# Why is $\int_{s} \mathbf{h}\cdot \mathbf{n} da = - \dfrac{dQ}{dt}$ & not $\int_{s} \mathbf{h} \cdot\mathbf{n} da = - \dfrac{dQ}{dt} .{dt}$?

I was reading the Lectures of Feynman about surface integral where a situation in which heat is conserved has been dealt. Let there be $Q$ heat energy present inside a body. Now, if there is net heat flow from the surface, the flux of heat flow should be proportional to the time rate of decrease of $Q$ inside the body ie. $$\int_{s} \mathbf{h}\cdot\mathbf{n} da \propto - \dfrac{dQ}{dt}$$ for a certain time $dt$. But, why should the relation be equalized; & if it is done, there must be a constant. Does that mean that constant is 1? Why? How can the rate of change become equal to the change itself?

• The equation you wrote is just a continuity equation As long as there is no additional heat being generated or lost within the surface, the equation will hold. – MonkeysUncle Mar 28 '15 at 19:49
• Seems to be effectively a duplicate of this question – Kyle Kanos Mar 28 '15 at 19:53
• both sides must be the same "size". In your equation the left is finite and the right an infinitesimal – user66432 Mar 28 '15 at 20:20

This is the general form in which a conservation law can be expressed, when the quantity being conserved can't be converted in other forms, or disappear and reappear anywhere else.

The flux density $\mathbf h$ is defined as that vector field that satisfies the equation you wrote, for every closed surface. More precisely, $\mathbf h \cdot \mathbf n$ represents the amount of $Q$ that flows in a unit time through a unit surface in the direction $\mathbf n$. In particular $\mathbf h$ has units: $$[h]=[Q]T^{-1}L^{-2}.$$

In general, a continuity equation has the form: $$\dfrac{\partial \rho}{\partial t} + \nabla\cdot \mathbf j=\sigma.$$ Here $\rho$ is the density of the quantity, let's call it $Q$, being conserved. $\mathbf j$ is the flux density, $\sigma$ is the density of “sources” and “sinks”, i.e., terms that account for the local generation/destruction (or, if we require locality, simply conversion) of $Q$. Some examples from EM include: $$\text {Conservation of electric charge:}\qquad \dfrac{\partial \rho _\text{e}}{\partial t}+\nabla \cdot {\mathbf j _\text{e}} =0,$$ $$\text {Conservation of energy}: \qquad \dfrac{\partial u}{\partial t}+\nabla \cdot \mathbf S+\mathbf E \cdot \mathbf j_{\text e}=0.$$

The physical meaning of a continuity equation for the quantity $Q$ is that $Q$ is conserved and that the conservation is local: the quantity can not disappear here and reapper there. Feynman was enligthening to me about this point, and I really suggest to read his Chapter 27 (II) when you go on with your study of electromagnetism.

Consider the conservation of energy above: apart from the actual expressions of $u$, $\mathbf S$ and $\mathbf j _\text{e}$, the equation says that the negative change of electromagnetic energy in a given volume equals the work (conversion) done inside the volume (the $\mathbf E \cdot \mathbf j _{\text e}$ term) plus the flow of energy through the surface (the $\nabla \cdot \mathbf S$ term). In principle, energy could disappear from a point inside the volume and reapper somewhere outside, in the form of electromagnetic energy or kinetic energy or what else, and this would not violate conservation of energy. However the conservation would not be local. The continuity equation is the way we express the locality of the conservation: all the energy differences depend on what goes on inside the volume (conversion in kinetic energy) or at the boundary (flow of energy).

• +1. Sir, can you tell what continuity equation actually is? And in the wiki article, it is written it deals with local conservation; why is it so?? Can you help? – user36790 Mar 29 '15 at 2:42
• I have added a brief discussion about continuity equation. If you don't understand what I wrote, I suggest to not waste your time with it. Go on reading Feynman and you will find some important examples of continuity equations - everything will then appear natural. – pppqqq Mar 29 '15 at 8:51