I am starting to learn the Fourier's law, but I am getting confused with the following sentence on Wikipedia:

The above differential equation, when integrated for a homogeneous material of 1-D geometry between two endpoints at constant temperature, gives the heat flow rate as:

I am wondering what "1-D geometry" means (isn't all the thing we deal with 3D geometry?) and curious what we could use this formula for (which shape?):

$$\frac{Q}{\Delta t} = -kA\frac{\Delta T}{\Delta x}$$

Also, could I use this formula for determining the heat transfer for a bottle that looks something like this?

enter image description here

If not, what should be the equation I use for the heat transfer for that bottle? I tried to search for the three dimensional form of Fourier's law, but found limited results and no useful information.


2 Answers 2


One dimension here means that the problem is homogeneous in the dimensions perpendicular to the dimension of interest. This means that the equations to be analyzed depend only on one variable, i.e., they are one-dimensional equations.

This is different from one-dimensional as understood, e.g., when applied to a thin rod, or a pipe (or one-dimensional structures on nanoscale, such as quantum wires) - where we would be talking about a system that is physically one-dimensional, in the sense that its length is much bigger than its width (transversal size.)

  • $\begingroup$ Thank you very much for the answer. So here the one dimension just mean change one variable in the equation, is that correct? Sorry as a non native English speaker the language is a little bit complex for me $\endgroup$
    – James
    Jan 23, 2023 at 8:48
  • $\begingroup$ Actually, I am doing an experiment to see how the change in the thickness of the bottle will affect the time it cools down. For this experiment, do you think I could use the above equation as my theoretical result? Thank you! $\endgroup$
    – James
    Jan 23, 2023 at 8:50
  • 1
    $\begingroup$ Sorry, to confirm, the one dimension just mean the heat all one from one direction, is it correct. Sorry and thank you soooo much! $\endgroup$
    – James
    Jan 23, 2023 at 8:59
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    $\begingroup$ @James if you think about it: a small piece of a cylinder is almost exactly like a cube. So you may choose to calculate it like that - as long as the thickness of the cylinder is a lot smaller than its radius. $\endgroup$ Jan 23, 2023 at 20:28
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    $\begingroup$ @James yes, from inside to outside. You're pretending the cylinder is unrolled to make a flat sheet. As long as the radius is much bigger than the thickness, this is a reasonable approximation. In physics, approximations are used all the time to make calculations simple. If the cylinder was very thick to its radius, this wouldn't be a good idea, because a thick cylinder has more area on one side than the other. But a thin cylinder has almost the same area, just like a sheet. $\endgroup$ Jan 23, 2023 at 20:44

what does "1 D geometry" means (isn't all the thing we deal with 3D geometry?)

Yes, our world is 3D. But if you had a very long item, insulated along the sides, the temperature variation throughout the cross-section might be minimal and uninteresting compared to the temperature variation along the length. Thus, you might model the heat flux as $q=-k\frac{dT}{dx}$ for simplicity.

More generally, you’d model the heat flux as $q=-k\nabla T$, the 3D version.

  • $\begingroup$ Thank you for the answer. So the 3D version seems to be the same from the 1D gemoetry, but the 3D version one has not be simpified (is that right)? In this case, should I use 3D version or 1D for the bottle $\endgroup$
    – James
    Jan 23, 2023 at 16:17
  • $\begingroup$ The picture above shows rotational symmetry, so you could operate in cylindrical coordinates and consider only the radial direction. Any introductory heat transfer book (e.g., Incropera & DeWitt) will discuss how to do this. $\endgroup$ Jan 23, 2023 at 16:44
  • $\begingroup$ Sorry I am getting a little bit confused here. What does the cylindrical coordinates related to the equation we are talking about and how could I embed it into Fourier's law? Thank you $\endgroup$
    – James
    Jan 23, 2023 at 16:52
  • $\begingroup$ In the heat equation, the Laplacian is different in cylindrical coordinates. The heat equation is what one solves to obtain a temperature distribution, given the boundary and initial conditions. Ignore this if it's not relevant. $\endgroup$ Jan 23, 2023 at 17:25
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    $\begingroup$ It is the heat equation with a cylindrical Laplacian, as described by the Wikipedia links in this comment chain. $\endgroup$ Jan 23, 2023 at 18:10

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