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In my plasma physics course, when studying the effects of magnetic mirrors, if we consider a magnetic field that primarily points in the $z$ direction as shown below:

Picture source: F. Chen - Introduction to Plasma Physics and Controlled Fusion

Picture source: F. Chen - Introduction to Plasma Physics and Controlled Fusion

Then the component of the force parallel to the principle direction of the magnetic field acting on a gyrating particle with guiding center on the $z$ axis is given by: $$ F_z = F_{\parallel}= - \mu \frac{\partial B_z}{\partial z} $$

where $\mu$ is the magnetic moment of the gyrating particle. My lecturer then gave the following argument:

In our argument, the direction of the B field along the $z$ axis was an arbitrary choice. Hence, we can generalize it to be: $$ F_{\parallel} = - \mu \frac{\partial B}{\partial s} = - \mu \nabla_{\parallel} B $$ where $s$ is the line element in the direction of $B$

It is not the first time I had seen this argument, where we can directly make a change from $d \rightarrow \nabla$. For example, when studying the grad B drift in plasmas, if we consider $B$ to point in the $z$ direction and $\nabla B$ to be in the $y$ direction, then the grad B drift is: $$ \mathbf{v}_{\nabla B} = \frac{1}{qB^2} \left(\pm \frac{q v_\perp r_L}{2} \frac{\partial B_z}{\partial y}\right) \hat{\mathbf{e}}_y \times \mathbf{B} $$

If we were to now consider the an arbitrary $\mathbf{B}$ and $\nabla B$, then the above expression becomes: $$ \mathbf{v}_{\nabla B} = \frac{1}{qB^2} \left(\pm \frac{q v_\perp r_L}{2} \nabla B \right) \times \mathbf{B} $$

where we again see the "derivative upgrade" $d \rightarrow \nabla$. Again, the explanation was:

The choice of the $y$ axis was arbitrary and hence should generalize to a general direction

I find these arguments incredibly "hand wavy" and would want to find out if there is some deep insight on why we can just swap $d \rightarrow \nabla$ in these equations?

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  • $\begingroup$ Changing to $\nabla$ means changing from a single-variable derivate to a multiple-variable gradient over several dimensions. What the arguments indicate without giving full detail is that there is symmetry in the derivation across all dimensions. The derivation along one axis thus applies along more axes. $\endgroup$
    – Steeven
    May 17 '21 at 11:55
  • $\begingroup$ @Steeven Is this always the case? I find it quite hand wavy, it seems to me that this argument will allow us to change the derivative to $\nabla$ for ANY physics equation. Are there any conditions needed before this switch can be made? $\endgroup$
    – D. Soul
    May 17 '21 at 12:25
  • $\begingroup$ Technically, these things are derived from first principles where the $\nabla$ operator is used first, not later. The intro text books tend to start with simple, 1D cases to illustrate the point and then often skip over the more general derivation by just saying that the 1D idealistic scenario can be generalized to arbitrary dimensions. With the $\nabla$B-drift case, you can see this by starting with the gradient along y then rotating to a new (poorly chosen) coordinate basis. $\endgroup$ May 17 '21 at 13:21
  • $\begingroup$ Have you heard of a directional derivative? $\endgroup$ May 17 '21 at 16:16
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I want to try to lift the confusion. First, I want to introduce two important aspects.

  1. Physics should not change if you change your coordinate system.

What does the real life care about the coordinate system you use in you calculations? Further, there are two facts to realize, because we generally choose an easy path in calculations: we usually align our coordinate system along the magnetic field. Thus,

  1. Direction parallel to the magnetic field $\mathbf B$ is distinguishable, direction perpendicular is not. Physics in parallel and perpendicular direction might be different.

Both of your examples are interesting, because they need us to use both of these facts. In the following, I want to use the notation $\mathbf B = B \, \mathbf b$, where $\mathbf b$ is the unit vector along $\mathbf B$ and $B=|\mathbf B|$ is the absolute value.

Your 1st example

You first example has the magnetic field along the z-direction, thus $\mathbf B = B_z \, \mathbf z$ and $B=B_z$. The force you wrote was, $$ F_{z} = - \mu \frac{\partial B_z}{\partial z} = - \mu \frac{\partial B}{\partial s} . $$ The last equal sign comes from the fact you mentioned, the z-direction was chosen arbitrarily. Why? Remember fact #1 above, you can always rotate your coordinate system without the physics to change. The z-direction was defined as the direction where the magnetic field points along. Why call it $\mathbf z$? It is more general to just call it $\mathbf b$. The operator $\nabla_{||}$ is nothing more than a projection of the gradient of $B$ along this direction, $$ \nabla_{||}B = \mathbf b \cdot \nabla B = b_x \frac{\partial B}{\partial x} + b_y \frac{\partial B}{\partial y} + b_z \frac{\partial B}{\partial z} = \frac{\partial B}{\partial s} . $$ The 3rd expression is just the definition of the derivative along the field line. Try using $\mathbf B = B_z \mathbf z$, you end up by at the "arbitrary" version. First, $\nabla_{||}$ is just a shorthand symbol, not an "upgrade" of "$\partial$". Second, it may look a bit hand-wavy to infere the general equation from the simple one, but how did you arive at the simple one? By exactly the same arguments as you would if you want to arive at the general one!

Your 2nd example

Now again, $B$ is pointing towards the z-direction. The perpendicular derivate is $\frac{\partial B_z}{\partial y} \, \mathbf y$, and points in this case in the $\mathbf y$ direction, which is again as you wrote arbitrary. Now we can make use of the point #2 above to make life more convenient. We decompose the gradient of B into a perpendicular and a parallel part,

$$ \nabla B = \frac{\partial B}{\partial x} \, \mathbf x + \frac{\partial B}{\partial y} \, \mathbf y + \frac{\partial B}{\partial z} \, \mathbf z = \nabla_\perp B + \nabla_{||} B. $$ If you cross it with $\mathbf B$ you get, $$ \nabla B \times \mathbf B = \nabla_\perp B \times \mathbf B + \nabla_{||} B \times \mathbf B = \nabla_\perp B \times \mathbf B + \mathbf b \cdot \nabla B \mathbf \times B \mathbf b = \nabla_\perp B \times \mathbf B. $$ where the second term is zero due to a vector triple product with two same vectors. This shows, that it does not matter if you use $\nabla$ or $\nabla_\perp$ in the vector product. This shows us, that the physics of gradB drift do not depend on the parallel gradient. This is the derivation of the most general form. How to connect to the simple one? Due to point #1 we can rotate the coordinate system as we want, e.g. we choose $\mathbf z$ to be the parallel direction, this leaves us with $$ \nabla_\perp B = \frac{\partial B_z}{\partial x} \, \mathbf x + \frac{\partial B_z}{\partial y} \mathbf y.$$ Note that this is general, we could choose not the letters $(\mathbf x, \mathbf y, \mathbf z)$ as the coordinate system but $(\mathbf \perp_1, \mathbf \perp_2, \mathbf ||)$ and just rename. If you now assume that $B_z$ only depends on $y$ (or $\nabla B$ along $\mathbf y$), you get your "arbitrary" expression.

Summary

Due to the special properties of the magnetic field (and lazy plasma physicists), one usually aligns one coordinate parallel to the magnetic field and the other two perpendicular. This gives a connection between parallel derivative $\partial/\partial_s$ and the operator $\nabla_{||}$, that comes from the decomposition into parallel and perpendicular, because you can write $\nabla = \nabla_\perp + \nabla_{||}$. Those are only shorthand symbols for expressions that directly link to partial derivatives $\nabla_{||} B= \mathbf b \partial_{||} B$ and $\nabla_\perp B= \perp_1 \partial_{\perp,1} B + \perp_2 \partial_{\perp,2} B$. Usually you then choose nice names like $\perp_1\rightarrow x, \perp_2\rightarrow y, ||\rightarrow z$. Therefore it appears, that $\partial$-terms are often replaced by $\nabla$-terms in the general expressions.

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  • $\begingroup$ Hi Sorry! How did you get: $$ b_x \frac{\partial B_x}{\partial x} + b_y \frac{\partial B_y}{\partial y} + b_z \frac{\partial B_z}{\partial z} = \frac{\partial B}{\partial s} $$ Why do we have $b_x = \frac{\partial x}{\partial s}$, $b_y = \frac{\partial y}{\partial s}$ and $b_z = \frac{\partial z}{\partial s}$? $\endgroup$
    – D. Soul
    May 18 '21 at 3:58
  • $\begingroup$ Sorry I did a mistake with the indices, it should be only $B$ in all terms. $\endgroup$
    – P. U.
    May 18 '21 at 6:02
  • $\begingroup$ Consider the equation of a magnetic field line (you can derive this assuming that a magnetic field line is a curve that is always tangential to the magnetic field), $\frac{B}{ds} = \frac{B_x}{dx} = \frac{B_y}{dy} = \frac{B_z}{dz}$. Those are all individual equations. Take the first, $\frac{B}{ds} = \frac{B_x}{dx}$ or $\frac{dx}{ds} = \frac{B_x}{B} = b_x$. The right hand side is just $b_x$. You can do this with the other ones as well. $\endgroup$
    – P. U.
    May 18 '21 at 6:25
  • $\begingroup$ I recommend the textbook of D'haeseleer "flux coordinates and magnetic field structure" if you are interested in all the details (you can find a pdf if you google for it). For example the definition of a magnetic field line, or the identity $\mathbf b \cdot \nabla = \frac{\partial}{\partial s}$ can be found inside. $\endgroup$
    – P. U.
    May 18 '21 at 6:28
  • $\begingroup$ Thank you! Please give me awhile before i accept your answer! $\endgroup$
    – D. Soul
    May 18 '21 at 10:01
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The gradient $\nabla$ can be thought of as the "components" of the differential operator d. More concretely, if $\phi$ is a scalar function, then $$\text{d}\phi = \frac{\partial\phi}{\partial x}\text{d}x + \frac{\partial\phi}{\partial y}\text{d}y + \frac{\partial \phi}{\partial z}\text{d}z = \nabla \phi\cdot (-),$$ where $(-)$ is a placeholder for a (unit) vector. The unit differentials $\text{d}x_i$ are defined to satisfy $\text{d}x_i(\hat{x}_j) = \delta_{ij},$ so for example if we use $(-) = \hat{y}$, then this simplifies to $$\text{d}\phi(\hat{y}) = \nabla\phi \cdot \hat{y} = \nabla_y \phi = \frac{\partial \phi}{\partial y}.$$

In your first example, $F_\parallel$ is the force along the $s$ direction, so we have $$F_\parallel = F\cdot s = (-\mu \nabla B)\cdot s = -\mu (\nabla B\cdot s) = -\mu \nabla_\parallel B.$$ If $s$ is the $z$ direction, this equation reads $$F_z = F\cdot \hat{e}_z = (-\mu \nabla B)\cdot \hat{e}_z= -\mu (\nabla B\cdot \hat{e}_z) = -\mu \nabla_z B = -\mu\frac{\partial B}{\partial z}.$$

The second example is really using the fact that the gradient can be expanded in components as $$\nabla B = \frac{\partial B}{\partial x} \hat{e}_x + \frac{\partial B}{\partial y} \hat{e}_y + \frac{\partial B}{\partial z} \hat{e}_z.$$ If $\nabla B$ is nonzero only in the $y$-direction, then $$\nabla B = \frac{\partial B}{\partial y} \hat{e}_y \implies \nabla B \cdot \hat{e}_y = \frac{\partial B}{\partial y}.$$

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  • $\begingroup$ Hi Thank you! Yes i can understand the argument when you go from $\nabla$ downwards to $\del$, I just am not comfortable with the argument that because "the axis chosen was arbitrary", we can simply change the derivative to $\nabla$, is there some insight for this? $\endgroup$
    – D. Soul
    May 17 '21 at 12:23
  • $\begingroup$ This sort of "engineer's induction" is not very rigorous I agree. I think there are two ways to think about this: 1. the formula using $\nabla$ is the postulate, and the case with the derivative is a special case of this formula; the presentation by the author to reverse this exposition is purely pedagogical. $\endgroup$
    – jsborne
    May 17 '21 at 13:16
  • $\begingroup$ 2. Since we have the formula given in terms of a standard basis, rewriting the formula to accommodate a general direction can be done by some kind of orthogonal transformation, under which expressions like $\nabla B$ are invariant. In this line of reasoning, the use of the derivative is equivalent to the use of $\nabla$; it's merely conventional. The author could've used $\nabla$ in both cases without loss of information. $\endgroup$
    – jsborne
    May 17 '21 at 13:17
  • $\begingroup$ I'd like to recommend reading this: en.wikipedia.org/wiki/Matrix_calculus. Maybe this will make it clear why the gradient is a generalization of the derivative. $\endgroup$
    – jsborne
    May 17 '21 at 13:20
  • $\begingroup$ Thank you! I will check it out and try to digest $\endgroup$
    – D. Soul
    May 18 '21 at 10:02

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