Let the electromagnetic field has $u$ as its energy density (amount of energy per unit volume in the field) and let $\bf S$ represents the energy flux- the amount of energy per unit time flowing across a unit area perpendicular to the flow).
Now electromagnetic field can interact with matter and do work on them; so this energy interaction must be considered when discussing energy conservation.
The field energy inside a volume $V$ is $\displaystyle \int_V u\,\mathrm dV\;.$ Amount of energy flowing out of the volume $V$ is given by $\displaystyle \int_\Sigma \mathbf S\cdot n\,\mathrm da \;.$
Now, the work done per unit time by the field on the matter inside the volume $V$ is given by $\displaystyle \int _V Nq(\mathbf E + \mathbf v\times \mathbf B)\cdot \mathbf v\,\mathrm dV$ where $N$ is the number of particles per unit volume; this can be written as \begin{align}\int _V Nq(\mathbf E + v\times \mathbf B)\,\mathrm dV& = \int_V Nq\mathbf E\cdot v\,\mathrm dV\\ &= \int_V \mathbf E\cdot \underbrace{(Nq\mathbf v)}_\textrm{current-density}\,\mathrm dV\\ &= \int_V E\cdot \mathbf J\,\mathrm dV \end{align}
So, the continuity equation is written thus: $$\underbrace{-\frac{\partial}{\partial t}\int_V u\,\mathrm dV}_\textrm{rate of change of energy inside volume $V$}=\underbrace{\int_\Sigma \mathbf S\cdot \mathbf n\,\mathrm da}_\textrm{amount of $\mathbf{field\, energy}$ flowing out of volume $V$ per unit time} + \underbrace{\int_V \mathbf E\cdot \mathbf J\,\mathrm dV}_\textrm{work done per unit time by the field on the matter inside volume $V$} \;. $$ This definitely implies the conservation of energy and moreover, the equation which OP wrote in his query is the offshoot of this same continuity equation.
Energy density and Poynting Vector:
We can write the differential form of the continuity equation above as:
$$-\frac{\partial u}{\partial t}= \textrm{div}\,\mathbf S + \mathbf E\cdot \mathbf J$$ (since, all the derivation is done in Cartesian coordinates, then $\textrm{div}\equiv \mathbf \nabla$).
Or, we can write $$\mathbf E\cdot \mathbf J= -\frac{\partial u}{\partial t}-\mathbf E\cdot \mathbf J\tag 1$$
In order to find out what $u$ and $\bf S$ actually are, it is assumed that they solely depend on the fields $\bf E$ and $\bf B\;.$
Now, using $$\mathbf J = \epsilon_oc^2\,\mathbf \nabla\times\mathbf B- \epsilon_0\frac{\partial\mathbf E}{\partial t}\,,$$ we can write $\mathbf E\cdot \mathbf J$ as $$\mathbf E\cdot \mathbf J= \epsilon_oc^2\,\mathbf E\cdot (\mathbf \nabla\times\mathbf B)- \epsilon_0\mathbf E\cdot\frac{\partial\mathbf E}{\partial t}\;.$$
Now, \begin{align}\mathbf E\cdot (\mathbf \nabla\times\mathbf B)&= (\mathbf \nabla\times\mathbf B)\cdot \mathbf E\\ &= \mathbf \nabla\cdot (\mathbf B\times\mathbf E)+ \mathbf B\cdot (\mathbf \nabla \times \mathbf E)\;.\end{align}
Therefore, \begin{align}\mathbf E\cdot \mathbf J&=\epsilon_oc^2\,\mathbf \nabla\cdot (\mathbf B\times\mathbf E)+\epsilon_oc^2\,\mathbf B\cdot (\mathbf \nabla \times \mathbf E)-\epsilon_0\mathbf E\cdot\frac{\partial\mathbf E}{\partial t}\\ &=\epsilon_oc^2\,\mathbf \nabla\cdot (\mathbf B\times\mathbf E)+\epsilon_oc^2\,\mathbf B\cdot (\mathbf \nabla \times \mathbf E)-\frac{\partial}{\partial t}\,\left(\frac12 \epsilon_0\,\mathbf E\cdot \mathbf E\right)\\ &=\mathbf \nabla\cdot (\epsilon_oc^2\,\mathbf B\times\mathbf E)+ \epsilon_oc^2 \mathbf B\cdot \left(-\frac{\partial\mathbf B}{\partial t}\right)-\frac{\partial}{\partial t}\,\left(\frac12 \epsilon_0\,\mathbf E\cdot \mathbf E\right)\\ &=\mathbf \nabla\cdot (\epsilon_oc^2\,\mathbf B\times\mathbf E) - \frac{\partial}{\partial t}\left(\frac{\epsilon_oc^2}{2}\,\mathbf B\cdot \mathbf B+ \frac{\epsilon_0}2 \,\mathbf E\cdot \mathbf E\right) \tag 2\end{align}
Comparing $(1)$ and $(2)\,,$ we get $$u = \frac{\epsilon_0}2 \,\mathbf E\cdot \mathbf E+\frac{\epsilon_oc^2}{2}\,\mathbf B\cdot \mathbf B\\ \mathbf S=\epsilon_oc^2\,\mathbf E\times\mathbf B \;.$$
This vector $\bf S\,,$ that is the energy flux vector we considered at the beginning of the making up of the continuity equation, is called the Poynting Vector.
The whole derivation is based on the continuity equation, which is the mathematical expression of conservation of energy.
OP can conclude the equation he wrote in the question by converting the differential form of equation $(2)$ into integral form; using the definition of $u$ and $\bf S$ derived above and lastly using Gauss's theorem.
References:
$\bullet$ Lectures on Physics by Feynman, Leighton, Sands.
you're saying that energy conservation is needed as an independent equation, in order to deduce that the Poynting vector gives energy flux. I don't think this is true. I think the energy continuity equation can be shown as a consequence of Maxwell's equations.
No! Never, ever I've made such statement. I've answered to your query I know the books say it represents energy flux, but how do you prove it represents energy flux?; that's it. What I've done is to clearly present the continuity equation before OP and have tried to clearly interpret each term and how they together imply conservation of energy. The continuity equation can indeed be derived from Maxwell's equation and this is what I've done to define $u$ and $\bf S\;.$ I'm really pretty upset on the allegation OP made.
But this is not a proof of conservation of energy.
It's indeed a proof of conservation of energy.
Contrast this with conservation of charge within electromagnetism. Conservation of charge is a consequence of Maxwell's equations, it does not have to be assumed independently.
Okay, let's fix this thing. Conservation of charge implies this continuity equation(this can be derivable from Maxwell's equations):
$$-\frac{\partial \rho_V}{\partial t}= \mathbf \nabla \mathbf J\;.$$
This suggests conservation of energy would look like:
$$-\frac{\partial u}{\partial t}= \mathbf \nabla \mathbf S\;.$$
But that is incomplete as it is the total energy and not just field energy which is conserved; the interaction of field with matter has to be taken into consideration.
is energy conservation a consequence of Maxwell's equations or something assumed independently from it?
Yes. The continuity equation meant for conservation of energy is derived from Maxwell's equation; that's how the definition of $u$ and $\bf S$ have been computed. OP must be confused to see in my answer first the continuity equation as if I'm saying continuity equation should be considered independently in order to consider conservation of energy. No! That was intentionally done in order to conceive that the Poynting Theorem indeed implies conservation of energy.
Best Answer
Poyntings theorem can be derived with 0 assumptions of energy conservation.It can be entirely derived from maxwell equations and work = $f.dr$
Consider I have a charge distribution.
Then the force element on a volume dv is given by
$df= (\vec{E}+\vec{v}×\vec{B})\rho dv$
$df= \vec{E}\rho dv+(\vec{v}×\vec{B})\rho dv$
an infinitesimal work element, is then given b
$df \cdot dr = \vec{E} \cdot \vec{dr} \rho dv+(\vec{v}×\vec{B})\cdot \vec{dr} \rho dv$
$dw = \vec{E} \cdot \vec{dr} \rho dv+(\vec{v}×\vec{B})\cdot \vec{dr} \rho dv$
dr isn't a line element in the form r(t), it is instead a position vector field line element representing the paths that each charge $\rho dv$ takes,
this is equivelent to $ dr = \vec{V} dt$ thus,
$dw = \rho\vec{E} \cdot \vec{V}dt dv+\rho(\vec{v}×\vec{B})\cdot \vec{V}dt dv$
the second term is zero since a vector by definition perpendicular to V, dotted with V is zero ( aka, magnetic fields do no work)
So the qauntity
$dw = \rho\vec{E} \cdot \vec{V}dt dv$
Represents the infinitesimal amount of work done by a single DQ of charge moving a tiny infinitesimal distance dr, ( or the distance moved by that charge with a velocity "V" in time "dt")
Meaning the rate at which work is done on a charge DQ(divided by dt) must be given by
$\frac{dw}{dt} = \rho\vec{E} \cdot \vec{V} dv$
Which, substituting in the definition of J is equivelant to
$\frac{dw}{dt} = \vec{E} \cdot \vec{J} dv$
This is the rate at which work is being done on an infinitesimal charge DQ,
lastly, to find the TOTAL rate at which work is being on charges in a volume, you simply integrate this about some volume
$\frac{dW}{dt} = \iiint \vec{E} \cdot \vec{J} dv$
I have chosen to work with rate of work on a single dq and then integrate last, you could have done the same integrating first
From here you can then derive poyntings theorem by eliminating J in favour of the fields using amperes law
To answer those last questions, E dot J intiluatively is the component of the velocity of my charge in the direction of the field, which is indicative of the work expression. There is also no relationship between J and B since magnetic fields do no work