A Conflicting results about Stress tensor symmetry on EM field

[Klaus3]

Consider the balance of momentum and angular momentum on a continuum body

$$ \int \frac{dp}{dt}dV = \int TndA + \int bdV $$
$$ div(T) + b = \frac{dp}{dt}$$
$$ \int r \times \frac{dp}{dt}dV = \int r \times TndA + \int r \times bdV $$

$p$ is the volumetric momentum density
$T$ is the stress tensor
$b$ is the body forces
$r$ is the position vector

Now, assume that the body forces is due to EM fields taking the general form as:

$$ b = div(T_{mx}) - \frac{dp_{EM}}{dt} $$

$T_{mx}$ is the maxwell stress tensor
$p_{EM}$ is the electromagnetic momentum density

The local balance of momentum becomes:
$$div(T_{mx} + T) - \frac{dp_{EM}}{dt }= \frac{dp}{dt}$$

Using this, the balance of angular momentum becomes:

$$ \int r \times \frac{dp}{dt}dV = \int r \times TndA + \int r \times (div(T_{mx}) - \frac{dp_{EM}}{dt})dV $$

By the divergence theorem :

$$ \int r \times \frac{dp}{dt}dV = \int e_{ijk} TdV + \int r \times (div(T_{mx} + T) - \frac{d(p_{EM} +p)}{dt})dV $$

$$\int e_{ijk} TdV =0$$
$$ e_{ijk} T =0$$

Thus the stress tensor is symmetric.

However, if i go back to the global balance of momentum and use the divergence theorem on divT_{mx} the resulting balance is:


$$ \int \frac{dp}{dt}dV = \int (T+T_{mx})ndA - \int \frac{dp_{EM}}{dt})dV $$

This gives the same local balance of momentum, however the global balance of angular momentum becomes:

$$ \int r \times \frac{dp}{dt}dV = \int r \times (T+T_{mx})ndA - \int r \times \frac{dp_{EM}}{dt})dV $$



By the divergence theorem :



$$ \int r \times \frac{dp}{dt}dV = \int e_{ijk} (T+T_{mx})dV + \int r \times (div(T_{mx} + T) - \frac{d(p_{EM} +p)}{dt})dV $$



$$\int e_{ijk} (T+T_{mx})dV =0$$

$$ e_{ijk} T = -e_{ijk} T_{mx}$$

Thus the stress tensor is not necessarily symmetric and its antisymmetric part is equal to the negative of the antisymmetric part of the maxwell stress tensor. For example, if the body in question has polarization moments, the maxwell stress tensor is not symmetric, this would cause the stress tensor to not be symmetric aswell.

Since one formulation says that the stress tensor is symmetric no matter what, but the other doesnt, what is the problem here? They cant both be correct. Which one it is correct and why?

 

[anuttarasammyak]

Explicit expression of Maxwell stress tensor

shows obvious symmetry. I do not understand your situation introducing other stress tensor and angular momentum for saying it does not have symmetry.

 

[Klaus3]

Explicit expression of Maxwell stress tensor
View attachment 363866
shows obvious symmetry. I do not understand your situation introducing other stress tensor and angular momentum for saying it does not have symmetry.

This expression for the maxwell stress tensor only works for linear dielectrics or monopolar bodies. It is not necessarily symmetric for polarizable materials/ferromagnetic and so on, you can see alternate versions of it in

Truesdell and Toupin "The classical field theories" . Sect 284, Page 690. Eq 284.11-13

In that case the explicit expression is

$$ \sigma_{ij} = D_iE_j + B_iH_j - \frac{1}{2} \delta_{ij} \left(\textbf{D \cdot E + B \cdot H}\right) $$

Which is not symmetric.

You also have other non-symmetric definitions in
Toupin : The elastic dielectric, link :https://hal.science/file/index/docid/930219/filename/The_Elasttic_Dielectric_Toupin_2.pdf

Equation 9.1

Or here
The energy-momentum tensor of electromagnetic fields in matter
Rodrigo Medina and J Stephany
https://arxiv.org/pdf/1703.02109

Equation 135

There are also other forms that represent the electromagnetic force as:

$$ div(T) + \frac{dp}{dt} $$

But $T$ is not called explicitly "The maxwell stress tensor" but rather, for example, the "Einstein-Laub" stress tensor. But mathematically they function the same way in my question.

 

[anuttarasammyak]

According to post #3 Maxwell stress tensor in media, especially anisotropic media, seems to be your issue.

Let me say about a tiny findings. My old textbook Panofsky-Phylips second edition https://www.amazon.co.jp/-/en/Wolfgang-K-H-Panofsky/dp/0486439240 eq.(10-37) states that

The suffix order is different from your quote so its antisymmetric part has a different sign. So there seems to exist two ways of the definition at least.

Let us think only electro not magnetic part for simplicity. The medium body has anisotropic dielectric coefficient s $\epsilon_1 > \epsilon_2 > \epsilon_3$.
$$T_{12}=\epsilon_2 E_1E_2=[\frac{\epsilon_1+\epsilon_2}{2} - \frac{\epsilon_1-\epsilon_2}{2}]E_1E_2$$
$$T_{21}=\epsilon_1 E_1E_2=[\frac{\epsilon_1+\epsilon_2}{2} + \frac{\epsilon_1-\epsilon_2}{2}]E_1E_2$$
etc. They are not the same and have symmetric and antisymmetric parts. I imagine antisymmetric part generates torque to media body so that the largest # \epsilon_1## has same direction with applied E so that free energy of the system be minimum. I would be glad if corrected.

The paper in arxiv you mentioned that your problem may concern the long history controverse. It may have something of your interest.

 

[Klaus3]

According to post #3 Maxwell stress tensor in media, especially anisotropic media, seems to be your issue.

Let me say about a tiny findings. My old textbook Panofsky-Phylips second edition https://www.amazon.co.jp/-/en/Wolfgang-K-H-Panofsky/dp/0486439240 eq.(10-37) states that
View attachment 363930
The suffix order is different from your quote so its antisymmetric part has a different sign. So there seems to exist two ways of the definition at least.

Let us think only electro not magnetic part for simplicity. The medium body has anisotropic dielectric coefficient s $\epsilon_1 > \epsilon_2 > \epsilon_3$.
$$T_{12}=\epsilon_2 E_1E_2=[\frac{\epsilon_1+\epsilon_2}{2} - \frac{\epsilon_1-\epsilon_2}{2}]E_1E_2$$
$$T_{21}=\epsilon_1 E_1E_2=[\frac{\epsilon_1+\epsilon_2}{2} + \frac{\epsilon_1-\epsilon_2}{2}]E_1E_2$$
etc. They are not the same and have symmetric and antisymmetric parts. I imagine antisymmetric part generates torque to media body so that the largest # \epsilon_1## has same direction with applied E so that free energy of the system be minimum. I would be glad if corrected.

The paper in arxiv you mentioned that your problem may concern the long history controverse. It may have something of your interest.

I am aware of the controversy, there is no (at least to my knowing) a general specific form of the maxwell stress tensor without disagreements. However, my issue isn't with the exact for of it, rather, with how it mathematically functions in the overall balances.

Let's forget we are talking about The electromagnetic force here, instead, just imagine a body force in the form of:

$$ b = div(T_1) + k $$

In the momentum balance, there are two ways of expressing this, either direct substitution or using the divergence theorem:

$$ \int \frac{dp}{dt}dV = \int T n dA + \int (div(T_1) + k) dV $$

$$ \int \frac{dp}{dt}dV = \int (T+T_1) n dA + \int k dV $$

These two equations are equivalent, and give the same local balance of momentum:

$$ \frac{dp}{dt}dV = div(T+T_1) + k $$

So far, no problems, however, when you do the angular momentum balance, the two equations possible are:

$$ \int r \times \frac{dp}{dt}dV = \int r \times T n dA + \int r \times (div(T_1) + k) dV $$

$$ \int r \times \frac{dp}{dt}dV = \int r \times (T+T_1) n dA + \int r \times k dV $$

And here's the problem: These two equations are NOT equivalent, the first equation gives a local balance in the form

$$ e_{ijk} T = 0 $$

Meaning T is symmetric

The second one gives:

$$ e_{ijk} T = -e_{ijk} T_1 $$

Meaning T is NOT necessarily symmetric, and may be asymmetric if $T_1$ is asymmetric

So, the problem is: Two equivalent forms of the momentum balance gives two distinct forms of the momentum balance. And i wonder which one is the correct one.

 

[anuttarasammyak]

And here's the problem: These two equations are NOT equivalent, the first equation gives a local balance in the form

The two equations are equation of motion and equation of angular momentum. They evolve with time. What do you mean by "local balance" ? What A is balancing with what B ?

 

[Klaus3]

The two equations are equation of motion and equation of angular momentum. They evolve with time. What do you mean by "local balance" ? What A is balancing with what B ?

Local balance is simply the statement that results when you drop the integral signs, by saying that the integral equation is valid for arbitrary volumes in the body, which leaves an equation relating the forces/torques densities.

When you drop the integral signs in the angular momentum equation, and apply the local form of the momentum balance, what is left are the two equations about the symmetry or asymmetry of the stress tensor.

Its simply a nomenclature, continuum mechanics books calls dynamical equations as "balance" equations.

 

[anuttarasammyak]

I see that "local balance" means taking a look at dyamical equations not in coordinate integral forms. Thanks.

For simplicity let body force $k=0$ and writing $t_{ik}=T_{ik}+T_{1\ ik}$ ,whole stress tensor, i-component of torque generated by $t_{ik}$ is
$$\epsilon_{ijk}\int_V x_j\frac{\partial t_{kl}}{\partial x_l}dV=\epsilon_{ijk}\int_A x_j\ t_{kl}\ n_l \ dA - \epsilon_{ijk}\int_V t_{kj} dV$$
stress tensor applies through surface so the second term must vanish, i.e. stress tensor is symmetric.I am not sure how you divide the whole stress tensor to the two, anyway, antisymmetric part of $T$ + antisymmetric part of $T_1$=0. Is this investigation agreeable to yours ?　

Therefore now I become skeptical to my post #4 about antisymmetric part of Maxwell stress tensor in anisotoropic media.

 

[Klaus3]

I see that "local balance" means taking a look at dyamical equations not in coordinate integral forms. Thanks.

For simplicity let body force $k=0$ and writing $t_{ik}=T_{ik}+T_{1\ ik}$ ,whole stress tensor, i-component of torque generated by $t_{ik}$ is
$$\epsilon_{ijk}\int_V x_j\frac{\partial t_{kl}}{\partial x_l}dV=\epsilon_{ijk}\int_A x_j\ t_{kl}\ n_l \ dA - \epsilon_{ijk}\int_V t_{kj} dV$$
stress tensor applies through surface so the second term must vanish, i.e. stress tensor is symmetric.I am not sure how you divide the whole stress tensor to the two, anyway, antisymmetric part of $T$ + antisymmetric part of $T_1$=0. Is this investigation agreeable to yours ?　

Therefore now I become skeptical to my post #4 about antisymmetric part of Maxwell stress tensor in anisotoropic media.

I am ok with the Total stress tensor being symmetric, however, the sum $T+T_1$ being symmetric doesn't imply that $T$ and $T_1$ are each symmetric.

However, if you keep the tensors separated, one of the derivations implies that $T$ is symmetric no matter what.

assuming $k = 0$

$$ \int TndA + \int div(T_1)dV = \int \frac{dp}{dt}dV$$

now, you don't do the divergence theorem here, you keep the equation as is, and formulate the angular momentum balance

$$ \int r \times TndA + \int r \times div(T_1)dV = \int r \times \frac{dp}{dt}dV$$

Now you bring in the divergence theorem, and put the angular momentum density to the LHS

$$ \int e_{ijk}TdV + \int r \times (div(T+T_1) - \frac{dp}{dt})dV = 0$$

now back to the linear momentum equation, dropping the integrals and using the divergence theorem:

$$div(T+T_1) - \frac{dp}{dt} = 0 $$

With this, make a substitution on the angular momentum equation's second term, it vanishes.

$$\int e_{ijk}TdV = 0 $$

Thus, $T$ is symmetric, regardless of $T_1$

However, things go differently if you formulate the linear momentum balance, in integral form, with the divergence theorem:

$$ \int (T+T_1)ndA = \int \frac{dp}{dt}dV$$

This implies an angular momentum with this form:

$$ \int r \times (T+T_1)ndA = \int r \times \frac{dp}{dt}dV$$

Then with the divergence theorem:

$$ \int e_{ijk}(T+T_1)dV + \int r \times (div(T+T_1) - \frac{dp}{dt})dV = 0$$

$$ e_{ijk}(T+T_1) = 0$$
The total stress $T+T_1$ is symmetric, however

$$e_{ijk}T = - e_{ijk}T_1$$

They are not individually symmetric, necessarily. One might be asymmetric as long as the other one is also asymmetric and opposite in sign.

 

[anuttarasammyak]

Let's forget we are talking about The electromagnetic force here, instead, just imagine a body force in the form of:

b=div(T1)+k

I am not sure how you divide the whole stress tensor to the two, anyway, antisymmetric part of T + antisymmetric part of T1=0. Is this investigation agreeable to yours ?

I am ok with the Total stress tensor being symmetric, however, the sum T+T1 being symmetric doesn't imply that T and T1 are each symmetric.

Can a body force have stress tensor part ? There seems to be contradiction in words. Maybe a contradiction causes another contradiction. Would you show some examples of b, T_1, k ?

 

[Klaus3]

I mean, electromagnetic force is the example. You can always use the maxwell equations to transform it into a divergence.

if you go onto the truesdell reference I said above. You will see that he makes this transformation and uses the maxwell stress tensor to reformulate the body forces as a surface.

And with two examples he shows one with a symmetric EM tensor and the other, with an asymmetric.

whether it is a genuine stress or simply a mathematical device, it still behaves differently if you use it one way or another, this is the whole issue for me.

 

[anuttarasammyak]

Say $T_1$ comes from EM, where does $T$, which you say symmetric, come from ? I am not a specialist of materials at all so take it a question from beginners please.

a. $T + T_1$ is symmetric
b. $T$ is symmetric
c. $T_1$ is not symmetric
We agree with a. I am not sure such division b and c could take place.

Coulomb force between the charges is a body force. Same situation is treated by Maxwell stress tensor in space. They are equivalent in effect. We can choose body force or tensor, but not both body force and tensor at a time. In our case "any forces which can be expressed by tensor form should not be counted in a body force" seems to be an appropriate approach. Otherwise the distinction of stress tensor and volume force would become meaningless.

 

[Klaus3]

Say $T_1$ comes from EM, where does $T$, which you say symmetric, come from ? I am not a specialist of materials at all so take it a question from beginners please.

a. $T + T_1$ is symmetric
b. $T$ is symmetric
c. $T_1$ is not symmetric
We agree with a. I am not sure such division b and c could take place.

Coulomb force between the charges is a body force. Same situation is treated by Maxwell stress tensor in space. They are equivalent in effect. We can choose body force or tensor, but not both body force and tensor at a time. In our case "any forces which can be expressed by tensor form should not be counted in a body force" seems to be an appropriate approach. Otherwise the distinction of stress tensor and volume force would become meaningless.

Microscopically speaking the distinction doesnt exist.

$T$ is supposed to account for contact forces. So forces between particles that are touching each other. On the micro scale, atoms and molecules never touch each other, the interaction they have is long range (i.e, body force). And these body forces are essentially electromagnetic, whether quantum or classical.

However, in the macro scale, the molecular and atomic distance where adjacent molecules interact with each other is regarded as negligible, so they are very much touching one another. So the electromagnetic force which is on the micro scale is long range, gets replaced by part long range part short range. This is a mere mathematical distinction to approximate the phenomena.

Specifically, $T_1$ represents what we would consider the force due to classical electromagnetic effects, you can understand it macroscopically as long range. However, once the (macroscopic) particles get closer and closer, the distinction becomes muddy, and you may have to account for surface effects, such as forces caused by surface charges.

$T$ will represent anything else. Forces of quantum nature, friction, pressure. Anything that cannot be described by maxwell equations, but can via other types of phenomena. It is normally specified via some experimental relation, such as newton's law of viscosity, hooke's law, etc.

In a nutshell, the distinction is mathematicsl, not necessarily physical. It gets even worse if you start accounting for the EM fields themselves to support "pressure", then the charged particles "stresses" the field which then carries this stress at the speed of light to other charged particles, like a rope connecting two bodies.

 

[anuttarasammyak]

Thank you so much for very kind explanation.

b. T is symmetric
c. T1 is not symmetric

I still do not find the evidence of this division.

Microscopically speaking the distinction doesnt exist.

The microscopic view denies the division to symmetric and non-symmetric tensors.

 

[Klaus3]

Thank you so much for very kind explanation.


I still do not find the evidence of this division.

The microscopic view denies the division to symmetric and non-symmetric tensors.

$#T$is going to be symmetric per the first derivation. It is not necessarily symmetric per the second derivation.

The character of $T_1$ here doesnt matter. The only thing it does is, in the second derivation, making $T$ asymmetric if it is asymmetric. The results are conflicting because of this.

 

[anuttarasammyak]

We know the whole tensor is symmetric. In mathematics no symmetric tensor is expressed as sum of symmetric tensor and non symmetric tensor. We can conclude that such division cannot take place.

 

[anuttarasammyak]

assuming k=0

∫TndA+∫div(T1)dV=∫dpdtdV.

You express one of stress tensor in surface integral and the other in volume integral.
Then in angular momentum calculation you make the two in volume integral and find T , originally in surface integral, symmetric.
Further if we make the two in surface integral we will confirm that T1, originally in volume integral, is symmetric.

For any stress tensor t, generated angular momentum is

$$\epsilon_{ijk}\int_V x_j\frac{\partial t_{kl}}{\partial x_l}dV=\epsilon_{ijk}\int_A x_j\ t_{kl}\ n_l \ dA - \epsilon_{ijk}\int_V t_{kj} dV$$

RHS second term is the difference of the two ways of the integration, which should be zero.

 

[Klaus3]

We know the whole tensor is symmetric. In mathematics no symmetric tensor is expressed as sum of symmetric tensor and non symmetric tensor. We can conclude that such division cannot take place.

But it can if both are non symmetric

 

[anuttarasammyak]

But it can if both are non symmetric

Now we know that the stress tensors T, T1, and T+T1 are all symmetric in post #17.