# Maxwell Stress Tensor in the absence of a magnetic field

• pafcu
In summary, the conversation discusses the derivation of the stress tensor in the case of a static electric field without a magnetic field. The stress tensor is derived using Maxwell's laws and assuming a flat surface with a perpendicular electric field. The final result is given as Eq. (5) and some textbooks argue that terms from the initial Lorentz force term can be removed without affecting the final result. However, the process of removing these terms is not clear and the conversation suggests using dyadics to clarify the derivation.
pafcu
I'm having some trouble calculating the stress tensor in the case of a static electric field without a magnetic field. Following the derivation on Wikipedia,

$$\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})$$

2. Get force density
$$\mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B}$$

3. Substitute using Maxwell's laws
$$\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}$$

4. Replace some curls and combine
$$\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} + (\mathbf{B}\cdot\boldsymbol{\nabla}) \mathbf{B} \right] - \frac{1}{2} \boldsymbol{\nabla}\left(\epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)$$

5. Get the tensor
$$\sigma_{i j} = \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)$$

6. Assuming B=0:
$$\sigma_{i j} = \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right)$$

7. Assume flat surface with perpendicular field (z-direction)
$$\sigma_{z z} = \epsilon_0 \left(E^2 - \frac{1}{2} E^2\right)=\frac{\epsilon_0}{2} E^2$$

This is the formula given in e.g. The Feynman Lectures in Physics Vol. 2 (Page 31-14), and some other textbooks.

However, this derivation seems to assume a magnetic field until the final steps. Since most terms in eq. 4 result from the initial v x B term (even those that depend only on E, $(\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E}$ and $\frac{1}{2} \boldsymbol{\nabla}\epsilon_0 E^2$ ), these should not be present in my case, and in fact eq 4 should be as simple as
$$\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E}\right]$$

Tensor calculus is not my strong point. To me it is not clear how to get from eq 4 to eq 5, and how modifying eq 4 alters the resulting stress tensor. Will it really still be the same as eq 6? To me it seems strange that removing terms would not affect the result, yet this seems to be what many textbooks claim. Or is there some reason why the initial v x B term can not be removed, even when there is no magnetic field?

The first two terms of Eq. (4) should be
$$\mathbf{f} = \epsilon_0\left[ \boldsymbol{\nabla}\cdot( \mathbf{E}\mathbf{E}) + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right]$$.
Then use $$(\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} =(1/2)\nabla({\bf E\cdot E })$$.
Writing these two terms with indices gives Eq. (5).
The other terms don't enter with no B field. dE/dt can't enter because this would imply a B field.

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I don't understand how you get
$$\mathbf{f} = \epsilon_0\left[ \boldsymbol{\nabla}\cdot( \mathbf{E}\mathbf{E}) + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right]$$

Looking at eq. 3
$$\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}$$

and assuming B = 0 gives me
$$\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E}\right]$$

Can you clarify?

pafcu said:
I don't understand how you get
$$\mathbf{f} = \epsilon_0\left[ \boldsymbol{\nabla}\cdot( \mathbf{E}\mathbf{E}) + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right]$$

Looking at eq. 3
$$\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}$$

and assuming B = 0 gives me
$$\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E}\right]$$
Can you clarify?
$$(\boldsymbol{\nabla}\cdot \mathbf{E} ){\bf E}= \nabla\cdot({\bf EE})-{(\bf E\cdot\nabla)E}$$.
If you put in indices for that, you get what you want.
The + sign was wrong in your rirst post.

Last edited:

I understand your confusion and concerns regarding the calculation of the Maxwell Stress Tensor in the absence of a magnetic field. Let me try to provide some clarification on the derivation and its assumptions.

Firstly, the derivation you have followed from Wikipedia assumes a general case where both electric and magnetic fields are present. This is why terms involving both fields are present in equation 4. However, as you correctly pointed out, in the absence of a magnetic field, these terms should not be present. This means that equation 4 should simplify to equation 7:

\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}

Now, to address your question about how this affects the resulting stress tensor, let's look at the definition of the stress tensor:

\sigma_{ij} = \rho\frac{d v_i}{d t} + f_i

In the absence of any forces acting on the system, the stress tensor reduces to:

\sigma_{ij} = \rho\frac{d v_i}{d t}

This means that the stress tensor is solely dependent on the acceleration of the particles in the system. In the case of a static electric field, the particles are not accelerating, so the stress tensor will be zero.

In other words, in the absence of a magnetic field, the stress tensor will only have non-zero components when there is a time-varying electric field, as seen in equation 7. This is consistent with the formula given in Feynman's lectures.

To summarize, the initial v x B term can be removed in the absence of a magnetic field, and it will affect the resulting stress tensor by making it zero, unless there is a time-varying electric field present. I hope this explanation helps to clarify the derivation and its assumptions.

## 1. What is the Maxwell Stress Tensor?

The Maxwell Stress Tensor is a mathematical expression that describes the stress and pressure exerted by an electromagnetic field on a material or object. It is derived from Maxwell's equations, which govern the behavior of electric and magnetic fields.

## 2. How is the Maxwell Stress Tensor calculated?

The Maxwell Stress Tensor is calculated by taking the cross product of the electric and magnetic field vectors at a specific point in space, and then multiplying by a constant known as the permittivity of free space. This calculation results in a matrix with six components, representing the stress and pressure in different directions.

## 3. What does it mean when the Maxwell Stress Tensor is in the absence of a magnetic field?

When the Maxwell Stress Tensor is in the absence of a magnetic field, it means that there is no magnetic field present in the system or that its effects are negligible. In this case, the stress and pressure are solely determined by the electric field.

## 4. How is the Maxwell Stress Tensor used in science?

The Maxwell Stress Tensor is used in various fields of science, such as electromagnetism, engineering, and materials science. It is often used to calculate the forces and stresses on objects caused by electromagnetic fields, and it plays a crucial role in understanding the behavior of materials under the influence of such fields.

## 5. What are some real-world applications of the Maxwell Stress Tensor?

The Maxwell Stress Tensor has numerous practical applications, including in the design and analysis of electrical and electronic devices, such as motors, generators, and transformers. It is also used in the development of materials for use in electromagnetic applications, such as antennas and shielding materials.

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