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?

 

[Rob2024]

That's a typo from me. It's Maxwell's equation. It's not from a textbook. It's from lecture notes. I wonder if the solution is wrong. But I can't be sure. I thought if we could prove the two vectors are perpendicular in the case of EM wave, it's sufficient to show the solution is wrong. i.e. if there is a case where the two vectors are perpendicular, then the solution is incomplete. For the Poynting vector, it usually discussed in the context of electric power transmission in a cable where $J, E, B$ are present. @kuruman no specific relationship between $E$ and $J$ are given.

 

[kuruman]

. @kuruman no specific relationship between $E$ and $J$ are given.

I thought one version of Ohm's law says $\mathbf E=\rho ~\mathbf J$ where $\rho$ is the resistivity. This should hold here unless there is no conductor but then there would be no $\mathbf J$ either.

 

[Rob2024]

@kuruman That's right. I still don't see how that allows us to show the two vectors are perpendicular. In fact it made it less like for those vectors to be perpendicular. It's not clear how $- \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t}$ can be either in the $B$ direction or $S$ direction. Then it would be perpendicular to $E$. Perhaps intuitively they are not perpendicular, but a proof would be nice.

 

[kuruman]

Maybe I misunderstand what you are trying to show. You have the equation $$\vec E \cdot \left( - \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t} \right)=0.$$ A dot product is zero when one of three possibilities is the case: (a) the first vector is zero or (b) the second vector is zero or (c) the two vectors are non-zero but perpendicular to each other.

Here
(a) is not the case because if $\mathbf E = 0$, there is no Poynting vector.
(c) is not the case because Ohm's law $\mathbf E=\rho ~\mathbf J$ ensures that the second vector has a component in the direction of the first vector so the two cannot be perpendicular to each other.
Therefore (b) is the case and $$\left( - \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t} \right)=0.$$Q.E.D.

 

[TSny]

Maybe I misunderstand what you are trying to show. You have the equation $$\vec E \cdot \left( - \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t} \right)=0.$$ A dot product is zero when one of three possibilities is the case: (a) the first vector is zero or (b) the second vector is zero or (c) the two vectors are non-zero but perpendicular to each other.

Here
(a) is not the case because if $\mathbf E = 0$, there is no Poynting vector.

The Poynting theorem holds even for a point in space and time for which $\vec S = 0$. So I don't see why we have to exclude $\vec E = 0$.

(c) is not the case because Ohm's law $\mathbf E=\rho ~\mathbf J$ ensures that the second vector has a component in the direction of the first vector so the two cannot be perpendicular to each other.

We could have a situation where Ohm's law doesn't apply. Consider the somewhat artificial example where we have an infinitely long plastic rod with a uniform charge density $\rho_e$. The rod slides along its length at a constant velocity $\mathbf v$. The current density inside the rod is then $\mathbf J = \rho_e \mathbf v$. But the electric field inside and outside the rod is radially outward and time-independent. At points within the rod, $\mathbf J$ and $\mathbf E$ are nonzero and perpendicular.

Of course, the Maxwell equation $- \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t} = 0$ holds in this example because $\dfrac{\partial \vec E}{\partial t} = 0$ and the current density produces a magnetic field satisfying Ampere's law $\nabla \times \vec B = \mu_0 \vec J$.

I still don't see how the author of the solution deduces $- \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t} = 0$ from $\vec E \cdot \left( - \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t} \right) = 0$ .

 

[kuruman]

We could have a situation where Ohm's law doesn't apply.

Perhaps, but not here. The term $\mathbf J\cdot \mathbf E$ in the starting equation for Poynting's theorem is the density of electric power dissipated by the Lorentz force acting on charge carriers. I don't think it can be omitted from the starting power density equation without loss of generality.

This so called derivation of Faraday's law from Poynting's theorem bothers me because it is begging the question. Faraday's law was obtained from and anchored on experimental observation. Poynting's theorem is derived on the assumption that Faraday's law is valid (see here.) This question is putting the cart before the horse.

 

[Rob2024]

The Poynting theorem holds even for a point in space and time for which $\vec S = 0$. So I don't see why we have to exclude $\vec E = 0$.


We could have a situation where Ohm's law doesn't apply. Consider the somewhat artificial example where we have an infinitely long plastic rod with a uniform charge density $\rho_e$. The rod slides along its length at a constant velocity $\mathbf v$. The current density inside the rod is then $\mathbf J = \rho_e \mathbf v$. But the electric field inside and outside the rod is radially outward and time-independent. At points within the rod, $\mathbf J$ and $\mathbf E$ are nonzero and perpendicular.

Of course, the Maxwell equation $- \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t} = 0$ holds in this example because $\dfrac{\partial \vec E}{\partial t} = 0$ and the current density produces a magnetic field satisfying Ampere's law $\nabla \times \vec B = \mu_0 \vec J$.

I still don't see how the author of the solution deduces $- \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t} = 0$ from $\vec E \cdot \left( - \vec J + \frac{1}{\mu_0} \nabla \times \vec B - \epsilon_0 \frac{\partial \vec E}{\partial t} \right) = 0$ .

This seems like a good counter example where the $E$ field is perpendicular to both $J$ and $\nabla \times B$ terms. I also agree with kuruman that the proof is backward.

 

[TSny]

Perhaps, but not here. The term $\mathbf J\cdot \mathbf E$ in the starting equation for Poynting's theorem is the density of electric power dissipated by the Lorentz force acting on charge carriers. I don't think it can be omitted from the starting power density equation without loss of generality.

I agree that the term $\mathbf J\cdot \mathbf E$ is essential in the general statement of the Poynting theorem. But, this term can be zero in some cases. It is not necessary for $\mathbf J$ and $\mathbf E$ to be proportional; that is, Ohm's law need not hold.

This so called derivation of Faraday's law from Poynting's theorem bothers me because it is begging the question. Faraday's law was obtained from and anchored on experimental observation. Poynting's theorem is derived on the assumption that Faraday's law is valid (see here.) This question is putting the cart before the horse.

Yes, I agree.

 

[PhDeezNutz]

I can't fathom the logic that

$\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \Rightarrow \vec{a} = \vec{c}$

One can easily come up with counter examples.

 

[LightPhoton]

I can't fathom the logic that

$\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \Rightarrow \vec{a} = \vec{c}$

One can easily come up with counter examples.

$\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \Rightarrow \vec a\cdot(\vec{b} - \vec{c})=0$ From where What Kuruman says applies