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Home , Body force, Equations of motion

EOSC 512 2019

4 Coordinate Systems and the of

4.1 Rates of change revisited

We have now derived the Navier-Stokes equations in an inertial (non-accelerating) for which Newton’s third law is valid. However, in oceanography and meteorology it is more natural to put ourselves in an earth-fixed coordinate frame that is rotating with the planet (Only natural – this is the frame from which we observe in the ocean and the atmosphere). However, rotating reference frames are not inertial. It is therefore nexessary to examine how the equations of motion must be altered to take this into account. Consider two observers. one fixed in inertial and one fixed in our rotating frame. Consider how the rate of change of a vector fixed in length will appear to both observers if the vector rotates with the rotation of the Earth.

Where ⌦is the angular of the rotating frame. The earth has a rotational period of one day, so ⌦= 5 1 7.29 10 s . ⇥ This introduces three interesting questions:

ˆ Dˆi 1. What rate of change of i ( Dt ) is seen by the inertial observer? By the rotating observer?

2. Now let ˆi be an arbitrary vector A~ = Ajˆij. How are the rates of change of this vector seen by the inertial and DA~ DA~ rotating observer related? That is, how does Dt relate to Dt ? inertial rotating ⇣ ⌘ ⇣ ⌘ ~ 3. Now choose A to be the velocity seen in the inertial frame (~u )inertial. How is the rate of change of the inertial velocity as seen in inertial space related to the as seen in the rotating coordinate frame?

We will answer these questions indiviually below:

1. Consider the motion of the ˆi in a t. It will swing around the rotation axis under the unfluence of the rotation, moving at right andles to itself and to the rotation vector as shown:

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The di↵erence between the two vectors, ~i can be approximated – for small t – as the chord generated in

that time between the perpendicular from the rotation axis to the tip of the unit vector at t = t0 &

the perpendicular distance from the rotation axis to the tip of the unit vector at t = t0 +t. Top-down view:

Here, the radius of the circle traced out by the tip of the unit vector is r =sin ˆi and the of swing is the product of the of rotation and the “swinging time”, ✓ =⌦t. Thus, the chord length is r✓ = ˆi sin ⌦t. ~ The direction of i is perpendicular to both the unit vector and the rotation vector since the change can only alter its direction and not its length. Thus, the change in the unit vector in time t is:

~ ⌦~ ˆi ˆi = ˆi sin ⌦t ⇥ (4.1) ⌦~ ˆi ⇥ Where ˆi sin ⌦= ⌦~ ˆi by the definition of the cross product. This leaves ~ˆi =t⌦ ˆi.Dividingbyt ⇥ ⇥ and taking the limit t 0, ! Dˆi = ⌦~ ˆi (4.2) Dt ⇥ !inertial

This is the rate of change of the unit vector ˆi seen by the inertial observer! Notice that if the observer is rotating with ⌦, the see no change at all, so:

Dˆi = 0 (4.3) Dt !rotating

2. Let A~ be represented as three components along the coordinate axes with unit vectors ˆi1, ˆi2, and ˆi3. Assuming the coordinate axes are still rotating with angular velocity ⌦.

The individual component of A~ along each coordinate axis is the shadow of the vector cast along that axis. It

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is a whose value and rate of change is seen the same by both the inertial and rotating observers. The inertial observer also sees the rate of change of the unit vectors. Thus,

DA~ DA Dˆi = i ˆi + A Dt Dt i i Dt !inertial DAi (4.4) = ˆi + A ⌦ ˆi Dt i i ⇥ DA ⇣ ⌘ = i ˆi +⌦ A~ Dt i ⇥

DAi ˆ ~ Physically, Dt ii is the rate of change of A seen by the rotating obeserver. He/she is only aware of the increase or decrease of the individual scalar coordinates along the coordinate unit vectors that the observer sees as constant. This can be interpreted as a stretching of the components of A~.⌦ A~ is an additional component ⇥ of the rate of change of A~ seen only by the inertial observer. This is the swinging of A~ due to the rotation. Together, DA~ DA~ = +⌦ A~ (4.5) Dt Dt ⇥ !inertial !rotating

3. Taking the previous and using ~x as the of the fluid element,

D~x DAx~ = +⌦ ~x (4.6) Dt Dt ⇥ ✓ ◆inertial !rotating The of position with respect to time is velocity, so:

~u = ~u +⌦ ~x (4.7) inertial rotating ⇥

Similarly, if we choose A~ to be the velocity seen in the inertial frame ~u inertial, D~u D~u inertial = inertial + ⌦~ ~u Dt Dt ⇥ inertial ✓ ◆inertial ✓ ◆rotating D = ~u rotating + ⌦~ ~x + ⌦~ ~u rotating + ⌦~ ~x Dt ⇥ rotating ⇥ ⇥ ✓ ⇣ ⌘◆ ⇣ ⌘ D~u rotating D~x = + ⌦~ + ⌦~ ~u rotating + ⌦~ ⌦~ ~x (4.8) Dt rotating ⇥ Dt rotating ⇥ ⇥ ⇥ ✓ ◆ ✓ ◆ ⇣ ⌘ D~u inertial D~u rotating = +2⌦~ ~u rotating + ⌦~ ⌦~ ~x Dt inertial Dt rotating ⇥ ⇥ ⇥ ✓ ◆ ✓ ◆ ⇣ ⌘ 2 3 1 | {z } | {z } | {z } 1 The acceleration that an observer in the rotating frame would see. ⌘ 2 The coriolis acceleration. ⌘ 3 The (familiar?) centripetal accerlation. This can be written as the gradient of a scalar, ⌦~ ⌦~ ~x = ⌘ ⇥ ⇥ 1 (⌦ ~x )2. This is the acceleration an object feels resulting from the change of direction (but⇣ not⌘ the 2 r ⇥ change in !) associated with uniform . Has magnitude ⌦ 2r where r is the distance | | between the rotation axis and the tip of the vector. It is directed toward the centre of the circle (hence centripetal!).

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4.2 The Coriolis acceleration/

We have just seen that the acceleration in the inertial frame is related to the acceleration that an observer would see in the rotating frame with two additional contributions: The centripetal accerlation (acceleration directed towards the centre of the circle traced out by the tip of the rotating vector) & the Coriolis acceleration. The so-called Coriolis acceleration acts perpendicular to the velocity as seen in the rotating frame. It represents the acceleration due to the swinging of the velocity vector by the rotation vector. A factor of two enters because the rotation vector swings the velocity ~u around, but also the velocity ⌦~ ~x which is not seen in the rotating rotating ⇥ frame. This term will increase if the position vector ~x increases, giving rise to a second factor of ⌦~ ~u in the ⇥ rotating acceleration. As we will see, the Coriolis acceleration is a dominant (lead order) term in the of large-scale, low frequency motion in the atmospheric and oceans. This includes general banded atmospheric circulations such as trade and westerlies and large-scale ocean gyre circulations. The Coriolis acceleration, like all , is produced by a force in the direction of the acceleration.

First we assume we have uniform velocity seen in the rotating system (which you will note is counterclockwise rotation). This implies that there will be no acceleration perceived in the system. In order for the flow to continue on a straight path as in the figure above, a force must be applied from the right. If that force is removed, the fluid element can no longer continue in a straight line and will veer to the right. This gives rise to the sensation that a “force” is acting to push the fluid to the right. Of course, since there are no in the inertial frame, then there are no acceleration in the inertial frame. For this reason, both the centripedal force and the are sometimes called pseudoforces as they are experienced only in the frame of reference of the body undergoing circular motion. It should be emphasized that there is nothing fictitious about these forces or accelerations – they are real – they are just not recognized as acceleration by an observer in the rotating frame.

An alternative, more heurisitic way to think of the Coriolis e↵ect is the tendency for objects in motion (viewed in a rotating frame of reference) to follow curved paths. They deflect to the right for counter-clockwise rotation. Consider an observer on Earth watching the motion of a fluid element in the atmosphere.

Because the Earth is rotating, points on the Earth’s surface travel eastward with a speed given by the Earth’s

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angular velocity. Further, because the Earth is spherical, the distance a point travels in one day at low latitudes is bigger compared to higher latitudes. As a results, the eastward velocity of a point on the Earth’s surface varies with latitude: Fast at the equator and slower as you move towards the poles.

Example 4.1. If a fluid parcel at the equator has a northward inertial velocity and no other forces acting, it also has an eastward component of velocity of approximately 444 m/s. To a rotating observer at 49°, it also has an eastward component of velocity due to di↵erence in the angular velocity between the equator and 49°.

To the observer at 49°, in some time t, the parcel has moved north, but also eastwards by an amount (444

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300) m/st. That is, the observer at 49° sees the parcel travel a curved path that is deflected to the right of its initial northwards. We can heuristically think of this as arising because the parcel is moving eastward faster than the ground over which it is flying as it moves northward (in a sense, it gets ahead of the planet). You can extend this argument to arrive at the rule of thumb results that due to the Coriolis e↵ect:

• In the northern hemisphere, an object in motion will get deflected to the right.

• In the souther hemisphere, an object in motion will get deflected to the left.

You can remember this with the following picture. denotes the starting location, the dashed arrow in the initial ⇥ trajectory and the solid arrows are the curved observed in the rotating frame. It is this e↵ect that

results in the deflection of the large-scale winds on Earth from N-S (that would results from the di↵erential heating at the equator vs the poles in the case of a non-rotating planet) to the E-W associated with the NE/SE trade winds, and the mid-latitude westerlies and the polar easterlies. In the absence of rotation:

When the earth starts to rotate:

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4.3 The Navier-Stokes Equations in a rotating frame

We are now in a position to rewrite the Navier-Stokes equations in a rotating frame and we will rarely look back. Going forward, we will always use the rotating frame, and (unless otherwise stated) the we use will be those observed from the rotating frame. And so we can dispense with the cumbersome (quantity)rotating demarcation. We we ever wish to use an inertial frame, we need only to set ⌦= 0 in our set of rotating equations. We will return to the general vector(-ish) representation of the Navier-Stokes equations for a fluid (equation 3.44).

D~u D~u inertial We will substitute Dt (remembering that this is now equivalent to our clunky Dt ) in terms of the inertial position, velocity, and acceleration vectors, seen in the rotation frame. This will⇣ require a⌘ bit of massaging.

2 2 ⌦~ ~x D~u ~ r ⇥ 2 ˆ @µ ⇢ + ⇢ (2⌦ ~u )=⇢ 0F + 1 P + µ ~u +( + µ) ( ~u ) +( )( ~u ) + ii2eij (4.9) Dt ⇥ ⇣ 2 ⌘ r r r r· r r· @xj B C a b c @ A d | {z } | {z } | {z } | {z } D~u Where Dt is the acceleration that an observer in the rotating frame would see, 2⌦ ~u is the Coriolis acceleration, 2 ⇥ ⌦~ ~x r( ⇥ ) and 2 is the centripedal acceleration moved to the RHS. Some notes:

• The spatial gradient are the same in both frames. Notice that the gradient in unaltered!

• The viscous terms a , b , c ,& d are each proportional to gradient of the rate of strain or the divergence of the velocity –each of which is independent of rotation. So these terms are also invariant between the two frames. We often assume that is constant, so we can set c = d = 0.

• We’ve moved the centripetal acceleration to the RHS where it can now be interpreted as the cenripetal force. This is convenient because the body force we most frequently encounter is due to , which can also be written as the gradient of a scalar (F~ = where is the gravitational potential). g r

Assuming this is the only body force acting (a good assumption for the types of flows we will consider) and assuming constant viscosity coecients (also usually a good assumption), the Navier-Stokes equations in the rotating frame simplify to: D~u 1 2 ⇢ + ⇢ (2⌦ ~u )= ⇢ ⌦~ ~x P + µ 2~u +( + µ) ( ~u ) (4.10) Dt ⇥ r 2 ⇥ r r r r· ✓ ⇣ ⌘ ◆

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Where the combination of the gradient of the gravitation and centrifual potential is what we experience as gravity. 2 (So ~g = 1 ⌦~ ~x ). This gives us a working form of the Navier-Stokes equations valid for an observer r 2 ⇥ ✓ ◆ in the rotating frame!⇣ ⌘ D~u ⇢ + ⇢ (2⌦ ~u )=⇢~g P + µ 2~u +( + µ) ( ~u ) (4.11) Dt ⇥ r r r r· We find that this will be applicable to most atmospheric and oceanic flows. Note that our equation for the conservation of is the same in the rotating frame, since it involves only the divergence of the velocity and the rate of change of a scalar.

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