<p>As far as the question "What can be causing this?," there are a variety of reasons. Let's look at the <a href="http://glossary.ametsoc.org/wiki/First_law_of_thermodynamics" rel="nofollow noreferrer">First law of Thermodynamics</a> (including molecular diffusion): <span class="math-container">$$\frac{DU}{Dt}=c_p\frac{DT}{Dt}=Q-p\frac{D\alpha}{Dt}+\nu \nabla^2 U$$</span>, where <span class="math-container">$U$</span> is the <a href="http://glossary.ametsoc.org/wiki/Internal_energy" rel="nofollow noreferrer">internal energy</a> of the ideal gas, <span class="math-container">$Q$</span> is diabatic heating, <span class="math-container">$p$</span> is pressure, and <span class="math-container">$\alpha$</span> is the specific volume, <span class="math-container">$\nu$</span> is the molecular diffusivity, and <span class="math-container">$c_p$</span> is the isobaric specific heat capacity.</p>

<p>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 <span class="math-container">$-p\frac{D\alpha}{Dt}$</span> term as <span class="math-container">$\alpha \frac{Dp}{Dt}\approx \alpha \frac{\partial p}{\partial z}\approx\alpha\rho g=g=-\Gamma_d c_p$</span>, <a href="http://glossary.ametsoc.org/wiki/Dry-adiabatic_lapse_rate" rel="nofollow noreferrer"><span class="math-container">$\Gamma_d$</span> is the dry adiabatic lapse rate</a>). Using that relationship, (1) may be rewritten as <span class="math-container">$$\frac{\partial T}{\partial t}=\frac{Q}{c_p}+\nu\nabla^2T-\frac{g}{c_p}-\Gamma_d- \vec{u} \cdot\nabla T$$</span> 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.</p>

<p>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. </p>

CC BY-SA 4.0
2020-06-08T08:28:41.460Z
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 ?
duff18
CC BY-SA 4.0
2020-06-08T11:44:49.680Z
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 $\partial p / \partial t$
Joscha Fregin
CC BY-SA 4.0
2020-06-05T15:04:10.137Z
2020-06-05T15:04:10.137Z
BarocliniCplusplus
<p>I would rewrite your equation in a more natural order, at least to me:</p>
<p><span class="math-container">$$\frac{DT}{Dt} = \frac{\partial T}{\partial t} + \mathbf{U} \cdot \nabla T $$</span></p>
<p>The term <span class="math-container">$\frac{DT}{Dt} $</span> is the total derivative of <span class="math-container">$T$</span> with respect to time, i.e. how temperature changes (in one particular place on Earth) with time passing, which can have different reasons. The term <span class="math-container">$\mathbf{U} \cdot \nabla T$</span>, as you point out, is the advective term. It expresses that winds can bring warmer or cooler air onto this place, making it cooler or warmer there, such as cold fronts.</p>
<p>The term  <span class="math-container">$\frac{\partial T}{\partial t} $</span> expresses the <span class="math-container">$local$</span> change of temperature with time. If no winds are blowing, temperature may change as well, for example due to insolation (cooler at night, warmer during the day), seasonality and so on.</p>
<p>Therefore, the term <span class="math-container">$\frac{DT}{Dt}$</span> on the left-hand side of the equation makes it maybe it more clear. It is not a component of the derivate, it is <strong>the total</strong> derivative, decomposed in local and advective terms. So advection <strong>is</strong> one of the components of <span class="math-container">$\frac{DT}{Dt}$</span>.</p>

CC BY-SA 4.0
2020-07-10T15:50:04.507Z
2020-07-10T15:50:04.507Z
ouranos
<p>The local derivative is linked to the total derivative by:</p>

<p><span class="math-container">$\frac{\partial T}{\partial t} = \frac{DT}{Dt} - \mathbf{U} \cdot \nabla T$</span></p>

<p>and the -<span class="math-container">$ \mathbf{U} \cdot \nabla T$</span> term represent the change by advection.</p>

<p>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.</p>

<p>I have trouble interpreting the <span class="math-container">$\frac{DT}{Dt}$</span> 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 ?</p>

CC BY-SA 4.0
2020-06-18T23:18:22.257Z
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.
JeopardyTempest
CC BY-SA 4.0
2020-06-18T23:21:13.203Z
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.
JeopardyTempest
<meteorology>
Total differentiation
duff18
# Total differentiation

(Asked by: username_0 on 2020-06-05)

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 ?

**Tags:** meteorology

> **Comments:**
> **username_1**: 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.  
> **username_1**: 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.  

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## 2 answers

### Answer by username_2 on 2020-06-05. Score: 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](http://glossary.ametsoc.org/wiki/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](http://glossary.ametsoc.org/wiki/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](http://glossary.ametsoc.org/wiki/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.

> **Comments:**
> **username_0**: 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 ?  
> **username_3**: 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 $\partial p / \partial t$  

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### Answer by username_4 on 2020-07-10. Score: 0

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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