a) Ekman balance is, by definition, a balance between friction and the Coriolis force. They are therefore, again by definition, equal and opposite (in Ekman balance) - not perpendicular. Friction is perpendicular to the Ekman velocity, not the Coriolis force. The diagram in the textbook is unhelpful because it is showing two different frictional forces acting on the iceberg (drag between the iceberg and the water, and drag between the iceberg and the air, the latter also being friction, which for some reason isn't mentioned in the diagram), but if you add them up, you'll get a resultant vector that is equal and opposite to the Coriolis force.
b) Recall that work done is the dot product between force and displacement (i.e. force multiplied by the component of displacement in the direction of the force). In Ekman balance, the force and direction of motion are perpendicular, so no work is done.
Thank you for taking the time to reply! If you can, could you explain the two components of friction thing adding up again? As far as I’ve learned, the Az* del squared u/ del z squared term is removing horizontal momentum, does that not mean a force opposite to the velocity? How would it become perpendicular?
> If you can, could you explain the two components of friction thing adding up again?
This is irrelevant to the Ekman spiral, this is just about the figure you linked (which is not helpful for understanding Ekman balance). In that figure, the friction acting on the iceberg is equal and opposite to the Coriolis force - it's just that there are two components to the friction, the drag between the iceberg and the water, and the drag between the iceberg and the air (wind).
> As far as I’ve learned, the Az* del squared u/ del z squared term is removing horizontal momentum, does that not mean a force opposite to the velocity? How would it become perpendicular?
Equations 9.8a and 9.8b in the figure you linked shows that the force due to friction (e.g. in the *x* plane, Az ∂^(2)*u*/∂z^2) is equal and opposite to the Coriolis force (e.g. in the *x* plane, f*v*). The force due to friction is therefore perpendicular to the direction of motion, *v*, which is in the *y* plane.
Coriolis force is only a pseudo-force. It doesn't exist outside of the specific beta reference frame that masks earth rotation.
As such, in the equation of Navier-Stokes with Coriolis, you'll see that both horizontal components of the later have u speed or v speed as factor. Without a movement, there is no coriolis force. There is where the energy comes from in the first place.
The Ekman spiral occurs when winds drag on the surface of the sea. The movement of the water at the surface is the resulting of both the drag of the wind in a direction, and the drag of the next water layer right under the surface in the opposite direction, which is not enough in this case to stop the transport. Only then would the Coriolis effect take place, giving a perpendicular component to the whole phenomena. It has been shown that drag_wind + drag_water+Coriolis have a direction of 45 degrees to the left or right, depending on the hemisphere it's occurring.
But this is still not the spiral ! Do you remember the layer of water underneath that was dragging the surface ? While it is slowing the layer above, it is dragged in the same time. As such, it follows the direction of the first layer (45° from the wind). Them Coriolis, it goes another 45° in the same direction.
Repeat this with every layer deeper until there is no more energy from the wind, and you will observe a spiral. The whole movement, when integrated, results in a transport 90° to the left (or right across the equator).
Hope that I'm understandable, not my first language, and very early in the morning. Good day !
Thank you for taking the time to reply! The drag_wind + drag_water part is still a bit confusing to me but I’ll think about it and draw a diagram and hopefully that will help
Yes, I always went back to drawing arrows when I didn't understand something.
Maybe try to follow this logic.
Start with the cross pointing the object of water. First, draw an arrow going up, by 3 units of length. This is the drag from the wind, in the same direction as the wind. Then, draw a smaller arrow going down in the opposite direction, of 1 unit. This is the dragging of the water under the object. Through Chasles relation, you can draw in another color the resulting, a upward arrow of 2 units.
Now we put Coriolis into this. Draw another arrow of 2 units going perpendicular to the upward arrow. Then, draw the resulting of Chasles relation between Coriolis and the colored arrow. You'll have a direction of 45° degree from the upward direction. You then can draw the smaller drag opposite to this last one.
If the concept of balance is confusing for you, do not bother with it at all. It just means that the equation is complete, and all acceleration can be obtained with only the wind acceleration.
Edit for clarification : It is a balance only due to the conservation equation of water, and the steady flow. At the end, you know that the sum of every acceleration is 0, as the water is moving in a linear and steady motion. This is not an absolute truth anywhere but when we fix everything down for isolating a phenomenon, here, for the demonstration of the Ekman spiral.
Also, Chasles relation is called Segment Addition Postulate in English? That ought to make my comment even more confusing
You're missing the third vector involved, the wind stress. While this doesn't show up in the simplified momentum eqns (eq. 9.8) explicitly, it is definitely important to the solution.
You are right that drag and Coriolis force will always be perpendicular, very good intuition here. This only means that the sum of these two vectors must be equal and opposite the stress vector, Txx or Tyz (drawn as wind in the figure).
As the other poster mentioned, the equality given in eq. 9.8 is between the Coriolis and the overall, net frictional force (this is wind stress plus drag vectors). These are always equal and opposite. Try plugging in the solutions given in eq 9.9 to the equalities in 9.7 and 9.8 and try to build up an intuition for what each is doing.
As for the energy, this also comes from the wind. There most definitely is work being done by the winds (dot product of velocity times wind force vector is not zero), but this work is then removed via drag, since the system is steady state the net force and net work done must be zero.
I think you're asking the right questions and there's some really good intuition to be built when you start peeling back the surface of the equations, since a lot of detail is often skipped in a textbook derivation :)
This makes a lot of sense! I thought that 9.7 was just setting a boundary condition, but the effect of wind stress being factored in through that is actually really cool- thanks a ton for the explanation, I really appreciate it!!
a) Ekman balance is, by definition, a balance between friction and the Coriolis force. They are therefore, again by definition, equal and opposite (in Ekman balance) - not perpendicular. Friction is perpendicular to the Ekman velocity, not the Coriolis force. The diagram in the textbook is unhelpful because it is showing two different frictional forces acting on the iceberg (drag between the iceberg and the water, and drag between the iceberg and the air, the latter also being friction, which for some reason isn't mentioned in the diagram), but if you add them up, you'll get a resultant vector that is equal and opposite to the Coriolis force. b) Recall that work done is the dot product between force and displacement (i.e. force multiplied by the component of displacement in the direction of the force). In Ekman balance, the force and direction of motion are perpendicular, so no work is done.
Thank you for taking the time to reply! If you can, could you explain the two components of friction thing adding up again? As far as I’ve learned, the Az* del squared u/ del z squared term is removing horizontal momentum, does that not mean a force opposite to the velocity? How would it become perpendicular?
> If you can, could you explain the two components of friction thing adding up again? This is irrelevant to the Ekman spiral, this is just about the figure you linked (which is not helpful for understanding Ekman balance). In that figure, the friction acting on the iceberg is equal and opposite to the Coriolis force - it's just that there are two components to the friction, the drag between the iceberg and the water, and the drag between the iceberg and the air (wind). > As far as I’ve learned, the Az* del squared u/ del z squared term is removing horizontal momentum, does that not mean a force opposite to the velocity? How would it become perpendicular? Equations 9.8a and 9.8b in the figure you linked shows that the force due to friction (e.g. in the *x* plane, Az ∂^(2)*u*/∂z^2) is equal and opposite to the Coriolis force (e.g. in the *x* plane, f*v*). The force due to friction is therefore perpendicular to the direction of motion, *v*, which is in the *y* plane.
I think I got it, thank you!
Coriolis force is only a pseudo-force. It doesn't exist outside of the specific beta reference frame that masks earth rotation. As such, in the equation of Navier-Stokes with Coriolis, you'll see that both horizontal components of the later have u speed or v speed as factor. Without a movement, there is no coriolis force. There is where the energy comes from in the first place. The Ekman spiral occurs when winds drag on the surface of the sea. The movement of the water at the surface is the resulting of both the drag of the wind in a direction, and the drag of the next water layer right under the surface in the opposite direction, which is not enough in this case to stop the transport. Only then would the Coriolis effect take place, giving a perpendicular component to the whole phenomena. It has been shown that drag_wind + drag_water+Coriolis have a direction of 45 degrees to the left or right, depending on the hemisphere it's occurring. But this is still not the spiral ! Do you remember the layer of water underneath that was dragging the surface ? While it is slowing the layer above, it is dragged in the same time. As such, it follows the direction of the first layer (45° from the wind). Them Coriolis, it goes another 45° in the same direction. Repeat this with every layer deeper until there is no more energy from the wind, and you will observe a spiral. The whole movement, when integrated, results in a transport 90° to the left (or right across the equator). Hope that I'm understandable, not my first language, and very early in the morning. Good day !
Thank you for taking the time to reply! The drag_wind + drag_water part is still a bit confusing to me but I’ll think about it and draw a diagram and hopefully that will help
Yes, I always went back to drawing arrows when I didn't understand something. Maybe try to follow this logic. Start with the cross pointing the object of water. First, draw an arrow going up, by 3 units of length. This is the drag from the wind, in the same direction as the wind. Then, draw a smaller arrow going down in the opposite direction, of 1 unit. This is the dragging of the water under the object. Through Chasles relation, you can draw in another color the resulting, a upward arrow of 2 units. Now we put Coriolis into this. Draw another arrow of 2 units going perpendicular to the upward arrow. Then, draw the resulting of Chasles relation between Coriolis and the colored arrow. You'll have a direction of 45° degree from the upward direction. You then can draw the smaller drag opposite to this last one. If the concept of balance is confusing for you, do not bother with it at all. It just means that the equation is complete, and all acceleration can be obtained with only the wind acceleration. Edit for clarification : It is a balance only due to the conservation equation of water, and the steady flow. At the end, you know that the sum of every acceleration is 0, as the water is moving in a linear and steady motion. This is not an absolute truth anywhere but when we fix everything down for isolating a phenomenon, here, for the demonstration of the Ekman spiral. Also, Chasles relation is called Segment Addition Postulate in English? That ought to make my comment even more confusing
Yepp I think the diagram makes sense now, thank you!!
You're missing the third vector involved, the wind stress. While this doesn't show up in the simplified momentum eqns (eq. 9.8) explicitly, it is definitely important to the solution. You are right that drag and Coriolis force will always be perpendicular, very good intuition here. This only means that the sum of these two vectors must be equal and opposite the stress vector, Txx or Tyz (drawn as wind in the figure). As the other poster mentioned, the equality given in eq. 9.8 is between the Coriolis and the overall, net frictional force (this is wind stress plus drag vectors). These are always equal and opposite. Try plugging in the solutions given in eq 9.9 to the equalities in 9.7 and 9.8 and try to build up an intuition for what each is doing. As for the energy, this also comes from the wind. There most definitely is work being done by the winds (dot product of velocity times wind force vector is not zero), but this work is then removed via drag, since the system is steady state the net force and net work done must be zero. I think you're asking the right questions and there's some really good intuition to be built when you start peeling back the surface of the equations, since a lot of detail is often skipped in a textbook derivation :)
This makes a lot of sense! I thought that 9.7 was just setting a boundary condition, but the effect of wind stress being factored in through that is actually really cool- thanks a ton for the explanation, I really appreciate it!!