Before getting to question, some background. Let $u(x,t)$ be the temperature in a laterally insulated rod of length $L$, at position $x$ and time $t$. The temperature satisfies the heat equation

$\partial_t u = \alpha \, \partial^2_x u$,

where $\alpha > 0$, thermal diffusivity of the rod, with Dirichlet (zero) boundary conditions say.

There are at least 2 math.stackexchange.com questions that involve the so-called Energy integral

$E(t) = \int_0^L u(x,t)^2 dx$.

Here they are:

Energy for the 1D Heat Equation


Heat equation and energy

Now my question: Why does $E(t)$ represent energy? It does not have the units of energy! It's well-known if you look at any derivation of heat equation or if you know just basic thermodynamics that

$c \rho u(x,t)$

does equal energy per unit length, where $c=$ specific heat, $\rho=$ density (so $c\rho =$ heat capacity of material). $\therefore$ the true energy of the bar equals (assuming $c$ and $\rho$ are constants)

$c \rho \int_0^L u(x,t)\, dx =$ true energy of rod.

Omitting constants, it would be OK to say

$\int_0^L u(x,t)\, dx=$ energy.

But, it is false to call $E(t) = \int_0^L u(x,t)^2 dx$ the energy of the bar.

I know that people introduce $E(t)$ to prove uniqueness of heat equation (I've seen this proof many times) but why do teachers call $E(t)$ the energy when it is not? I do see that $E(t)$ does have resemblance to Kinetic energy (like $(1/2) m v^2$ with a squared term) but in the context of heat, $E(t)$ certainly does not equal energy, so why do teachers call it energy? Are they disingenuous? My final question then is, since $E(t)$ is not the true energy, what physical quantity does it represent?

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    $\begingroup$ Good question. The energy I would associate with the heat equation is the Dirichlet energy, $\int u_x^2.$ $\endgroup$ – user7530 Feb 13 '15 at 20:57
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    $\begingroup$ I think your question belongs more to physics SE $\endgroup$ – Mister Benjamin Dover Feb 14 '15 at 14:40
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    $\begingroup$ They use the term "Energy" simply because (1) it involves integrating something squared, and (2) because you can prove it is non-increasing using similar mathematical techniques that show energy is non-increasing in other physical systems. It should not be thought of in terms of the standard physics notion of "Energy." $\endgroup$ – Stephen Montgomery-Smith Feb 16 '15 at 0:38
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    $\begingroup$ @Curiosity Without knowing the precise context, I don't see why they would have called it this. Also, I am not saying they are right - I am just trying to come up with some plausible argument as to why they may have come up with this terminology. $\endgroup$ – Stephen Montgomery-Smith Feb 16 '15 at 5:19
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    $\begingroup$ One halfway decent reason to call such a quantity an "energy" is that its minimizer should represent a solution to its corresponding PDE. (Not this particular quantity though, I'm also not sure why it's not the Dirichlet energy $\int u_x^2$.) A minimizer corresponds to a stationary point of the functional, so one might think of the statement "solutions to the PDE are minimizers of the energy" as a sort of conservation law. More concretely, looking for minimizers of the "energy" leads you to the Euler-Lagrange equations. $\endgroup$ – Gyu Eun Lee Feb 19 '15 at 23:30

Nearly a year ago, I gave in the comments a non-answer that would justify calling the quantity $$ \int u(x,t)^2~dx = E(t) $$ the energy of the rod. Namely, one can think of minimizing the energy functional as satisfying a conservation law, and thus minimizing the energy leads you to solutions of the associated partial differential equations.

One year out, I think I have a more concrete, dimensional-analysis based answer. Let us examine the equation $$ \partial_t u = -\alpha\partial_{xx}u $$ for its dimensions. $u = u(x,t)$ represents temperature at the spacetime coordinate $(x,t)$, so it has units of temperature $T$. $\alpha$ here is the thermal diffusivity, which has units of length squared over time, $L^2/\tau$. Finally, not in the equation but lurking in the background is the kinetic energy, which is related to temperature by the equation $$ E_k = \frac{3}{2}kT, $$ where $k$ is the Boltzmann constant and has units of energy over temperature, $E/T$.

Let me use $[unit]$ to represent the dimensions of any units in any expression. So for example, the units in the heat equation check out: $\partial_t u$ has units of temperature over time, while $\partial_xx u$ has units of temperature over length squared, which gives us $$ [T\tau^{-1}] = [L^2\tau^{-1}][TL^{-2}]. $$ Now, in our scenario the total length of the rod is fixed, so without loss of generality we may treat it as a constant. Dimensionally, this is saying we will take $[L] = [1]$, so it drops out of any of our equations. This is all fine, as we see in the heat equation the units of length cancel each other anyway.

So let us turn to the energy functional. In terms of units, $$ \int u(x,t)^2~dx = \int [T^2]~d[L] = [T^2L] = [T^2]. $$ So the energy functional has units $[T^2]$, which isn't where we want to be (we want units of energy). However, the primary thing we do with taking the energy functional is minimize it. Minimizing the energy functional $E(t)$ is equivalent to minimizing the square root of the energy functional, $\sqrt{E(t)}$, which has units of $[T]$.

Still not there. But aha! Energy and temperature, $[E]$ and $[T]$, are related by a proportionality law! That is,the equation $$ E_k = \frac{3}{2}kT $$ tells us precisely how to convert from temperature to energy, and the conversion preserves order. So if we can minimize the square root of the "energy functional" $\sqrt{E(t)}$, which has units of $T$, then we automatically know how to minimize the actual energy, which according to our dimensional analysis must be something like $$ \sqrt{\int \frac{9}{4}k^2 u(x,t)^2~dx } = \frac{3k}{2}\sqrt{E(t)}, $$ where the right-hand side is actually integrating the physical notion of energy squared. And now, the units do indeed check out: these are quantities with units of energy. There are surely other ways to define a natural notion of a physical energy functional using an integral, but the units of this definition work and it has nice mathematical properties (pretty much exactly those of $\sqrt{E(t)}$, which we know has an excellent mathematical theory). This revolves around the observation that since $k$ is a constant, minimizing the "energy" functional defined using units of heat is equivalent to minimizing what you would expect to be the "physical energy" functional.

Morally, the fact that $k$ is a constant (despite having units of $[ET^{-1}]$) gives us a temperature-energy equivalence law, which tells us that for dimensional analysis purposes $[E]$ and $[T]$ are indistinguishable. This is what the physicist in me needs to say to satisfy himself. The mathematician in me chooses units for temperature and energy so that $k = 2/3$, and then I just have the simple equivalence $E_k = T$, and now I happily exchange temperature and energy at will. And presumably this is what mathematicians of the past have done.

Treating $L$ like a constant also needs some justification, but I think if we think of $L$ as very small and argue that computing the energy of large homogeneous bodies consists of computing the energy on small pieces and summing, then we can justify that assumption as well.

  • $\begingroup$ But, if we use dimensional analysis, then according to this reasoning, any quantity whatsoever that has dimensions of T would be considered energy. For example, any L^p norm of u(x,t) would be considered energy. I agree that we can think of any L^p norm of u(x,t), or any quantity involving u so that the result has dimensions of T, as a quantity representing "energy", not in the standard sense of thermodynamics, but in a mathematical sense. However, my original question was the interpretation of E(t) as a geniune physical quantity. $\endgroup$ – Curiosity Jan 22 '16 at 12:44
  • $\begingroup$ @Curiosity I agree, I'm not suggesting that the $L^2$ norm of $u(x,t)$ should be regarded as the actual energy. In fact, the energy functional is well defined for any $L^2$ function $u$, not just those that solve the heat equation, so it stands to reason that it won't represent a physical energy in general. continued... $\endgroup$ – Gyu Eun Lee Jan 22 '16 at 23:51
  • $\begingroup$ Rather than claiming that the energy functional genuinely represents the energy of a body, I think we should interpret the terminology "energy functional" as an auxiliary functional that is closely related to the physical energy. Dimensional analysis supports the idea that this relation exists, but of course we shouldn't always expect dimensional analysis to actually give us meaningful physical answers. $\endgroup$ – Gyu Eun Lee Jan 22 '16 at 23:51

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