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It is a question vector calculus and Maxwell's laws. I put it this way. Let's say, we are working in a $3$-Dimensional space ( e.g $x\cdot y\cdot z = 4\cdot3\cdot2$, a certain room/class of that size ) .

Within this room, the heat obey a certain equation ( for e.g. $T = 25 + 5z$ ) .We know that heat flows from higher temperature regions to lower temperature regions. With this information in mind

How could I be able to determine the amplitude and the direction of travel of the thermal energy with the Del operator ( gradient ) ? .

I'm not looking for a definitive response, but an equation that could give me potentially the result for the amplitude and the direction. I also want to know if thermal energy follows a loop pattern within my room (and be able to explain it mathematically by using the del operator once again of course)?

You can find the lecture source here.

Thank you.

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Instead of lowering my question, maybe you could point me out what am I missing... –  fneron Jul 9 '12 at 15:31
Dear fneron , I think you are new to Math.SE. Welcome to Math.SE. The problem with your question is this : 1) You didn't state what you are looking for precisely. 2) You didn't follow punctuation and formatting tips while writing. 3) Please state where you are stuck. Thank you, BTW +1 for your question. –  Iyengar Jul 9 '12 at 16:23
Thank you very much. Is that better? –  fneron Jul 9 '12 at 16:49
Wait, I edit this one for you. –  Iyengar Jul 9 '12 at 17:01
perfect thx! Overall question is clear? –  fneron Jul 9 '12 at 17:06

2 Answers 2

Hello I try to present what I know the best.

We know that heat flow takes place in three forms :

  1. Conduction.
  2. Convection.
  3. Radiation.

For one dimensional heat flow, we have $q= k \dfrac{T_2-T_1}{L}$ , where $T_1$ and $T_2$ are terminal temperatures and $L$ is the length of the material. Switching to $3$-D spaces as you said, we need to look at the fourier system of the heat transfer before proceeding further. $\dfrac{q_x}{A}= - K \dfrac{dT}{dx}$. So integrating we obtain $$\dfrac{q_x}{A}\large \int_{0}^{L} \ dx= -k \int_{T_1}^{T_2} \ dT.$$

So given that the heat flow ( rate of conduction ) in the $3$-D space $(x,y,z)$ is given by the following equation $$ q= - k \nabla T = - k \bigg( \hat{i} \dfrac{\partial T}{\partial x}+\hat{j} \dfrac{\partial T}{\partial y}+\hat{k} \dfrac{\partial T}{\partial y}\bigg).$$The negative sign, there indicates the transfer of heat from one place to another. So you can simply substitute the heat equation in the place of $T$ and compute the partial derivatives to get the rate of flow of heat.

I can write up the cylindrical and spherical coordinate version too . It can be like this .

  1. For cylindrical coordinate system $(r,\phi)$ we have $$ \large q= - k \nabla T = - k \bigg(\hat{i} \dfrac{\partial T}{\partial r}+\hat{j} \dfrac{1}{r} \dfrac{\partial T}{\partial y}+\hat{k} \dfrac{\partial T}{\partial Z}\bigg).$$ and individual heat flows are given by $$\large q_r = - K \dfrac{dT}{dx}, \\ \large q_{\phi} = - \dfrac{K}{r}\dfrac{dT}{d \phi},\\ \large q_{Z} = - K\dfrac{dT}{dZ}$$ where $\large q_r$ represents the flow in $r$-th direction. Here we include an extra dimension $Z$ to make $2$-D into $3$-D.
  2. For Spherical coordinate system $(r,\theta,\phi)$ we have $$ \large q= - k \nabla T = - k \bigg(\hat{i} \dfrac{\partial T}{\partial r}+\hat{j} \dfrac{1}{r} \dfrac{\partial T}{\partial \theta}+\dfrac{\hat{k}}{r\sin \theta} \dfrac{\partial T}{\partial \phi}\bigg).$$

I hope this will be of some use to you. Feel free to ask if you want to hear more on anything.

Thank you.

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Great Answer! Great explanation... Unfortunately, I already had this in mind. I have read this part in my lecture source on page 9-10 (section 2.42, 2.43 and 2.44). +1 for answering a well written answer. –  fneron Jul 9 '12 at 18:09
@fneron : Thank you, I did nothing. Oh, BTW I didn't see your lecture notes. But I think the spherical and cylindrical versions are not there in that article you kept there. BTW are you a software engineer, what is the meaning of your name fneron ? –  Iyengar Jul 9 '12 at 18:17
Yes. I'm a student in computer/software engineering. It's a short term for my real name (first letter of my first name + last name)... –  fneron Jul 9 '12 at 18:24
@fneron : Well, it was very nice meeting you. I hope you have understood how to post the questions, don't get down from the bus that you caught ;). I faced lot many situations like that and learnt formatting. When somebody behave rudely with you think them as a sand-paper, they may be hard, but they will end up making you shine. That is my philosophy. All the best. –  Iyengar Jul 9 '12 at 18:33
I like your metaphor ;) Thank you for you help and nice meeting you too! –  fneron Jul 9 '12 at 18:40

(Thermal) Energy is scalar so it does NOT have orientation. Thermal field(If such a field is defined in Physics) has both orientation and magnitude.

In Electromagnetics :

Orientation: $E= \nabla\cdot V$

Magnitude (If energy is stored in field) :

$$W= \dfrac{1}{2}\large\int_{\mathbb{R}^{3}} |E|^2 \ dt$$

I think you may plug $T$ for $V$ in Thermodynamics.

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thank you. I'm looking into Maxwell equations to confirm, but I am stubbling on the heat diffusion equation: grad(grad(T))... –  fneron Jul 9 '12 at 15:35

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