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In his MIT OCW course, Professor Kleitman explores the derivative of temperature when it depends on position and time:

$T(x, y, z, t)$ ... is a function of position $(x, y, z)$ and time $t$... What is the derivative with respect to time of the temperature experienced by a body [under trajectory] $r(t)$? $$\begin{align*} dT &= \frac {\partial T}{\partial x} dx + \frac {\partial T}{\partial y} dy + \frac {\partial T}{\partial z} dz + \frac {\partial T}{\partial t} dt \\ &\quad ... \\ \frac {dT} t &= \nabla T \cdot \frac {dr}{dt} + \frac {\partial T}{\partial t} \end{align*}$$ (edited)

The issue I have with this is that $T$ means three different things:

  • A function depending on time only: $\frac {dT} t$
  • A function depending on position only: $\nabla T$
  • A function depending on both: $\frac {\partial T}{\partial t}$

Is there a simpler way to write this? I tried to use functional notation to simplify both the notation and the derivation:

Let $$\begin{align} r \in \mathbb R^3 &\text{ (position)}\\ t \in \mathbb R &\text{ (time)}\\ U(r, t): \mathbb R^4 \to \mathbb R &\text{ (temperature at an arbitrary position and time)} \\ R(t): \mathbb R \to \mathbb R^3 &\text{ (a trajectory)} \\ T_R(t): \mathbb R \to \mathbb R = U(R(t), t) &\text{ (temperature along trajectory }R\text{ at time } t \text{)}. \end{align}$$

Then $$\begin{align} DT_R(t) &= \underbrace{\underbrace{DU(R(t), t)}_{1 \times 4} \underbrace{\begin{bmatrix}DR(t) \\ 1\end{bmatrix}}_{4 \times 1}}_{1 \times 1}\\ \end{align}$$

I'd like to conclude that $$\require{enclose}\enclose{horizontalstrike}{DT_R(t) = (DU(t))_{xyz} \cdot DR(t) + (DU(t))_t}$$ $$DT_R(t) = (DU(R(t),t))_{xyz} \cdot DR(t) + (DU(R(t),t))_t$$

but I'm not sure that's any clearer. The problem is that, while functional notation simplifies many things, including concepts, it gets very cumbersome when you have to talk about a subset of dimensions.

(Besides arguably simplifying notation, I believe this approach makes the actual derivation much clearer.)

Is my derivation correct? Is there a clearer way to write it? Is there a clearer way to handle this? See also How to write clear proofs involving curves and surfaces: The path of steepest descent along an ellipsoid for a similar issue.

SRobertJames
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  • The derivation is mostly fine, except that $DU(t)$ makes no sense. What you want are the spatial/time derivatives of $U$ evaluated at the point $(R(t),t)$. See here and the sublinks for more. Anyway let me assure you that writing things out correctly is not a bad thing, and clarifying this notational issue now (earlier the better) will only save you from lots of confusion later on. – peek-a-boo Aug 15 '23 at 19:53
  • @peek-a-boo Thanks. Fixed now? But that notation is still very opaque. The gradient/Leibnitz notation used at the quote at the top, while being inconsistent, still makes the intent clear. – SRobertJames Aug 15 '23 at 20:17
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    in my experience, it seems easy, then sometime afterwards things start looking confusing (especially if you start learning thermodynamics/Lagrangian mechanics), until you clarify the foundational question, and then everything looks simple again, i.e if we plot an understanding vs time graph, it looks like a $U$ shape :) The point is that it’s fine to use the abuse of notation as long as you (and whoever you’re writing for) know about it. – peek-a-boo Aug 15 '23 at 20:28

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