我理解压力坐标以正确的方式吗?- 江南体育网页版- - - - -地球科学堆江南电子竞技平台栈交换 最近30从www.hoelymoley.com 2023 - 06 - 27 - t09:55:32z //www.hoelymoley.com/feeds/question/25028 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/25028 4 我理解压力坐标以正确的方式吗? MichaelW //www.hoelymoley.com/users/21047 2023 - 03 - 14 - t19:42:33z 2023 - 03 - 16 - t17:12:33z < p >问题很简单:我仍然有大气压力问题坐标:< / p > < p > < a href = " https://i.stack.imgur.com/FDpSF.png " rel = " nofollow noreferrer " > < img src = " https://i.stack.imgur.com/FDpSF.png " alt = "在这里输入图像描述" / > < / > < / p > < p >照片中的压力依赖性,什么是空运的速度坐标移动与< span class = " math-container " > $ \ vec v = (u, v = 0 w = 0) $ < / span > <强>压力沿纬向方向坐标> < /强大?< / p > < p >我理解正确的话,在压力坐标没有而是ω?< / p > < p >我想说我们有那么元组<跨类= " math-container " > $ $ (u, v = 0,ω)$ $ < / span >与< span class = " math-container " > $ $ \ω= Dp / Dt = (p / \ \部分部分x) _z \ cdot Dx / Dt = (p / \ \部分部分x) _z \ cdot u $ $ < / span > < / p > < p >但我不确定,因为它看起来有点奇怪,在第一个系统我们有0 w和压力系统-垂直“speed"。< / p > < p >尽管我们通常只考虑水平分量u, v的风,由于压力坐标我们有一个非零“vertical"组件(因为压力减少包裹右移的例子)。比更容易压力坐标如何处理当有一个额外的组件类< span = " math-container " > \ω< / span >美元,我们有三个速度分量来考虑,而不只是两个(当然我知道其他方程变得更简单密度消失…)?< / p > < p >我有一个错误在我的考虑吗? < / p > //www.hoelymoley.com/questions/25028/-/25031 # 25031 3 答案由Joscha Fregin我理解压力坐标以正确的方式吗? Joscha Fregin //www.hoelymoley.com/users/18833 2023 - 03 - 15 - t16:40:50z 2023 - 03 - 16 - t17:12:33z < ol > <李> < p >压力坐标通常使用< span class = " math-container " > \ω< / span >美元的垂直速度分量,而不是< span class = " math-container " > w < / span >美元如笛卡尔坐标。李李< / p > < / > < > < p >获得压力坐标系中的速度组件:< / p > < /李> < / ol > < p >使用物质导数可以与垂直速度分量和我将在下面描述怎么做。< / p > < p >开头两个坐标系统类< span = " math-container " >美元(x, y, z, t) $ < / span >和<跨类= " math-container " >美元(x, y, p、t) $ < / span >与垂直坐标<跨类= " math-container " > z z =美元美元(t, x, y, p) < / span >和<跨类= " math-container " > $ p = p (t, x, y, z) < / span >美元。一些变量的导数< span class = " math-container " > < / span >美元美元对其他变量类< span = " math-container " > < / span >加元美元(可以<跨类= " math-container " > $ x $ < / span >, < span class = " math-container " > $ y $ < / span >或<跨类= " math-container " > t < / span >)美元在一个系统中其他相关如下:< / p > < p > <跨类= " math-container " > \开始{方程}\离开(\压裂{\部分}{部分c \} \右)_z = \离开(\压裂{\部分}{部分c \} \右)_p + \压裂{\部分}{\部分p} \离开(\压裂{\部分p}{部分c \} \右)_z。结束\{方程}< / span >的垂直坐标认为< span class = " math-container " > \开始{方程}\压裂{\部分}{\部分p} = \压裂{\部分}{部分z \} \压裂{\部分z}{\部分p}, \{方程}< / span >结束,因此,我们发现<跨类= " math-container " > \开始{方程}\离开(\压裂{\部分}{\部分c} \右)_z = \离开(\压裂{\部分}{\部分c} \右)_p + \压裂{\部分}{部分z \} \压裂{\部分z}{\部分p} \离开(\压裂{\部分p}{\部分c} \右)_z。结束\{方程}< / span > < / p > < p >这个(2 d)梯度写道:< / p > < p > <跨类= " math-container " > \开始{方程}\ nabla_z = \ nabla_p + \压裂{\部分z}{\部分p} \压裂{\部分}{部分z \} \ nabla_z p。\{方程}< / span >结束时间导数是微不足道的,因为它遵循从上面的一般形式。Now we can write the material derivative in pressure coordinates as follows: \begin{equation} \begin{split} \left( \frac{\text{D}}{\text{D} t} \right)_p &= \left( \frac{\partial}{\partial t} \right)_p+ \vec{u} \cdot \nabla_p + \omega \frac{\partial}{\partial p}\\ &= \left( \frac{\partial }{\partial t} \right)_z - \frac{\partial z}{\partial p}\left( \frac{\partial p}{\partial t} \right)_z \frac{\partial }{\partial z} + \vec{u} \cdot \left[ \nabla_z - \frac{\partial z}{\partial p} \nabla_z p \frac{\partial }{\partial z} \right] + \omega \frac{\partial z}{\partial p}\frac{\partial}{\partial z} \\ &= \left( \frac{\partial }{\partial t} \right)_z + \vec{u} \cdot \nabla_z + \left[\omega - \left(\frac{\partial p}{\partial t}\right)_z - \vec{u} \cdot \nabla_z p \right] \frac{\partial z}{\partial p} \frac{\partial}{\partial z} . \end{split} \end{equation} Hence, \begin{equation} w = \left[\omega - \left(\frac{\partial p}{\partial t}\right)_z - \vec{u} \cdot \nabla_z p \right] \frac{\partial z}{\partial p} \end{equation} and \begin{equation} \omega =\left( \frac{\partial p }{\partial t}\right)_z + \vec{u} \cdot \nabla_z p + w \frac{\partial p}{\partial z}. \end{equation} What typically follows is the use of the hydrostatic approximation in the last term. If you do a (large) scale analysis you will find that $-w\rho g = w \partial p/\partial z$ is the dominating term and thus, you can approximate $\omega = -w\rho g$. However, based on your sketch I think it would not add to an understanding to just say $\omega = 0$, since $w = 0$.

Why it makes sense to use pressure coordinates I tried to answer here, although I want to add that in a theoretical context it can be much more useful to apply e.g. sigma coordinates (terrain following coordinates) to simplify the treatment of boundary conditions.

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