# Why does this equation refer to magnetic tension?

The force density in a plasma, according to MHD, is $$\mathbf{F} = -\nabla P -\nabla(\frac{B^2}{2\mu_0}) + \frac{1}{\mu_0}(\mathbf{B}\cdot\nabla)\mathbf{B}$$

where first term is the thermal pressure gradient, the second term is the magnetic pressure gradient and the third term is the magnetic tension

The magnetic tension tends to straighten out the field lines thus reducing their curvature.

How do I see this from the form of the equation $\frac{1}{\mu_0}(\mathbf{B}\cdot\nabla)\mathbf{B}$?

• The name tension comes from the form of the Stress tensor that arises from this force density. The 'pressures' are the scalar part of this tensor, while the last term comes from a dyadic part. – Hydro Guy Apr 14 '14 at 14:26

I think the easiest way is to start off with the conservation of momentum in the conservation-law form: $$\frac{\partial\boldsymbol\pi}{\partial t}=\nabla\cdot\mathbb T=0$$ where $\boldsymbol\pi=\rho\mathbf u$ and $$\mathbb T=\rho\mathbf u\mathbf u+\left(p+\frac12B^2\right)\mathbb I - \mathbf B\mathbf B$$ is the stress tensor and, as discussed before, $\mathbf{aa}$ is the dyad product. The magnetic and pressure terms can be combined to a single tensor in "matrix" notation (assuming a flow along the $z$ axis) $$\left(\begin{array}{ccc}p+\frac12B^2 & 0 & 0 \\ 0 & p+\frac12B^2 & 0 \\ 0 & 0 & p-\frac12B^2\end{array}\right)\begin{array}{c}\perp \\ \\ \perp\\ \\ \parallel\end{array}$$ with the $\perp,\,\parallel$ symbols indicate that those diagonal components are the component along perpendicular or parallel to the flow. Visually, this is akin to the following image (taken from Page 149 of the book Principles of Magnetohydrodynamics by Hans Goedblood & Stefaan Poedts)

You can see that the magnetic field terms parallel to the tube are pulling it outwards, causing the tube to reduce its curvature (straightening it).

In general a term $(\mathbf{v} \cdot \mathbf{\nabla}) \mathbf{A}$ is the directional derivative of $\mathbf{A}$ in the direction of $\mathbf{v}$. It tells you how $\mathbf{A}$ is changing when you move along $\mathbf{v}$.

Here, we have $(\mathbf{B} \cdot \mathbf{\nabla}) \mathbf{B}$, which is telling how much $\mathbf{B}$ changes as we move in the direction of $\mathbf{B}$. This term will be zero if $\mathbf{B}$ is constant in magnitude and direction as you move along a stream line of $\mathbf{B}$.

What if $\mathbf{B}$ is not constant as you move along a flow line? Let's look at the example where $\mathbf{B}$ points in the $\hat{x}$ direction, but gets bigger as you move from the origin. So let's say $\mathbf{B} = B_0 x^2/L^2 \hat{x}$. Then $(\mathbf{B} \cdot \mathbf{\nabla}) \mathbf{B} = 2 B_0^2 x^3 / L^4 \hat{x}$. Since this comes in with a relative negative sign compared to the pressure, this term in this case is like a pressure pointing in the $-\hat{x}$ direction and will work to smooth out the magnetic field so it gets bigger for small $x$ and smaller for big $x$. You can convince yourself that for other field configurations it will work to make the field constant along stream lines.

• Sorry why should I consider it with a negative sign? – SuperCiocia Apr 14 '14 at 14:54
• because when you look at $\mathbf{F}$, the term with pressure has a negative sign and the term you were asking about doesn't. So the term you are looking at is actually like a negative pressure. – Brian Moths Apr 14 '14 at 15:43
• should I take the minus sign even when I have a constant B field (constant in magnitude) but with bending field lines? – SuperCiocia Apr 14 '14 at 16:31
• Also, naive question: if, as you say, $(\mathbf{v}\cdot\mathbf{A})$ is the directional derivative of $\mathbf{A}$ in the direction of $\mathbf{v}$, if $\mathbf{v}$ and $\mathbf{A}$ are both equal to $\mathbf{B}$, shouldn't the magnetic field $\mathbf{B}$ be constant along the direction of $\mathbf{B}$? – SuperCiocia Apr 14 '14 at 16:34
• Also, if I integrate $(\mathbf{B} \cdot \mathbf{\nabla}) \mathbf{B} = 2 B_0^2 x^3 / L^4 \mathbf{\hat{x}}$ between some $x_1$ and $x_2$ and times it by the area $A$ so that I get a force density$\times$volume = force, I get a force in the $+\mathbf{\hat{x}}$ direction...? – SuperCiocia Apr 14 '14 at 16:37