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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...2012-07-09
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    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.2012-07-09
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    Thank you very much. Is that better?2012-07-09
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    Wait, I edit this one for you.2012-07-09
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    perfect thx! Overall question is clear?2012-07-09
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    I think the most appropriate forum to ask this things is http://physics.stackexchange.com/ , try posting it there, if you don't get a response here.2012-07-09
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    Please add Physics tag to your question in addition to differential-equations and vector-analysis tags.2012-07-09
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    thank you. Appreciated this formatting help! I'm a programmer, most of the time I just copy/paste my code into those forums with a brief explanation. But physics and math are different ;)2012-07-09
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    Sorry for removing all the stuff you said. The people in this community are only interested in the question, if you say the background in writing up the question they will simply down-vote. Another issue is that, don't use BOLD letters unless for headings. That will make your question look tidy. But I need to remove all the unnecessary part and kept the quintessence. You are free to roll-back if you don't like my edit. Thank you .2012-07-09
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    Yes physics and mathematics are different. But this forum is only for mathematics. So it offers an adhoc relaxation for the physics questions , if you can keep the "physics" tag. See the matter present below the tag.2012-07-09
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    No you are right. I trust you ;)2012-07-09

2 Answers 2

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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.2012-07-09
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    @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 ?2012-07-09
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    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)...2012-07-09
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    @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.2012-07-09
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    I like your metaphor ;) Thank you for you help and nice meeting you too!2012-07-09
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(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))...2012-07-09