# Maxwell stress tensor in electrodynamics

• Petar Mali
In summary, the conversation discusses the use of dyadics and their notation in the derivation of the MST. It is mentioned that \{\vec{E},\vec{D}\} is not equal to \{\vec{D},\vec{E}\} in general, but is equal when epsilon and mu are symmetric. The conversation also touches on the use of different symbols for anticommutators and the importance of remembering the \cdot symbol. Finally, a mathematical proof is given to show that \nabla(\vec{a}\cdot \vec{b})=\frac{1}{2}\nabla(\vec{a}\cdot \vec{b}) when \hat{\epsilon} and \hat{\mu
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

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

Just to remind you

definition

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

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

Any idea?

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

$$\hat{1}$$ - 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

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

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.

Last edited:
Meir Achuz said:
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?

Petar Mali said:
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?
$$\{\vec{E},\vec{D}\}$$

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.

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

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

For anticommutator you can use symbol

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

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

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

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

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

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

## Question 1: What is the Maxwell stress tensor?

The Maxwell stress tensor is a mathematical concept in electrodynamics that describes the distribution of electromagnetic forces within a given region of space. It is a 4x4 matrix that contains information about the electric and magnetic fields and how they interact with each other.

## Question 2: How is the Maxwell stress tensor used in electrodynamics?

The Maxwell stress tensor is used to calculate the force and torque exerted on a charged particle by an electric and magnetic field. It is also used to calculate the stress and strain in a material due to the presence of electromagnetic fields.

## Question 3: What are the components of the Maxwell stress tensor?

The components of the Maxwell stress tensor are the electric field, magnetic field, and their cross-products. The diagonal terms represent the normal stresses caused by the electric and magnetic fields, while the off-diagonal terms represent the shear stresses.

## Question 4: How is the Maxwell stress tensor related to the electromagnetic stress-energy tensor?

The Maxwell stress tensor is a part of the electromagnetic stress-energy tensor, which describes the energy and momentum density of electromagnetic fields. The stress tensor is the spatial components of the stress-energy tensor, while the time component represents the energy density.

## Question 5: What are some applications of the Maxwell stress tensor in real-world situations?

The Maxwell stress tensor has numerous applications in various fields such as engineering, physics, and materials science. It is used to study the behavior of materials under the influence of electromagnetic fields, design electromagnetic devices, and understand the dynamics of charged particles in particle accelerators and plasma physics experiments.

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