Deriving Maxwell's equation from Poynting Theorem

[Rob2024]

Homework Statement

Poynting Theorem states $$- \vec J \cdot \vec E - \nabla \cdot {\vec S} = \frac{\partial u}{\partial t}$$ where $\vec J$ is current density, $\vec S$ is the Poynting vector and $u = \frac{1}{2} ( \epsilon_0 E^2 + \frac{B^2}{\mu_0})$ is the electromagnetic energy density at a point. What is $\vec J$ at a point where $\vec B = 0$ according to Poynting Theorem? (Hint: you may need the identity: $\nabla (\vec A \times \vec B) = \vec B \cdot \nabla \times A - \vec A \cdot \nabla \times \vec B$)

Relevant Equations

$$- \vec J \cdot \vec E - \nabla \cdot {\vec S} = \frac{ \partial u}{\partial t}$$
$$\nabla (\vec A \times \vec B) = \vec B \cdot \nabla \times A - \vec A \cdot \nabla \times \vec B$$

Here is the solution:

From vector identity,
$$\nabla (\vec A \times \vec B) = \vec B \cdot \nabla \times A - \vec A \cdot \nabla \times \vec B $$
If $\vec B = 0$, then $$\nabla \vec S = \nabla \frac{\vec E \times \vec B}{\mu_0} = \frac{1}{\mu_0} (\nabla \times E \cdot \vec B - \cdot E \cdot \nabla \times \vec B) = - \frac{1}{\mu_0} \cdot E \cdot \nabla \times\vec B$$
Then the Poynting Theorem becomes,
$$- \vec J \cdot \vec E + \frac{1}{\mu_0} \vec E \cdot \nabla \times \vec B = \frac{\partial u}{\partial t} $$
$$ \vec E \cdot ( - \vec J + \frac{1}{\mu_0} \nabla \times \vec B) = \epsilon_0 \vec E \cdot \frac{\partial \vec E}{\partial t} + \frac{\vec B}{\mu_0} \cdot \frac {\partial \vec B}{\partial t}$$
$$ \vec E \cdot ( - \vec J + \frac{1}{\mu_0} \nabla \times \vec B) = \epsilon_0 \vec E \cdot \frac{\partial \vec E}{\partial t}$$
$$ - \vec J + \frac{1}{\mu_0} \nabla \times \vec B = \epsilon_0 \cdot \frac{\partial \vec E}{\partial t}$$
$$\nabla \times \vec B = \mu_0 \vec J + \epsilon_0 \mu_0 \cdot \frac{\partial \vec E}{\partial t}$$
Therefore,
$$\vec J = \frac{1}{\mu_0} \nabla \times\vec B - \epsilon_0 \cdot \frac{\partial \vec E}{\partial t}$$

When I looked through the solution I realized there is a caveat in the above proof, such that if the vector $(- \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t})$ is perpendicular to $\vec E$, the resulting dot product can still be 0,
$$\vec E \cdot (- \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t}) = 0$$

Is there a way to show, those two vectors cannot be perpendicular in an electromagnetic wave not knowing the Faraday's law (in addition to the condition $\vec B = 0$)? Thanks,

 

[kuruman]

How are $\mathbf E$ and $\mathbf J$ related?

 

[TSny]

When I looked through the solution I realized there is a caveat in the above proof, such that if the vector $(- \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t})$ is perpendicular to $\vec E$, the resulting dot product can still be 0,
$$\vec E \cdot (- \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t}) = 0$$

I'm with you in that I also don't see how they get from $$ \vec E \cdot ( - \vec J + \frac{1}{\mu_0} \nabla \times \vec B) = \epsilon_0 \vec E \cdot \frac{\partial \vec E}{\partial t}$$ to
$$ - \vec J + \frac{1}{\mu_0} \nabla \times \vec B = \epsilon_0 \frac{\partial \vec E}{\partial t}$$

Is there a way to show, those two vectors cannot be perpendicular in an electromagnetic wave not knowing the Faraday's law (in addition to the condition $\vec B = 0$)?

The Maxwell equation $\nabla \times \vec B = \mu_0 \vec J +\mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t}$ is not the Maxwell equation that represents Faraday's law.

Also, the homework statement doesn't mention that we are dealing with an electromagnetic wave.

Is this a problem from a textbook?