<p>You're confusing time-dependent and time independent equations: The time-dependent model tells you how <span class="math-container">$\vec U$</span> will evolve, it doesn't have to evolve into a steady-state. The steady, time-independent model tells you which <span class="math-container">$\vec U$</span> fulfills that equation as a criterion.<br />
Therefore that <span class="math-container">$\vec U$</span> usually gets a special name, the geostrophic velocity <span class="math-container">$\vec U_g$</span>, and the time-dependent equation also can be reformulated into a form of <span class="math-container">$\vec U =\vec U' -\vec U_g$</span>.</p>
<p>So when the current <span class="math-container">$\vec U_g$</span> changes, we can see how <span class="math-container">$d_t \vec U$</span> generates non-perpendicular terms until <span class="math-container">$\vec U = \vec U_g$</span> and balance is restored.</p>
<blockquote>
<p>In the moment I regard this as a discrepancy.</p>
</blockquote>
<p>I don't presume you regard the existence of satellites as fantasy, to which your argument would equally apply? For a circular orbit, the centrifugal force is perpendicular to the velocity, so it's all a hoax?</p>

CC BY-SA 4.0
2022-12-19T15:19:37.057Z
I thoght that the total change of U with time is
 
dU = ∂U/∂t * dt + ∂U/dx * dx + ∂U/dy * dy + ∂U/dz * dz
 
and 

Du/Dt =  ∂U/∂t  + ∂U/dx * dx/dt + ∂U/dy * dy/dt + ∂U/dz * dz/dt = 
          ∂U/∂t  + ∇U * U
    
So DU/dt gives me the rate of change of speed of a parcel as it flows along, as the "total" derivative.

In contrast  ∂U/∂t is the local rate of change, regarding U in a fixed point.

Isn't it DU/Dt which is to be used in Newton's Law? If not, what else?

I thought that I understood it, but you make me complete unsure now.
MichaelW
CC BY-SA 4.0
2022-12-19T15:24:49.037Z
For a circular orbit a satellite doesn't change its absolute value of speed, only direction. But in the flow of air not only direction is changing, but also absolute value.
MichaelW
CC BY-SA 4.0
2022-12-19T15:28:55.037Z
What do you mean with "time dependent"? If I regard U as a field and consider one point in space, the U can change with time on that point. This is a non-stationary state, but this I don't mean: I mean that even in stationary flow an air parcel must feel some force if it has U1 at one point and a moment later U2 on a second point.
MichaelW
CC BY-SA 4.0
2022-12-19T15:41:01.003Z
To be sure we talk about the same thing I added a drawing to my question.
MichaelW
CC BY-SA 4.0
2022-12-19T15:43:12.070Z
@MichaelW Your image is wrong. $dU/dt$ is not zero at all those three points, when you introduce a pressure gradient.
AtmosphericPrisonEscape
CC BY-SA 4.0
2022-12-19T15:45:11.333Z
@MichaelW Satellite: You're confusing quite many things here. The satellite equation is also a vectorial one. If you write it down as scalar (after projection with some radial operator) then your wind direction remains constant all the same, if you leave the pressure gradient constant.
AtmosphericPrisonEscape
CC BY-SA 4.0
2022-12-19T15:51:58.907Z
Sorry, I still cannot follow. You say "dU/dt  is not zero at all those three points."
I meant ∂U/∂t. Why not: The speed of the fluid at the location of the three points does not change with time. What changes is the speed of the parcel itself relative to the frame.

And yes, there is a pressure gradient, but only perpendicular to the line of flow, which is  on a contour line of constant pressure. Along the black lines there is no pressure change, they shall depict isobaric surfaces.
I feel like an idiot now...
MichaelW
CC BY-SA 4.0
2022-12-19T16:08:24.893Z
@MichaelW The situation you depicted is physically inconsistent with a constant magnitude of velocity. Either, initially, $\vec u =const$ everwhere, then due to the change in pressure gradient $d_t \vec U \neq 0$. Or, $\vec U(x,y)$ has some nontrivial profile, which fulfills the pressure gradient balance. In the latter case, the Coriolis force provides the required acceleration to keep $d_t \vec U = 0$ everywhere.
AtmosphericPrisonEscape
CC BY-SA 4.0
2022-12-19T16:09:52.457Z
What you seem confused about is how the steady state is achieved: As already said, when $d_t \vec U \neq 0$, then the coriolis acts to accelerate against the pressure gradient, until vectorial balance is achieved, in which case $\vec U \perp \vec \nabla P$.
AtmosphericPrisonEscape
CC BY-SA 4.0
2022-12-19T16:21:13.600Z
Sorry ..., but I talk only about a steady state, where wind velocity and direction is constant at every place. Is this unphysically? - I don't think so: Why shouldn't it be possible to measure wind at ten different places and it is constant there (with different values at the ten places but constant at each place)? Therefore I have drawn the red vectors with different size. At the same time geostrophic wind goes along lines of constant pressure, so the pressure for the left, middle and right point is the same. This means the parcel moves along constant pressure, but changes its velocity.
MichaelW
CC BY-SA 4.0
2022-12-19T16:26:29.367Z
I appreciate really your help but I'm obviously an idiot. Sorry, but I'm totally frustrated now...
MichaelW
CC BY-SA 4.0
2022-12-19T16:37:00.840Z
Other way around: When you say the picture incorrect, how must it be corrected in order to give a consistent view? Saying ∂U/∂t cannot be the solution, because I want to describe the stationary flow. But what screws to turn I have left then? Is it wrong, that the velocity changes from left to right? Might flow maybe not along constant pressure surfaces?  But this is what I learned...
MichaelW
CC BY-SA 4.0
2022-12-19T17:20:30.277Z
What I cannot understand additionally: There is Bernoulli's Law: It says essentially, that the sum of static and dynamic pressure doesn't change along a stream line: ϱv²/2+p = const. But when v changes, p cannot be constant (assuming constant temperature, p is proportional to ϱ). Isn't there a flaw from the beginning when I say, that along the streamline p=const?
MichaelW
CC BY-SA 4.0
2022-12-20T11:35:30.160Z
@MichaelW I am not sure i am able to help you more than what I've already tried to explain: The perpendicular velocity 'knows' about the pressure gradient in steady state, because it is accelerated towards the steady solution, while being in non-steady state. Once ss is reached, that velocity profile is kept. However even in ss, with dP being non-constant, accelerations, but steady ones, must exist (which then act along the v-vector). $v\nabla v$ is a good candidate for that. As for Bernoulli, the theorem (rederive it yourself) is unable to capture 2-D effects. It's useful for pipes.
AtmosphericPrisonEscape
CC BY-SA 4.0
2022-12-20T16:11:34.540Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/141440/discussion-between-michaelw-and-atmosphericprisonescape).
MichaelW
CC BY-SA 4.0
2022-12-21T07:30:38.583Z
I came finally to the conclusion, that geostrophic wind motion is only a rough estimation of balance and not applicable to curved flow. It is not a mathematically rigid solution for flow, only a first order approximation how flow could be roughly.
MichaelW
CC BY-SA 4.0
2022-12-19T13:34:00.030Z
2022-12-19T13:34:00.030Z
AtmosphericPrisonEscape
<p>Equation for acceleration of an air parcel is</p>
<p><span class="math-container">$$D \vec U/Dt = -2 \vec \omega \times \vec U -\frac{1}{\rho} \vec \nabla p + \vec g$$</span></p>
<p>Now, for geostrophic wind, U is parallel to the contour lines of constant p.</p>
<p>But if this is the case, then all forces above (beside g) are perpendicular to U. So how can U change when the lines of constant pressure get closer: in this case velocity should rise because larger gradient of p. In the moment I regard this as a discrepancy. How to resolve it?</p>
<p>EDIT:</p>
<p>This is what I mean in a picture:</p>
<p><a href="https://i.sstatic.net/aSFgP.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/aSFgP.png" alt="enter image description here" /></a></p>
<p>As the the blue air parcel moves from left to right on a line of constant pressure it becomes slower (red arrow), although the local change of speed is zero (stationary flow). It means, there must be a force on it which directs along its direction of flow (in fact its energy is decreased, because it is slower). The question is: where is this force coming from, since both pressure force and Coriolis force are perpendicular to the parcel's speed in each moment.</p>

<meteorology><wind><atmosphere-modelling><fluid-dynamics>
How can speed of geostrophic wind change when forces are always perpendicular to speed?
MichaelW
# How can speed of geostrophic wind change when forces are always perpendicular to speed?

(Asked by: username_0 on 2022-12-19)

Equation for acceleration of an air parcel is

$$D \vec U/Dt = -2 \vec \omega \times \vec U -\frac{1}{\rho} \vec \nabla p + \vec g$$

Now, for geostrophic wind, U is parallel to the contour lines of constant p.

But if this is the case, then all forces above (beside g) are perpendicular to U. So how can U change when the lines of constant pressure get closer: in this case velocity should rise because larger gradient of p. In the moment I regard this as a discrepancy. How to resolve it?

EDIT:

This is what I mean in a picture:

[![enter image description here](https://i.sstatic.net/aSFgP.png)](https://i.sstatic.net/aSFgP.png)

As the the blue air parcel moves from left to right on a line of constant pressure it becomes slower (red arrow), although the local change of speed is zero (stationary flow). It means, there must be a force on it which directs along its direction of flow (in fact its energy is decreased, because it is slower). The question is: where is this force coming from, since both pressure force and Coriolis force are perpendicular to the parcel's speed in each moment.

**Tags:** meteorology, wind, atmosphere-modelling, fluid-dynamics


---

## 1 answers

### Answer by username_1 on 2022-12-19. Score: 2 | Accepted answer

You're confusing time-dependent and time independent equations: The time-dependent model tells you how $\vec U$ will evolve, it doesn't have to evolve into a steady-state. The steady, time-independent model tells you which $\vec U$ fulfills that equation as a criterion.  
Therefore that $\vec U$ usually gets a special name, the geostrophic velocity $\vec U\_g$, and the time-dependent equation also can be reformulated into a form of $\vec U =\vec U' -\vec U\_g$.

So when the current $\vec U\_g$ changes, we can see how $d\_t \vec U$ generates non-perpendicular terms until $\vec U = \vec U\_g$ and balance is restored.

> In the moment I regard this as a discrepancy.

I don't presume you regard the existence of satellites as fantasy, to which your argument would equally apply? For a circular orbit, the centrifugal force is perpendicular to the velocity, so it's all a hoax?

> **Comments:**
> **username_0**: I thoght that the total change of U with time is   dU = ∂U/∂t * dt + ∂U/dx * dx + ∂U/dy * dy + ∂U/dz * dz   and   Du/Dt =  ∂U/∂t  + ∂U/dx * dx/dt + ∂U/dy * dy/dt + ∂U/dz * dz/dt =            ∂U/∂t  + ∇U * U      So DU/dt gives me the rate of change of speed of a parcel as it flows along, as the "total" derivative.  In contrast  ∂U/∂t is the local rate of change, regarding U in a fixed point.  Isn't it DU/Dt which is to be used in Newton's Law? If not, what else?  I thought that I understood it, but you make me complete unsure now.  
> **username_0**: For a circular orbit a satellite doesn't change its absolute value of speed, only direction. But in the flow of air not only direction is changing, but also absolute value.  
> **username_0**: What do you mean with "time dependent"? If I regard U as a field and consider one point in space, the U can change with time on that point. This is a non-stationary state, but this I don't mean: I mean that even in stationary flow an air parcel must feel some force if it has U1 at one point and a moment later U2 on a second point.  
> **username_0**: To be sure we talk about the same thing I added a drawing to my question.  
> **username_1**: @username_0 Your image is wrong. $dU/dt$ is not zero at all those three points, when you introduce a pressure gradient.  
> **username_1**: @username_0 Satellite: You're confusing quite many things here. The satellite equation is also a vectorial one. If you write it down as scalar (after projection with some radial operator) then your wind direction remains constant all the same, if you leave the pressure gradient constant.  
> **username_0**: Sorry, I still cannot follow. You say "dU/dt  is not zero at all those three points." I meant ∂U/∂t. Why not: The speed of the fluid at the location of the three points does not change with time. What changes is the speed of the parcel itself relative to the frame.  And yes, there is a pressure gradient, but only perpendicular to the line of flow, which is  on a contour line of constant pressure. Along the black lines there is no pressure change, they shall depict isobaric surfaces. I feel like an idiot now...  
> **username_1**: @username_0 The situation you depicted is physically inconsistent with a constant magnitude of velocity. Either, initially, $\vec u =const$ everwhere, then due to the change in pressure gradient $d_t \vec U \neq 0$. Or, $\vec U(x,y)$ has some nontrivial profile, which fulfills the pressure gradient balance. In the latter case, the Coriolis force provides the required acceleration to keep $d_t \vec U = 0$ everywhere.  
> **username_1**: What you seem confused about is how the steady state is achieved: As already said, when $d_t \vec U \neq 0$, then the coriolis acts to accelerate against the pressure gradient, until vectorial balance is achieved, in which case $\vec U \perp \vec \nabla P$.  
> **username_0**: Sorry ..., but I talk only about a steady state, where wind velocity and direction is constant at every place. Is this unphysically? - I don't think so: Why shouldn't it be possible to measure wind at ten different places and it is constant there (with different values at the ten places but constant at each place)? Therefore I have drawn the red vectors with different size. At the same time geostrophic wind goes along lines of constant pressure, so the pressure for the left, middle and right point is the same. This means the parcel moves along constant pressure, but changes its velocity.  
> **username_0**: I appreciate really your help but I'm obviously an idiot. Sorry, but I'm totally frustrated now...  
> **username_0**: Other way around: When you say the picture incorrect, how must it be corrected in order to give a consistent view? Saying ∂U/∂t cannot be the solution, because I want to describe the stationary flow. But what screws to turn I have left then? Is it wrong, that the velocity changes from left to right? Might flow maybe not along constant pressure surfaces?  But this is what I learned...  
> **username_0**: What I cannot understand additionally: There is Bernoulli's Law: It says essentially, that the sum of static and dynamic pressure doesn't change along a stream line: ϱv²/2+p = const. But when v changes, p cannot be constant (assuming constant temperature, p is proportional to ϱ). Isn't there a flaw from the beginning when I say, that along the streamline p=const?  
> **username_1**: @username_0 I am not sure i am able to help you more than what I've already tried to explain: The perpendicular velocity 'knows' about the pressure gradient in steady state, because it is accelerated towards the steady solution, while being in non-steady state. Once ss is reached, that velocity profile is kept. However even in ss, with dP being non-constant, accelerations, but steady ones, must exist (which then act along the v-vector). $v\nabla v$ is a good candidate for that. As for Bernoulli, the theorem (rederive it yourself) is unable to capture 2-D effects. It's useful for pipes.  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/141440/discussion-between-username_0-and-username_1).  
> **username_0**: I came finally to the conclusion, that geostrophic wind motion is only a rough estimation of balance and not applicable to curved flow. It is not a mathematically rigid solution for flow, only a first order approximation how flow could be roughly.  

---

