# What's the most general form of Maxwell's stress tensor for EM fields in matter?

In Griffith's Introduction to Electrodynamics we find a derivation of the relation $$\mathbf f + \frac{\partial \mathbf g}{\partial t} = \nabla \cdot \overleftrightarrow{\mathbf T} ,$$ where $$\mathbf f$$ is the Lorentz force per unit volume, $$\mathbf g$$ is the electromagnetic linear momentum per unit volume, and $$\overleftrightarrow{\mathbf T}$$ is Maxwell's stress tensor. However, this derivation employs Maxwell's equations in a vacuum, or at least in media that exhibit no polarization or magnetization.

My aim is to derive a similar relation for Maxwell's tensor, but starting from the more general set of equations that include the fields $$\mathbf D$$, $$\mathbf E$$, $$\mathbf B$$, and $$\mathbf H$$, with the general relations $$\mathbf D = \varepsilon_0 \mathbf E + \mathbf P$$ and $$\mathbf B = \mu_0(\mathbf H + \mathbf M)$$. Starting again from the local expression of the Lorentz force on free charges and current densities $$\mathbf f = \rho_0 \mathbf E + \mathbf J_0 \times\mathbf B,$$ I've managed to show that $$\mathbf f = -\frac{\partial \mathbf g}{\partial t} + (\nabla \cdot \mathbf D)\mathbf E + (\nabla \cdot \mathbf B) \mathbf H - \mathbf D \times (\nabla \times \mathbf E) - \mathbf B \times (\nabla \times \mathbf H),$$ but I cannot seem to be able to continue. Intuitively, I think I should get something of the form $$T_{ij} = E_i D_j + H_i B_j - \frac 1 2 \delta_{ij} (\mathbf E \cdot \mathbf D + \mathbf H \cdot \mathbf B),$$ which does revert to Griffith's $$T_{ij} = \varepsilon_0 E_i E_j + \frac 1 {\mu_0} B_i B_j - \frac 1 2 \delta_{ij} \left( \varepsilon_0 E^2 + \frac 1 {\mu_0} B^2\right)$$ as soon as $$\mathbf P$$ and $$\mathbf M$$ both vanish, and also happens to contain the expression for the electromagnetic energy density $$u = \frac 1 2 \mathbf E \cdot \mathbf D + \frac 1 2 \mathbf H \cdot \mathbf B.$$ Maybe I'm missing some relevant vector calculus identities, or relations between the fields. Otherwise, I guess the only way out is to express $$\mathbf D$$ and $$\mathbf B$$ in terms of $$\mathbf E$$ and $$\mathbf H$$ respectively, through the usual relations (valid in all homogeneous isotropic linear media) $$\mathbf D = \varepsilon \mathbf E, \qquad \mathbf B = \mu \mathbf H,$$ with $$\varepsilon = \varepsilon_0 \varepsilon_r = \varepsilon_0 (1+\chi_e)$$ and $$\mu = \mu_0 \mu_r = \mu_0(1+\chi_m)$$, which would yield $$T_{ij} = \varepsilon E_i E_j + \frac 1 {\mu} B_i B_j - \frac 1 2 \delta_{ij} \left( \varepsilon E^2 + \frac 1 {\mu} B^2\right),$$ but this means we lose generality, and we won't be able to treat e.g. ferroelectric or ferromagnetic materials, or anisotropic linear media.

Any thoughts?

Your guessed answer is almost correct. The right order must be $$T_{ij}=D_iE_j+B_iH_j$$+... A derivation with the necessary assumptions is in pp. 283-284 of my EM textbook.

The venerable Griffiths answers your question himself (indirectly because he doesn't use tensors) in his paper with Hnizdo What's the Use of Bound Charge (arXiv:1506.02590).

They derive two results that are relevant to your question, which in the form of your equation are:

$$\mathbf f + \frac \partial {\partial t} \left( \varepsilon_0 \mathbf E \times \mathbf B \right) = \nabla \cdot \overleftrightarrow{\mathbf T}$$

and

$$\mathbf f_o + \frac \partial {\partial t} \left( \frac 1 {c^2} \mathbf E \times \mathbf H \right) = \nabla \cdot \overleftrightarrow{\mathbf T}$$

where

$$T_{ij} = E_i D_j + H_i B_j - \frac 1 2 \delta_{ij} \left[ \varepsilon_0 E^2 + \mu_0 \left( H^2 - M^2 \right) \right]$$

$$\mathbf f$$ is the rate of change of total momentum density and $$\mathbf f_o$$ is the ‘overt’ force density, which is equal to $$\mathbf f$$ less the rate of change of the ‘hidden’ momentum density:

$$\mathbf f_o = \mathbf f - \frac \partial {\partial t} \left( \frac 1 {c^2} \mathbf M \times \mathbf E \right)$$

• Indeed, Griffiths shall be enshrined in the pantheon of great physics teachers! But only Jackson is worthy of our veneration! Commented Jan 10 at 3:20

After much consideration I now think that some of the force densitities that Griffiths and Hnizdo propose in their paper are wrong, so I’m writing to modify my previous answer.

1. It is incomplete in that his force density only accounts for the forces on free charges and currents and ignores the forces on dipoles.

2. It makes the simplifying assumption that the material is both linear and homogeneous.

3. His force density is the same as that arising from Minkowski’s stress-energy tensor for the electromagnetic field in matter.

This last point is of particular importance because Minkowski’s tensor is Lorentz covariant, which means that the resulting four-force density is also Lorentz covariant.

Griffiths and Hnizdo’s force densities do not arise from a Lorentz-covariant stress-energy tensor, so that they cannot be correct. However, Griffiths and Hnizdo’s Equation 73 is equal to the divergence of the stress component of a stress-energy tensor for the electromagnetic field in matter due to Medina and Stephany (https://arxiv.org/abs/1703.02109):

$$\Theta ^ {\mu \nu} + \eta _ {\alpha \beta} F ^ {\mu \alpha} P ^ {\nu \beta}$$

where $$\Theta ^ {\mu \nu}$$ is the (symmetric) electromagnetic stress-energy tensor, $$F ^ {\mu \alpha}$$ is the electromagnetic field tensor, and $$P ^ {\nu \beta}$$ is the polarization-magnetization tensor.

Medina and Stephany’s tensor is also Lorentz covariant, so it, too, yields a four-force density that is Lorentz covariant. In fact, the corresponding force density differs from that in Jerrold Franklin’s textbook by a total derivative: $$\frac 1 2 \nabla ( \mathbf P \cdot \mathbf E + \mathbf M \cdot \mathbf B )$$, which appears to be the force density on dipoles in linear and homogeneous matter. The quantity $$\mathbf P \cdot \mathbf E + \mathbf M \cdot \mathbf B$$ is a Lorentz scalar, as would be expected.

Griffiths and Hnizdo err in two ways.

Firstly, they calculate that the force on an electric dipole is $$( \mathbf p \cdot \nabla ) \mathbf E + \dot {\mathbf p} \times \mathbf B$$ instead of \begin {align} \nabla ( \mathbf p \cdot \mathbf E ) & = ( \mathbf p \cdot \nabla ) \mathbf E + \mathbf p \times ( \nabla \times \mathbf E ) \\ & = ( \mathbf p \cdot \nabla ) \mathbf E + \dot {\mathbf p} \times \mathbf B - \frac \partial {\partial t} ( \mathbf p \times \mathbf B ) \end {align}

This doesn’t mean that they ignore the contribution of dipoles to the momentum of the electromagnetic field. They account for this by including an adjustment for the rate of change of the hidden-momentum density (Equation 80). This is their second error: my final equation above shows that if their adjustment had been for the rate of change of $$\mathbf P \times \mathbf B$$ instead it would have offset their original error.

Does this mean that the concept of hidden momentum is unnecessary? No, it appears in the momentum density component of the stress-energy tensor for polarized and magnetized matter:

$$T ^ {\mu \nu} + \eta _ {\alpha \beta} G ^ {\mu \alpha} M ^ {\nu \beta}$$

where $$T ^ {\mu \nu}$$ is the stress-energy tensor for ‘bare’ matter (i.e. matter that is neither polarized nor magnetized) and $$G ^ {\mu \alpha}$$ and $$M ^ {\nu \beta}$$ are the duals of $$F ^ {\mu \alpha}$$ and $$P ^ {\nu \beta}$$ respectively.

Like $$\eta _ {\alpha \beta} F ^ {\mu \alpha} P ^ {\nu \beta}$$, $$\eta _ {\alpha \beta} G ^ {\mu \alpha} M ^ {\nu \beta}$$ is Lorentz covariant, and their sum is symmetric, as Abraham desired. They are complementary in that, like two sides of a coin, it’s not possible to have one without the other.

This implies that, in addition to the canonical four-force density arising from Medina and Stephany’s tensor,

$$- \partial _ \nu ( \Theta ^ {\mu \nu} + \eta _ {\alpha \beta} F ^ {\mu \alpha} P ^ {\nu \beta} )$$

there is also a kinetic four-force density,

$$- \partial _ \nu ( \Theta ^ {\mu \nu} + \eta _ {\alpha \beta} F ^ {\mu \alpha} P ^ {\nu \beta} + \eta _ {\alpha \beta} G ^ {\mu \alpha} M ^ {\nu \beta} )$$

which accounts for the hidden momentum of magnetic dipoles.

However, my answer to the question of what's the most general form of Maxwell's stress tensor for EM fields in matter stands: it’s (the negative of) the stress component of Medina and Stephany’s tensor (Equation 135), which does not depend on the linearity and homogeneity of the matter. The difference between it and Jerrold Franklin’s Maxwell stress dyadic is due to the force density on dipoles (in linear and homogeneous matter).

PS Jerrold Franklin appears to have adopted a different convention for the divergence of a tensor from mine: $$\partial _ i T ^ {i j}$$ rather than $$\partial _ j T ^ {i j}$$.

Using the formula (https://en.wikipedia.org/wiki/Vector_calculus_identities#Cross_product_rule) $$\mathbf{V}\times \left(\nabla \times \mathbf{W} \right)= \left(\nabla \mathbf{W} \right) \cdot \mathbf{V}-\left(\mathbf{V} \cdot \nabla \right) \mathbf{W}$$ That, put in the formula gives $$\mathbf{f}+ \frac{\partial \mathbf{g}}{\partial t}= \left(\nabla \cdot \mathbf{D} \right) \mathbf{E}+\left(\nabla \cdot \mathbf{B} \right) \mathbf{H} - \left(\nabla \mathbf{E} \right) \cdot \mathbf{D}+\left(\mathbf{D} \cdot \nabla \right) \mathbf{E} - \left(\nabla \mathbf{H} \right) \cdot \mathbf{B}+\left(\mathbf{B} \cdot \nabla \right) \mathbf{H}$$ Expressed in terms of components, limited just to the terms with $$\mathbf{E}$$ and $$\mathbf{D}$$, $$f_i+\frac{\partial g_i}{\partial t}=\left(\frac{\partial D_j}{\partial x_j} \right) E_i+ D_j \frac{\partial E_i}{\partial x_j}-\left(\frac{\partial E_i}{\partial x_j} \right) D_i+...$$ Now, the first 2 terms can be expressed as $$\frac{\partial}{\partial x_j} (E_i \, D_j)$$ but, to add something like $$-\delta_{ij} \mathbf{E}\cdot \mathbf{D}$$, we need to also have a term $$-E_i \left(\frac{\partial D_i}{\partial x_j} \right)$$.
Honestly, I don't know how to fix this.
By the way, $$\mathbf{g}=\mathbf{D}\times \mathbf{B}$$.