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When physicists say energy is conserved, do they mean that energy satisfies the continuity equation:

$$\triangledown \cdot j+\dot{\rho}=0$$

On the internet there is plenty of talk of how the continuity equation applies to conservation of charge, fluid dynamics, and so forth, but I can't find any mention of how it applies to the conservation of energy. Why? Is it because it is problematic to talk about energy current density ($j$)?

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The fundamental reason why energy is conserved, is invariance of the physical laws by a time translation $t \to t + t_0$, in a Lagrangian formulation. This is a particular case of the Noether theorem, which states, that if a Lagrangian has a continuous symmetry, there is a corresponding conserved quantity.

Now, if we look in detail, conservation of a charge $Q$ may be expressed as a local conservation law implying a (electric) current $j_\mu$, that is $\partial^\mu j_\mu =0$

This is the relativistic notation, with $\mu =0$ is the time coordinate ($j_0=\rho$), and $\mu =1,2,3$ are the spatial coordinates ($j_1 = -j^x, j_2=-j^y, j_3=-j^z$). Upper and lower indices are related by the metric matrices $\eta^{\mu\nu}$ and $\eta_{\mu\nu}$ which are simply diagonal matrices ($1, -1,-1, -1$). So $\partial^\mu j_\mu =0$ is simply $\frac{\partial \rho}{\partial t} + \vec \nabla .\vec j =0$

Now, for the conservation of momentum/energy $P_\nu$, you see that (comparing to $Q$) you have a supplementary indice $\nu$ (for instance the energy is $P_0$, while the momentum coordinates are $P_1, P_2, P_3$), so the corresponding local "current" has now $2$ indices : $T_{\mu \nu}$, and the local conservation law is simply $\partial^\mu T_{\mu\nu} =0$

The "current" $T_{\mu\nu}$ is more known as the stress-energy tensor.

So, the current for the energy $E=P_0$ is $T_{\mu0}= T_{0\mu}$ (the stress-energy tensor is symmetric in its indices), and the local law conservation for energy may be written : $\partial^\mu T_{\mu0} =0$

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Usually, when physicists talk about energy being conserved, they mean Energy being a Noether charge on the fundamental level, c.f. wikipedia.

As a very general result, one can derive that the time derivative of Energy is zero $$ \frac{d}{dt} E = 0.$$ This result is only true if the Lagrangian description of your system does not explicitly depend on time though. Therefore for a non-isolated system one might find a formulation that allows energy dissipation.

In this picture, the continuity equation follows from the invariance of the Lagrangian of electromagnetics under gauge transformations.

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On the internet there is plenty of talk of how the continuity equation applies to conservation of charge, fluid dynamics, and so forth, but I can't find any mention of how it applies to the conservation of energy. Why? Is it because it is problematic to talk about energy current density (j)?

The continuity equation is fine for energy, and sometimes appears, for example in the theory of incompressible liquid without gravity we can focus just on kinetic energy and then we have $$ \rho = \frac{1}{2}\rho_{liq} v^2, $$ $$ \mathbf j = (\rho + p)\mathbf v, $$ where $\rho_{liq}$ is density and $p$ is pressure of the liquid. In general continuum mechanics and electromagnetic theory other expressions may be used. If you did not find much about this on the Internet, then perhaps it is time to study some good books on hydrodynamics and EM theory:-) The 2nd volume of the Feynman lectures is very good for these topics.

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