# How can speed of geostrophic wind change when forces are always perpendicular to speed?

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:

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.

## 1 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?

• 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. Dec 19, 2022 at 15:19
• 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. Dec 19, 2022 at 15:24
• @MichaelW Your image is wrong. $dU/dt$ is not zero at all those three points, when you introduce a pressure gradient. Dec 19, 2022 at 15:43
• @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. Dec 19, 2022 at 15:45
• 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... Dec 19, 2022 at 15:51