Maxwell stress tensor in electrodynamics

[Petar Mali]

$$\hat{N}=\{\vec{E},\vec{D}\}+\{\vec{H},\vec{B}\}-\frac{1}{2}(\vec{D}\cdot\vec{E}+\vec{B}\cdot\vec{H})\hat{1}$$

$$\hat{1}$$ - unit tensor

If I look $$\{\vec{E},\vec{D}\}$$. I know that

$$\{\vec{E},\vec{D}\}=\{\vec{D},\vec{E}\}^*$$

But when I can say that

$$\{\vec{E},\vec{D}\}=\{\vec{D},\vec{E}\}$$?

and when can I say that

$$\{\vec{H},\vec{B}\}=\{\vec{B},\vec{H}\}$$?

Thanks for your answer.

Just to remind you

definition

$$\{\vec{A},\vec{B}\}\cdot \vec{C}=\vec{A}(\vec{B}\cdot \vec{C})$$

$$\vec{C}\cdot \{\vec{A},\vec{B}\}=(\vec{C}\cdot\vec{A})\vec{B}$$

 

[Petar Mali]

Any idea?

 

[Meir Achuz]

$$\hat{N}=\{\vec{E},\vec{D}\}+\{\vec{H},\vec{B}\}-\frac{1}{2}(\vec{D}\cdot\vec{E}+\vec{B}\cdot\vec{H})\hat{1}$$

$$\hat{1}$$ - unit tensor

If I look $$\{\vec{E},\vec{D}\}$$. I know that

$$\{\vec{E},\vec{D}\}=\{\vec{D},\vec{E}\}^*$$

But when I can say that

$$\{\vec{E},\vec{D}\}=\{\vec{D},\vec{E}\}$$?

and when can I say that

$$\{\vec{H},\vec{B}\}=\{\vec{B},\vec{H}\}$$?

Thanks for your answer.

Just to remind you

definition

$$\{\vec{A},\vec{B}\}\cdot \vec{C}=\vec{A}(\vec{B}\cdot \vec{C})$$

$$\vec{C}\cdot \{\vec{A},\vec{B}\}=(\vec{C}\cdot\vec{A})\vec{B}$$

You seem to be using dyadics, which is fine with me, but I am not familiar with your notation
$$\{\vec{E},\vec{D}\}$$.
I just use $$\vec{E}\vec{D}$$

I don't know what your star notation means. Star usually means cc,
but you seem to be using it as matrix transpose.

$$\{\vec{E},\vec{D}\$$ is not equal to
$$\{\vec{D},\vec{E}\}$$ for two general vectors, but the derivation of the MST requires that epsilon and mu be symmetric.
In this case, the equality holds.

 

[Petar Mali]

You seem to be using dyadics, which is fine with me, but I am not familiar with your notation
$$\{\vec{E},\vec{D}\}$$.
I just use $$\vec{E}\vec{D}$$

I don't know what your star notation means. Star usually means cc,
but you seem to be using it as matrix transpose.

$$\{\vec{E},\vec{D}\$$ is not equal to
$$\{\vec{D},\vec{E}\}$$ for two general vectors, but the derivation of the MST requires that epsilon and mu be symmetric.
In this case, the equality holds.

I like more

$$\{\vec{E},\vec{D}\}$$

for example

$$\nabla\cdot \{\vec{E},\vec{D}\}$$

You write this like

$$\nabla\cdot(\vec{E}\vec{D})$$

Is that true?

This is OK but people sometimes forget $$\cdot$$ and that can make confusion!

How can I show that $$\hat{\epsilon}$$ and $$\hat{\mu}$$ must be symmetrical in that case? Are you sure?

 

[Born2bwire]

I like more

$$\{\vec{E},\vec{D}\}$$

for example

$$\nabla\cdot \{\vec{E},\vec{D}\}$$

You write this like

$$\nabla\cdot(\vec{E}\vec{D})$$

Is that true?

This is OK but people sometimes forget $$\cdot$$ and that can make confusion!

How can I show that $$\hat{\epsilon}$$ and $$\hat{\mu}$$ must be symmetrical in that case? Are you sure?

What is this operator?
$$\{\vec{E},\vec{D}\}$$

 

[Meir Achuz]

In the derivation of the MST, one step is to show that grad(D.E) with D held constant is equal to (1/2)grad(D.E). This requires that epsilon_ij be symmetric and constant.

 

[Meir Achuz]

By the way, in physics, {D,E} is used to denote the anticommutator.
That is why I use DE.

 

[Petar Mali]

Well $$\{,\}$$ is also in some books symbol for Poisson bracket, and somewhere people use $$[,]_{PB}$$

For anticommutator you can use symbol

$$[,]_{+}$$.

Now for your answer

It's used that

$$(\vec{D}\cdot \nabla)\vec{E}+\vec{D}\times rot\vec{E}=\nabla(\frac{1}{2}\vec{D}\cdot \vec{E})=\nabla \cdot (\frac{1}{2}(\vec{D}\cdot \vec{E})\hat{1})$$

If $$\hat{\epsilon}$$ and $$\hat{\mu}$$ are symmetric. They can be time functions? Right?

Using that I get

$$\hat{N}=\{\vec{E},\vec{D}\}+\{\vec{H},\vec{B}\}-\frac{1}{2}(\vec{D}\cdot\vec{E}+\vec{B}\cdot\vec{H })\hat{1}$$

 

[Petar Mali]

$$\nabla(\vec{a}\cdot \vec{b})=\nabla(\vec{a}^*\cdot \vec{b})+\nabla(\vec{a}\cdot \vec{b}^*)$$

If $$\vec{b}=\hat{\alpha}\vec{a}$$

$$\hat{\alpha}$$ -symmetric tensor

$$\vec{a}^*\cdot\vec{b}=\vec{a}^*\cdot \hat{\alpha}\vec{a}=\vec{a}\cdot \hat{\alpha} \vec{a}^*$$

$$\vec{a} \cdot \vec{b}^*=\vec{a} \cdot \hat{\alpha}\vec{a}^*$$

So

$$\nabla(\vec{a}\cdot \vec{b})=2\nabla(\vec{a}\cdot \vec{b}^*)$$

$$\vec{a}\times rot\vec{b}^*=\vec{a}\times (\nabla \times \vec{b}^*)=\nabla(\vec{a}\cdot \vec{b}^*)-(\vec{a}\cdot \nabla)\vec{b}^*$$

so

$$(\vec{a}\cdot \nabla)\vec{b}+\vec{a} \times rot\vec{b}=\frac{1}{2}\nabla(\vec{a}\cdot \vec{b})$$

Thanks for your answer!