In a certain system of units the electromagnetic stress tensor Mij is given by Mij = EiEj + BiBj − 1/2 δij(EkEk + BkBk), where the electric and magnetic fields, E and B, are first-order tensors. Show that Mij is a second-order tensor. Consider a situation in which |E| = |B| but the directions of E

Question

In a certain system of units the electromagnetic stress tensor [latex]{M}_{ij}[/latex] is given by

[latex]{M}_{ij} = {E}_{i}{E}_{j} + {B}_{i}{B}_{j} −\frac { 1 }{ 2 } {δ}_{ij} ({E}_{k}{E}_{k} + {B}_{k}{B}_{k})[/latex],

where the electric and magnetic fields, E and B, are first-order tensors. Show that [latex]{M}_{ij}[/latex] is a second-order tensor.

Consider a situation in which |E| = |B| but the directions of E and B are not parallel. Show that E ± B are principal axes of the stress tensor and find the corresponding principal values. Determine the third principal axis and its corresponding principal value.

Accepted Answer

In the calculation of the transformed RHS, [latex]\mathbf{E}[/latex] and [latex]\mathbf{B}[/latex] transform with a single 'L-matrix', but [latex]\delta_{i j}[/latex], being a second-order tensor, requires two. It may simply be noticed that [latex]E_{k} E_{k}[/latex] and [latex]B_{k} B_{k}[/latex] are scalars and therefore unaltered in the transformation; but, if not, then the orthogonal properties of [latex]L, L_{i k} L_{j k}=\delta_{i j}[/latex] and [latex]L_{k i} L_{k j}=\delta_{i j}[/latex], are needed:

[latex]\begin{aligned}M_{i j}= & E_{i} E_{j}+B_{i} B_{j}-\frac{1}{2} \delta_{i j}\left(E_{k} E_{k}+B_{k} B_{k} ight) \M_{i j}^{\prime}= & L_{i m} E_{m} L_{j n} E_{n}+L_{i m} B_{m} L_{j n} B_{n} \& \quad-\frac{1}{2} L_{i p} L_{j q} \delta_{p q}\left(L_{k r} E_{r} L_{k s} E_{s}+L_{k r} B_{r} L_{k s} B_{s} ight) \= & L_{i m} L_{j n}\left(E_{m} E_{n}+B_{m} B_{n} ight)-\frac{1}{2} L_{i p} L_{j q} \delta_{p q}\left(\delta_{r s} E_{r} E_{s}+\delta_{r s}B_{r} B_{s} ight) \= & L_{i m} L_{j n}\left[E_{m} E_{n}+B_{m} B_{n}-\frac{1}{2} \delta_{m n}\left(E_{r} E_{r}+B_{r} B_{r} ight) ight] \= & L_{i m} L_{j n} M_{m n} .\end{aligned}[/latex]

To obtain the penultimate line we relabelled the dummy suffices p and q as m and n. Thus [latex]M_{i j}[/latex] transforms as a second-order tensor; it is real and symmetric and will therefore have orthogonal eigenvectors.

For the case [latex]|\mathbf{E}|=|\mathbf{B}|[/latex], i.e. [latex]E^{2}=B^{2}[/latex], denote [latex]E_{i} \pm B_{i}[/latex] by [latex]v_{i}[/latex] and consider

[latex]\begin{aligned}M_{i j} v_{j}= & M_{i j}\left(E_{j} \pm B_{j} ight) \= & E_{i} E_{j}\left(E_{j} \pm B_{j} ight)+B_{i} B_{j}\left(E_{j} \pm B_{j} ight)-\frac{1}{2} \delta_{i j}\left(E^{2}+B^{2} ight)\left(E_{j} \pm B_{j} ight) \= & E_{i} E^{2} \pm E_{i}(\mathbf{E} \cdot \mathbf{B})+B_{i}(\mathbf{B} \cdot \mathbf{E}) \pm B_{i} B^{2} \& \quad-\frac{1}{2}\left(E^{2}+B^{2} ight)\left(E_{i} \pm B_{i} ight) \= & \left(E_{i} \pm B_{i} ight)\left[E^{2} \pm(\mathbf{E} \cdot \mathbf{B})-\frac{1}{2} 2 E^{2} ight], ext { using } E^{2}=B^{2}, \= & \pm(\mathbf{E} \cdot \mathbf{B})\left(E_{i} \pm B_{i} ight) \= & \pm(\mathbf{E} \cdot \mathbf{B}) v_{i} .\end{aligned}[/latex]

This shows that [latex]\mathbf{E} \pm \mathbf{B}[/latex] are eigenvectors of [latex]M_{i j}[/latex] (i.e. its principal axes) with principal values [latex]\pm(\mathbf{E} \cdot \mathbf{B})[/latex].

The third principal axis is orthogonal to both of these and is therefore in the direction

(E + B) × (E − B) = 0 + (B × E) − (E × B) − 0 = 2(B × E)

To determine its principal value, consider

[latex]\begin{aligned}M_{i j}(\mathbf{B} imes \mathbf{E})_{j} & =M_{i j} \epsilon_{j l m} B_{l} E_{m} \& =E_{i} E_{j} \epsilon_{j l m} B_{l} E_{m}+B_{i} B_{j} \epsilon_{j l m} B_{l} E_{m}-\frac{1}{2} \delta_{i j} 2 E^{2} \epsilon_{j l m} B_{l} E_{m} \& =0+0-E^{2}(\mathbf{B} imes \mathbf{E})_{i}, ext { since } \epsilon_{j l m} X_{l} X_{j}=0 .\end{aligned}[/latex]

Thus, the third principal value is [latex]-E^{2}[/latex] (or [latex]-B^{2}[/latex] ). This value could have been deduced from the trace of [latex]M_{i j}=E^{2}+B^{2}-\frac{3}{2}\left(E^{2}+B^{2} ight)=-E^{2}[/latex], since the two eigenvalues found previously are [latex]\pm \mathbf{E} \cdot \mathbf{B}[/latex], which sum to zero. The three eigenvalues together must add up to the trace; hence, the third one is [latex]-E^{2}[/latex].

Question 26.15: In a certain system of units the electromagnetic stress tens...

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