Published on *METEO 300: Fundamentals of Atmospheric Science* (https://www.e-education.psu.edu/meteo300)

Scientists like things that are conserved. There are good reasons for this. First, if something is conserved, that means we can always count on it being the same no matter what happens. Second, when we write down the equation for the conserved quantity, we can use that equation to understand how the equation’s variables will change with differing conditions. For example, in Lesson 2, we were able to use the First Law of Thermodynamics (a.k.a., conservation of energy) along with the Ideal Gas Law to derive the equation for potential temperature, which is very useful for understanding and calculating the vertical motion of air parcels.

In atmospheric dynamics, we like three conservation laws:

**conservation of energy**(The 1^{st}Law of Thermodynamics [1])**conservation of mass****conservation of momentum**(Newton’s Second Law, but really three equations—one in each direction)

So, let’s step back and look at the mass of an air parcel, which equals the density times the volume of the parcel:

$m=\rho V$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaad2gacqGH9aqpcaGGGcGaeqyWdiNaamOvaaaa@3897@

In a parcel, the mass is conserved, and since *m = ρV*,

$\frac{D}{Dt}\left(m\right)=\frac{D}{Dt}\left(\rho V\right)=0$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaalaaapaqaa8qacaWGebaapaqaa8qacaWGebGaamiDaaaadaqadaWdaeaapeGaamyBaaGaayjkaiaawMcaaiabg2da9maalaaapaqaa8qacaWGebaapaqaa8qacaWGebGaamiDaaaadaqadaWdaeaapeGaeqyWdiNaamOvaaGaayjkaiaawMcaaiabg2da9iaaicdaaaa@4235@

Apply the product rule to Equation [10.2]:

$V\frac{D\rho}{Dt}+\rho \frac{DV}{Dt}=0$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaadAfadaWcaaWdaeaapeGaamiraiabeg8aYbWdaeaapeGaamiraiaadshaaaGaey4kaSIaeqyWdi3aaSaaa8aabaWdbiaadseacaWGwbaapaqaa8qacaWGebGaamiDaaaacqGH9aqpcaaIWaaaaa@406A@

Divide both sides by *ρV*:

$\frac{1}{\rho}\frac{D\rho}{Dt}+\frac{1}{V}\frac{DV}{dt}=0$

Recall that the specific rate of change in parcel volume is equal to the divergence (Equation 9.4) and so we can write:

$\frac{1}{V}\frac{DV}{dt}=\overrightarrow{\nabla}\u2022\overrightarrow{U}$ $$

Rearranging the equation gives us an expression for the conservation of mass:

$\frac{1}{\rho}\frac{D\rho}{Dt}+\overrightarrow{\nabla}\u2022\overrightarrow{U}=0$

This equation is for the conservation of mass in a continuous fluid (i.e., the fluid particles are so small that the air parcel behaves like a fluid). It is also called the **Equation of Continuity**. Physically, this equation means that if the flow is converging ($\overrightarrow{\nabla}\u2022\overrightarrow{U}<0$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaaFiaabaGaey4bIenacaGLxdcacaGGIaYaa8HaaeaacaWGvbaacaGLxdcacqGH8aapcaaIWaaaaa@3B2C@ ), then the density must increase ($\frac{D\rho}{Dt}$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaalaaapaqaa8qacaWGebGaeqyWdihapaqaa8qacaWGebGaamiDaaaaaaa@3779@ >0). Note that in Lesson 9.5 [3]we said that density doesn’t change much at any fixed pressure level, which is how we were able to relate horizontal divergence/convergence with vertical ascent/descent. What *did* change was the vertical size of the air parcel as the horizontal size increased or decreased. The total mass, however, remained the same.

Credit: Licensed under Public Domain via Wikimedia Commons [4]

Newton’s 2^{nd} Law, **F** = *m***a**, applies to a mass with respect to the inertial coordinate system of space. But we are interested in motion with respect to the rotating Earth. So, to apply Newton’s 2^{nd} Law to Earth’s atmosphere, our mathematics will need to account for the forces of Earth’s rotating coordinate system:

$$\overrightarrow{F}=m\overrightarrow{a}={{\displaystyle \sum}}^{\text{}}\text{(realforces)}+{{\displaystyle \sum}}^{\text{}}\text{(apparentforces)}=m\frac{D\overrightarrow{U}}{Dt}$$[2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@60E8@

where the first set of forces are real forces and the second set are apparent (or effective) forces that will be used to correct for using a coordinate system attached to a rotating Earth.

When we use the word “specific” as an adjective describing a noun in science, we mean that noun divided by mass. So, specific force is **F**/m = **a**, acceleration. In what follows, we will use the terms “force” and “acceleration” interchangeably, assuming that if we say “force,” we mean “force/mass,” which is acceleration. At this point, you should be able to check the units—if there is no “kg,” then obviously we are talking about accelerations.

$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{D\overrightarrow{U}}{Dt}$

Here is another chance to earn **one point of extra credit**: **Picture of the Week**!

- You take a picture of some atmospheric phenomenon—a cloud, wind-blown dust, precipitation, haze, winds blowing different directions—anything that strikes you as interesting.
- Add a short description of the processes that you think are causing your observation. A Word file is a good format for submission.
- Use your name as the name of the file. Upload it to the
**Picture of the Week Dropbox**in this week's lesson module. To be eligible for the week, your picture must be submitted by 23:59 UT on Sunday of each week. - I will be the sole judge of the weekly winners. A student can win up to three times.
- There will be a Picture of the Week dropbox each week through Lesson 11. Keep submitting entries!

**Links**

[1] https://www.e-education.psu.edu/meteo300/8

[2] mailto:MathType@MTEF

[3] https://www.e-education.psu.edu/meteo300/node/726

[4] http://commons.wikimedia.org/wiki/File:Isaac_Newton_woodcut,_frontispiece_to_Mach.jpg#/media/File:Isaac_Newton_woodcut,_frontispiece_to_Mach.jpg