^*}\pm\sqrt{\sigma^{*2}+\frac{1}{4\sigma^{*2}}-(2j+1)} \end{gather} > $\lambda_e = \sqrt{c_n/\beta}$赤道變形半徑 ![image](https://hackmd.io/_uploads/B1tB7kU0p.png =80%x) ![image](https://hackmd.io/_uploads/BkuwHk80a.png =80%x) ### 13.6陸地邊界對赤道陷波之效應 #### 能量方程式: ![image](https://hackmd.io/_uploads/Syc5U1URp.png =50%x) ##### 假設水平方向無阻礙 第三項$=0$,反向。積分: ![image](https://hackmd.io/_uploads/ry60Lk8Ap.png =60%x) 平均總能量密度、能量通量密度: ![image](https://hackmd.io/_uploads/HJpEvJL06.png =60%x) ![image](https://hackmd.io/_uploads/Hk6PDk806.png =60%x) ![image](https://hackmd.io/_uploads/HJltD1LRa.png =40%x) ##### 假設水平方向有阻礙 $\Rightarrow$垂直海岸方向上之流速分量必須為0、海岸陷波可能產生,只能用完整方程式 ###### 西岸東西流速分量$u=u_I+u_R=0$ 入射: $$ u_1=\frac{iA^I_{nJ}\sqrt{c_n\beta}}{2}\left\{\frac{\sqrt{2(J+1)}D_{J+1}(\xi)}{(\sigma-k^I_Jc_n)}+\frac{\sqrt{2J}D_{J-1}(\xi)}{(\sigma+k^I_Jc_n)}\right\}\exp(-i\sigma t) $$ 反射加總 \begin{align} u_R(x,y,t)=&\frac{i\sqrt{c_n\beta}}{\sqrt{2}}\sum^J_{j=1}A^R_{nj}\left\{\frac{\sqrt{(j+1)}D_{j+1}(\xi)}{\sigma-k^R_jc_n}+\frac{\sqrt{j}D_{J-1}(\xi)}{\sigma+k^R_jc_n}\right\}\exp(i(k^R_jx-\sigma t))\\ -& \frac{-\sigma}{\sqrt{2c_n\beta}}A_YD_1(\xi)\exp\left\{i\left(\frac{\sigma}{c_n}-\frac{\beta}{\sigma}\right)x-\sigma t\right\} \\ +& iA_KD_0(\xi)exp\left\{i\frac{\sigma}{c_n}x-\sigma t\right\} \end{align} 需要解降階遞迴: ![image](https://hackmd.io/_uploads/SyBRllI0T.png =40%x) **解出:** 柳井波:$A^R_{n2}=\frac{\sigma(\sigma+k^R_2c_n)}{\sqrt{2}\beta c_n}A_Y$ 赤道凱爾文波:$A^R_{n1}=-\frac{\sqrt{2}(\sigma+k^R_1c_n)}{\sqrt{\beta c_n}}A_K$ - 入射模態數$J \in$odd $\rightarrow$反射波水位即東西流速分量模態數$\in$even(反射不包含柳井波) - 入射模態數$J \in$even $\rightarrow$反射波水位即東西流速分量模態數$\in$odd(南北反對稱、反射不包含赤道Kelvin waves) ###### 東岸 \begin{align} u_I=&\frac{iA^I_{nJ}\sqrt{c_n\beta}}{2}\left\{\frac{\sqrt{2(J+1)}D_{J+1}(\xi)}{(\sigma-k^I_Jc_n)}+\frac{\sqrt{2J}D_{J-1}(\xi)}{(\sigma+k^I_Jc_n)}\right\}\exp(-i\sigma t)\\ u_R(x,y,t)=&\frac{i\sqrt{c_n\beta}}{\sqrt{2}}\sum^{\infty}_{j=J}A^R_{nj}\left\{\frac{\sqrt{(j+1)}D_{j+1}(\xi)}{\sigma-k^R_jc_n}+\frac{\sqrt{j}D_{J-1}(\xi)}{\sigma+k^R_jc_n}\right\}\exp(i(k^R_jx-\sigma t)) \end{align} 反射波中無赤道凱爾文波與柳井波 需要解升階遞迴: ![image](https://hackmd.io/_uploads/rJEchJLCT.png =80%x) ![image](https://hackmd.io/_uploads/rJZonyLCa.png =40%x) 柳井波:$A^R_{n2}=\frac{\sigma(\sigma+k^R_2c_n)}{\sqrt{2}\beta c_n}A_Y$ 赤道凱爾文波:$A^R_{n1}=-\frac{\sqrt{2}(\sigma+k^R_1c_n)}{\sqrt{\beta c_n}}A_K$ 固定頻率入射波升階計算中,模態每次增2,會達到$(2j+1)(\beta_n)>\left(\dfrac{\sigma}{c_n}\right)^2 +\left(\dfrac{\beta}{2\sigma}\right)^2$,出現複數(不具向西傳播能力的震盪,引發南北向傳播) 最後,如果為封閉海盆且無摩擦損耗可形成不衰退的自由震盪的自由振盪。 <!-- \begin{gather*} \partial_t \bar{E_{nj}}+\partial_x(g\bar{U_{nj}\eta_{nj}})+\partial_y(g\bar{V_{nj}\eta_{nj}})=0\\ \frac{\partial\eta_n}{\partial t}+H_n\left(\frac{\partial U_n}{\partial x}+\frac{\partial V_n}{\partial y}\right)=0 \end{gather*} --> ## CH 14 熱帶海洋對風力作用之反應 > **typo:** > pp467: 赤道海域 H 與 g'典型數值約為 H = 100 m (**g'未給**) > ### 14.1 熱帶風場以及海洋環流 - 南北信風交會,形成ITCZ (雲、降水) ![Screenshot 2024-03-24 at 02.33.19](https://hackmd.io/_uploads/HJYNZi2Ra.jpg =60%x) ![Screenshot 2024-03-24 at 02.33.26](https://hackmd.io/_uploads/BytNZonAa.jpg =60%x) ![Screenshot 2024-03-24 at 02.33.36](https://hackmd.io/_uploads/B1YEWjnA6.jpg =60%x) - ITCZ季移 ![Screenshot 2024-03-24 at 02.33.43](https://hackmd.io/_uploads/By_V-i2Ca.png =60%x) - 全球熱帶海洋風場變化最劇烈之處為印度洋,當地風向會發生季節性顛轉(5 月至 10 月盛行西南季風,12 月至 3 月盛行東北季風, 4 月與 11 月則為轉換期 - 西南季風最強勁處則發生在東非沿岸的低層大氣噴射氣流區 ![Screenshot 2024-03-24 at 02.33.51](https://hackmd.io/_uploads/HytE-o2RT.jpg =60%x) #### 熱帶太平洋海面溫度分布 - 赤道海溫較低、赤道北方有一暖水帶外。海盆溫度為東低於西(湧升作用、海流平流) - 熱帶海域盛行偏東風。科氏力造成艾克曼搬運->湧升流。 - 美洲海岸之盛行偏南與偏北風造成離岸方向艾克曼搬運->沿岸湧升流 ![Screenshot 2024-03-24 at 02.34.03](https://hackmd.io/_uploads/HkK4-o3Ra.jpg =60%x) ![Screenshot 2024-03-24 at 02.34.10](https://hackmd.io/_uploads/ByKVWsnA6.jpg =60%x) ![Screenshot 2024-03-24 at 02.34.19](https://hackmd.io/_uploads/HytVZin06.jpg =60%x) > NEC < 20 cm/s、SEC > 100 cm/s、NECC 50 cm/s、SECC(換日線以西) ![Screenshot 2024-03-24 at 02.34.26](https://hackmd.io/_uploads/H1F4-i2Ap.png =60%x) ![Screenshot 2024-03-24 at 02.34.35](https://hackmd.io/_uploads/HJtN-shAa.jpg =60%x) ### 14.2 赤道流與赤道潛流 #### Revisit Equatorial $\beta$ plane > $f = 2\Omega \sin\psi \sim 2\Omega\psi$ > \begin{gather} > U = 0.5\ m/s,\ L = 1000\ km\\ > R_o = \frac{U}{|f|L} > \end{gather} > - 緯度介於±2º,科氏力可忽略 > - 只要不過於接近赤道,大尺度海流可以維持地轉特性 > #### 剛蓋假設 > 靜水平衡、海面為剛蓋、恆定風場[風應力$\tau=(\tau_x, \tau_y)$為常數] > 斜溫層的平均厚度(H)遠大於斜溫層起伏(η), $\eta \ll H$ ) > 海洋上層為微小振幅運動、下層無限深且無運動 > 不考慮降水、蒸發作用 > 若上層密度為ρ、上下水層密度差 Δρ > (赤道海域H = 100 m,Δρ/ρ = 0.002) ##### **減重力 1 1⁄2層淺水模型** $$ \begin {gather} -fv=-\frac{g'}{\rho_0}\frac{\partial \eta}{\partial x} + F_x + \nu_z\frac{\partial^2 u}{\partial z^2}\\ fu=-\frac{g'}{\rho_0}\frac{\partial \eta}{\partial y} + F_y + \nu_z\frac{\partial^2 v}{\partial z^2}\\ \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0 \end{gather} $$ ##### 剛蓋BC \begin{align} z = 0:&\ \nu_z \frac{\partial u}{\partial z} = \frac{\tau_x}{\rho_0},\ \nu_z \frac{\partial v}{\partial z} = \frac{\tau_y}{\rho_0}\\ z = -H:&\ \frac{\partial u}{\partial z} = \frac{\partial v}{\partial z} =0\\ &\frac{\partial U}{\partial x}+\frac{\partial V}{\partial y}=0 \end{align} ##### 垂直積分 $$ \begin {gather} -\frac{fV}{H}=-\frac{g'}{\rho_0}\frac{\partial \eta}{\partial x} - \frac{rU}{H} + \frac{\tau_x}{\rho_0 H}\\ \frac{fU}{H}=-\frac{g'}{\rho_0}\frac{\partial \eta}{\partial y} - \frac{rV}{H} + \frac{\tau_y}{\rho_0 H}\\ \Rightarrow \beta\frac{\partial \Psi}{\partial x}+r\nabla^2_H\Psi = \nabla\times \left(\frac{\tau}{\rho_0}\right) \end{gather} $$ > $U=-\dfrac{\partial \Psi}{\partial y}$, $V=\dfrac{\partial \Psi}{\partial x}$. > $(U,V)=\int^0_{-H}(u,v)dz$ ##### 折減流 $$ \begin {gather} -fv'=- \frac{\tau_x}{\rho_0 H}-ru' + \nu_z\frac{\partial^2 u'}{\partial z^2}\\ fu'=- \frac{\tau_y}{\rho_0 H}-rv'+ \nu_z\frac{\partial^2 v'}{\partial z^2}\\ \frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}+\frac{\partial w}{\partial z}=0 \end{gather} $$ > $(u',v')=(u-\dfrac{U}{H},=v-\dfrac{V}{H}) = (u_{interior}+u_{Ekman},v_{interior}+v_{Ekman})$. > $0=\int^0_{-H}(u',v')dz$ ###### 內部不隨深度改變 ![image](https://hackmd.io/_uploads/SkIH_HyyR.png) $$ \begin {gather} -fv_I=- \frac{\tau_x}{\rho_0 H}-ru_I\\ fu_I=- \frac{\tau_y}{\rho_0 H}-rv_I\\ \frac{\partial u_I}{\partial x}+\frac{\partial v_I}{\partial y}+\frac{\partial w_I}{\partial z}=0\\ \Rightarrow (u_I,v_I) = \left(\frac{-r\tau_x-\beta y\tau_y}{\rho_0H(r^2+\beta^2y^2)}, \frac{-r\tau_y+\beta y\tau_x}{\rho_0H(r^2+\beta^2y^2)}\right) \end{gather} $$ 帶入連續方程可得$w$。($u_I$, $v_I$只與$y$有關) ###### Ekman layer $$ \begin {gather} -fv_E=-ru_E + \nu_z\frac{\partial^2 u_E}{\partial z^2}\\ fu_E=-rv_E+ \nu_z\frac{\partial^2 v_E}{\partial z^2}\\ \frac{\partial u_E}{\partial x}+\frac{\partial v_E}{\partial y}+\frac{\partial w_E}{\partial z}=0\\ \Rightarrow (U_E,V_E) = \left(\frac{r\tau_x+\beta y\tau_y}{\rho_0(r^2+\beta^2y^2)}, \frac{r\tau_x-\beta y\tau_y}{\rho_0(r^2+\beta^2y^2)}\right) = -H(u_I,v_I) \end{gather} $$ ##### 赤道上 ###### 折減流(無科氏力) $$ \begin {gather} 0=- \frac{\tau_x}{\rho_0 H}-ru' + \nu_z\frac{\partial^2 u'}{\partial z^2}\\ 0=- \frac{\tau_y}{\rho_0 H}-rv'+ \nu_z\frac{\partial^2 v'}{\partial z^2}\\ \frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}+\frac{\partial w}{\partial z}=0 \end{gather} $$ if $v'=0$, then $u'=\dfrac{\tau_x}{\rho_0Hr}\left(\dfrac{(\alpha H)\cosh \alpha(z+H)}{\sinh \alpha H}-1\right)$, $\alpha = \sqrt{r/\nu_z}$ ![image](https://hackmd.io/_uploads/Sy6XpHJJ0.png =40%x) - 看見赤道潛流的存在(補償風成流) #### 史佛卓流 ![image](https://hackmd.io/_uploads/SyT2pryyR.png =80%x) 忽略水平摩擦$\rho_0\beta V = \nabla\times \tau$ $$ \frac{\partial U}{\partial x}=-\frac{\partial V}{\partial y} = \dfrac{1}{\rho_0\beta}\dfrac{\partial(\nabla\times \tau)}{\partial y} $$ ### 14.3突發性風場 equtorial jet / Yoshida jet:大氣西風引發表面海流。 #### **減重力 1 1⁄2層淺水模型** > 赤道β平面、靜水平衡、海面為剛蓋海盆在東西方向上無邊界限制、斜溫層平均厚度(H)遠大於斜溫層厚度起伏變化(η,以下改稱距平)、下層無限深、微小振幅運動、不考慮降水與蒸發以及省略。 水平摩擦作用 $$ \begin {gather} \frac{\partial u}{\partial t}-\beta yv = -g'\frac{\partial \eta}{\partial x} + \frac{\tau_x}{\rho_0 H}\\ \frac{\partial v}{\partial t}+\beta yu = -g'\frac{\partial \eta}{\partial y} + \frac{\tau_y}{\rho_0 H}\\ \frac{\partial \eta}{\partial t}+H\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0\\ \end{gather} $$ 赤道Kelvin waves 需要124天由西到東橫跨太平洋(15000 km)。 > $u=U/H$,$v=V/H$,$g'=(\delta \rho/\rho_0 )g$ > $H = 100$ m、$\delta \rho/\rho_0 = 0.002$ > $c_1 = \sqrt{gH_1}\Rightarrow H_1 = (g'/g)H = 20$ cm、$c_1 = 1.4$ m/s > 對$t$微分、消除$\eta$: \begin{align} \left(\frac{\partial^2}{\partial t^2}-c_1^2\frac{\partial^2}{\partial x^2}\right)u &= \left(\beta y\frac{\partial}{\partial t}+c_1^2\frac{\partial^2}{\partial x\partial y}\right)v+\frac{1}{\rho_0 H}\frac{\partial\tau_x}{\partial t}\\ \left(\frac{\partial^2}{\partial t^2}-c_1^2\frac{\partial^2}{\partial y^2}\right)v &= \left(-\beta y\frac{\partial}{\partial t}+c_1^2\frac{\partial^2}{\partial x\partial y}\right)v+\frac{1}{\rho_0 H}\frac{\partial\tau_y}{\partial t}\\ \Rightarrow\frac{\partial}{\partial t}\left(\frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2}\right)v &+\beta\frac{\partial v}{\partial x} -\frac{(\beta y)^2}{c_1^2}\frac{\partial v}{\partial t} -\frac{1}{c_1^2} \frac{\partial^3 v}{\partial t^3} = F\\ F = \frac{\beta y}{c_1^2}\frac{\partial}{\partial t}\frac{\tau_x}{\rho_0 H}&-\frac{1}{c_1^2}\frac{\partial^2}{\partial t^2}\frac{\tau_y}{\rho_0 H}+\frac{1}{\rho_0 H}\frac{\partial}{\partial x}\left(\frac{\partial\tau_y}{\partial x}-\frac{\partial\tau_x}{\partial y}\right) \end{align} 斜壓下,mod 1之赤道陷波為$F=0$之解,若$F\neq 0$為特解。 假設平靜海面($t=0$ when $u=v=\eta =0$)吹均勻西風: ![image](https://hackmd.io/_uploads/rkXVYf-xR.png =45%x) > $t$大第三項可以省略 > ![image](https://hackmd.io/_uploads/H1WItfZgA.png =40%x) 無因次化後利用Hermit Fuction可以迭代求解: \begin{align} \frac{d^2Q}{d\xi^2}-\xi^2Q&=-\xi\\ Q_1(\xi) &\approx -{\xi^3\over 3!}+a_1\xi+a_2=-{\xi^3\over 3!}+\left({dQ\over d\xi}\right)_{\xi=0}\xi \end{align} ![image](https://hackmd.io/_uploads/S1dd_fbgA.png =80%x) > v之最大值位於1.48變形半徑($\lambda_\beta$),以內輻合、以外輻散 > $\eta$為厚度變化,代表輻散輻合強度 只要風持續,混合層會不斷加厚;風停後,赤道線波方程組控制波的傳播(混合層也不加厚)。因嚴格遵守地轉平衡、無非地轉項,若無外力,赤道噴流可以一直持續、只有反向東風可以停止。另外,加速過高會產生不穩定;如果有南北向邊界,需要考慮自由震盪。 ### 14.4週期性風場 > 若靜水平衡、水平底,如果**流體密度不均勻**(穩定分層),運動可分解為不同的垂直模態組合、運動方程可以降維化簡。假設剛蓋、水平方向上無限制、赤道β平面,第 $n$ 垂直模態運動之控制方程可寫為: $$ \begin {gather} \frac{\partial u_n}{\partial t}-fv_n = -g\frac{\partial \eta_n}{\partial x} + \frac{\tau_n^x}{\rho_0 H}\\ \frac{\partial v_n}{\partial t}+fu_n = -g\frac{\partial \eta_n}{\partial y} + \frac{\tau_n^y}{\rho_0 H}\\ \frac{\partial \eta_n}{\partial t}+H_n\left(\frac{\partial U_n}{\partial x}+\frac{\partial V_n}{\partial y}\right)=0\\ \end{gather} $$ 無因次化: ![image](https://hackmd.io/_uploads/SJYu0G-e0.png) $$ \begin {gather} \frac{\partial u^*}{\partial t^*}-y^*v^* = -\frac{\partial \eta^*}{\partial x^*} + \tau_x^*\\ \frac{\partial v_n}{\partial t^*}+fu_n = -\frac{\partial \eta^*}{\partial y^*} + \tau_y^*\\ \frac{\partial \eta^*}{\partial t^*}+\left(\frac{\partial u^*}{\partial x^*}+\frac{\partial v^*}{\partial y^*}\right)=0\\ \end{gather} $$ if $q-\eta+u$ and $r=\eta-u$: ![image](https://hackmd.io/_uploads/H1FQFtQe0.png =40%x) 單一變數$v$之方程: ![image](https://hackmd.io/_uploads/ryoSFKQlA.png =45%x) ![image](https://hackmd.io/_uploads/BkqtYYXgC.png =40%x) 利用Hermite Function 展開,得出不同模態的關係式 ![image](https://hackmd.io/_uploads/r1Cl9FXlA.png =70%x) 若風應力為單一週期: ![image](https://hackmd.io/_uploads/Hkg8qY7gC.png =70%x) ### 14.5小風域、短延時風場 #### 流體調整 第一階段:重力波快速調整(類似f平面)、趨近地轉平衡。 第二階段:赤道陷波向準地轉調整 #### 西風事件 * 熱帶季內振盪(又稱為麥儒振盪,Madden Julian Oscillation,週期約 30-60 天,Zhang 2005) * 冷半球中緯度的寒潮事件(會引發流向赤道方向的氣流,後者跨越赤道後受柯氏力影響轉為偏西風) * 熱帶風暴(熱帶低壓靠赤道側為偏西風) * 熱帶海域中尺度過程 風驅艾克曼作用會促使位於西太平洋暖池之風力作用區內的海洋上層暖水堆積,之後堆積的暖水伴隨風驅赤道凱文波向東傳播、可能演化成為聖嬰事件 #### **減重力 1 1⁄2層淺水模型 (§14.3)** $$ \begin {gather} \frac{\partial u}{\partial t}-\beta yv = -g'\frac{\partial \eta}{\partial x} + \frac{\tau_x}{\rho_0 H}+A_H\nabla^2_Hu\\ \frac{\partial v}{\partial t}+\beta yu = -g'\frac{\partial \eta}{\partial y} + \frac{\tau_y}{\rho_0 H}+A_H\nabla^2_Hv\\ \frac{\partial \eta}{\partial t}+H\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0\\ \end{gather} $$ > $A_H$:水平方向渦黏係數 > 加入側向摩擦項: > 1. 簡便的非滑動條件處理邊界問題($u=v=0$), > 2. 濾波器(壓抑短波以去除運算誤差、長波較不受影響) ![image](https://hackmd.io/_uploads/HJ72Wm-xA.png) > 風場作用30天(南北對稱) > 波動分別向西(-)與向東傳(+)、其中向東移動較快 向東:赤道凱爾文波(主要)、赤道羅士培波短波、赤道重力波 (短波距離有限) 向西:赤道羅士培波長波、赤道重力波(快) ![image](https://hackmd.io/_uploads/rySg4XWeC.png) ### 14.6聖嬰事件 > 東風減弱、暖水東移、漁獲減少 > 南方震盪現象 #### ONI (Oceanic Nino Index) 透過SST距平值判斷聖嬰事件發生 聖嬰3.4海域 0.5~1:弱事件 1~1.5:中事件 1.5~2:強事件 \>2:極強事件 #### SOI (Southern Oscillation Index) 大溪地、達爾文SLP差值距平值 SOI+:強信風 SOI-:弱信風、西風 #### 相位平均 1. 前期(t-1年):SOI-、ONI增加 2. 中期(t=0年): **年初:** ONI很大、ITCZ北移停滯、東太平洋海溫將水異常增加(50~100 cm/s) **4月:** 西太平洋對流東移(印尼降雨減少、SOI 續降);東方異常強降雨以及風速距平出現北風分量 **7月:** ONI 正值最大(年底、次年年初次高峰)。7至9月印尼降雨少 **9月:** 信風衰落,西風東移,西太洋赤道附近的輻合帶東移(SOI-),ITCZ往赤道移,而南太平洋輻合帶則往北移 **年底(以及次年年初):** 異常區最大 3. 末期(t+1年):ONI減小或ONI-,反聖嬰開始。 #### Bjerknes mechanism 氣流輻合區改變位置,斜溫層構造無法維持。波動(陷波、Kelvinwave)調整過程中,使暖池範圍擴大。東太平洋冷卻後,即結束。接著冷海溫快速西傳使得反聖嬰肇始。 大氣經由風場驅動海水移動(快),海溫分佈改變影響大氣風場(慢)。 \begin{align} {\partial T\over\partial t}&+{\partial \over\partial x}(uT)+{\partial \over\partial y}(vT)+{\partial \over\partial z}(wT) = {Q \over\rho_0c_pH}\\ Q&=\mathbf{Q}_{SW}-Q_{LW}-\mathbf{Q}_{LH}-Q_{SH} \end{align} 假設水平變化小、$H_m = H+\eta$ (14.6.3、14.6.4): \begin{align} {\partial T'\over\partial t}&+u{\partial\bar{T}\over\partial x}+v{\partial\bar{T}\over\partial y}+ {Q_{S'}+Q_{W}\over\rho_oc_pH_m} = -\lambda_1(T-T_a)\\ {\partial T'\over\partial t}&=-u{\partial\bar{T}\over\partial x}-v{\partial\bar{T}\over\partial y}-\gamma_1{w_e\over H_m}(T-T_d)-\lambda_1(T-T_a) \end{align} > $Q_{S'}$:進入深水層、$Q_W$:垂直項進深水層 考慮動量方程(§14.3)、加入線性摩擦、源匯效應(降水、蒸發)、半地轉(赤道Kelvin、Rossby)、只有東西風應力: $$ \begin {gather} \frac{\partial u}{\partial t}-\beta yv = -g'\frac{\partial \eta}{\partial x} -ru+ \frac{\tau_x}{\rho_0 H}\\ \beta yu = -g'\frac{\partial \eta}{\partial y}\\ \frac{\partial \eta}{\partial t}+H\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=-r\eta\\ \end{gather} $$ 赤道上層垂直積分、邊界$u=0$(第二式為零): \begin{align} -g'H{\partial\eta\over\partial x}-t\int_{-H}^0 udz+{\tau_x\over\rho_0}=0\\ {\partial\eta\over\partial x}={\tau_x\over\rho_0g'H} \end{align} > 恆定均勻東風時,斜溫層西深東淺 > 簡化模型假設上、下層密度固定、不受海溫變化影響 > 實際,厚水處海溫高,薄水處海溫低 ##### 松野-基爾模型 Matsuno-Gill model > 近似淺水方程 > 無$\tau_x$、改用大氣壓、加入海面熱源 ![image](https://hackmd.io/_uploads/HyXutXWgR.png =60%x) - 若上升氣流越強(對流越強)或西風在 x 方向上跨越的範圍越大(即西風的風域越長),那麼上游的西風便會越強。 #### 震盪 > 季內(麥儒震盪)、季節、年際(ENSO) - 太平洋最顯著,太平洋沃克環流亦為三大洋中變動程度最大者 - 季節震盪:太陽輻射年週期變化有關,東西擺盪 - ENSO是一種減幅震盪,當代有兩種主要理論: - 1. 自給振盪(self-sustained oscillation):海洋與大氣耦合產生不穩定自然振盪 - 2. 外力激發:由隨機的外在驅動力或擾動激發 - 正回饋因子:聖嬰時暖水東傳、反聖嬰時冷水冬傳 - 負回饋因子:聖嬰時冷水東傳、反聖嬰時暖水冬傳 #### 自給震盪機制: ##### **延時振盪器**(delayed oscillator) 赤道凱爾文波(西風驅動),推動上層海水向東平流。 $$ {\partial \left\langle T'(t) \right\rangle \over \partial t } = a_1\left\langle T'(t) \right\rangle-b_1\left\langle T'(t-\Delta) \right\rangle $$ >$a_1$、$b_1$比例係數 還有其他形式,例如考慮中太平洋深對流、不可慮海溫 分界點移動(與緯項海流有關): $$ {dx\over dt}=u'(t)=u_1(t)+u_2(t-\Delta)=a_2x(t)-b_2x(t-\Delta) $$ > $\Delta$:西傳Rossby wave反射之Rossby wave回到發源地的時間差 > 令解為:$r+i\sigma=a_2-b_2\exp(-(r+i\sigma)\Delta)$ 展開可得: $(\sigma\Delta)\approx \sqrt{2}(\ln(b_2\Delta)-a_2\Delta+1)^{1/2}$與$(r\Delta)\approx {1\over 3}(2\ln(b_2\Delta)+a_2\Delta-1)$ > $r>0$:不穩定;$r\leq 0$:穩定 > if $r=0$: $$ \begin{gather} 1-a_2\Delta=2\ln(b_2\Delta)\\ T={2\pi\over\sigma}={\sqrt{2}\pi\Delta\over\sqrt{3\ln(b_2\Delta)}}\approx {10.3\over\sqrt{\ln(b_2\Delta)}} \end{gather} $$ ##### **排放-重注振盪器**(discharge-recharge oscillator) 風應力旋度的作用 積分赤道周圍條狀海面(14.6.14): \begin{align} \int_{5^{\circ}N}vdx-\int_{5^{\circ}S}vdx+{1\over H}\left({\partial\over\partial t}+r\right)\int_{S}\eta dS=0 \end{align} 計算v的近似(14.6.15): ![image](https://hackmd.io/_uploads/Sy1rd4GlR.png =65%x) ![Screenshot 2024-04-09 at 11.39.48](https://hackmd.io/_uploads/S1Zq_Vzg0.png =73%x) 與14.6.15代入14.6.14: ![Screenshot 2024-04-09 at 11.40.08](https://hackmd.io/_uploads/Skrou4zlR.png =75%x) 考慮平均厚度($\left\langle\eta\right\rangle=\left(\int_S\eta dS\right)/S$)、$\left\langle T'\right\rangle$與$\left\langle\tau_x\right\rangle$為正比: ![Screenshot 2024-04-09 at 11.40.22](https://hackmd.io/_uploads/HJXh_NGeC.png =30%x) 取時間平均,得平均場之平衡關係式: ![Screenshot 2024-04-09 at 11.40.44](https://hackmd.io/_uploads/rkFa_Nzl0.png =60%x) 14.6.4扣除平均場: ![Screenshot 2024-04-09 at 11.40.59](https://hackmd.io/_uploads/B1P0uVMeR.png =90%x) 條狀海面積分: ![Screenshot 2024-04-09 at 11.41.11](https://hackmd.io/_uploads/Bym1FNfeR.png =30%x) **微分方程**: ![Screenshot 2024-04-09 at 11.41.23](https://hackmd.io/_uploads/SJklFVMe0.png =60%x) 代入$\left\langle T'\right\rangle=e^{\omega t}$: ![Screenshot 2024-04-09 at 11.41.41](https://hackmd.io/_uploads/BJ-ZF4MgC.png =50%x) 解出微分方程: ![Screenshot 2024-04-09 at 11.41.56](https://hackmd.io/_uploads/BkZMt4MxC.png =50%x) ##### 西太平洋振盪器(Western Pacific oscillator) 西太平洋西風區因**距平東風**造成的修正作用,距平東風導致西岸出現向東傳的湧升型(混合層厚度距平 η 為負)赤道凱爾文波,**聖嬰事件的負回饋因子** ##### 平流-反射振盪器(advective-reflective oscillator) 聖嬰事件時 1. **東向的海流**會推動西太平洋暖池海水向東太平洋平流(正回饋因子) 2. 伴隨之距平西風卻會產生向東傳的下沉型赤道凱爾文波以及向西傳的湧升型赤道羅士培波長波(西傳長波反射後形成湧升型赤道凱爾文波、東傳下沉型赤道 凱爾文波反射形成下沉型赤道羅士培波長波[非頻散波、流速向西])(負回饋因子) > 平均流在暖池也在東緣輻合,使暖池海水向西 > ENSO不干擾氣候旱澇出現時間、影響到旱澇強度與範圍