# Demonstration of magnetic field lines “tension” and “pressure” using the Maxwell stress tensor

I would like to show that the magnetic fields lines are under "tension" along the lines and exert "pressure" perpendicularly to the lines, using the Maxwell tensor only. I have a sign ambiguity.

In Special Relativity, the electromagnetic field stress-energy-momentum tensor is defined with these components (using cartesian coordinates in an inertial frame. I'm using $$c = 1$$ and $$\mu_0 = 1$$ for simplicity. Metric signature is the "mostly negative" : $$\eta = (1, -1, -1, -1)$$): $$$$\tag{1} T^{ab} = {F^a}_c \, F^{cb} + \frac{1}{4} \, \eta^{ab} \, F_{cd} \, F^{cd}.$$$$ The Maxwell stress tensor is defined by the space part: $$$$\tag{2} T^{ij} = {F^i}_c \, F^{c j} - \frac{1}{4} \, \delta^{ij} \, F_{cd} \, F^{cd}.$$$$ Explicitely, assuming a magnetic field only ($$F_{0 i} = 0$$ and $$F_{ij} = -\, \varepsilon_{ijk} \, B_k$$): $$$$\tag{3} T_{ij} = \frac{1}{2} \, B^2 \, \delta_{ij} - B_i \, B_j.$$$$ Now, I introduce an unit vector $$\boldsymbol{\mathrm{n}}$$ oriented in the direction of the magnetic field : $$\boldsymbol{\mathrm{B}} \cdot \boldsymbol{\mathrm{n}} = B$$, and a unit vector $$\boldsymbol{\mathrm{u}}$$ perpendicular to the local field line: $$\boldsymbol{\mathrm{B}} \cdot \boldsymbol{\mathrm{u}} = 0$$. Contracting these vectors to the components (3) gives these (notice the index position and the sign change, because of the Minkowski metric, since $$\eta_{ij} = -\, \delta_{ij}$$): \begin{align}\tag{4} {T^i}_j \, n^j &= \frac{1}{2} \, B^2 \, n^i, &{T^i}_j \, u^j &= -\, \frac{1}{2} \, B^2 \, u^i, \end{align} According to these equations, the eigenvalue associated to the eigenvector vector $$\boldsymbol{\mathrm{n}}$$ is the positive magnetic energy density ($$\mathcal{U} = \frac{1}{2} \, B^2 > 0$$), while it is negative for the eigenvector $$\boldsymbol{\mathrm{u}}$$: \begin{align}\tag{5} {T^i}_j \, n^j &= \mathcal{U} \, n^i, &{T^i}_j \, u^j &= -\, \mathcal{U} \, u^i,. \end{align} Tension should be a "negative pressure". The sign difference between both eigenvalues shows that the magnetic field line is under tension in some direction and exert a "pressure" in the orthogonal direction. But how can I show that "tension" (i.e negative pressure) is along the field line (i.e in the direction of vector $$\boldsymbol{\mathrm{n}}$$), and not in its perpendicular direction? It is not clear that the two equations of (5) are telling that there is negative pressure (i.e tension) in the $$\boldsymbol{\mathrm{n}}$$ direction and positive pressure in the $$\boldsymbol{\mathrm{u}}$$ direction. By lowering all the indices of these equations, I could also write these: \begin{align}\tag{6} T_{ij} \, n_j &= -\, \mathcal{U} \, n_i, &T_{ij} \, u_j &= +\, \mathcal{U} \, u_i. \end{align}

• I'm not certain what you're asking. You already showed that the tensor has opposite sign in the $n$ and two independent $u$ directions. So are you just asking why the eigenvalue of $T^i_{\,\,j}$ has the interpretation as a negative pressure rather than positive? – octonion Dec 19 '18 at 22:11
• @octonion, no, I'm asking what is the proper way to show that the field lines are in tension along themselves. Tension is a negative pressure, but the first equation of (5) has a positive eigenvalue. This equation alone isn't sufficient to show that it's actually representing a tension. – Cham Dec 20 '18 at 0:56
• So although you said "no" it sounds like you just restated what I suspected your question was. Compare your result with the energy momentum tensor of a perfect fluid in the mostly negative signature. You will see that the sign you are getting is consistent with it being interpreted as a negative pressure. – octonion Dec 20 '18 at 10:42
• @octonion, can you be more specific? Can you offer a detailed answer to the question? I don't see clearly how the Maxwell 3-tensor shows that the field lines are under tension, while exerting a transverse pressure. – Cham Dec 20 '18 at 13:48
• @octonion, a prefect fluid is locally isotropic. The space part of its energy-momentum is $$T_{ij} = (\rho + p) \, \gamma^2 \, v_i \, v_j + p \, \delta_{ij}.$$ The Maxwell tensor (3) can be written as this:$$T_{ij} = - 2 \, \mathcal{U} \, n_i \, n_j + \mathcal{U} \, \delta_{ij},$$ and doesn't represent a kind of isotropic fluid. – Cham Dec 20 '18 at 13:56