The local derivative is linked to the total derivative by:

$\frac{\partial T}{\partial t} = \frac{DT}{Dt} - \mathbf{U} \cdot \nabla T$

and the -$ \mathbf{U} \cdot \nabla T$ term represent the change by advection.

So, for example cold air blowing in a region of warm temperatures will bring cold advection and it will contribute negatively to the local derivative.

I have trouble interpreting the $\frac{DT}{Dt}$ term. If it represents how the temperature is changing withing the flow regardless of the flow moving to cooler/warmer regions, what could this be caused by ? Say expansion and consequent cooling of the air parcels ?

  • $\begingroup$ If I remember right, I picture it like a cold front blowing across a large continent. At each location (the local derivative) will see the temperature drop as the front passes, (then eventually perhaps rise back up as that cold airmass pushes away... unless another cold air mass comes in). On the other hand, the airmass itself will moderate (warm) as it reaches lower latitudes and gets heated by the sun... the total derivative. $\endgroup$ Commented Jun 18, 2020 at 23:18
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    $\begingroup$ Now in terms of the equation written this way, focusing on the local change... the temperature change during the day at a location will be the combination of the advection and the airmass change itself. So in other words, during the day, the temperature change will be a combination of what air is blowing in, plus how the air mass is being changed by effects like heating. ... During the day the temperature at a spot usually goes up because of the DT/Dt term... solar heating. But if enough cold air is blown in (the advection term), the temperature may manage not to rise, or even drop. $\endgroup$ Commented Jun 18, 2020 at 23:21

2 Answers 2


As far as the question "What can be causing this?," there are a variety of reasons. Let's look at the First law of Thermodynamics (including molecular diffusion): $$\frac{DU}{Dt}=c_p\frac{DT}{Dt}=Q-p\frac{D\alpha}{Dt}+\nu \nabla^2 U$$, where $U$ is the internal energy of the ideal gas, $Q$ is diabatic heating, $p$ is pressure, and $\alpha$ is the specific volume, $\nu$ is the molecular diffusivity, and $c_p$ is the isobaric specific heat capacity.

So, without diabatic heating, one way is to change the volume or pressure, which varies the most with changes in altitude (you can rewrite the $-p\frac{D\alpha}{Dt}$ term as $\alpha \frac{Dp}{Dt}\approx \alpha \frac{\partial p}{\partial z}\approx\alpha\rho g=g=-\Gamma_d c_p$, $\Gamma_d$ is the dry adiabatic lapse rate). Using that relationship, (1) may be rewritten as $$\frac{\partial T}{\partial t}=\frac{Q}{c_p}+\nu\nabla^2T-\frac{g}{c_p}-\Gamma_d- \vec{u} \cdot\nabla T$$ Another way is through molecular heat exchange, which is so small it is usually neglected, unless it is in a micrometeorological context. And then there is diabatic heating, perhaps the most loaded term of the equation.

Diabatic heating is any heating source that requires entropy to change. This includes latent heat exchange, radiative heating, etc. Turbulent mixing may arguably be considered in this for large scales, but not small scales.

  • $\begingroup$ Thanks for taking your time to write such a detailed answer ! Of course, the $DT/Dt$ change (absorption of radiation, change in volume/pressure, etc) is happening at the same local point where the $\partial T/\partial t$ is computed, correct ? $\endgroup$
    – duff18
    Commented Jun 8, 2020 at 8:28
  • $\begingroup$ No. $D/Dt$ gives you the change following a parcel. You can imagine it like this: You are traveling with a hot air balloon (the parcel). Lets say you are interested in the pressure. In one time step your balloon moved from location a to location b. Calculating $Dp/Dt$ gives you the change of pressure as you move from location a to b. The change in pressure at the specific location a (or b) may be different. You get this local change of pressure by calculating $\partial p / \partial t$ $\endgroup$ Commented Jun 8, 2020 at 11:44

I would rewrite your equation in a more natural order, at least to me:

$$\frac{DT}{Dt} = \frac{\partial T}{\partial t} + \mathbf{U} \cdot \nabla T $$

The term $\frac{DT}{Dt} $ is the total derivative of $T$ with respect to time, i.e. how temperature changes (in one particular place on Earth) with time passing, which can have different reasons. The term $\mathbf{U} \cdot \nabla T$, as you point out, is the advective term. It expresses that winds can bring warmer or cooler air onto this place, making it cooler or warmer there, such as cold fronts.

The term $\frac{\partial T}{\partial t} $ expresses the $local$ change of temperature with time. If no winds are blowing, temperature may change as well, for example due to insolation (cooler at night, warmer during the day), seasonality and so on.

Therefore, the term $\frac{DT}{Dt}$ on the left-hand side of the equation makes it maybe it more clear. It is not a component of the derivate, it is the total derivative, decomposed in local and advective terms. So advection is one of the components of $\frac{DT}{Dt}$.


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