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

<br /> \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}<br />

 

[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!